Precalculus

🧭 Overview

🧠 One-sentence thesis

Careful tracking of units and rates is essential for solving real-world problems, because numbers always come with units attached and rates precisely describe how quantities change over time.

📌 Key points (3–5)

• Units must be consistent: when applying formulas, all quantities must use the same unit system (e.g., all distances in feet or all in miles).
• Unit conversion is algebraic: units behave like numbers—multiply by conversion factors and cancel matching units in numerator and denominator.
• Rate definition: rate equals change in quantity divided by change in time; the Greek letter Δ (delta) is shorthand for "change in."
• Common confusion: Δ quantity is always (final value minus initial value), not just "any difference"—order matters.
• Density as a ratio: density equals mass divided by volume; given any two of density/mass/volume, you can solve for the third.

📏 Working with units

📏 Why units matter

• Numbers in real problems don't occur in isolation; each number comes with a unit attached (feet, seconds, grams, etc.).
• The excerpt emphasizes: you must be VERY CAREFUL to use consistent units throughout a problem.
• Example: A marathon runner's speed is given in feet/second, but the race distance is in miles—you must convert one to match the other before applying formulas.

🔄 Unit conversion mechanics

Unit conversion: multiply by a conversion factor (a ratio equal to 1) and cancel matching units algebraically.

• Treat units like algebraic symbols: cancel common units in numerator and denominator.
• Example from the excerpt:
  • 26.2 miles × (5,280 feet per mile) = (26.2)(5,280) feet
  • The "mile" units cancel, leaving feet.
• You can chain conversions: seconds → minutes → hours by multiplying successive conversion factors.

🧮 Applying formulas with units

The excerpt uses the distance-speed-time relationship:

(total distance traveled) = (constant speed) × (elapsed time)

Written with units: (ft) = (ft/sec) × (sec)

• Plug in values with their units and verify that units on both sides match.
• Example: 138,336 ft = 18 ft/sec × t
  Solve: t = 138,336 ft ÷ (18 ft/sec) = 7,685.33 sec
• The division of units: ft ÷ (ft/sec) = ft × (sec/ft) = sec.

⚖️ Density as a unit ratio

Density = mass / volume

• Pure water has density 1 gram per cubic centimeter (1 g/cm³).
• Given any two of {density, mass, volume}, solve for the third.
• Example from the excerpt: 857 g of a substance with volume 2.1 liters
  Density = (857 g) / (2.1 L) = 408 g/L
  Convert liters to cm³: 1 L = 1,000 cm³, so density = 0.408 g/cm³.

🌐 Multi-step unit problems

The gold sphere example shows a chain of reasoning:

• Given: mass = 100 kg, density of gold = 19.3 g/cm³
• Convert mass: 100 kg × (1,000 g/kg) = 100,000 g
• Volume of sphere: V = (4/3) π r³
• Density equation: 19.3 g/cm³ = 100,000 g / V
• Solve for V, then for radius r.
• Don't confuse: the formula connects three quantities; you must isolate the unknown algebraically while keeping units consistent.

⏱️ Rates and change over time

⏱️ What a rate measures

Rate (or rate of change) = (change in the quantity) / (change in time)

• A rate describes how fast a quantity is changing, not the quantity itself.
• Common shorthand: Δ (delta) means "change in," so
  rate = Δ quantity / Δ time

🔢 Calculating Δ (change)

The excerpt defines change precisely:

Δ quantity = (value at final time) − (value at initial time)
Δ time = (final time) − (initial time)

• Order matters: always subtract initial from final.
• Example: Temperature at 8:00 am is 65°F, at 10:00 am is 71°F.
  • Δ temperature = 71 − 65 = 6 degrees
  • Δ time = 10:00 − 8:00 = 2 hours
  • Rate = 6 deg / 2 hr = 3 deg/hr (temperature is increasing).

🔄 Positive vs negative rates

• If the final value is greater than the initial value, Δ is positive → the quantity is increasing.
• If the final value is less than the initial value, Δ is negative → the quantity is decreasing.
• Example from the excerpt: On June 5, temperature at 8:00 am is 71° and at 10:00 am is 65°.
  • Δ temperature = 65 − 71 = −6 degrees
  • Rate = −6 deg / 2 hr = −3 deg/hr (temperature is decreasing).
• Don't confuse: a negative rate means the quantity is going down, not that you made a calculation error.

📐 Rate formula structure

Component	Meaning	Units example
Δ quantity	Final value minus initial value	degrees Fahrenheit
Δ time	Final time minus initial time	hours
Rate	Δ quantity / Δ time	degrees per hour

• The units of the rate are always "quantity units per time unit."
• Example: speed in feet per second, temperature change in degrees per hour, cost increase in dollars per year.

🧭 Overview

🧠 One-sentence thesis

When a quantity changes at a constant rate over time, the total change equals the rate multiplied by the elapsed time, providing a fundamental tool for predicting how quantities evolve.

📌 Key points (3–5)

• What a rate measures: the change in a quantity divided by the change in time, capturing how fast something is changing.
• How to calculate rate: subtract the initial value from the final value of the quantity, then divide by the time elapsed (final time minus initial time).
• Sign matters: rates can be positive (increasing) or negative (decreasing); mixing up initial and final values produces the wrong sign.
• Common confusion: the formula "Total Change = Rate × Time" only works for constant rates—situations where the rate stays the same throughout the time period.
• Why it matters: this principle allows us to predict total changes in real-world quantities like distance, temperature, or area when the rate of change is steady.

📐 Understanding rates

📏 What a rate is

Rate (or rate of change): change in the quantity divided by change in time.

• A rate describes how fast something is changing, not just how much it has changed.
• The excerpt uses the shorthand notation Δ (delta) to mean "change in."
• Written as: rate = Δ quantity / Δ time
• Example: temperature changing from 65°F to 71°F over 2 hours gives a rate of 3 degrees per hour.

🧮 How to calculate a rate

The calculation requires comparing values at two specific moments:

• Δ quantity = (value at final time) − (value at initial time)
• Δ time = (final time) − (initial time)
• Then divide: rate = Δ quantity / Δ time

Don't confuse: You must be careful about which time is "initial" and which is "final." Swapping them accidentally will flip the sign of your rate.

➕➖ Positive vs negative rates

The excerpt gives two temperature examples on different days:

Day	Initial (8:00 am)	Final (10:00 am)	Rate	Meaning
June 4	65°F	71°F	+3 deg/hr	Temperature increasing
June 5	71°F	65°F	−3 deg/hr	Temperature decreasing

• Same temperatures, same time interval, but different order → opposite signs.
• A positive rate means the quantity is increasing; a negative rate means it is decreasing.

🔄 Constant rates and total change

🎯 What "constant rate" means

Constant rate: a rate that stays the same for all time periods.

• Example: driving at a steady 60 miles per hour means your distance is changing at a constant rate.
• The speedometer needle staying steady indicates the rate is constant.
• Most real-world situations have varying rates, but constant rates are simpler and the focus of this course.

🧩 The total change formula

When the rate is constant, the excerpt gives the key principle:

Total Change in some Quantity = Rate × Time

• This formula only works when the rate is constant throughout the time period.
• It allows you to predict how much a quantity will change over a given time.
• Example: A puddle's surface area grows at 4 square centimeters per hour. After 84 minutes (which is 84/60 hours), the total surface area is: (4 cm²/hr) × (84/60 hr) = 5.6 cm².

Don't confuse: This formula does not work for non-constant rates. The excerpt notes that one main goal of calculus is to develop a version that handles changing rates.

🌊 Worked example: the leaking pipe

💧 Problem setup

The excerpt provides a detailed example:

• A water pipe drips onto the floor, creating a circular puddle.
• The puddle's surface area increases at a constant rate of 4 cm²/hour.
• Question: What is the surface area and radius after 84 minutes?

🔢 Solution steps

1. Identify the pieces: quantity changing = surface area; rate = 4 cm²/hr; time = 84 minutes
2. Convert units: 84 minutes = 84/60 hours = 1.4 hours
3. Apply the formula: Total Surface Area = Rate × Time = (4 cm²/hr) × (1.4 hr) = 5.6 cm²
4. Find dimensions: Using the circle area formula (area = π times radius squared), solve for radius: r = square root of (5.6/π) = 1.335 cm

This example shows how the total change formula connects to geometric properties when combined with other formulas.

🧭 Overview

🧠 One-sentence thesis

Mathematical modeling is a problem-solving technique that describes and predicts by isolating key features through equations, and while initial models may be crude, they can be refined to become more informative and adaptable.

📌 Key points (3–5)

• What modeling does: it is used to both "describe" and "predict" across many disciplines by adding mathematical structure.
• Models as caricatures: a model picks out certain features and focuses on those at the expense of others, requiring experience to know which features matter.
• Progression from crude to refined: initial models may be qualitative categories, but adding numerical structure makes them more informative and modifiable.
• Common confusion: a "good" model is not necessarily perfect—it isolates the right features and can be easily modified, even if some details (like variation in homework time) are not captured.
• Expect initial difficulty: modeling leads to frustration and confusion at first, but comfort increases with practice.

🎨 What modeling is and why it matters

🎨 Core definition and purpose

Modeling: a method used to both "describe" and "predict" by adding mathematical structure to a problem.

• It is not just writing equations—mathematics arises only after preliminary work.
• The excerpt emphasizes that "clean equations and formulas only arise after some (or typically a lot) of preliminary work."
• Modeling appears "any time we are problem solving and consciously trying" to understand and forecast.

🖼️ The caricature analogy

• A model is like a caricature: it picks out certain features (like a nose or face) and focuses on those at the expense of others.
• This means models are selective, not exhaustive.
• Don't confuse: a model is not meant to capture every detail; it deliberately simplifies.

🧗 Learning curve

• The excerpt warns that "in the beginning, modeling will lead to frustration and confusion."
• Experience is needed to distinguish "good" models (which isolate the right features) from "bad" ones.
• By the end of a course, "comfort level will dramatically increase."

🔄 From crude to refined models

🔄 The example: estimating study time

The excerpt walks through Example 1.3.1: "How much time do you anticipate studying precalculus each week?"

Stage 1: Crude qualitative model

• Initial response: "a little" or "way too much!"
• This is already modeling—categorizing total study time into qualitative bins based on past experience.
• It is a "crude model" but still a model.

Stage 2: Numerical model

• The excerpt refines the estimate by breaking down total time T into components:
  • T = (hours in class) + (hours reading text) + (hours doing homework)
• Known: 5 hours in class per week.
• Reading: if one page takes 15 minutes on average and there are r pages, then hours reading = (15/60) × r.
• Homework: if one problem takes 25 minutes on average and there are h problems, then hours doing homework = (25/60) × h.
• Final model: T = 5 + (15/60)r + (25/60)h hours.

✅ Evaluating the model

Is this a good model?

• Yes: it is "certainly more informative" than the crude categorical model.
• Key strength: "it clearly isolates the features being used to make our estimated time commitment and it can be easily modified as the amount of reading or homework changes."
• Not perfect: "some homework problems will take a lot more than 25 minutes!"—the model uses averages and does not capture variation.

Takeaway: A "pretty good model" is one that isolates the right features and is adaptable, even if it is not perfect in every detail.

🧩 Key characteristics of useful models

🧩 Informative and transparent

• A refined model is more informative than a crude one.
• It makes explicit which features (variables) are being used.
• Example: the study-time model shows exactly how reading pages and homework problems contribute to total time.

🔧 Modifiable and adaptable

• A good model "can be easily modified" when conditions change.
• Example: if the number of homework problems h changes week to week, the formula T = 5 + (15/60)r + (25/60)h can be recalculated without starting over.

⚖️ Trade-offs and imperfection

• Models simplify reality; they do not capture every detail.
• The excerpt acknowledges that "some homework problems will take a lot more than 25 minutes," but the model uses an average.
• Don't confuse: "good" does not mean "perfect"—it means the model isolates the right features and is useful for the problem at hand.

🧭 Overview

🧠 One-sentence thesis

These exercises apply unit conversions, rate reasoning, geometric formulas, and algebraic modeling to real-world scenarios, reinforcing the modeling process introduced earlier in the chapter.

📌 Key points (3–5)

• Unit conversion practice: problems require converting between different units (time, speed, area, volume) to compare quantities or verify equivalences.
• Rate and proportion reasoning: many problems involve constant rates (pizza orders, dripping water, temperature change) and require finding totals or future values.
• Geometric and physical modeling: several problems use formulas for area, volume, density, and mass to answer questions about shapes, objects, and physical scenarios.
• Common confusion: speed vs. pace—pace is measured in minutes per mile (inverse of speed), so higher pace numbers mean slower running.
• Algebraic manipulation: later problems require solving equations, simplifying fractions, and translating word statements into mathematical expressions ("pseudo-equations").

🔄 Unit conversion and comparison

🔄 Time and speed conversions

• Problem 1.1(a): verify that 7685.33 seconds equals 2 hours 8 minutes 5.33 seconds by converting hours and minutes to seconds and summing.
• Problem 1.1(b): compare 100 mph vs. 150 ft/s by converting both to the same unit (e.g., both to ft/s or both to mph).
  • Don't confuse: different units can hide which quantity is larger; always convert to a common unit before comparing.

💰 Salary comparison

• Problem 1.1(c): Gina earns 1 cent/second for a 40-hour work week; Tiare earns $1400 for the same period.
  • Convert Gina's rate to total weekly pay: (cents per second) × (seconds in 40 hours) → compare to Tiare's $1400.
  • Example: calculate total seconds in 40 hours, multiply by 0.01 dollars, then compare.

🎓 Credit hours and study time

• Problem 1.1(d): a degree requires 180 credits; each credit corresponds to one hour of class per week for a 10-week quarter, plus 2.5 hours of outside study per class hour.
  • Total time per credit = (1 hour in class + 2.5 hours study) × 10 weeks.
  • Total hours for degree = 180 credits × (time per credit).
  • This models total investment (class + study) needed to graduate.

🏃 Rate and pace problems

🏃 Marathon pace vs. speed

Pace = minutes per mile (the inverse of speed).

• Problem 1.5: runners use pace (min/mile) instead of speed (distance/time).
  • (a) Convert Lee's speed (16 ft/sec) to pace (min/mile): first convert ft/sec to miles/hour, then invert to get min/mile.
  • (b) Convert Allyson's pace (6 min/mile) to speed: invert pace to get miles/min, then convert to desired units.
  • (c) Compare Adrienne's pace (5.7 min/mile) to Dave's speed (10.3 mph) by converting both to the same unit.
  • Don't confuse: lower pace number = faster runner; higher speed number = faster runner.

🚴 Bicycle loop problem

• Problem 1.2: Sarah completes a loop in 2 hours 40 minutes; if her speed increased by 1 km/hr, time would drop by 6 minutes.
  • Let loop length = L km, original speed = v km/hr.
  • Original time: L/v = 2 hours 40 min.
  • New time: L/(v+1) = 2 hours 34 min.
  • Solve the system to find L.

🍕 Pizza order rate

• Problem 1.9: Pagliacci receives 18 orders in 4 minutes (constant rate).
  • Orders in 4 hours = (18 orders / 4 min) × (240 min).
  • Profit rate: $11 per 10 orders.
  • Find when cumulative profit exceeds $1,000 by setting up a time equation.

📐 Geometry and physical quantities

📐 Density and volume

• Problem 1.3: density of lead = 11.34 g/cm³, aluminum = 2.69 g/cm³; find radius of spheres each with mass 50 kg.
  • Use mass = density × volume and volume of sphere = (4/3)π r³.
  • Solve for r in each case.

🗼 Eiffel Tower and air mass

• Problem 1.4: Tower mass = 7.3 million kg, height = 324 m, square base side = 125 m, steel volume = 930 m³, air density = 1.225 kg/m³.
  • Imagine a cylinder containing the Tower; find air mass in that cylinder.
  • Compare air mass to Tower mass.
  • Example: cylinder volume = π × (base radius)² × height; air mass = volume × density.

🍕 Pizza value comparison

• Problem 1.7: 10-inch diameter pizza for $8 vs. 15-inch diameter pizza for $16.
  • Compare price per area: area = π × (diameter/2)².
  • Calculate cost per square inch for each pizza.
  • Don't confuse: diameter vs. radius—use radius in the area formula.

🍕 Pizza slice geometry

• Problem 1.10: 16-inch diameter pie sliced into 8 equal pieces; Lee cuts off an edge, John takes a triangular slice.
  • Compare areas of Lee's edge piece vs. John's triangular slice.
  • Requires geometric calculation of each region's area.

🔬 Scientific and relativistic modeling

🔬 Relativity and mass

• Problem 1.8: Einstein's relativity predicts mass increase at high speeds.
  • Formula: m = m₀ / √(1 - v²/c²), where c = 3×10⁸ m/sec (speed of light).
  • (a) Calculate Dave's mass (m₀ = 66 kg) at 90%, 99%, 99.9% of light speed.
  • (b) Find speed needed for Dave to reach 500 kg.
  • Example: as v approaches c, the denominator approaches zero, so mass grows very large.

🧬 DNA length

• Problem 1.11: a single cell contains ~2 meters of DNA (if straightened); human body has 10¹⁴ cells.
  • Total DNA length = 2 m × 10¹⁴.
  • How many times does this wrap around Earth's equator?
  • Divide total length by Earth's circumference.

💧 Changing rates and areas

💧 Dripping water puddle

• Problem 1.12: circular puddle area increases at constant rate of 11 cm²/hour.
  • (a) Find area and radius after 1 minute, 92 minutes, 5 hours, 1 day.
    • Area at time t = 11 × t (in hours).
    • Radius: solve A = π r² for r.
  • (b) Is radius increasing at a constant rate?
    • Since A = π r², r = √(A/π); radius grows as the square root of time, not linearly.
  • Don't confuse: constant rate of area increase ≠ constant rate of radius increase.

🌡️ Temperature rate of change

• Problem 1.16: temperature changes over time intervals.
  • (a) 44°F at 7:00 am, 50°F at 10:00 am → rate = (50 - 44) / (3 hours) = 2°F/hour.
  • (b) Use the same rate to predict temperature at 2:00 pm.
  • (c) 54°F at 4:30 pm, 26°F at 6:15 pm → calculate rate over 1.75 hours.
  • Example: rate of change = (final temp - initial temp) / (time elapsed).

🧮 Algebraic expressions and equations

🧮 Pseudo-equations

• Problem 1.6: translate word statements into mathematical expressions.
  • Example: "John's salary is at most $56,000 a year" → (John's salary) ≤ $56,000.
  • "More than 28% of his salary in taxes" → (Taxes) > 0.28 × (Salary).
  • "Between 1500 and 1800 students" → 1500 ≤ (Number of students) ≤ 1800.
  • "Twice the number of happy math students exceeds five times the number of happy chemistry students" → 2×(Math) > 5×(Chem).
  • Don't confuse: "at most" (≤) vs. "at least" (≥) vs. "more than" (>).

🧮 Solving equations

• Problem 1.17: solve for the unknown variable.
  • (a) 3t - 7 = 11 + t → isolate t.
  • (b) √(1 + 1/a) = 3 → square both sides, solve for a.
  • (c) √(x² + a²) = 2a + x → square and simplify.
  • (d) 1 - t > 4 - 2t → solve the inequality.
  • (e) Combine fractions: 2/x - 1/(x+1) → find common denominator.

🌳 Apple orchard yield model

• Problem 1.14: 60 trees, each yielding 12 bushels; removing a tree increases yield per tree by 0.45 bushels.
  • Let x = number of trees, N = yield per tree.
  • (a) Find formula for N in terms of x: start with x=60, N=12; if x=59, N=12.45; pattern: N = 12 + 0.45×(60 - x).
  • (b) Why does yield increase when trees are removed? Possible reasons: less competition for sunlight, water, nutrients.

📉 Tax rate formula

• Problem 1.15: proposed tax rate after x years: r / (1 + 1/(1 + 1/x)).
  • Simplify the nested fraction to a single fraction.
  • Calculate new rate after 1, 2, 5, 20 years.
  • Does the rate increase or decrease over time?
  • Congress claims a 75% cut eventually—verify by examining the limit as x grows large.

🌍 Large-scale environmental problems

🌍 Sewage volume

• Problem 1.13: Seattle dumped 20 million gallons/day into Lake Washington in the 1950s.
  • (a) Weekly total = 20 million × 7; yearly total = 20 million × 365.
  • (b) Visualize daily sewage as a rectangular prism with football field base (100 yards × 50 yards); find height h.
    • Convert gallons to cubic feet (7.5 gallons = 1 ft³), then to cubic yards.
    • Volume = base area × height → solve for h.

🧭 Overview

🧠 One-sentence thesis

The xy-coordinate system solves the problem of uniquely locating any point in a plane by assigning each point a unique pair of real numbers (x, y) through perpendicular axes.

📌 Key points (3–5)

• The core problem: finding a foolproof method to reconstruct the location of points on a plane using pairs of real numbers.
• How it works: two perpendicular axes (x-axis and y-axis) intersect at the origin, dividing the plane into four quadrants.
• Bidirectional mapping: the system is reversible—you can go from a point to a pair of numbers, and from a pair of numbers back to a unique point.
• Common confusion: the axes are not just lines; they are copies of the real number line with positive, negative, and zero portions.
• Uniqueness guarantee: different points always produce different coordinate pairs, and different coordinate pairs always produce different points.

📐 The basic setup

📐 What a coordinate system is

Coordinate system: Every point P in the xy-plane corresponds to a unique pair of real numbers (x, y), where x is a number on the horizontal x-axis and y is a number on the vertical y-axis.

• The excerpt motivates this with a real-world analogy: finding a restaurant on a city map by using a grid of streets (addresses).
• The coordinate system is like a city map grid—it catalogs points using pairs of real numbers.

✏️ The two axes

• Start by drawing two perpendicular lines: the horizontal axis (x-axis) and the vertical axis (y-axis).
• Each axis is a copy of the real number line, divided into three parts: positive numbers, negative numbers, and zero.
• The origin is the intersection point of these two axes.

🧭 Positive and negative directions

• Positive x-axis: numbers to the right of the origin on the x-axis.
• Positive y-axis: numbers above the origin on the y-axis.
• The negative portions are on the opposite sides (left for x-axis, below for y-axis).
• Don't confuse: the axes are not just geometric lines; they carry the structure of the real number line with signed directions.

🔄 From points to numbers

🔄 The forward procedure: point P → pair (x, y)

The excerpt describes a three-step process:

1. Draw parallel lines: Through point P, draw one line parallel to the x-axis (call it ℓ) and one line parallel to the y-axis (call it ℓ*).
2. Find intersections:
   • Line ℓ crosses the y-axis at exactly one point; call that point "y."
   • Line ℓ* crosses the x-axis at exactly one point; call that point "x."
3. Uniqueness: If you start with two different points P and Q, you get two different pairs of numbers (x, y) and (x*, y*), meaning either x ≠ x* or y ≠ y*.

• This procedure assigns a unique pair of real numbers to each point.
• Example: If P is located somewhere in the plane, drawing the parallel lines will pinpoint exactly which x-value and y-value correspond to P.

🔁 From numbers to points

🔁 The reverse procedure: pair (x, y) → point P

The excerpt emphasizes that the procedure is reversible:

1. Locate the numbers: Find the number x on the x-axis and the number y on the y-axis.
2. Draw parallel lines:
   • Draw a line ℓ parallel to the x-axis passing through y on the y-axis.
   • Draw a line ℓ* parallel to the y-axis passing through x on the x-axis.
3. Find the intersection: The two lines ℓ and ℓ* intersect at exactly one point in the plane; call it P.

• If you start with two different pairs of real numbers, the corresponding points in the plane will be different.
• This bidirectional mapping is the key feature: you can go back and forth between points and coordinate pairs without ambiguity.

🔑 Why reversibility matters

• The excerpt states: "The great thing about the procedure we just described is that it is reversible!"
• This means the coordinate system provides a one-to-one correspondence: each point has exactly one coordinate pair, and each coordinate pair corresponds to exactly one point.
• Don't confuse: this is not just about labeling points; it's about establishing a foolproof reconstruction method.

🗺️ Quadrants and location

🗺️ The four quadrants

The plane is divided into four regions based on the signs of the coordinates:

Quadrant	First coordinate (x)	Second coordinate (y)	Location
First	Positive	Positive	Upper right
Second	Negative	Positive	Upper left
Third	Negative	Negative	Lower left
Fourth	Positive	Negative	Lower right

• Every point in the plane lies in one of these four quadrants or on one of the two axes.
• The excerpt notes that quadrant terminology is useful to give a rough sense of location (the sentence is incomplete but implies this is a coarse locator).

📍 Points on the axes

• Points that lie exactly on the x-axis or y-axis do not belong to any quadrant.
• Example: A point on the positive x-axis has coordinates (x, 0) where x > 0; a point on the negative y-axis has coordinates (0, y) where y < 0.

🧭 Overview

🧠 One-sentence thesis

A coordinate system requires three design choices—scaling, labeling, and units on each axis—and changing the aspect ratio (the relative scale between axes) can alter how the same data appears visually without changing the underlying mathematical relationships.

📌 Key points (3–5)

• What a coordinate system involves: scaling, labeling, and units on each of the axes.
• Scaling and aspect ratio: the aspect ratio is the length of one unit on the vertical axis divided by the length of one unit on the horizontal axis; different aspect ratios change the visual appearance of the same set of points.
• How aspect ratio affects graphs: the same points (e.g., on a parabola) look different when the aspect ratio changes, even though the mathematical relationship is unchanged.
• Common confusion: aspect ratio is not about the absolute size of the graph; it is about the relative length of one unit on the vertical axis compared to one unit on the horizontal axis.
• Units can differ between axes: the two axes may use different types of units (labels), which is useful for real-world applications like plotting performance over time.

📏 Scaling and aspect ratio

📏 What scaling means

• Scaling refers to how "one unit" on each axis is represented by a physical length on the graph.
• The excerpt emphasizes that you can choose different scales for the horizontal and vertical axes.

📐 Aspect ratio definition

Aspect ratio = (length of one unit on the vertical axis) / (length of one unit on the horizontal axis).

• It is a fraction that compares the physical lengths representing one unit on each axis.
• Example: if one unit on the x-axis is drawn as 2 cm and one unit on the y-axis is drawn as 1 cm, the aspect ratio is 1/2.
• Don't confuse: aspect ratio is not the range of values on each axis; it is the ratio of the physical lengths used to represent one unit.

🎨 How aspect ratio changes appearance

The excerpt provides an exercise:

• Plot the same set of points (including (1,1), (−1,1), (−4/5, 16/25), etc.) on two coordinate systems.
• In the first system, the scale on each axis is the same (aspect ratio = 1).
• In the second system, "one unit" on the x-axis has the same length as "two units" on the y-axis (aspect ratio = 1/2).
• Both pictures show the points lying on a parabola, but the visual shape changes because of the different aspect ratio.

Key insight: The mathematical relationship (the parabola) is the same, but the visual appearance depends on the aspect ratio.

Example: A circle with aspect ratio 1 may look like an ellipse with aspect ratio 1/2.

🛠️ Practical use

• In problem solving, you often need to make a rough assumption about the relative axis scaling based on the information given.
• Most graphing devices let you specify the aspect ratio.

🏷️ Axes units and labels

🏷️ Different units on different axes

• Sometimes the two axes involve different types of units (labels).
• The excerpt mentions this is useful for real-world applications and begins an example about a marketing director presenting product performance data (the example is incomplete in the excerpt).

📊 Why this matters

• Using different units on each axis allows you to plot relationships between different kinds of quantities (e.g., time on one axis, sales on another).
• The excerpt emphasizes "the power of using pictures" to communicate information effectively.

Don't confuse: having different units on each axis does not mean the coordinate system is invalid; it is a deliberate design choice to represent real-world data.

🧭 Overview

🧠 One-sentence thesis

Imposing a coordinate system—choosing where to place the origin and axes—is a critical first step in all modeling problems, and although many choices are possible, experience helps identify the most natural one.

📌 Key points (3–5)

• What imposing a coordinate system means: deciding where to draw the xy-coordinate system (where to place the origin and axes) in a real-world scenario.
• Multiple valid choices exist: there is often more than one way to impose a coordinate system; any may work, but some are more natural than others.
• Natural choice criteria: a good choice often keeps motion or key features in one quadrant (e.g., all coordinates non-negative) or simplifies the algebra.
• Common confusion: there is no single "correct" coordinate system—the choice depends on the problem context and what makes the math simpler or more intuitive.
• Why it matters: the coordinate system choice directly affects how you write equations for positions, distances, and other quantities; a poor choice can complicate the problem unnecessarily.

🎯 What imposing a coordinate system means

🎯 The core decision

Imposing a coordinate system: the initial problem-solving step of deciding on a choice of xy-coordinate system.

• It is not just drawing axes; it is deciding where to place the origin and how to orient the axes relative to the physical scenario.
• The excerpt emphasizes that this is "a key step in all modeling problems."
• Example: in a ball-toss scenario, you must decide whether the origin is at the top of the cliff, the bottom, the launch point, or the landing point.

🔄 Reversibility and uniqueness

• The excerpt (from earlier sections) notes that the coordinate system procedure is reversible: every point corresponds to a unique pair of real numbers, and every pair corresponds to a unique point.
• This means once you impose a coordinate system, you can translate back and forth between geometric locations and numeric coordinates.

🧩 Multiple valid choices and how to pick

🧩 Many possible choices

• The excerpt states: "There will often be many possible choices."
• No single choice is universally "right"—different origins and orientations can all describe the same physical situation.
• Don't confuse: "many possible choices" does not mean "arbitrary"—each choice has trade-offs in simplicity and clarity.

🌟 What makes a choice "natural"

• The excerpt says: "it takes problem solving experience to develop intuition for a 'natural' choice."
• A natural choice often:
  • Keeps motion or key features in one quadrant (so coordinates are non-negative).
  • Simplifies the equations you need to write.
  • Aligns with the problem's symmetry or starting conditions.

Criterion	Why it helps
Motion in one quadrant	All coordinates non-negative; easier to interpret
Origin at a key point	Simplifies initial conditions (e.g., starting position is (0,0))
Axes aligned with motion	Reduces the number of coordinates that change

🏀 Example: tossed ball scenario

🏀 Four logical choices

• The excerpt revisits the tossed ball scenario (from page 1) and shows four possible coordinate systems (Figure 2.7):
  1. Origin at the top of the cliff.
  2. Origin at the bottom of the cliff.
  3. Origin at the landing point.
  4. Origin at the launch point.
• All four choices "will work," but they differ in convenience.

✅ The most natural choice

• The excerpt identifies Figure 2.7(b)—origin at the bottom of the cliff—as "probably the most natural choice."
• Why: in this coordinate system, the motion of the ball takes place entirely in the first quadrant, so both x and y coordinates are non-negative throughout the flight.
• This avoids negative numbers and makes the path easier to visualize and describe.

🏃 Example: runners colliding on a runway

🏃 The scenario

• Michael and Aaron run toward each other from opposite ends of a 10,000 ft. airport runway.
• The problem asks: where and when will they collide?

📍 Imposing the coordinate system

• The excerpt chooses:
  • Michael starts at the origin: M = (0, 0).
  • Aaron starts at the far end: A = (10,000, 0).
• This choice places both runners on the x-axis (y = 0 throughout) and makes the initial positions simple.

🧮 Writing position as a function of time

• Michael's position: moving at 15 ft/second to the right.
  • After t seconds, he has traveled 15t feet.
  • Position: M(t) = (15t, 0).
• Aaron's position: moving at 8 ft/second to the left from (10,000, 0).
  • After t seconds, he has traveled 8t feet to the left.
  • Position: A(t) = (10,000 − 8t, 0).
• Example: after 1 second, Michael is at (15, 0) and Aaron is at (9,992, 0).

🔍 Finding the collision

• The key observation: collision occurs when the coordinates of Michael and Aaron are equal.
• Because both move along the x-axis, set their x-coordinates equal:
  • 15t = 10,000 − 8t
  • 23t = 10,000
  • t ≈ 434.8 seconds
• Plug t back in to find the location:
  • M(434.8) = (15 × 434.8, 0) = (6,522, 0).
• Conclusion: they collide at approximately (6,522, 0) after about 434.8 seconds.

🎓 Why this coordinate choice worked

• Placing Michael at the origin and Aaron at (10,000, 0) made the initial conditions simple (no offsets).
• Both runners stayed on the x-axis, so the problem reduced to a one-dimensional algebra problem.
• A different choice (e.g., origin at the midpoint) would still work but might require more bookkeeping.

🔗 Connection to visualization and modeling

🔗 From coordinates to pictures

• The excerpt (from earlier sections) emphasizes that "mathematical modeling is all about relating concrete phenomena and symbolic equations."
• Visualization typically involves plotting points in the plane, either from a list or a prescription.
• Imposing a coordinate system is the bridge: it lets you translate a physical scenario into numeric coordinates that you can plot and analyze.

🔗 Why this step is key

• Without a coordinate system, you cannot write equations for positions, velocities, or other quantities.
• The choice of coordinate system affects:
  • The form of your equations.
  • The ease of solving the problem.
  • The clarity of your visualization.
• The excerpt states this is "a key step in all modeling problems"—it is foundational, not optional.

🧭 Overview

🧠 One-sentence thesis

The distance formula uses the Pythagorean theorem to compute straight-line distance between two points in the plane, relying on the changes in x- and y-coordinates (Δx and Δy), which are directed distances that depend on which point is chosen as the starting point.

📌 Key points (3–5)

• Distance formula: the straight-line distance d between two points P = (x₁, y₁) and Q = (x₂, y₂) is the square root of (Δx)² + (Δy)², derived from the Pythagorean theorem.
• Directed distances: Δx and Δy represent changes in coordinates from a beginning point to an ending point; they can be negative if the ending coordinate is smaller than the beginning coordinate.
• Order matters for Δx and Δy: reversing the beginning and ending points flips the sign of Δx and Δy, but distance d remains the same because squaring eliminates the sign.
• Common confusion: do NOT distribute the square root over addition—√(a² + b²) is not equal to √a² + √b²; you must add first, then take the square root.
• Application: the distance formula applies to real-world motion problems where objects move in perpendicular directions, forming a right triangle.

📐 The distance formula and its foundation

📐 How the formula is derived

• The excerpt starts with two points P = (x₁, y₁) and Q = (x₂, y₂) in the xy-plane, with the same units on both axes (e.g., both in feet).
• Imagine an arrow from P (the starting position) to Q (the ending position).
• A right triangle is formed: the horizontal leg has length |Δx| and the vertical leg has length |Δy|; the hypotenuse is the straight-line distance d.
• By the Pythagorean theorem: d² = (Δx)² + (Δy)², so d = √((Δx)² + (Δy)²).

📏 The formal distance formula

Distance formula: If P = (x₁, y₁) and Q = (x₂, y₂) are two points in the plane, then the straight-line distance between them (in the same units as the axes) is d = √((Δx)² + (Δy)²) = √((x₂ − x₁)² + (y₂ − y₁)²).

• This formula gives the distance you would travel if you flew along the straight line segment connecting the two points.
• Example from the excerpt: if P = (1, 1) and Q = (5, 4), then Δx = 5 − 1 = 4, Δy = 4 − 1 = 3, and d = 5.

🧭 Directed distances: Δx and Δy

🧭 What Δx and Δy represent

• Δx and Δy track the change in coordinates as you move from a beginning point to an ending point.
• The excerpt defines them as:
  • Δx = (x-coordinate of ending point) − (x-coordinate of beginning point) = x₂ − x₁
  • Δy = (y-coordinate of ending point) − (y-coordinate of beginning point) = y₂ − y₁
• These are called directed distances in the x and y directions.

🔄 Order matters: beginning vs ending point

• You must specify which point is the beginning and which is the ending.
• If you reverse the choice, Δx and Δy both change sign (become negative).
• Example: if P = (5, 4) and Q = (1, 1), then Δx = 1 − 5 = −4 and Δy = 1 − 4 = −3.
• However, the distance d remains the same because (Δx)² = (|Δx|)² and (Δy)² = (|Δy|)²; squaring eliminates the sign.
• The excerpt emphasizes: d = √((Δx)² + (Δy)²) = √((|Δx|)² + (|Δy|)²).

🔮 Why directed distances matter

• The notion of directed distance is important for later topics: lines (Chapter 4), vectors, and calculus.
• It captures not just "how far" but also "in which direction" the coordinates change.

⚠️ Common algebraic mistake

⚠️ Do NOT distribute the square root

• The excerpt includes a CAUTION about a very common mistake:
  • Wrong: √(3² + 4²) = √3² + √4² = 3 + 4 = 7
  • Right: √(3² + 4²) = √(9 + 16) = √25 = 5
• You must add the squared terms before taking the square root.
• Don't confuse: the square root of a sum is not the sum of the square roots.

🚗 Application: two departing cars

🚗 Problem setup

• Two cars depart from a four-way intersection at the same time.
• One heads East, the other heads North; both travel at 30 ft/sec.
• The excerpt asks: (1) how far apart are they after 1 hour 12 minutes? (2) when are they exactly 1 mile apart?

🚗 Building the model

• Impose a coordinate system with the intersection at the origin.
• After t seconds, the Eastbound car is at (a, 0) where a = 30t, and the Northbound car is at (0, b) where b = 30t.
• Distance between them: d = √(a² + b²) = √((30t)² + (30t)²) = √(2t²(30)²) = 30t√2.

🚗 Solving part 1: distance after 1 hour 12 minutes

• Convert time to seconds: 1 hr 12 min = 1.2 hr = 1.2 × 60 × 60 = 4,320 sec.
• Substitute t = 4,320: d = 30 × 4,320 × √2 = 129,600√2 feet ≈ 183,282 feet.
• Convert to miles (1 mile = 5,280 feet): d ≈ 34.71 miles.

🚗 Solving part 2: when are they 1 mile apart?

• Set d = 1 mile = 5,280 feet: 30t√2 = 5,280.
• Solve for t: t = 5,280 / (30√2) ≈ 124.45 seconds ≈ 2 minutes 4 seconds.
• Example: the two cars will be 1 mile apart in 2 minutes and 4 seconds.

🧩 Interpreting the distance formula geometrically

🧩 The right triangle interpretation

• The excerpt emphasizes using the right triangle in the figure.
• The horizontal leg has length |Δx|, the vertical leg has length |Δy|, and the hypotenuse has length d.
• This visual model makes the Pythagorean theorem connection clear.

🧩 Units and consistency

• The excerpt assumes "the units on each axis are the same" (e.g., both in feet).
• The resulting distance d is in the same units.
• Example: in the car problem, coordinates are in feet, so d is in feet; convert to miles as needed.

🧭 Overview

🧠 One-sentence thesis

These exercises apply the distance formula and coordinate geometry to solve motion problems involving two moving objects, requiring careful unit conversion, coordinate setup, and algebraic manipulation.

📌 Key points (3–5)

• Core skill: compute distance between two points using the Pythagorean theorem and track changing positions over time.
• Motion problems: set up coordinate systems for objects moving in perpendicular directions (north/south, east/west) and find when they reach specific distances apart.
• Unit conversion: convert between feet/second, miles/hour, hours/minutes/seconds to ensure consistent units in formulas.
• Common confusion: distinguish between Δx and Δy depending on which point is "beginning" vs "ending"; the distance formula works regardless of order, but the signs of Δx and Δy change.
• Algebraic manipulation: solve for time or position by setting distance expressions equal to target values and isolating the variable.

📐 Distance formula fundamentals

📏 Computing distance and deltas

Distance between points P = (x₁, y₁) and Q = (x₂, y₂): d = square root of [(x₂ − x₁)² + (y₂ − y₁)²]

• Δx = change in x-coordinate = (ending x) − (starting x)
• Δy = change in y-coordinate = (ending y) − (starting y)
• The distance formula is symmetric: switching which point is "beginning" vs "ending" changes the signs of Δx and Δy but does not change d (because squaring eliminates sign differences).

Example (Problem 2.2): For M = (a, b) and N = (s, t), all three formulas are equivalent:

• d = √[(a − s)² + (b − t)²]
• d = √[(a − s)² + (t − b)²]
• d = √[(s − a)² + (t − b)²]

🔄 Order matters for deltas, not distance

• If M is beginning and N is ending: Δx = s − a, Δy = t − b
• If N is beginning and M is ending: Δx = a − s, Δy = b − t
• Don't confuse: the distance d is always positive and the same regardless of order; Δx and Δy can be positive, negative, or zero and their signs depend on direction.

📍 Special cases

• Δx = 0: the two points have the same x-coordinate, so they lie on a vertical line.
• Δy = 0: the two points have the same y-coordinate, so they lie on a horizontal line.

🚗 Motion in perpendicular directions

🧭 Setting up coordinates

• Typical setup: place the origin at the starting point or intersection; one object moves along the x-axis, the other along the y-axis.
• Position as a function of time: if an object moves at constant speed v in one direction, its coordinate changes by v·t after time t.

Example (Problem 2.3): Steve walks north at 3 mph starting at 6 AM; Elsie walks west at 3.5 mph starting at 8 AM. Place origin at the starting point. After Elsie starts, Steve has already walked for 2 hours (6 miles north). At time t hours after Elsie starts:

• Steve's position: (0, 6 + 3t)
• Elsie's position: (−3.5t, 0)
• Distance between them: d = √[(3.5t)² + (6 + 3t)²]

🕒 Time offsets and unit conversion

• When objects start at different times, adjust the time variable accordingly.
• Convert all speeds to the same units before computing distance.

Example (Problem 2.4): Erik's speed is given in feet/second; the ferry's speed is in miles/hour. Convert one to match the other:

• 1 mile = 5,280 feet
• 1 hour = 3,600 seconds
• Erik at 10 ft/sec = (10 ft/sec) · (1 mile / 5,280 ft) · (3,600 sec / 1 hr) ≈ 6.82 mph

📊 Tracking positions over time

Problems often ask for a table of positions at specific times (e.g., 0 sec, 30 sec, 7 min, t hours). Fill in by substituting the time into each object's position formula.

Time	Object 1 position	Object 2 position	Distance between
0 sec	starting point	starting point	0
t sec	(x₁(t), y₁(t))	(x₂(t), y₂(t))	√[(x₂−x₁)² + (y₂−y₁)²]

🎯 Solving for specific conditions

⏱️ When does distance equal a target value?

Set the distance expression equal to the target and solve for t.

Example (Problem 2.5c): Two cars leave an intersection; one goes north at 30 ft/sec, the other west at 58 ft/sec. When are they 1 mile apart?

• After t seconds: north car at (0, 30t), west car at (−58t, 0)
• Distance: d = √[(58t)² + (30t)²] = √[3,364t² + 900t²] = √[4,264t²] = t·√4,264
• Set t·√4,264 = 5,280 feet (1 mile) and solve for t.

📍 When do coordinates coincide or satisfy conditions?

• x-coordinates equal: set x₁(t) = x₂(t) and solve for t.
• y-coordinates equal: set y₁(t) = y₂(t) and solve for t.
• One coordinate reaches a value: set x(t) = target value and solve for t.

Example (Problem 2.8b): Spider at s(t) = (1 + 2t, 2 + t), ant at a(t) = (15 − 2t, 2t).

• Spider's x-coordinate equals 5: 1 + 2t = 5 → t = 2 minutes.
• Ant's y-coordinate equals 5: 2t = 5 → t = 2.5 minutes.

🐜 Which object reaches a point first?

Solve for the time each object reaches the point, then compare.

Example (Problem 2.8f): Sugar cube at (9, 6). For the spider: 1 + 2t = 9 and 2 + t = 6 both give t = 4. For the ant: 15 − 2t = 9 and 2t = 6 both give t = 3. The ant reaches the sugar cube first.

🧮 Algebraic techniques

✅ Verifying solutions

Problem 2.11 asks you to check whether given algebraic steps are correct, even if the final answer happens to be right.

• Check each step: does the manipulation follow algebra rules?
• Common errors: incorrect sign handling, invalid cancellation, misapplying exponent rules.

Example (Problem 2.11b): The claim (x + y)² − (x − y)² = (x² + y²) − (x² − y²) = 0 is wrong. The correct expansion is:

• (x + y)² = x² + 2xy + y²
• (x − y)² = x² − 2xy + y²
• Difference = 4xy, not 0.

🔢 Solving with parameters

When constants (α, β) appear, treat them as fixed numbers and isolate x.

Example (Problem 2.12a): Solve αx + β = 1 / (αx − β).

• Multiply both sides by (αx − β): (αx + β)(αx − β) = 1
• Expand: α²x² − β² = 1
• Solve: x² = (1 + β²) / α², so x = ±√[(1 + β²) / α²]

🧩 Simplifying expressions

• Expand squares: (a + b)² = a² + 2ab + b²
• Combine like terms under a square root: √[A² + B²] cannot be simplified to A + B.
• Combine fractions: find a common denominator.

Example (Problem 2.13c): 1/(t − 1) − 1/(t + 1) = [(t + 1) − (t − 1)] / [(t − 1)(t + 1)] = 2 / (t² − 1).

🚤 Optimization and path problems

🛶 Minimizing travel time

Problem 2.7 involves Brooke paddling at 2 mph and walking at 4 mph to reach a destination. The total time depends on where she beaches the kayak.

• Set up: let x = distance along shore from point A to where she lands.
• Paddle distance: use Pythagorean theorem with the 5-mile offshore distance and x.
• Walk distance: remaining distance along shore.
• Total time = (paddle distance / paddle speed) + (walk distance / walk speed).
• Compare times for different landing points (directly to A, directly to destination, halfway, etc.) to find the minimum.

🕷️ Speed along a line

Problem 2.8g asks for the speed of each bug. If position is (x(t), y(t)), speed = distance traveled per unit time.

• Compute the distance between s(0) and s(1) for the spider, and between a(0) and a(1) for the ant.
• The bug with the larger distance per minute is moving faster.

🚗 Collision problems

Problem 2.9: find the Ferrari's speed so that it reaches the intersection at the same time as the Mercedes.

• Mercedes time to intersection = 400 feet / (32 mph converted to ft/sec).
• Ferrari time to intersection = 624 feet / (Ferrari speed in ft/sec).
• Set the two times equal and solve for Ferrari speed.

🧭 Overview

🧠 One-sentence thesis

Horizontal and vertical lines are the simplest curves in the plane, and each can be described by a single equation that tests whether a point lies on the line.

📌 Key points (3–5)

• Fundamental question for curves: Can we give a condition (an equation) that tells us precisely when a point lies on a curve?
• Horizontal lines: A horizontal line through k on the y-axis is the graph of the equation y = k; x can be any real number.
• Vertical lines: A vertical line through h on the x-axis is the graph of the equation x = h; y can be any real number.
• Common confusion: Horizontal lines constrain only y (not x), while vertical lines constrain only x (not y).
• Graph of an equation: The set of all solution points (x, y) that make the equation true when plotted in the xy-coordinate system.

📐 The core idea: curves and equations

🎯 The fundamental question

• When confronted with a curve in the plane, we always try to answer: Can we give a condition that tells us precisely when a point lies on the curve?
• The condition typically involves an equation in two variables (like x and y).
• This chapter focuses on the three simplest situations: horizontal lines, vertical lines, and circles.

🔗 The link between geometry and algebra

• The connection is achieved by plotting all solutions (x, y) of an equation in the xy-coordinate system.
• If you take any point on the curve, plug its coordinates into the equation, and you get a true statement, then that point is a solution.
• The set of all solutions is called the graph of the equation.

↔️ Horizontal lines

📏 What a horizontal line looks like

• A horizontal line is parallel to the x-axis.
• Example: A line passing through 2 on the y-axis is a horizontal line ℓ that passes through the point (0, 2).

🧪 The equation test for horizontal lines

A horizontal line ℓ passing through k on the y-axis is precisely a plot of all solutions (x, y) of the equation y = k.

• Set notation: ℓ = {(x, 2) | x is any real number} means "the set of all points (x, 2) where x equals any real number."
• The equation y = 2 does not involve the variable x and only constrains y to equal 2.
• Therefore, x can take on any real value, while y is fixed at 2.
• Example: Points (−1, 2), (0, 2), (2, 2), and any (x, 2) all lie on the line because they satisfy y = 2.

✅ How the test works

• Take any point on the horizontal line.
• Plug its coordinates into the equation y = k.
• You get a true statement, confirming the point is on the line.

↕️ Vertical lines

📏 What a vertical line looks like

• A vertical line is perpendicular to the x-axis (or parallel to the y-axis).
• Example: A line passing through 3 on the x-axis is a vertical line m that passes through the point (3, 0).

🧪 The equation test for vertical lines

A vertical line m passing through h on the x-axis is precisely a plot of all solutions (x, y) of the equation x = h.

• Set notation: m = {(3, y) | y is any real number} means "the set of all points (3, y) where y equals any real number."
• The equation x = 3 does not involve the variable y and only specifies that x = 3.
• Therefore, y can take on any real number value, while x is fixed at 3.
• Example: Points (3, −2), (3, 0), (3, 3), and any (3, y) all lie on the line because they satisfy x = 3.

✅ How the test works

• Take any point on the vertical line.
• Plug its coordinates into the equation x = h.
• You get a true statement, confirming the point is on the line.

🔄 Comparing horizontal and vertical lines

Line type	Passes through	Equation	Which variable is free?	Which variable is fixed?
Horizontal	k on the y-axis	y = k	x (any real number)	y (equals k)
Vertical	h on the x-axis	x = h	y (any real number)	x (equals h)

⚠️ Don't confuse

• Horizontal lines (y = k) constrain only the y-coordinate; the x-coordinate can be anything.
• Vertical lines (x = h) constrain only the x-coordinate; the y-coordinate can be anything.
• The variable that appears in the equation is the one that is fixed; the other variable is free.

🧭 Overview

🧠 One-sentence thesis

A circle of radius r centered at (h, k) is precisely the graph of all solution points (x, y) that satisfy the equation (x − h)² + (y − k)² = r².

📌 Key points (3–5)

• Core idea: A circle is the set of all points at a fixed distance r from a center point, and this geometric definition translates directly into an algebraic equation.
• Standard form equation: (x − h)² + (y − k)² = r² describes a circle with center (h, k) and radius r.
• Common confusion: Watch the minus signs—an equation like (x + 3)² means the center x-coordinate is −3, not +3.
• Connection between geometry and algebra: Plotting all solutions (x, y) of the circle equation gives the geometric circle; conversely, every point on the geometric circle satisfies the equation.

🔗 From geometry to algebra

🔗 Distance formula foundation

The excerpt starts with the distance formula applied to the origin:

• For any point Q = (x, y) and the origin P = (0, 0), the distance is √(x² + y²).
• A point (x, y) is at distance r from the origin if and only if r = √(x² + y²).
• Squaring both sides gives x² + y² = r².

This shows:

{ (x, y) | distance from (x, y) to origin is r } = { (x, y) | x² + y² = r² }

📐 Visualizing the circle

The excerpt uses a physical analogy:

• Fasten a pencil to one end of a non-elastic string of length r.
• Tack the other end to the origin.
• Holding the string tight and moving the pencil around traces all points at distance r from the origin.
• This physical process produces the same set of points as the equation x² + y² = r².

Example: The excerpt verifies that (−r/√2, r/√2) is a solution by calculating (−r/√2)² + (r/√2)² = r²/2 + r²/2 = r².

📦 Standard form of a circle

📦 General center and radius

The same reasoning extends to any center point:

Definition 3.2.1 (Circles): Let (h, k) be a given point in the xy-plane and r > 0 a given positive real number. The circle of radius r centered at (h, k) is precisely all of the solutions (x, y) of the equation (x − h)² + (y − k)² = r²; i.e., the circle is the graph of this equation.

• This equation is called the standard form of the equation of a circle.
• From standard form, you can immediately read off both the center (h, k) and the radius r.

⚠️ Minus sign caution

Don't confuse: The standard form uses (x − h), not (x + h).

• If you see (x + 3)², rewrite it as (x − (−3))² to identify the center x-coordinate as −3.
• Example from the excerpt: (x + 3)² + (y − 1)² = 7 is NOT in standard form.
  • Rewrite: (x − (−3))² + (y − 1)² = (√7)²
  • This describes a circle of radius √7 centered at (−3, 1).

The excerpt emphasizes: Be very careful with the minus signs "−" in the standard form for a circle equation.

🧮 Worked examples

🧮 Unit circle

The circle of radius 1 centered at the origin is the graph of x² + y² = 1.

• This is called the unit circle and will be used extensively (according to the excerpt).

🧮 Circle with center (1, −1)

A circle of radius 3 centered at (h, k) = (1, −1):

• Standard form: (x − 1)² + (y − (−1))² = 3²
• Equivalently: (x − 1)² + (y + 1)² = 9
• Equivalently (expanded): x² + y² − 2x + 2y = 7

All three forms describe the same circle; standard form is most useful for reading center and radius directly.

🧮 Testing whether a point lies on a circle

The circle of radius √5 centered at (2, −3) does not pass through the origin.

• Why? Because (0, 0) is not a solution of (x − 2)² + (y + 3)² = 5.
• Checking: (0 − 2)² + (0 + 3)² = 4 + 9 = 13 ≠ 5.

Example: To verify a point lies on a circle, plug its coordinates into the equation and check whether you get a true statement.

🔄 Graph of an equation

🔄 What "graph" means

The excerpt defines:

The set of all solutions of the equation is the graph of the equation.

• For a circle equation, the graph is the geometric circle.
• For any equation in x and y, plotting all solution points (x, y) in the xy-coordinate system gives the graph.
• This is the fundamental link between algebra (equations) and geometry (curves).

🔄 Equation ↔ Geometry correspondence

Geometric object	Equation	What it means
Circle, radius r, center origin	x² + y² = r²	All points at distance r from (0, 0)
Circle, radius r, center (h, k)	(x − h)² + (y − k)² = r²	All points at distance r from (h, k)

The excerpt shows that these two descriptions—geometric (distance) and algebraic (equation)—are equivalent via Equation 3.1.

🧭 Overview

🧠 One-sentence thesis

Finding where two curves intersect means solving their equations simultaneously to determine the exact coordinates where the curves cross each other.

📌 Key points (3–5)

• What "intersect" means: the place where two curves cut into or cross one another.
• Visual vs algebraic: pictures help decide whether curves intersect; algebra finds the exact coordinates of intersection points.
• Intersection patterns: two horizontal lines (or two vertical lines) never intersect; a horizontal and vertical line always intersect once; a line and a circle can intersect in zero, one, or two points.
• Common confusion: don't confuse the circle's equation with the line's equation—you must impose the line's condition (e.g., x = 0) on the circle's equation to find where they meet.
• The core procedure: replace the simpler equation's constraint into the more complex equation, then solve for the remaining variable.

🔍 What intersection means

🔍 Etymology and definition

• The word "intersect" comes from Latin: "inter" (within or in between) + "sectus" (to cut).
• Intersection: the place where two curves cut into each other; where they cross one another.

🖼️ Visual vs algebraic approaches

• If pictures of two curves are given, you can often visually decide whether they intersect.
• Drawing a picture is a good first step for any physical problem.
• However, to find the explicit coordinates of intersection points, you need algebra.

📐 Intersection patterns for simple curves

📐 Lines intersecting lines

Type of lines	Intersection behavior
Two different horizontal lines	Never intersect
Two different vertical lines	Never intersect
One horizontal + one vertical	Always intersect exactly once

• Example: A horizontal line y = k and a vertical line x = h intersect at the single point (h, k).

🔵 Lines intersecting circles

• A horizontal or vertical line may or may not intersect a given circle.
• Three possibilities:
  • Two points (line crosses through the circle)
  • One point (line is tangent to the circle)
  • No points (line misses the circle entirely)
• Don't confuse: the number of intersections depends on the relative position of the line and circle—you cannot tell without calculation or a careful diagram.

🛠️ The procedure for finding intersection coordinates

🛠️ Set up a system of equations

• Each curve has its own equation.
• To find intersections, you must simultaneously solve both equations.
• Example: To find where a circle intersects the y-axis, you have:
  • Circle equation: (x − h)² + (y − k)² = r²
  • y-axis equation: x = 0

🔄 Impose one equation's condition on the other

• Take the simpler equation (often a line like x = 0 or y = 0) and replace that variable in the more complex equation.
• This "imposes the conditions of the second equation on the first equation."
• Example: Replace x = 0 in the circle equation to get (0 − h)² + (y − k)² = r², then solve for y.

✅ Solve for the remaining variable

• After substitution, you have one equation in one variable.
• Solve it (often by isolating a squared term, then taking ± square root).
• You may get zero, one, or two solutions, corresponding to the number of intersection points.
• Example: (y − 40)² = 3,200 gives y = 40 ± √3,200, yielding two y-values and thus two intersection points.

🌆 Worked example: street light illumination

🌆 Problem setup

• Scenario: A halogen street light on a 50-foot pole creates a circular illuminated area 120 feet in diameter.
• The pole is located 20 feet east and 40 feet north of the intersection of Parkside Ave. (north/south) and Wilson St. (east/west).
• Question: What portion of each street is illuminated?

🗺️ Coordinate system and circle equation

• Impose a coordinate system with the street intersection at the origin; units are feet.
• The illuminated disc is centered at (20, 40) with radius r = 60 feet (diameter 120 feet).
• Circle equation in standard form:

  (x − 20)² + (y − 40)² = 3,600

📍 Finding intersection with the y-axis (Parkside Ave.)

• The y-axis has equation x = 0.
• System of equations:
  • (x − 20)² + (y − 40)² = 3,600
  • x = 0
• Substitute x = 0 into the circle equation:
  • (0 − 20)² + (y − 40)² = 3,600
  • 400 + (y − 40)² = 3,600
  • (y − 40)² = 3,200
  • y − 40 = ±√3,200
  • y = 40 ± √3,200 ≈ 96.57 or −16.57
• Result: Two intersection points P = (0, 96.57) and Q = (0, −16.57).

📍 Finding intersection with the x-axis (Wilson St.)

• The x-axis has equation y = 0.
• System of equations:
  • (x − 20)² + (y − 40)² = 3,600
  • y = 0
• Substitute y = 0 into the circle equation:
  • (x − 20)² + (0 − 40)² = 3,600
  • (x − 20)² + 1,600 = 3,600
  • (x − 20)² = 2,000
  • x − 20 = ±√2,000
  • x = 20 ± √2,000 ≈ 64.72 or −24.72
• Result: Two intersection points S = (64.72, 0) and R = (−24.72, 0).

🎯 Interpretation

• The illuminated portion of Parkside Ave. (y-axis) runs from Q to P.
• The illuminated portion of Wilson St. (x-axis) runs from R to S.
• Notice: two solutions for each axis correspond to the circle crossing each street at two points.

🧭 Overview

🧠 One-sentence thesis

Horizontal lines, vertical lines, and circles each have standard algebraic equations that allow us to describe their geometry and find intersection points systematically.

📌 Key points (3–5)

• Standard forms: horizontal lines are y = c, vertical lines are x = c, and circles are (x − h)² + (y − k)² = r².
• Circle parameters: (h, k) is the center and r is the radius in the circle equation.
• Finding intersections: to find where a circle meets a line, substitute the line's equation into the circle's equation and solve simultaneously.
• Common confusion: don't confuse the center coordinates (h, k) with the radius r—the center locates the circle, the radius sizes it.

📐 Standard equation forms

📏 Horizontal lines

Every horizontal line has equation of the form y = c.

• c is a constant value.
• The line runs parallel to the x-axis at height c.
• Example: y = 5 is a horizontal line 5 units above the x-axis.

📏 Vertical lines

Every vertical line has equation of the form x = c.

• c is a constant value.
• The line runs parallel to the y-axis at position c.
• Example: x = −3 is a vertical line 3 units to the left of the y-axis.

⭕ Circles

Every circle has equation of the form (x − h)² + (y − k)² = r² where (h, k) is the center of the circle, and r is the circle's radius.

• The center (h, k) tells you where the circle is located in the coordinate plane.
• The radius r tells you how large the circle is (distance from center to any point on the circle).
• Example: (x − 20)² + (y − 40)² = 3600 describes a circle centered at (20, 40) with radius 60 (since r² = 3600 means r = 60).

Don't confuse: the center coordinates (h, k) appear with subtraction in the equation; the radius r appears squared on the right side.

🔍 Finding intersection points

🔍 General approach

The excerpt demonstrates a systematic method for finding where a circle intersects horizontal or vertical lines:

1. Write down both equations (the circle equation and the line equation).
2. Simultaneously solve by substituting the line's condition into the circle equation.
3. Solve the resulting equation for the remaining variable.
4. Interpret the solutions as intersection points.

🔢 Circle with the y-axis

To find where a circle meets the y-axis:

• The y-axis has equation x = 0.
• Substitute x = 0 into the circle equation (x − h)² + (y − k)² = r².
• Solve for y.

Example from the excerpt: For the circle (x − 20)² + (y − 40)² = 3600 and the y-axis (x = 0):

• Substitute: (0 − 20)² + (y − 40)² = 3600
• Simplify: 400 + (y − 40)² = 3600
• Solve: (y − 40)² = 3200, so y − 40 = ±√3200
• Results: y = 40 + √3200 ≈ 96.57 or y = 40 − √3200 ≈ −16.57
• Intersection points: P = (0, 96.57) and Q = (0, −16.57)

Why two solutions: A circle typically crosses a line at two points (unless it's tangent or doesn't intersect at all).

🔢 Circle with the x-axis

To find where a circle meets the x-axis:

• The x-axis has equation y = 0.
• Substitute y = 0 into the circle equation.
• Solve for x.

Example from the excerpt: For the same circle and the x-axis (y = 0):

• Substitute: (x − 20)² + (0 − 40)² = 3600
• Simplify: (x − 20)² + 1600 = 3600
• Solve: (x − 20)² = 2000, so x − 20 = ±√2000
• Results: x = 20 + √2000 ≈ 64.72 or x = 20 − √2000 ≈ −24.72
• Intersection points: S = (64.72, 0) and R = (−24.72, 0)

📊 Summary comparison

Curve type	Standard equation	Key parameters
Horizontal line	y = c	c = height
Vertical line	x = c	c = position
Circle	(x − h)² + (y − k)² = r²	(h, k) = center, r = radius

Why this matters: These standard forms let you quickly identify the curve type and extract geometric information; they also provide the starting point for finding intersections with other curves.

🧭 Overview

🧠 One-sentence thesis

These exercises apply the equations of circles, horizontal lines, and vertical lines to solve geometric problems involving real-world scenarios like moving sprinklers, Ferris wheels, and radar zones.

📌 Key points (3–5)

• Core skill practiced: writing and manipulating circle equations in the form (x − h)² + (y − k)² = r², identifying centers and radii, and finding intersection points.
• Real-world modeling: problems involve time-dependent motion (sprinklers moving, ferries traveling) combined with circular regions (puddles, radar zones, wheat fields).
• Coordinate system setup: many problems require imposing a coordinate system on a physical scenario before writing equations.
• Common confusion: distinguishing the boundary (the circle itself), interior (inside the disk), and exterior (outside the circle) when describing circular regions.
• Integration of concepts: exercises combine circles with lines (horizontal, vertical, and general), time-based motion, and geometric reasoning about intersections.

📐 Mechanical circle equations

📐 Writing circle equations from geometric data

Problems 3.1(a)–(c) focus on translating geometric descriptions into algebraic form:

• Given a center (h, k) and radius r, write (x − h)² + (y − k)² = r².
• Example: A circle of radius 3 centered at (−3, 4) becomes (x − (−3))² + (y − 4)² = 9, or (x + 3)² + (y − 4)² = 9.
• Diameter vs radius: Problem 3.1(b) gives diameter 1/2, so radius is 1/4; the equation uses r² = (1/4)² = 1/16.
• Multiple solutions: Problem 3.1(c) asks for four different circles of radius 2 through (1, 1)—any center (h, k) satisfying (1 − h)² + (1 − k)² = 4 works.

🔍 Testing points on a circle

Problem 3.1(d) gives the equation (x − 1)² + (y + 1)² = 4 and asks which points lie on the graph:

• Substitute each point's coordinates into the equation and check if the equality holds.
• Example: For (1, 1), compute (1 − 1)² + (1 + 1)² = 0 + 4 = 4 ✓; for (0, 0), compute (0 − 1)² + (0 + 1)² = 1 + 1 = 2 ≠ 4 ✗.
• This reinforces that the equation describes exactly the set of points at distance r from the center.

🔄 Completing the square to find center and radius

Problem 3.2 presents circle equations in expanded form (e.g., x² − 6x + y² + 2y − 2 = 0) and asks for the center and radius:

• Group x-terms and y-terms separately, then complete the square for each variable.
• Example: x² − 6x becomes (x − 3)² − 9; y² + 2y becomes (y + 1)² − 1.
• Rewrite the equation in standard form to read off (h, k) and r.
• This skill is essential for converting between algebraic and geometric representations.

🚰 Time-dependent motion and circular regions

🚰 Expanding circular puddle (Problem 3.3)

A broken water main creates a circular puddle with radius r = 5t feet (t in seconds):

• The puddle's boundary is a circle centered at the intersection with time-varying radius.
• Part (a): A runner 6 miles away runs toward the center at 17 ft/sec. Find when the runner's distance from the center equals the puddle radius.
  • Convert 6 miles to feet, write the runner's distance as a function of time, set it equal to 5t, and solve for t.
• Part (b): The runner starts 6 miles east and 5000 feet north, runs due west at 17 ft/sec. Impose coordinates with the intersection at the origin, write the runner's position (x(t), y(t)), and find when (x(t))² + (y(t))² = (5t)².
• Key idea: Combine the circle equation with parametric motion equations.

🎡 Ferris wheel and dropped objects (Problem 3.4)

A Ferris wheel of radius 60 feet is mounted on a 62-foot tower:

• Part (a): Impose a coordinate system (e.g., origin at ground level below the tower center).
• Part (b): A rider at 100 feet above ground drops an ice cream cone straight down—find where it lands (same x-coordinate, y = 0).
• Part (c): The operator stands 24 feet to one side on the ground. Find rider positions so a dropped cone lands on the operator.
  • The rider is on the circle (x − 0)² + (y − 62)² = 60²; the cone lands at (x, 0). Set x = ±24 and solve for y to find two rider positions.
• This problem practices translating a physical setup into coordinates and using the circle equation to answer geometric questions.

🚜 Moving tractor sprinkler (Problem 3.5)

A sprinkler starts 100 feet south of a 10-foot-wide sidewalk, waters a 20-foot radius, and moves north at 1/2 inch/second:

• Part (a): Impose coordinates (e.g., origin at the sprinkler's starting point or at the sidewalk). Write equations for the sidewalk's north and south boundaries (horizontal lines).
• Part (b): Find when the circle (x − 0)² + (y − position(t))² = 20² first intersects the south boundary line.
  • The sprinkler's y-coordinate is y(t) = initial y + (1/2 inch/sec × t) converted to feet.
  • Solve for t when the circle's southern edge (y(t) − 20) reaches the sidewalk's south boundary.
• Parts (c)–(d): Find when the circle's northern edge clears the sidewalk's north boundary, then compute the duration water falls on the sidewalk.
• Parts (e)–(f): Sketch the watered region after a given time and calculate the total grass area watered (excluding the sidewalk).
• Don't confuse: The sprinkler's center moves, so the circle equation's center (h, k(t)) changes with time; the radius stays constant.

⛵ Ferry, sailboat, and radar zones (Problem 3.6)

⛵ Setting up coordinates and ferry path

Erik's sailboat is 3 miles east and 2 miles north of Kingston; a ferry travels from Kingston toward Edmonds (6 miles due east), then turns south toward Ballard:

• Part (a): Impose coordinates with Ballard as the origin. Translate all locations into (x, y) coordinates, then write equations for the ferry's path (two line segments).
• The ferry's motion is piecewise: first along a horizontal line (constant y), then along a vertical line (constant x).

📡 Radar zone as a circular disk

The sailboat's radar detects objects within 3 miles (a circular disk of radius 3 centered at the sailboat):

• Part (b): Write the boundary equation (x − sailboat x)² + (y − sailboat y)² = 3².
  • Interior: (x − sailboat x)² + (y − sailboat y)² < 9 (inside the disk).
  • Exterior: (x − sailboat x)² + (y − sailboat y)² > 9 (outside the circle).
• Sketch the radar zone by finding where the line from Kingston to Edmonds intersects the circle.
• Parts (c)–(e): Determine when the ferry enters and exits the radar zone (solve for intersections of the ferry's path with the circle), then compute the time spent inside.
• Key distinction: The boundary is the circle (equality); interior and exterior use inequalities.

🌾 Combine harvesting a wheat field (Problem 3.7)

🌾 Cutting a swath through a circular field

Nora drives a combine east at 10 ft/sec toward a circular wheat field of radius 200 feet; the combine cuts a 20-foot-wide swath:

• The combine's corner "a" starts 60 feet north and 60 feet west of the field's western edge.
• Part (a): Find when corner "a" first enters the circular field (solve for when the corner's position satisfies the circle equation).
• Part (b): Find when the combine first cuts a full 20-foot swath (when both edges of the swath are inside the field).
• Part (c): Calculate the total time wheat is being cut (from first entry to last exit of the swath).
• Part (d): Estimate the area of the swath cut (a rectangular strip intersected with the circular field).
• This problem combines time-based motion, circle intersections, and geometric area estimation.

🧮 Algebraic skills review (Problem 3.8)

🧮 Solving equations and conditional solutions

Problem 3.8 reviews algebraic techniques needed for the geometric problems:

• Parts (a)–(b): Solve rational equations (clear denominators, check for extraneous solutions).
• Parts (c)–(d): Given a fixed value for one variable (e.g., x = −2 or y = 3), substitute into a circle or ellipse equation and solve for the other variable.
  • Example: If x = −2 in (x + 1)² + (y − 1)² = 10, then (−2 + 1)² + (y − 1)² = 10 → 1 + (y − 1)² = 10 → (y − 1)² = 9 → y = 4 or y = −2.
• These skills are foundational for finding intersection points of circles with horizontal or vertical lines, as mentioned in the summary from section 3.4.

🧭 Overview

🧠 One-sentence thesis

By assuming that earning power changes linearly over time, we can construct mathematical models (straight-line equations) that predict future or past income for women and men based on two historical data points.

📌 Key points (3–5)

• The problem: predict future earning power of women and men using historical income data from 1970 and 1987.
• The modeling assumption: earning power follows a straight line—if someone earns $y in year x, the point (x, y) lies on the line connecting the two known data points.
• Why "linear model": the model demands that predicted data points lie on a straight line in a coordinate system.
• Common confusion: this is an assumption that requires statistical validation in the real world; the model is not automatically correct just because we draw a line.
• The mathematical task: find equations that describe when a point (x, y) lies on the line through the given data points.

📊 Setting up the problem

📊 The data

The excerpt provides average annual income for full-time workers:

Group	1970	1987
Women	$5,616	$18,531
Men	$9,521	$28,313

• Source: 1990 Statistical Abstract of the U.S.
• Each pair of numbers gives two points in an xy-coordinate system.

🗺️ Visualizing the data

• Axes: x-axis = year, y-axis = dollars.
• Women's data points: P = (1970, 5616) and Q = (1987, 18531).
• Men's data points: R = (1970, 9521) and S = (1987, 28313).
• The excerpt plots these points on two identical coordinate systems (Figure 4.1).

🧮 The linear modeling assumption

🧮 What the assumption says

Linear model assumption: For women, if the earning power in year x is $y, then the point (x, y) lies on the line connecting P and Q. Likewise, for men, if the earning power in year x is $y, then the point (x, y) lies on the line connecting R and S.

• This means we are demanding that all predicted income values fall on a straight line.
• The excerpt emphasizes that in the real world, validating this assumption would require statistical analysis.

🎯 Why this approach

• The assumption simplifies the problem: instead of guessing arbitrary future values, we use the pattern suggested by the two known points.
• Example: if we want to predict women's income in 1995, we find the y-value on the line through P and Q when x = 1995.
• Don't confuse: the line is not "the truth"—it is a model based on an assumption that may or may not hold.

🔧 The mathematical goal

• Find equations that describe when a point (x, y) lies on one of the two lines.
• The excerpt states that the next subsection will review the mathematics needed to show that the lines are the graphs of Equations 4.1 and 4.2.

📐 The resulting equations

📐 Women's earning power equation

The excerpt provides:

y_women = [(18,531 − 5,616) / (1987 − 1970)] × (x − 1970) + 5,616
= (12,915 / 17) × (x − 1970) + 5,616

• Numerator (18,531 − 5,616) = change in income.
• Denominator (1987 − 1970) = change in years.
• The fraction represents the rate of change (slope).
• The term (x − 1970) shifts the starting point to 1970.
• Adding 5,616 accounts for the income in the base year 1970.

📐 Men's earning power equation

The excerpt provides:

y_men = [(28,313 − 9,521) / (1987 − 1970)] × (x − 1970) + 9,521
= (18,792 / 17) × (x − 1970) + 9,521

• Same structure as the women's equation.
• Numerator (28,313 − 9,521) = change in men's income.
• The slope (18,792 / 17) is steeper than women's slope (12,915 / 17), reflecting faster income growth for men over the period.

🧩 Understanding lines and equations

🧩 Key geometric fact

Important Fact: Two different points completely determine a straight line.

• This means if you know two points on a line, you can reconstruct the entire line.
• In practice: line up a ruler with the two points.
• This fact is the foundation for constructing the linear models from the data.

📏 Vertical vs non-vertical lines

The excerpt distinguishes two types of simple lines:

Line type	Equation	Description
Vertical	x = h	All points have the same x-coordinate h
Horizontal	y = k	All points have the same y-coordinate k
Sloped (non-vertical)	Involves both x and y	Both coordinates change as you move along the line

• The earning power lines are non-vertical because both year (x) and income (y) change.
• Don't confuse: vertical lines cannot be used for this model because they would mean "all incomes occur in the same year," which makes no sense for prediction.

🔢 Slope of a non-vertical line

Slope (denoted m): the ratio of change in y to change in x, calculated as m = Δy / Δx = (y₂ − y₁) / (x₂ − x₁).

• Rise: Δy = change in y going from point P to point Q.
• Run: Δx = change in x going from point P to point Q.
• Alternative names: "rise over run" or "difference quotient."
• The slope is always defined for non-vertical lines because Δx ≠ 0 (no division by zero).
• Example from the excerpt: if P = (1, 1) and Q = (4, 5), then rise = 4, run = 3, so slope m = 4/3.

⚠️ Why non-vertical matters

• For a non-vertical line, Δx is never zero, so the slope calculation is always valid.
• Vertical lines would have Δx = 0, leading to division by zero (undefined slope).
• The earning power problem uses non-vertical lines because time (x) always increases.

🧭 Overview

🧠 One-sentence thesis

A non-vertical line is completely determined by its slope (the constant rate of change of y with respect to x) and a single point, and this relationship enables us to model real-world phenomena like motion, tuition growth, and earning power through linear equations.

📌 Key points (3–5)

• Slope as rate of change: The slope m = Δy/Δx measures how y changes per unit change in x, and this rate is constant everywhere on a non-vertical line.
• Three equivalent formulas: Any non-vertical line can be expressed using two-point, point-slope, or slope-intercept form depending on what data you have.
• Common confusion—order matters: When computing slope, you must subtract coordinates in the same order (destination minus starting point) for both x and y; reversing the order in only one coordinate gives the wrong answer.
• Minimal data requirement: A linear model needs only two data points, or one point plus a slope, or an intercept plus a slope—no more data is necessary.
• Units carry through: The slope inherits units from the variables (e.g., dollars/year), making it interpretable as a real-world rate like speed or cost growth.

📐 Understanding slope and its calculation

📏 What slope measures

Slope of a line ℓ: the ratio m = Δy/Δx = (change in y)/(change in x), where Δy = y₂ − y₁ and Δx = x₂ − x₁ for two points P = (x₁, y₁) and Q = (x₂, y₂) on the line.

• Slope is also called "rise over run" because Δy is the rise and Δx is the run.
• It is a difference quotient (ratio of two differences).
• For non-vertical lines, Δx is never zero, so the division is always defined.

⚠️ Order consistency requirement

Critical rule: When computing slope from two points, you must use the same order for both coordinates.

• Correct: m = (y₂ − y₁)/(x₂ − x₁) or m = (y₁ − y₂)/(x₁ − x₂) (both give the same result).
• Wrong: m ≠ (y₂ − y₁)/(x₁ − x₂) and m ≠ (y₁ − y₂)/(x₂ − x₁) (mixing orders gives incorrect slope).

Example: For P = (1, 1) and Q = (4, 5), the rise = Δy = 4 and run = Δx = 3, so m = 4/3.

🔄 Slope is independent of point choice

• You can pick any two distinct points on the line and always get the same slope.
• Why: Similar triangles formed by different point pairs have proportional sides, so the ratios are equal.
• This freedom lets you choose convenient points (e.g., where x = 0) to simplify calculations.

📝 Three formulas for line equations

🎯 Two-point formula

When you know two points P = (x₁, y₁) and Q = (x₂, y₂):

y = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) + y₁

• Plug in the coordinates of both points to compute the slope, then use one point to anchor the equation.
• Example: For P = (1, 1) and Q = (4, 5), the equation becomes y = (4/3)(x − 1) + 1.

🎯 Point-slope formula

When you know one point (x₁, y₁) and the slope m:

y = m(x − x₁) + y₁

• This is the most direct form when slope is already known.
• Interpretation: A line is determined by a point on it and the rate of change.

🎯 Slope-intercept formula

When you know the slope m and the y-intercept b (where the line crosses the y-axis at (0, b)):

y = mx + b

• Most compact form; easiest to read off slope and intercept directly.
• The y-intercept b is the value of y when x = 0.
• You can derive this from point-slope by setting x₁ = 0 and y₁ = b.

Relationship: All three formulas describe the same line; they are algebraically equivalent but use different starting information.

🎬 Slope as rate of change

📊 Interpreting slope with units

• Slope = Δy/Δx inherits units from the numerator and denominator.
• Example: If y is in dollars and x is in years, then m has units dollars/year.
• This makes slope a rate: "the value changes by m dollars per year."

When units are the same: If both x and y have the same units, the slope is dimensionless (just a number).

🏃 Motion along a line

For an object moving at constant speed m along a straight line:

s = b + mt

• s = location at time t
• b = initial location (at t = 0)
• m = speed (rate of change of position)
• This is a linear equation in slope-intercept form with s on the vertical axis and t on the horizontal axis.

Notation in physics: Often written as s = s₀ + vt, where s₀ is initial position and v is velocity.

Important distinction: The graph of s = mt + b (a line in the ts-coordinate system) is a visual aid for the equation, not a picture of the object's physical path.

Example: Mookie starts 10 meters ahead and runs at 5 m/sec. After t seconds, his distance from the starting point is s_M = 10 + 5t.

🔧 Building and using linear models

🛠️ Minimal data requirements

A linear model is completely determined by:

1. One data point and a slope, or
2. Two data points, or
3. An intercept and a slope.

• This tells experimenters how much data to collect: two measurements suffice if the relationship is known to be linear.
• Don't confuse: More data can improve accuracy in real-world noisy situations, but mathematically only two points are needed to define the line.

📈 Making predictions with linear models

Once you have a linear equation, you can:

1. Predict y for a given x: Plug x into the equation and compute y.
2. Predict x for a given y: Set the equation equal to the desired y value and solve for x.

Example (tuition model): Given y = 180(x − 1989) + 1,827 where x is year and y is tuition in dollars:

• To find tuition in year 2000: plug in x = 2000 → y = $3,807.
• To find when tuition reaches $10,000: solve 10,000 = 180(x − 1989) + 1,827 → x ≈ 2035.

🎯 Graphical interpretation

• Draw a vertical line at the known x-value to find the corresponding y on the model line.
• Draw a horizontal line at the known y-value to find the corresponding x on the model line.
• The intersection point gives the answer.

🔀 Parallel and perpendicular lines

↔️ Parallel lines

Two non-vertical lines are parallel if and only if they have the same slope.

• Same slope means same rate of change; the lines never meet.
• Example: If line ℓ has slope m = −2/5, any parallel line also has slope −2/5.

⊥ Perpendicular lines

Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.

• If one line has slope m, a perpendicular line has slope −1/m.
• Example: If line ℓ has slope −2/5, a perpendicular line has slope 5/2.

Application—shortest distance: To find the shortest distance from a point P to a line ℓ:

1. Construct a line through P perpendicular to ℓ (use the negative reciprocal slope).
2. Find where this perpendicular line intersects ℓ.
3. Compute the distance from P to the intersection point.

🚁 Uniform linear motion with parametric equations

🎯 Parametric equations of motion

When an object moves along a line at constant speed, its position (x, y) at time t is given by:

x = a + bt
y = c + dt

• These are parametric equations: both coordinates depend on the parameter t (time).
• Four constants (a, b, c, d) need to be determined.
• Knowing the object's location at two different times provides enough information (four equations: two for each time).

🏃 Finding equations from two positions

Example: Bob runs from (2, 3) to (5, −4) in 6 seconds.

Step 1: Set t = 0 when Bob is at (2, 3).

• At t = 0: x = 2 and y = 3.
• Plug into x = a + b(0) and y = c + d(0) → a = 2, c = 3.

Step 2: At t = 6, Bob is at (5, −4).

• Plug into x = 2 + 6b and y = 3 + 6d.
• Solve: 5 = 2 + 6b → b = 1/2; −4 = 3 + 6d → d = −7/6.

Result: x = 2 + (1/2)t and y = 3 − (7/6)t.

🏃 Finding equations from position, path, and speed

When you know the starting point, the line of travel, and the speed:

1. Use the starting point to get one time-position pair (usually t = 0).
2. Pick any other point on the line in the correct direction.
3. Calculate the distance to that point and divide by speed to get the time.
4. Now you have two time-position pairs; proceed as above.

Example: Olga starts at (3, 5), runs along y = −(1/3)x + 6 at 7 m/sec away from the y-axis. Pick another point like (6, 4), find the distance √10 ≈ 3.162 meters, so time = 3.162/7 ≈ 0.452 seconds. Then solve for the four constants.

🧭 Overview

🧠 One-sentence thesis

The slope of a non-vertical line—defined as the ratio of change in y to change in x—is constant everywhere on the line and determines the line's equation together with any point on it.

📌 Key points (3–5)

• What slope measures: the ratio of vertical change (rise, Δy) to horizontal change (run, Δx) between any two points on a line.
• Slope is independent of point choice: any two distinct points on the same line yield the same slope, thanks to similar triangles.
• Three equivalent formulas: two-point, point-slope, and slope-intercept forms all describe the same line depending on what data you have.
• Common confusion: you must subtract coordinates in the same order for both x and y; reversing order in only one coordinate gives the wrong slope.
• Slope as rate of change: slope represents the constant rate at which y changes with respect to x, and units matter (e.g., dollars per year).

📐 Defining slope

📐 The slope formula

Slope of a non-vertical line ℓ: m = Δy / Δx = (y₂ − y₁) / (x₂ − x₁), where P = (x₁, y₁) and Q = (x₂, y₂) are any two points on ℓ.

• Δx = change in x going from P to Q = x₂ − x₁
• Δy = change in y going from P to Q = y₂ − y₁
• The line must be non-vertical so that Δx ≠ 0 (avoiding division by zero).

🏔️ Rise over run terminology

• Rise: another name for Δy (vertical change).
• Run: another name for Δx (horizontal change).
• Slope is often called "rise over run."
• Slope is also called a difference quotient because both numerator and denominator involve differences.

Example: If P = (1, 1) and Q = (4, 5), then rise = Δy = 4 and run = Δx = 3, so m = 4/3.

⚠️ Order matters—but consistently

• Correct: You may reverse the order in both calculations and get the same slope:
  • m = (y₂ − y₁) / (x₂ − x₁) = (y₁ − y₂) / (x₁ − x₂) = −Δy / −Δx
• Incorrect: Reversing order in only one coordinate gives the wrong slope:
  • m ≠ (y₂ − y₁) / (x₁ − x₂)
  • m ≠ (y₁ − y₂) / (x₂ − x₁)
• Don't confuse: always subtract destination minus starting point for both x and y, or starting minus destination for both.

🔁 Slope is the same everywhere

🔁 Independence from point choice

• The slope calculation does not depend on which two points P and Q you pick on the line.
• You are free to choose your favorite two points to compute the slope.

🔺 Why: similar triangles

• If you pick two different pairs of points on the same line, you form two similar right triangles.
• Similar triangles have equal ratios of corresponding sides.
• Therefore: (y₂ − y₁) / (x₂ − x₁) = (y₂* − y₁*) / (x₂* − x₁*) for any two pairs of points.

Example: Using P = (1, 1) and Q = (4, 5) gives m = 4/3; using any other two points on the same line will also give m = 4/3.

📝 Three formulas for a line

📝 Point-slope formula

Point-slope formula: y = m(x − x₁) + y₁

• Data required: one point (x₁, y₁) on the line and the slope m.
• This formula says: a point (x, y) lies on ℓ if and only if its coordinates satisfy this equation.

📝 Two-point formula

Two-point formula: y = [(y₂ − y₁) / (x₂ − x₁)](x − x₁) + y₁

• Data required: two points P = (x₁, y₁) and Q = (x₂, y₂) on the line.
• This is just the point-slope formula with the slope written out explicitly.

📝 Slope-intercept formula

Slope-intercept formula: y = mx + b

• Data required: the slope m and the y-intercept b (the y-coordinate where the line crosses the y-axis).
• The point R = (0, b) is the y-intercept.
• Derivation: plug x = 0 into the point-slope formula to find b = −mx₁ + y₁, then rewrite y = m(x − x₁) + y₁ as y = mx + b.

Formula	Data needed	Form
Two-point	Two points (x₁, y₁) and (x₂, y₂)	y = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) + y₁
Point-slope	One point (x₁, y₁) and slope m	y = m(x − x₁) + y₁
Slope-intercept	Slope m and y-intercept b	y = mx + b

🧪 Example: verifying points

Example: For the line through P = (1, 1) and Q = (4, 5), the slope is m = 4/3 and the point-slope equation is y = (4/3)(x − 1) + 1.

• Plug in x = 0: y = (4/3)(0 − 1) + 1 = −1/3, so (0, −1/3) lies on the line.
• Plug in x = 1: y = (4/3)(1 − 1) + 1 = 1, so (1, 1) lies on the line (as expected).
• Plug in x = 6: y = (4/3)(6 − 1) + 1 = 23/3, so (6, 23/3) lies on the line.
• Plug in x = −1: y = (4/3)(−1 − 1) + 1 = −5/3, so (−1, −5/3) lies on the line.
• The point (0, 0) does not satisfy the equation, so it does not lie on the line.

As a set: ℓ = {(x, (4/3)(x − 1) + 1) | x is any real number}.

📊 Slope as rate of change

📊 Constant rate interpretation

Rate of change of y with respect to x: slope = Δy / Δx = (change in y) / (change in x).

• For a line, this rate is constant—it is the same no matter where you compute it on the line.
• A line is determined by a point on the line and the rate of change of y with respect to x.

🧮 Units in rate of change

• The units of slope are (units of y) / (units of x).
• Example: If y is house value in dollars and x is time in years, and the equation is y = 10,000x + 200,000, then:
  • Slope m = 10,000 dollars/year.
  • Interpretation: the house value is changing at a rate of 10,000 dollars per year.
• If x and y have the same units, the slope is dimensionless (a pure number).

Example: After 5 years (x = 5), the house value is y = 10,000(5) + 200,000 = 250,000 dollars. The slope-intercept form shows the slope is 10,000 dollars/year.

🔗 General lines and linear equations

🔗 General form

Linear equation: Any line in the plane is the graph of an equation of the form Ax + By + C = 0, for some constants A, B, C.

• Non-vertical lines are of most interest because they can be viewed as graphs of functions (discussed in Chapter 5).
• The excerpt notes that all lines—vertical or non-vertical—can be written in this general form.

🧭 Overview

🧠 One-sentence thesis

Any line in the plane can be expressed as a linear equation of the form Ax + By + C = 0, and non-vertical lines are especially useful because they represent functions and describe constant rates of change.

📌 Key points (3–5)

• Three equation forms for non-vertical lines: two-point formula, point-slope formula, and slope-intercept formula—each suited to different given data.
• General form: every line (vertical or non-vertical) can be written as Ax + By + C = 0, called a linear equation.
• Slope as rate of change: for non-vertical lines, slope equals the constant rate of change of y with respect to x (Δy/Δx).
• Units matter: when x and y have different units (e.g., years and dollars), the slope carries those units (e.g., dollars/year); when units are the same, slope is dimensionless.
• Common confusion: distinguish the graph of an equation (a visual aid in the coordinate system) from the physical path or situation it models (e.g., motion along a line).

📐 Three formulas for non-vertical lines

📐 Two-point formula

Two-point formula: y = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) + y₁

• When to use: you know two different points P = (x₁, y₁) and Q = (x₂, y₂) on the line.
• How it works: the fraction (y₂ − y₁)/(x₂ − x₁) is the slope; the rest adjusts for the starting point.
• Example: if you have points (1, 1) and some other point, plug them in to get the equation.

📐 Point-slope formula

Point-slope formula: y = m(x − x₁) + y₁

• When to use: you know one point P = (x₁, y₁) on the line and the slope m.
• Interpretation: a line is determined by a point on the line and the rate of change of y with respect to x.
• This is the most direct form when slope is already known.

📐 Slope-intercept formula

Slope-intercept formula: y = mx + b

• When to use: you know the slope m and the y-intercept (the point where the line crosses the y-axis, (0, b)).
• Derivation: plug x = 0 into the point-slope formula to find b = −mx₁ + y₁; rearrange to get y = mx + b.
• Why it matters: b is the y-intercept, the value of y when x = 0.
• Example: in the excerpt, the line ℓ = {(x, (4/3)(x − 1) + 1) | x is any real number} can be rewritten in slope-intercept form.

🧮 General form and linear equations

🧮 The general linear equation

Linear equation: Ax + By + C = 0 for some constants A, B, C.

• Key insight: any line in the plane—vertical or non-vertical—can be written this way.
• Non-vertical lines are of most interest because they can be viewed as graphs of functions (discussed in Chapter 5).
• Vertical lines (where x is constant) cannot be written as y = ... but still fit Ax + By + C = 0 (with B = 0).

📊 Slope as rate of change

📊 Constant rate of change

Rate of change of y with respect to x: slope = Δy/Δx = (change in y)/(change in x).

• Constant: the rate of change is the same no matter where you compute the slope on the line.
• This is what makes linear relationships special: the change is uniform.
• Point-slope interpretation: a line is determined by a point on the line and the rate of change of y with respect to x.

📊 Units and interpretation

• Different units: if x and y have different units, the slope carries those units.
  • Example: y = 10,000x + 200,000 relates house value (dollars) to years owned. The slope m = 10,000 has units dollars/year, meaning the house value increases at 10,000 dollars per year.
  • After 5 years (x = 5), the value is y = 10,000(5) + 200,000 = $250,000.
• Same units: if x and y have the same units, the units cancel out in Δy/Δx, so the slope is dimensionless (just a number).
  • This occurs in chemistry and physics when comparing quantities with the same dimension.

📊 Speed as a rate of change

• Motion example: an object moves along a straight line at constant speed m.
  • Let b = initial location (starting position).
  • Let s = location at time t.
  • The equation is s = b + mt (or s = mt + b in slope-intercept form).
• Graph vs. path:
  • The graph of s = mt + b in the ts-coordinate system is a visual aid for the equation.
  • The actual path of the object (Figure 4.7 in the excerpt) is the physical motion along a line.
  • Don't confuse: the graph is not the path; it is a tool to answer questions about the motion.
• Notation in physics: speed m is often written as v (velocity), and b as s₀ (initial position at time zero), giving s = s₀ + vt.

Concept	Symbol	Meaning
Initial location	b (or s₀)	Starting position of the object
Speed	m (or v)	Constant rate of motion
Location at time t	s	Position after time t
Equation	s = b + mt	Linear relationship between position and time

• Rate of change in this context: Δs/Δt = m, the slope of the graph, represents the speed of the object.
• The s-intercept (b) is the location when t = 0.

🧭 Overview

🧠 One-sentence thesis

The slope of a non-vertical line represents a constant rate of change of y with respect to x, which allows us to model real-world phenomena like house values and motion, and to make predictions using linear equations.

📌 Key points (3–5)

• Slope as rate of change: the slope of a line equals the rate of change of y with respect to x, and this rate is constant everywhere on the line.
• Units matter: the slope carries units from both variables (e.g., dollars/year), which gives the rate of change a real-world interpretation.
• Point-slope interpretation: a line is fully determined by one point on the line plus the rate of change of y with respect to x.
• Common confusion: distinguish between the graph of an equation (a visual aid in the coordinate system) and the actual physical path or scenario it models.
• Two types of predictions: linear models let you predict (i) the output value at a given input, or (ii) the input value needed to reach a desired output.

📐 Slope as constant rate of change

📏 What slope measures

Rate of change of y with respect to x: slope = Δy / Δx = (change in y) / (change in x).

• The excerpt emphasizes that for a non-vertical line, this rate of change is constant.
• No matter where you compute the slope on the line, you get the same value.
• This constancy is what makes linear relationships special.

🔢 Point-slope interpretation

• The excerpt states: "A line is determined by a point on the line and the rate of change of y with respect to x."
• In other words, if you know one point and the slope (rate of change), you can write the equation of the entire line.
• This is the foundation of the point-slope formula.

🏷️ Units and real-world meaning

🏷️ How units work in rate of change

• When you compute Δy / Δx, the units from the numerator and denominator combine.
• Example: If y is in dollars and x is in years, then slope = Δy / Δx has units of dollars/year.
• This tells you the rate at which y changes per unit of x.

🏠 House value example

• The excerpt gives the equation y = 10,000x + 200,000, where y is house value (dollars) and x is years owned.
• The slope m = 10,000 means the house value changes at a rate of 10,000 dollars/year.
• After 5 years, x = 5, so y = 10,000(5) + 200,000 = $250,000.
• The slope-intercept form makes it easy to read off the rate of change.

🔄 When units cancel

• If both x and y have the same units, the units cancel in Δy / Δx.
• The slope becomes a unit-less number.
• The excerpt notes this occurs in chemistry and physics applications.

🏃 Motion along a line

🏃 Speed as a rate of change

• Speed is an important type of rate: it measures how fast an object's location changes over time.
• If an object moves at constant speed m along a straight line, its location s at time t is given by a linear equation.

📍 Initial location and the motion equation

• Let b = initial location of the object (where it starts).
• Let m = constant speed.
• Let s = location of the object at time t.
• The equation is: s = b + mt.
• This is a linear equation in the variables s and t, with slope m and s-intercept b.

📊 Graph vs. physical path

• Don't confuse: the graph of s = mt + b (plotted in the ts-coordinate system) is a visual aid, not the actual path of the object.
• The graph helps answer questions about the equation, which in turn tells us about the motion.
• The excerpt emphasizes this distinction: the graph is attached to the equation, not to the physical scene.

🔤 Notation in physics

• In physics texts, speed m is often written as v (for velocity).
• Initial location b is often written as s₀ (subscript "0" meaning "time zero").
• The equation becomes s = s₀ + vt.
• Also, since t represents time, only the positive t-axis is drawn (time is non-negative).

🥏 Frisbee example

Scenario: Linda, Asia, and Mookie play frisbee. Mookie starts 10 meters in front of Linda and runs at 5 m/sec. Asia starts 34 meters in front of Linda and runs at 4 m/sec. Both run directly away from Linda after she yells "go!"

• Let s_M = distance between Linda and Mookie after t seconds.
  • s_M = (initial distance) + (distance Mookie runs in t seconds) = 10 + 5t.
• Let s_A = distance between Linda and Asia after t seconds.
  • s_A = (initial distance) + (distance Asia runs in t seconds) = 34 + 4t.
• Let s_MA = distance between Mookie and Asia after t seconds.
  • s_MA = s_A − s_M = (34 + 4t) − (10 + 5t) = 24 − t meters.
• All three distances are given by linear equations of the form s = b + mt.

💰 Earning power application

💰 Building the linear model

• The excerpt returns to the motivating problem: modeling men's earning power over time.
• Two data points are chosen: R (beginning) = (1970, 9,521) and S (ending) = (1987, 28,313).
• Slope calculation:
  • Δy / Δx = (28,313 − 9,521) / (1987 − 1970) = 18,792 / 17.
• Using the point-slope formula with point R:
  • y = (18,792 / 17)(x − 1970) + 9,521.
• This line passes through both R and S.

🔮 Two types of predictions

The excerpt explains that the linear model allows two kinds of predictions:

Prediction type	What you know	What you find	Method
(i) Predict earnings at a date	The year x	The earnings y	Plug x into the equation and solve for y
(ii) Predict when earnings reach a value	The desired earnings y	The year x	Set y equal to the desired value and solve for x

📅 Predicting earnings in 1995 (type i)

• Draw a vertical line at x = 1995 up to the graph; label the intersection U.
• Draw a horizontal line through U to the y-axis.
• The y-coordinate is (18,792 / 17)(1995 − 1970) + 9,521 = 37,156.
• Conclusion: men's earning power in 1995 is predicted to be $37,156.

📅 Predicting when earnings reach $33,000 (type ii)

• We want to find x such that y = 33,000.
• Set up the equation: (18,792 / 17)(x − 1970) + 9,521 = 33,000.
• Solve for x: x = 1991.24.
• Conclusion: men's earning power will reach $33,000 at the end of the first quarter of 1991.
• Graphically: draw a horizontal line at y = 33,000, fin

🧭 Overview

🧠 One-sentence thesis

A linear model built from two data points can predict both future values at a given time and when a desired value will occur, as demonstrated by modeling men's and women's earning power from 1970 to 1987.

📌 Key points (3–5)

• Building the model: Use the first and last data points to calculate slope and apply the point-slope formula to create a linear equation.
• Two types of predictions: (i) predict earnings at a specific date by plugging in the year, or (ii) predict when earnings will reach a target value by solving for the year.
• Graphical interpretation: Vertical lines find earnings at a given year; horizontal lines find the year for a given earnings level.
• Common confusion: Don't mix up the two prediction directions—vertical line → read y-value (earnings); horizontal line → read x-value (year).
• Model requirements: A linear model needs either one point + slope, two points, or an intercept + slope (Important Fact 4.7.1).

📊 Building the linear model from data

📍 Choosing endpoints

• The excerpt uses the first and last data points from the earning power dataset to define the line.
• For men's earning power:
  • Beginning point: R = (1970, 9,521)
  • Ending point: S = (1987, 28,313)
• This approach captures the overall trend across the entire period.

📐 Calculating the slope

The slope represents the rate of change in earnings per year:

• Change in y (earnings): 28,313 − 9,521 = 18,792
• Change in x (years): 1987 − 1970 = 17
• Slope m = 18,792 / 17

Interpretation: Men's earning power increases by approximately 18,792/17 dollars per year under this model.

✏️ Writing the equation

Apply the point-slope formula using point R and the calculated slope:

y = (18,792/17)(x − 1970) + 9,521

• This equation passes through both points R and S.
• x represents the year, y represents earnings in dollars.
• The model is now ready for making predictions.

🔮 Making predictions with the model

📅 Predicting earnings at a given year

Question: What are men's earnings in 1995?

Graphical method:

• Draw a vertical line at x = 1995 up to the graph.
• Label the intersection point U.
• Draw a horizontal line through U to the y-axis.
• Read the y-coordinate.

Algebraic method:

• Substitute x = 1995 into the equation:
• y = (18,792/17)(1995 − 1970) + 9,521 = 37,156
• Answer: Point U = (1995, 37,156), so earnings are $37,156 in 1995.

📆 Predicting when earnings reach a target

Question: When will men's earning power equal $33,000?

Setup: We want a point T on the line with y-coordinate = 33,000.

Algebraic method:

• Set the equation equal to 33,000:
• (18,792/17)(x − 1970) + 9,521 = 33,000
• Solve for x: x = 1991.24
• Answer: Earnings reach $33,000 at the end of the first quarter of 1991.

Graphical method:

• Draw a horizontal line at y = 33,000.
• Label the intersection point T on the model.
• Draw a vertical line through T to the x-axis.
• Read the x-coordinate: 1991.24.
• Point T = (1991.24, 33,000).

🔄 Don't confuse the two directions

Prediction type	Given	Find	Graphical tool	Algebraic approach
Earnings at a date	Year (x-value)	Earnings (y-value)	Vertical line → read y	Plug x into equation
Date for target earnings	Earnings (y-value)	Year (x-value)	Horizontal line → read x	Set equation = target, solve for x

👥 Comparing men's and women's models

👩 Women's earning power model

The excerpt states that the women's model is:

y = (12,915/17)(x − 1970) + 5,616

• This uses the same time period (1970–1987).
• The slope 12,915/17 is smaller than men's slope 18,792/17.
• The starting point (intercept adjusted to 1970) is lower: 5,616 vs. 9,521.

❓ Are women gaining on men?

The excerpt poses this question for the exercises but does not answer it directly.

What the slopes tell us:

• Men's rate: 18,792/17 ≈ 1,105 dollars/year
• Women's rate: 12,915/17 ≈ 760 dollars/year
• Men's earnings grow faster in absolute dollars per year under these models.

🛠️ What's needed to build a linear model

📋 Three equivalent ways

The excerpt provides Important Fact 4.7.1:

A linear model is completely determined by:

1. One data point and a slope (a rate of change), or
2. Two data points, or
3. An intercept and a slope (a rate of change).

Why this matters:

• In a laboratory or data collection context, knowing the situation follows a linear model tells you how much data you need.
• You don't need to collect many points—just enough to satisfy one of the three conditions above.

🔍 Which method was used?

In the earning power problem, method 2 (two data points) was used:

• The first data point (1970) and the last data point (1987) fully determined the linear model.
• From those two points, the slope was calculated, then the equation was written.

🧭 Overview

🧠 One-sentence thesis

A linear model is completely determined by knowing just two pieces of information from a small set of possibilities, which guides how much data must be collected in practical situations.

📌 Key points (3–5)

• Core conclusion: A linear model can be fully constructed from one data point plus a slope, two data points, or an intercept plus a slope.
• Why it matters: Knowing that a situation follows a linear model affects how much data needs to be collected (e.g., in a laboratory).
• Three equivalent ways: Each of the three combinations provides enough information to determine the entire line uniquely.
• Common confusion: Don't assume you need many data points—for linear models, two points (or equivalent information) are sufficient to define the entire relationship.

📐 The three ways to determine a linear model

📍 One data point and a slope

• If you know one specific point the line passes through and the rate of change (slope), you can construct the entire linear model.
• This uses the "point-slope formula" approach.
• Example: If you know tuition was $1,827 in 1989 and it increases by $180 per year, you can build the full model.

📍 Two data points

• Knowing two points the line passes through is sufficient to determine the line completely.
• You can calculate the slope from the two points, then use either point with that slope.
• Example: Knowing tuition in 1989 and 1995 allows you to construct the linear tuition model (as shown in the application example).

📍 An intercept and a slope

• If you know where the line crosses the y-axis (the intercept) and the rate of change (slope), the model is fully determined.
• This uses the "slope-intercept form" directly.
• Example: If a line crosses the y-axis at 5 and has slope negative two-fifths, the equation is immediately y = negative two-fifths times x plus 5.

🔬 Why this matters in practice

🔬 Data collection efficiency

Important Fact: A linear model is completely determined by (1) one data point and a slope, or (2) two data points, or (3) an intercept and a slope.

• In laboratory or research contexts, knowing the situation follows a linear model changes data collection strategy.
• You don't need to gather many data points—just enough to satisfy one of the three conditions.
• This saves time and resources when studying linear relationships.

🔬 Model construction strategy

• The excerpt emphasizes thinking about "the information needed to construct that particular kind of mathematical model."
• For linear models specifically, the minimum requirement is clear: two independent pieces of information (in any of the three combinations).
• Don't confuse: This is specific to linear models; other types of mathematical models discussed later in the text will have different requirements.

📊 Comparison of the three methods

Method	What you need	What you calculate	Form used
Point + slope	One coordinate pair + rate of change	The full equation directly	Point-slope formula
Two points	Two coordinate pairs	Slope first, then equation	Two-point formula
Intercept + slope	y-intercept value + rate of change	The full equation directly	Slope-intercept form

• All three methods are equivalent in the sense that they each provide exactly enough information to uniquely determine the line.
• The choice of method depends on what information is available in the problem context.

🧭 Overview

🧠 One-sentence thesis

Linear equations model real-world situations—from tuition growth to flight paths—by using slope, intercepts, and intersection techniques to answer practical questions about time, distance, and relationships between quantities.

📌 Key points (3–5)

• Parallel and perpendicular lines: parallel lines share the same slope; perpendicular lines have slopes that are negative reciprocals of each other.
• Intersection problems: finding where two curves meet requires solving a system of equations; the quadratic formula handles equations with squared terms.
• Uniform linear motion: objects moving at constant speed along a line can be described by parametric equations (x and y each as linear functions of time t).
• Common confusion: perpendicular slope is not "opposite" but "negative reciprocal"—if one slope is m, the perpendicular slope is negative 1 divided by m.
• Practical applications: these techniques solve real problems like predicting costs, finding closest distances, and calculating travel times.

📐 Parallel and perpendicular lines

📏 Parallel lines

Two non-vertical lines in the plane are parallel precisely when they both have the same slope.

• If line ℓ has slope m, any line parallel to ℓ must also have slope m.
• Example: if ℓ has slope negative 2/5, a parallel line through y-intercept 5 is y = (negative 2/5)x + 5.
• The y-intercept can differ, but the slope must match exactly.

⊥ Perpendicular lines

Two non-vertical lines are perpendicular precisely when their slopes are negative reciprocals of one another.

• If line ℓ has slope m, a perpendicular line has slope negative 1/m.
• Example: if ℓ has slope negative 2/5, the perpendicular slope is 5/2 (flip the fraction and change the sign).
• Don't confuse: "negative reciprocal" means both flip and negate, not just one operation.

🧮 Finding parallel and perpendicular equations

The excerpt shows a worked example:

• Given points (1, 1) and (6, −1), the slope is negative 2/5.
• For a parallel line through y-intercept 5: use slope-intercept form with the same slope.
• For a perpendicular line through (4, 6): use the negative reciprocal slope 5/2 and point-slope form.

🔗 Intersecting curves and the quadratic formula

🔍 What intersection problems involve

• Finding where two curves meet means solving their equations simultaneously.
• The excerpt focuses on cases where one equation is linear and the other involves squared terms (like a circle).

🧪 The quadratic formula

Quadratic Formula: Consider the equation az² + bz + c = 0, where a, b, c are constants. The solutions are given by z = (negative b plus-or-minus square root of (b² minus 4ac)) divided by 2a.

• Use this when you have an equation with a squared variable.
• The solutions are real numbers if and only if b² minus 4ac is greater than or equal to 0.
• Example: after substituting a line equation into a circle equation, you get a quadratic in x; apply the formula to find the x-coordinates of intersection points.

📍 Crop duster example walkthrough

The excerpt presents a detailed scenario:

• A circular irrigated field (radius 1 mile, center at origin) and a crop duster flying in a straight line.
• Step 1: Find the line equation using two points (the exit point Q and the spotting point S); slope is negative 0.8, so y = negative 0.8x minus 0.8.
• Step 2: Find entry point P by solving the line equation and the circle equation x² + y² = 1 simultaneously; substitution gives a quadratic, yielding x = negative 1 or 0.2195; P is at (0.2195, negative 0.9756).
• Step 3: Calculate distance from P to Q using the distance formula (1.562 miles), then divide by speed (120 mph) to get time (47 seconds).
• Step 4: Find shortest distance from the flight path to the center by constructing a perpendicular line through the origin (slope 1.25), intersecting it with the flight path, and using the distance formula (0.6263 miles).

🎯 Shortest distance technique

• To find the shortest distance from a point to a line, construct a perpendicular line through that point.
• The intersection of the perpendicular with the original line gives the closest point.
• Then use the distance formula to measure from the point to this closest point.

🚀 Uniform linear motion

🛤️ What uniform linear motion means

When an object moves along a line in the xy-plane at a constant speed, we say that the object exhibits uniform linear motion.

• The object's location is described by a pair of linear equations in time t.
• These are called parametric equations of motion: x = a + bt and y = c + dt, where a, b, c, d are constants.

🔢 Determining the equations

• There are four constants (a, b, c, d), so you need four pieces of information.
• Typically, knowing the object's location at two different times provides enough data.
• Example: if you know position at t = 0 and t = 1, you can solve for all four constants.

⏱️ Why parametric form is useful

• It separates horizontal and vertical motion into independent equations.
• Each coordinate (x and y) changes linearly with time, making predictions straightforward.
• Example: to find where the object is at a specific time, plug that t-value into both equations.

📊 Summary table: key techniques

Technique	When to use	Key formula or rule
Parallel lines	Need a line with same direction	Same slope as original line
Perpendicular lines	Need a line at right angles	Slope is negative reciprocal
Quadratic formula	Solving equations with squared terms	z = (−b ± √(b² − 4ac)) / 2a
Shortest distance to line	Finding closest point from a point to a line	Construct perpendicular, find intersection, use distance formula
Parametric motion	Describing moving objects	x = a + bt, y = c + dt

🧭 Overview

🧠 One-sentence thesis

The quadratic formula and perpendicular-line techniques enable solving real-world problems such as finding where moving objects intersect boundaries and calculating shortest distances from points to lines.

📌 Key points (3–5)

• Quadratic formula application: solves equations of the form az² + bz + c = 0, essential for finding intersection points between curves and lines.
• Shortest distance technique: construct a perpendicular line from the point to the original line, then find the intersection to locate the closest point.
• Uniform linear motion: an object moving at constant speed along a straight line can be described by parametric equations x = a + bt and y = c + dt.
• Common confusion: perpendicular lines have slopes that are negative reciprocals (if one line has slope m, the perpendicular has slope −1/m).
• Determining motion equations: knowing an object's location at two different times provides the four pieces of information needed to find all four constants in the parametric equations.

🧮 Quadratic Formula Essentials

🔢 The formula and when solutions are real

Quadratic Formula: For the equation az² + bz + c = 0 where a, b, c are constants, the solutions are z = (−b ± √(b² − 4ac)) / (2a). The solutions are real numbers if and only if b² − 4ac ≥ 0.

• The formula gives two solutions (using ± means "plus or minus").
• The expression b² − 4ac under the square root determines whether solutions are real.
• If b² − 4ac < 0, the square root of a negative number means no real solutions exist.

🌾 Example scenario: crop duster and circular field

Setup: A crop duster flies at 120 mph, spotted 2 miles South and 1.5 miles East of a circular irrigated field (radius 1 mile, center at origin). The flight path is a straight line.

Key steps:

1. Find the flight path equation: Using two points (the Western-most exit point Q = (−1, 0) and the spotting point S = (1.5, −2)), calculate slope m = −0.8, giving line equation y = −0.8x − 0.8.
2. Find entry point P: Solve the line equation simultaneously with the circle boundary x² + y² = 1 by substituting y = −0.8x − 0.8 into x² + y² = 1, yielding x² + (−0.8x − 0.8)² = 1, which simplifies to 1.64x² + 1.28x − 0.36 = 0. Apply the quadratic formula to get x = −1 or x = 0.2195; the entry point is P = (0.2195, −0.9756).
3. Calculate time over field: Distance from P to Q is 1.562 miles; at 120 mph, time = 1.562/120 hours = 0.01302 hours = 47 seconds.

📏 Shortest Distance from Point to Line

🔧 The perpendicular-line technique

Core idea: To find the shortest distance from a point to a line, construct a perpendicular line through the point, find where it intersects the original line, then measure the distance to that intersection.

Why perpendicular: The shortest path from a point to a line is always along the perpendicular direction.

🛤️ Step-by-step method

Using the crop duster example to find shortest distance from the origin (0, 0) to the flight path y = −0.8x − 0.8:

1. Find perpendicular slope: Original line has slope m = −0.8, so perpendicular slope = −1/(−0.8) = 1.25.
2. Write perpendicular line equation: Through origin with slope 1.25: y = 1.25x.
3. Find intersection: Set −0.8x − 0.8 = 1.25x and solve: x = −0.3902, giving closest point (−0.39, −0.49).
4. Calculate distance: Using distance formula: d = √((−0.39)² + (−0.49)²) = 0.6263 miles.

Don't confuse: The closest point on the line is not necessarily one of the known points on the line; it must be found by intersection with the perpendicular.

🏃 Uniform Linear Motion

🎯 What uniform linear motion means

Uniform linear motion: An object moves along a line in the xy-plane at constant speed.

• Location is described by parametric equations: x = a + bt and y = c + dt, where t represents time.
• Four constants (a, b, c, d) must be determined.
• The motion is "uniform" (constant speed) and "linear" (straight line).

🧩 Determining the four constants

Information needed: The object's location at two different times provides exactly four pieces of information (two coordinates at each time).

Example: Bob runs from (2, 3) to (5, −4) in 6 seconds.

Step	What to do	Result
Set time reference	Let t = 0 when Bob is at (2, 3)	t measures time since leaving (2, 3)
Use first location	At t = 0: x = 2 and y = 3	From x = a + b(0) and y = c + d(0), get a = 2 and c = 3
Use second location	At t = 6: x = 5 and y = −4	From 5 = 2 + 6b and −4 = 3 + 6d, solve to get b = 1/2 and d = −7/6
Final equations	x = 2 + (1/2)t and y = 3 − (7/6)t	Can now find location at any time

✅ Verification and application

• Check your work: Plug in the known times to verify you get the correct locations.
  • Example: At t = 0, x = 2 + (1/2)(0) = 2 and y = 3 − (7/6)(0) = 3 ✓
  • At t = 6, x = 2 + (1/2)(6) = 5 and y = 3 − (7/6)(6) = −4 ✓
• Find location at any time: At t = 30 seconds, x = 2 + (1/2)(30) = 17 and y = 3 − (7/6)(30) = −32, so Bob is at (17, −32).

🧭 Overview

🧠 One-sentence thesis

Uniform linear motion describes an object moving at constant speed along a straight line in the xy-plane using parametric equations that express position as linear functions of time.

📌 Key points (3–5)

• What parametric equations are: a pair of linear equations x = a + bt and y = c + dt that describe an object's location at any time t.
• What information is needed: four constants (a, b, c, d) require four pieces of information—typically the object's location at two different times.
• How to find the equations: use known positions at specific times to solve for the constants step-by-step.
• Common confusion: when given speed and direction instead of two positions, you must first calculate a second position using distance and speed before applying the standard method.
• Why it matters: once you have the equations, you can easily calculate the object's location at any time.

📐 Core concept: parametric equations

📐 What uniform linear motion means

Uniform linear motion: when an object moves along a line in the xy-plane at a constant speed.

• The motion is described by two linear equations involving time t as the variable.
• The general form is: x = a + bt and y = c + dt, where a, b, c, d are constants.
• These are called parametric equations of motion because they define motion in terms of the parameter t.

🔢 Why four constants

• Four constants (a, b, c, d) need to be determined.
• This requires four pieces of information.
• Knowing the object's location at two points in time provides exactly four pieces of information (two x-coordinates and two y-coordinates).

🛠️ Standard method: two known positions

🛠️ Setting up the time reference

• Choose t = 0 to represent a convenient starting instant (usually when the object is at the first known position).
• Then t represents the time elapsed since that starting instant.

🧮 Finding constants a and c

• Plug in t = 0 with the first known position (x₁, y₁).
• From x = a + b(0) = a, conclude that a = x₁.
• From y = c + d(0) = c, conclude that c = y₁.
• This determines two of the four constants immediately.

🧮 Finding constants b and d

• Use the second known position (x₂, y₂) at time t = t₂.
• Plug into x = a + bt₂ and y = c + dt₂.
• Solve for b and d:
  • b = (x₂ - a) / t₂
  • d = (y₂ - c) / t₂

✅ Verification tip

• Check your equations by plugging in both known times.
• At t = 0, you should get the first position.
• At t = t₂, you should get the second position.

Example: Bob runs from (2, 3) to (5, -4) in 6 seconds. Setting t = 0 at the start gives a = 2, c = 3. At t = 6, solving 5 = 2 + 6b and -4 = 3 + 6d yields b = 1/2 and d = -7/6. His equations are x = 2 + (1/2)t and y = 3 - (7/6)t.

🔄 Alternative scenario: speed and direction given

🔄 When you don't have two positions directly

• Sometimes you know only one position, plus the speed, direction, and path.
• The strategy: use the extra information to calculate a second position, then apply the standard method.

🎯 Step-by-step approach

1. Identify the starting position (this gives you the first position at t = 0).
2. Pick any convenient point on the line of motion in the correct direction.
3. Calculate the distance from the starting position to this new point using the distance formula.
4. Find the time to reach this point by dividing distance by speed.
5. Now you have two positions at two times—proceed with the standard method.

📍 Direction matters

• "Heading away from the y-axis" or similar phrases tell you which direction along the line to choose your second point.
• Any point in the correct direction will work; the choice is for convenience.

Example: Olga starts at (3, 5) on the line y = -(1/3)x + 6, running at 7 meters per second away from the y-axis. Choosing the point (6, 4) on the line, the distance is √((6-3)² + (4-5)²) = √10 meters. Time to reach it: √10 / 7 ≈ 0.452 seconds. Now with two positions—(3, 5) at t = 0 and (6, 4) at t ≈ 0.452—apply the standard method to find a = 3, c = 5, b ≈ 6.64, d ≈ -2.21.

🧰 Practical applications

🧰 Finding position at any time

• Once you have the parametric equations, simply plug in the desired time t.
• No need to recalculate from scratch.

Example: Using Bob's equations x = 2 + (1/2)t and y = 3 - (7/6)t, at t = 30 seconds his position is x = 2 + (1/2)(30) = 17 and y = 3 - (7/6)(30) = -32, so he is at (17, -32).

🔗 Connection to other concepts

• The excerpt shows this topic follows work on intersecting curves and perpendicular lines.
• The same coordinate geometry tools (distance formula, line equations) are used throughout.

🧭 Overview

🧠 One-sentence thesis

An object moving at constant speed along a straight line can be described by parametric equations that express both x and y coordinates as linear functions of time.

📌 Key points (3–5)

• Parametric equations structure: motion is described by x = a + bt and y = c + dt, where t is time and a, b, c, d are constants.
• Two approaches to finding equations: either know the object's position at two different times, or know one position plus speed and direction information.
• Converting speed and direction to a second point: when given speed and a line equation, pick any point on the line in the correct direction, calculate the distance from the starting point, and divide by speed to find the time.
• Common confusion: the constants a and c represent the starting position (when t = 0), while b and d represent the rates of change in the x and y directions.
• Why it matters: parametric equations let you predict an object's location at any future time, which is essential for motion problems in physics and engineering.

🎯 Setting up parametric equations

📍 The general form

Parametric equations of motion: x = a + bt and y = c + dt, where t represents time and (x, y) represents the object's location at time t.

• The constants a, b, c, and d must be determined from the given information.
• This form assumes uniform linear motion: constant speed along a straight line.
• The equations are "parametric" because both x and y depend on a third variable (the parameter t).

🔢 Finding the constants a and c

• These represent the starting position when t = 0.
• Substitute t = 0 into the equations: x = a + b(0) = a and y = c + d(0) = c.
• Example: if the object starts at point (3, 5), then a = 3 and c = 5.

🔢 Finding the constants b and d

• These represent the rates of change in the x and y directions.
• You need a second point in time to determine b and d.
• Substitute the second time and position into the equations and solve for b and d.
• Example: if at t = 0.45175395 the object is at (6, 4), and you already know a = 3 and c = 5, then:
  • 6 = 3 + b(0.45175395) → b = 3 / 0.45175395 = 6.64078311
  • 4 = 5 + d(0.45175395) → d = -1 / 0.45175395 = -2.21359436

🧮 Working with speed and direction

🛤️ When you know the path and speed

• Sometimes you only know one position, plus the line the object travels along, the direction, and the speed.
• Strategy: use this information to construct a second point in time, then solve as usual.
• The excerpt demonstrates this approach in Example 4.11.2.

📏 Picking a second point on the line

• Choose any point on the line in the correct direction from the starting point.
• The excerpt emphasizes "any will do"—the choice is arbitrary as long as it's on the correct line and in the correct direction.
• Example: Olga starts at (3, 5) and runs along the line y = -1/3 x + 6 away from the y-axis; the point (6, 4) is on this line and in the correct direction.

⏱️ Finding the time to reach the second point

• Calculate the distance from the starting point to the chosen second point using the distance formula: square root of (change in x squared plus change in y squared).
• Divide this distance by the speed to get the time.
• Example: distance from (3, 5) to (6, 4) is square root of ((6 - 3) squared + (4 - 5) squared) = square root of (9 + 1) = square root of 10; at 7 meters per second, time = square root of 10 / 7 = 0.45175395 seconds.
• Don't confuse: this is the time to reach the second point, not the total time of motion.

🔄 Reducing to the standard approach

• Once you have two points with their corresponding times, the problem becomes identical to the case where you knew two positions at two times from the start.
• The excerpt states: "At this point, we are now in the same situation as in the last example."
• Solve for a, b, c, and d using the method described earlier.

📐 Summary of the method

🗺️ Step-by-step process

Step	What to do	Example from excerpt
1. Identify starting position	This gives you a and c when t = 0	Olga starts at (3, 5), so a = 3, c = 5
2. Find a second point	Either given directly, or construct from speed/direction	Choose (6, 4) on the line y = -1/3 x + 6
3. Find the time for the second point	If not given, calculate distance and divide by speed	Distance = square root of 10, speed = 7, so t = 0.45175395
4. Solve for b and d	Use the second point and time in the parametric equations	b = 6.64078311, d = -2.21359436
5. Write the final equations	Substitute all constants into x = a + bt and y = c + dt	x = 3 + 6.64078311t, y = 5 - 2.21359436t

⚠️ Key assumptions

• The motion is uniform: constant speed.
• The motion is linear: along a straight line.
• Time t is measured from a chosen starting moment (often t = 0 at the starting position).

🧭 Overview

🧠 One-sentence thesis

This collection of exercises applies linear modeling to real-world scenarios—from real estate pricing and temperature conversion to motion problems—demonstrating how lines, slopes, and intersections solve practical geometry and rate-of-change questions.

📌 Key points (3–5)

• Core skill: translating word problems into linear equations (slope-intercept form, parametric form, or point-slope form) and using algebra or geometry to answer questions.
• Real-world contexts: property sales trends, temperature scales, moving objects (people, sprinklers, trains), and geometric constraints (triangles, circles, lines of sight).
• Common confusion: distinguishing when to use a single linear equation (e.g., price vs. year) versus parametric equations (e.g., position vs. time for a moving object).
• Intersection and distance: many problems require finding where two lines meet, or computing distances from a point to a line or between moving objects.
• Why it matters: linear models are the simplest tool for approximating trends, predicting future values, and solving optimization problems (e.g., "when is the ball closest to the cup?").

📐 Writing and interpreting linear equations

📐 Forms of a line

The exercises ask you to work with lines in multiple forms:

• Slope-intercept: ( y = mx + b ), where ( m ) is slope and ( b ) is the y-intercept.
• Standard form: ( Ax + By = C ).
• Point-slope: determined by a point and a slope, or by two points.

Example: Problem 4.1 asks you to write equations for lines given various pieces of information (slope and y-intercept, two points, x-intercept and slope, etc.).

🔍 Completing a table

Problem 4.4 presents a table with columns for equation, slope, y-intercept, and two points on the line. You must fill in missing entries.

• If you know the equation, read off the slope and y-intercept directly (if in slope-intercept form).
• If you know two points, compute the slope as (change in y) / (change in x), then use point-slope form to find the equation.
• Don't confuse: a point on the line satisfies the equation; the y-intercept is the special point where x = 0.

🏡 Real-world linear models

🏡 Property sales over time

Problem 4.5 gives sales prices for Seattle and Port Townsend in 1970 and 1990.

• You model price ( y ) as a linear function of year ( x ): find the slope (change in price / change in years) and use one data point to solve for the intercept.
• The problem then asks when the two cities' prices will be equal (set the two equations equal and solve for ( x )), when one price will be $15,000 more or less than the other, and when one will be double the other.
• Example: to find when Seattle's price is double Port Townsend's, set ( y_{\text{Seattle}} = 2 \times y_{\text{Port Townsend}} ) and solve for ( x ).

🌡️ Temperature conversion

Problem 4.9 asks you to derive the linear relationship between Celsius and Fahrenheit.

• You know two points: (0°C, 32°F) and (100°C, 212°F).
• Compute the slope: (212 − 32) / (100 − 0) = 180/100 = 9/5.
• Write the equation ( F = (9/5)C + 32 ) (or solve for C in terms of F).
• Don't confuse: the slope is not 1; the two scales have different "degrees per unit."

🏃 Motion and parametric equations

🏃 Constant-speed motion

Several problems involve objects moving in straight lines at constant speeds.

• Parametric form: if an object starts at point ((x_0, y_0)) and moves toward ((x_1, y_1)) at speed ( v ), you can write
  • ( x(t) = x_0 + (\text{horizontal component of velocity}) \times t )
  • ( y(t) = y_0 + (\text{vertical component of velocity}) \times t )
• Example: Problem 4.13 has Margot walking from a point 30 feet east of a statue toward a point 24 feet north of the statue at 4 ft/sec. You first find the direction vector, then scale it to speed 4, and write ( x(t) ) and ( y(t) ).

🎯 Distance and closest approach

Many problems ask "when is the moving object closest to a fixed point?"

• Write an expression for the distance from the object's position ((x(t), y(t))) to the fixed point.
• Distance is (\sqrt{(\Delta x)^2 + (\Delta y)^2}); to minimize, you can minimize the square of the distance (to avoid the square root).
• Set the derivative to zero (calculus) or complete the square (algebra) to find the time of closest approach.
• Example: Problem 4.6 asks where a golf ball (traveling in a straight line) is closest to the cup; Problem 4.10 asks how close Pam's train gets to Paris.

🤝 Two moving objects

Problem 4.14 has Juliet and Mercutio moving at constant speeds; you must find when Mercutio reaches the y-axis, given that he passes through the same x-axis point as Juliet but 2 seconds later.

• Write parametric equations for each person.
• Use the constraint (same x-intercept, different times) to find Mercutio's speed or direction.
• Don't confuse: "passing through the same point" vs. "passing through at the same time."

📏 Geometry and intersections

📏 Triangles and areas

Problem 4.3 asks for the area of triangles formed by lines and axes.

• Find the vertices by solving for intersections (set equations equal, or set x or y to zero).
• Use the formula: area = (1/2) × base × height.
• Example: if a line ( y = mx + b ) (with ( m < 0 ) and ( b > 0 )) cuts off a triangle from the first quadrant, the vertices are (0, b), (−b/m, 0), and the origin; area = (1/2) × (−b/m) × b.

🔵 Circles and tangency

Problem 4.8 involves a grain silo (a circle in top view) and lines of sight.

• A line tangent to a circle is perpendicular to the radius at the point of tangency.
• You can compute the slope of the tangent line two ways: (1) using two points (the tangent point and the observer's position), and (2) using the perpendicularity condition (slope of tangent × slope of radius = −1).
• Set these equal to find the coordinates of the tangent point.

🧵 Bungee cord and constraints

Problem 4.7 has Allyson and Adrian connected by a bungee cord that can stretch from 30 to 90 feet.

• Write parametric equations for each person's position.
• Compute the distance between them at any time ( t ): (\sqrt{(\Delta x)^2 + (\Delta y)^2}).
• The cord reaches maximum length (90 feet) when this distance equals 90; solve for the time or Allyson's position.

🧮 Algebraic techniques

🧮 Solving for unknowns

Problem 4.2 involves an unknown constant ( \alpha ).

• Find the intersection of two lines by setting their equations equal and solving for ( x ) (or ( y )); your answer will involve ( \alpha ).
• Then impose a condition (e.g., "the x-coordinate is 10" or "the point lies on the x-axis, so y = 0") to solve for ( \alpha ).

🧮 Non-linear equations

Problems 4.15 and 4.16 include equations with square roots, fourth powers, or fractions.

• For equations like ( \frac{1}{x} - \frac{1}{x+1} = 3 ), clear denominators by multiplying through.
• For equations like ( 2 = \sqrt{(1+t)^2 + (1-2t)^2} ), square both sides to eliminate the square root, then expand and simplify.
• For ( x^4 - 4x^2 + 2 = 0 ), substitute ( u = x^2 ) to get a quadratic in ( u ), solve for ( u ), then take square roots to find ( x ).
• Don't confuse: squaring can introduce extraneous solutions; always check your answers in the original equation.

🚿 Extended application: the crawling sprinkler

🚿 Setup and coordinate system

Problem 4.12 describes a sprinkler that moves north along a hose at 1/2 inch/second, watering a 20-foot radius circle.

• Impose a coordinate system: place the origin at a convenient point (e.g., the initial sprinkler position or a corner of the sidewalk).
• The hose is a line; find its equation using the given geometry (100 feet south of the sidewalk, which is 10 feet wide).
• The southern boundary of the sidewalk is a line parallel to the hose, shifted north by 100 feet.

🚿 Tracking the wet region

After 33 minutes (1980 seconds), the sprinkler has moved north by (1/2 inch/sec) × 1980 sec = 990 inches = 82.5 feet.

• At each moment, the sprinkler waters a circle of radius 20 feet.
• The "wet portion of the sidewalk" is the union of all circles traced out as the sprinkler moves.
• To find the length of the wet portion on the southern edge, determine where the circle first touches the sidewalk and where it last touches (as the sprinkler moves north).
• Use the Pythagorean theorem or the distance formula to find these points.

🚿 Northern boundary

The northern boundary of the sidewalk is 10 feet north of the southern boundary.

• If the southern boundary has equation ( y = mx + b ), the northern boundary has equation ( y = mx + (b + 10) ) (assuming the sidewalk runs east-west; adjust if the hose is not perpendicular).
• The hint mentions "properties of right triangles," suggesting you may need to use the Pythagorean theorem to relate horizontal and vertical distances.

🧭 Overview

🧠 One-sentence thesis

These exercises apply linear modeling techniques to a wide range of geometric, motion, and real-world problems, requiring students to translate scenarios into equations, find intersections, and interpret linear relationships.

📌 Key points (3–5)

• Core skill practiced: translating word problems and geometric scenarios into linear equations and coordinate systems.
• Types of problems: line equations and intersections, area calculations, motion problems with constant speed, real-world modeling (property prices, temperature conversion).
• Common confusion: distinguishing between finding an equation from geometric constraints vs. solving for unknowns in existing equations; also, parametric vs. standard form for motion problems.
• Key techniques: using slope and intercepts, setting up coordinate systems, finding intersection points by solving simultaneous equations, distance formulas.
• Why it matters: these exercises build fluency in moving between algebraic, geometric, and real-world representations of linear relationships.

📐 Line equations and geometric constraints

📐 Finding equations from conditions

Several problems ask you to write the equation of a line given various pieces of information:

• Slope and y-intercept (Problem 4.1b, 4.1g): use the form y = mx + b directly.
• Crossing points and slope (Problem 4.1h): if the line crosses the x-axis at x = 1, the point (1, 0) is on the line; combine with slope m = 1 to find the equation.
• Two points (implicit in many problems): use the two-point form or calculate slope first, then find intercept.

Example: A line with slope m = 1 passing through (1, 0) has equation y - 0 = 1(x - 1), or y = x - 1.

🔗 Intersections and unknown parameters

Problem 4.2 introduces an unknown constant α:

• You are given y = 2 - (1/2)x and y = 1 + αx.
• Find the intersection point in terms of α by setting the equations equal and solving for x and y.
• Then find specific values of α that make the intersection satisfy additional conditions (e.g., x-coordinate = 10, or lying on the x-axis).

Don't confuse: finding the intersection point (solve for x and y) vs. finding the parameter α that makes the intersection satisfy a constraint (solve for α after expressing the intersection in terms of α).

📏 Area of triangles formed by lines

Problem 4.3 asks for areas of triangles determined by lines and axes:

• Part (a): three lines (y = -(1/2)x + 5, y = 6x, and the y-axis) form a triangle; find vertices by solving pairs of equations, then compute area.
• Part (b): general formula—if b > 0 and m < 0, the line y = mx + b intersects the positive x-axis and positive y-axis, cutting off a triangle from the first quadrant; express area in terms of m and b.
• Part (c): given that a triangle has area 10, solve for the unknown slope m.

Problem part	Given	Find
4.3(a)	Three specific lines	Area of triangle
4.3(b)	General line y = mx + b	Area formula in terms of m, b
4.3(c)	Area = 10	The slope m

🏘️ Real-world linear modeling

🏘️ Property price modeling (Problem 4.5)

You are given property prices in Seattle and Port Townsend for 1970 and 1990:

• Task: find linear models relating year x and sales price y for each city.
• Use the two data points (1970, $38,000) and (1990, $175,000) for Seattle; similarly for Port Townsend.
• The model is a line through these two points.

Applications of the model:

• Predict prices in other years (1983, 1998).
• Find when prices in the two cities will be equal (solve for x when the two equations are equal).
• Find when one price is $15,000 more or less than the other (set up and solve equations with the given difference).
• Determine when one price is double the other (set one equation equal to twice the other and solve).

Don't confuse: interpolation (predicting within the data range, e.g., 1983) vs. extrapolation (predicting outside the range, e.g., 1998 or future years)—both use the same linear model, but extrapolation assumes the trend continues.

🌡️ Temperature conversion (Problem 4.9)

The relationship between Fahrenheit (°F) and Celsius (°C) is linear.

• Given: 32°F = 0°C and 212°F = 100°C.
• Task: find the equation C = (expression in F) and F = (expression in C).
• Use the two points (32, 0) and (212, 100) in the (F, C) plane to find the slope and intercept.
• Apply the formula to convert -23°C to Fahrenheit.

Example: The slope is (100 - 0)/(212 - 32) = 100/180 = 5/9, so C = (5/9)(F - 32).

🏃 Motion and distance problems

🏃 Constant-speed motion along a line

Many problems involve objects moving at constant speed in straight lines:

• Problem 4.6: a golf ball travels at 10 ft/sec along a straight path; find when and where it enters/exits a circular green, and when it is closest to the cup.
• Problem 4.7: Allyson and Adrian move at different speeds (10 ft/sec and 8 ft/sec) in specified directions, connected by a bungee cord; find positions and cord length at a given time.
• Problem 4.10: Pam travels by train from Rome to Florence; find the closest distance to Paris during the trip.
• Problem 4.11: Angela runs toward Tiff at 12 ft/sec; find when Angela is closest to Mary.

Key technique:

• Set up a coordinate system.
• Write parametric equations for position as a function of time t: x(t) and y(t).
• Use the distance formula to express distance to a fixed point as a function of t.
• Minimize distance by solving for t (often involves calculus or completing the square, but the excerpt does not detail methods).

🏃 Parametric equations (Problem 4.13, 4.14)

Parametric equations: express x and y coordinates separately as functions of time t.

• Problem 4.13: Margot walks from (30, 0) toward (0, 24) at 4 ft/sec; write x(t) and y(t), then find distance to the statue at time t and solve for when distance = 28 feet.
• Problem 4.14: Juliet and Mercutio move at constant speeds; given start points and destinations, write parametric equations and solve for when Mercutio reaches the y-axis.

Don't confuse: parametric form (x and y both functions of t) vs. standard form (y as a function of x)—parametric is more natural for motion problems.

🎯 Closest approach problems

Several problems ask "when is the moving object closest to a fixed point?"

• Method: express distance as a function of time (or position parameter), then find the minimum.
• The excerpt hints at using the distance formula: distance = square root of ((x-coordinate difference)² + (y-coordinate difference)²).
• Problem 4.6(d), 4.10, 4.11 all ask for closest distance.

Example: If a ball's position is (x(t), y(t)) and the cup is at (a, b), distance d(t) = √[(x(t) - a)² + (y(t) - b)²]; find t that minimizes d(t).

🧮 Geometric and applied setups

🧮 Coordinate system imposition

Problem 4.12 (crawling tractor sprinkler) and Problem 4.8 (grain silo line of sight) require:

• Imposing a coordinate system: choose an origin and axes that simplify the problem.
• Describing positions and boundaries: write equations for lines (e.g., sidewalk edges, lines of sight).
• Using geometric properties: right triangles, perpendicularity, tangency conditions.

Problem 4.8 hint: to find the tangent line, compute slope two ways (using two points, and using perpendicularity to a radius) and set them equal.

🧮 Table completion (Problem 4.4)

Problem 4.4 asks you to complete a table relating:

• Equation
• Slope
• y-intercept
• Points on the line

Skills tested: converting between different representations of a line (equation, slope-intercept form, point-slope form); checking consistency.

Don't confuse: a point on the line vs. the y-intercept (which is the specific point where x = 0).

🧪 Algebraic problem-solving

🧪 Solving equations (Problem 4.15, 4.16)

The excerpt includes algebraic exercises:

• Rational equations (4.15a): solve 1/x - 1/(x+1) = 3 by finding a common denominator.
• Equations with square roots (4.15b, c, d): solve for t in equations like 2 = √[(1+t)² + (1-2t)²]; square both sides and simplify.
• Polynomial equations (4.16a): solve x⁴ - 4x² + 2 = 0 by substituting u = x² (quadratic in u).
• Equations with square roots of variables (4.16b): solve y - 2√y = 4 by substituting u = √y.

Common technique: substitution to reduce a complicated equation to a simpler form (linear, quadratic).

Don't confuse: solving for a variable in a linear equation vs. solving a quadratic or higher-degree equation (may have multiple solutions or require factoring/quadratic formula).

🧭 Overview

🧠 One-sentence thesis

The same physical situation can be described in three interconnected ways—data tables, visual plots, and mathematical equations—with equations being the most powerful because they generate infinite data points.

📌 Key points (3–5)

• Three modes of description: any motion or relationship can be represented as tabulated data, a coordinate plot, or an equation.
• How to move between modes: data points can be plotted visually; plots can be read as data; equations generate both data and plots.
• More data reveals the pattern: increasing the number of data points (shorter time intervals) makes the underlying curve clearer.
• Common confusion: don't think of these as separate tools—they are three views of the same relationship; the equation is not "different information" but a complete prescription for all possible data points.
• Why equations are most powerful: a finite data table gives only discrete points, but an equation produces an infinite number of data points for any input value.

📊 The three descriptive modes

📋 Data tables

• A table records specific measurements at particular times.
• Example: the seagull scenario uses a table showing height (in feet) at specific times (in seconds).
• Limitation: tables are finite—they only show selected moments, not the complete picture.

📈 Visual plots

• Data points are plotted in a coordinate system (e.g., time on one axis, height on the other).
• Each point (t, s) means "at time t seconds, the height is s feet."
• The reverse process works too: any point on a plot can be read as a data entry.

🧮 Equations

An equation is a formula or prescription that tells how to produce a data point for any given input value.

• Example from the excerpt: s = 15/8 times (t minus 4) squared minus 10.
• Plugging in specific t values (0, 1, 2, etc.) produces the corresponding s values from the data table.
• Key advantage: equations generate all possible data points, not just a finite sample.

🔄 Moving between the three modes

➡️ From data to plot

• Take each pair of values from the table and mark the corresponding point in the coordinate system.
• The excerpt shows 11 data points initially, then more points at half-second and quarter-second intervals.

➡️ From equation to data and plot

• Substitute any input value into the equation to calculate the output.
• This produces as many data points as needed.
• Example: the equation in the excerpt generates all the tabulated data when t = 0, 1, 2, ..., 10.

🔍 From plot back to data

• Read coordinates directly from any point on the plot.
• Interpretation: a point (t, s) on the plot means "the quantity is s at time t."

🎯 Why more data points matter

📉 Revealing the underlying pattern

• With only a few data points, the pattern is unclear.
• As the number of points increases (by using shorter time intervals), the points "fill in" a recognizable shape.
• The excerpt shows how the seagull data reveals a parabola shape when more points are plotted.

⏱️ Discrete vs. continuous

• Discrete data: a finite collection of measurements at specific moments.
• Continuous description: an equation that works for any input value in a range (e.g., any time between 0 and 10 seconds).
• Don't confuse: more data points help visualize the pattern, but only an equation provides truly continuous information.

💪 The power of equations

♾️ Infinite data generation

• A table with 11 entries shows only 11 moments.
• An equation can produce the value for any time t between 0 and 10 (or beyond).
• This is described as "A LOT more powerful than a finite (discrete) collection of tabulated data."

🎨 Visualization without tedium

• Creating longer and longer tables is tedious.
• With an equation, you can generate as many points as needed for any desired level of detail.
• Example: the same equation produces data at 1-second, half-second, and quarter-second intervals without rewriting anything.

🧭 Overview

🧠 One-sentence thesis

A function is a procedure that assigns exactly one unique output to each allowable input, and it can be represented through equations, tables of data, or plotted curves.

📌 Key points (3–5)

• Core idea: A function is a "procedure" that produces a unique output for any acceptable input value.
• Three representations: Functions can be described using tables of data, plots of curves, or equations.
• Equation form: When using equations, a function requires three parts: a rule y = f(x), a domain (allowed inputs), and a range (resulting outputs).
• Common confusion: Not every equation is a function—if one input produces multiple outputs, it violates the function definition.
• Why it matters: Functions let us move from finite discrete data to infinite possibilities, making predictions and modeling real-world relationships far more powerful.

🌊 From data to formulas

🌊 The seagull example

The excerpt uses a seagull hovering above a cliff to illustrate how functions emerge from data:

• Initially, height s (feet above cliff level) is recorded at specific times t (seconds).
• As more data points are collected, they "fill in" a portion of a parabola.
• Creating longer tables is tedious; what we really want is a formula (a prescription) that produces the gull's height at any given time.

🔑 The power of an equation

The equation given is: s equals fifteen-eighths times the quantity (t minus 4) squared, minus 10.

• Plugging in t = 0, 1, 2, 9, 10 produces s = 20, 6.88, −2.5, 36.88, 57.5, matching the initial tabulated data.
• The same equation generates all data points for plots using half-second and quarter-second intervals.
• Key advantage: The equation produces an infinite number of data points (any t between 0 and 10), whereas a table is finite (discrete).
• Don't confuse: The excerpt acknowledges the equation comes from mathematical modeling but focuses on how the equation relates to data and plots.

🎯 Core definition and representations

🎯 Informal definition

Definition: A function is a procedure for assigning a unique output to any allowable input.

• The key word is "procedure".
• This broad definition applies to many real-world situations where two changing quantities are related.

📊 Three ways to represent a function

Representation	How it works	Example from excerpt
Table of data	Two columns list input and output; reading across each row is the "procedure"	Seagull height at specific times
Plot of a curve	For each x on the horizontal axis, the y coordinate of point P on the curve is the output	Figure 5.4 showing a curve
Equation	Plug in an x value and use algebra to produce a unique output y	s = (15/8)(t − 4)² − 10

🌍 Real-world examples

The excerpt lists everyday relationships that can be modeled as functions:

• Postage cost related to weight of an item.
• Investment value depending on time elapsed.
• Cell population in a growth medium related to time.
• Speed of a chemical reaction related to temperature.

In all cases, we want a procedure to assign a unique output value to any acceptable input value.

📐 Formal definition (equation viewpoint)

📐 Three-part package

Definition 5.2.2: A function is a package consisting of three parts:

1. An equation of the form y = "a mathematical expression only involving the variable x," written as y = f(x). Each x value produces exactly one (unique) y value. The mathematical expression f(x) is called "the rule."
2. A set D of x-values we are allowed to plug into f(x), called the "domain."
3. The set R of output values f(x), where x varies over the domain, called the "range."

🔤 Independent and dependent variables

• Independent variable (x): the "input data"—we have freedom in choosing these values.
• Dependent variable (y): the "output data"—once we specify an x value, the y value is uniquely determined by the rule f(x).

⚠️ Not every equation is a function

Example: The equation x + y² = 1 is not a function in the independent variable x.

• Plugging in x = 0 gives y² = 1, which has two solutions: y = ±1.
• This violates the "unique output" condition.
• However, solving for x in terms of y gives x = 1 − y², which does define a function x = g(y) in the independent variable y.
• Don't confuse: The same equation can be a function in one variable but not in another.

📚 Examples of functions (equation form)

📚 Linear and constant functions

Example (i): y = −2x + 3

• The rule is f(x) = −2x + 3.
• If the domain is all real numbers, this defines a function.

Example (ii): Same rule f(x) = −2x + 3, but domain is all non-negative real numbers.

• This is a different function from Example (i) because the domains differ.
• This illustrates a restricted domain: starting with a function on all real numbers, then restricting to a subset.

Example (iii): y = b (where b is a constant)

• The rule is f(x) = b.
• These are called constant functions.
• The graph is a horizontal line; if b = 0, it's the horizontal axis.

📚 Functions with restricted domains

Example (iv): y = 1/x

• The rule f(x) = 1/x defines a function as long as we do not plug in x = 0.
• Domain: all non-zero real numbers.

Example (v): y = square root of (1 − x²)

• Before plugging in x values, ensure the expression under the radical is non-negative (so the square root is a real number).
• Solve the inequality: 0 ≤ 1 − x², which gives −1 ≤ x ≤ 1.
• Domain: −1 ≤ x ≤ 1.

📏 Interval notation

The excerpt provides a table summarizing interval notation on the number line:

• All numbers between a and b, possibly equal to either: a ≤ x ≤ b
• All numbers between a and b, not equal to either: a < x < b
• All numbers between a and b, not equal to b, possibly equal to a: a ≤ x < b
• All numbers between a and b, not equal to a, possibly equal to b: a < x ≤ b

Typically, the domain is the entire number line, an interval, or a finite union of intervals.

🔧 Conceptual viewpoint: function as a process

🔧 The "black box" metaphor

Conceptually, think of a function as a process:

• An allowable input goes into a "black box."
• Out pops a unique new value denoted f(x).
• The excerpt compares this to a machine making hula-hoops: tubes go in, hoops come out.

This viewpoint is useful when a function is described in words rather than equations.

🔧 Word-based examples

Example (i): Total water used by a household since midnight

• Let y be the total gallons of water used between 12:00am and time t (hours).
• Given a time t, the household will have used a specific unique amount, call it S(t).
• Function: y = S(t), independent variable t, dependent variable y.
• Domain: 0 ≤ t ≤ 24.
• Range: 0 ≤ y ≤ S(24) (the largest possible value is at t = 24).

Example (ii): Height of a basketball center while dribbling

• Let s be the height at time t seconds after starting to dribble.
• At any frozen moment, the ball's center has a single unique height, call it h(t).
• Function: s = h(t).
• Domain: a given time interval, e.g., 0 ≤ t ≤ 2 (first 2 seconds).
• Range: all possible heights attained by the basketball center during those 2 seconds.

Example (iii): State sales tax on a taxable item

• Let T be the state tax (dollars) due on an item selling for z dollars.
• For a given price z, the tax due is a single unique amount, call it W(z).
• Function: T = W(z), independent variable z.
• (The excerpt cuts off here, but the pattern is clear.)

🧭 Overview

🧠 One-sentence thesis

The graph of a function is the visual plot of all solution points (x, f(x)) in the coordinate system, and the vertical line test determines whether a curve represents a function by checking that each x-value corresponds to exactly one y-value.

📌 Key points (3–5)

• What the graph is: the set of all points (x, f(x)) where x is in the domain, plotted in the xy-coordinate system.
• How to find/test points on the graph: given x in the domain, the point (x, f(x)) is on the graph; given a point (u, v), it is on the graph only if v = f(u) is true.
• The vertical line test: a curve is the graph of a function on domain D if and only if every vertical line through D intersects the curve exactly once.
• Common confusion: a point (u, v) being "near" the graph vs. actually on it—only v = f(u) guarantees the point is on the graph.
• Why it matters: the graph translates symbolic function equations into visual data, making patterns and relationships easier to see.

📊 From symbolic to visual representation

📊 What a solution point means

A point (x, y) is a solution to the function equation y = f(x) if plugging x and y into the equation gives a true statement.

• The function definition ensures that for each x in the domain, there is only one y-value: f(x).
• This means (x, f(x)) is the only solution with first coordinate x.
• Example: For y = -2x + 3, the point (2, -1) is a solution because -1 = -2·2 + 3 is true.

🗺️ Definition of the graph

Graph = {(x, f(x)) | x in the domain}

• The graph is the plot of all solutions of the equation y = f(x) in the xy-coordinate system.
• It is common to refer to this as either "the graph of f(x)" or "the graph of f."
• The excerpt shows that tabulated data (a table of x and y values) and visual data (plotted points) are two views of the same information.

🔢 Example: linear function

The excerpt uses f(x) = -2x + 3 on the domain of all real numbers:

x	y = -2x + 3	Point (x, y)
-1	5	(-1, 5)
0	3	(0, 3)
1	1	(1, 1)
2	-1	(2, -1)

• Plotting these points produces a line with slope m = -2 and y-intercept 3.
• The graph of the function f(x) = -2x + 3 is the same as the graph of the equation y = -2x + 3.

🔍 Procedure: finding and testing points on the graph

✅ How to produce a point on the graph

• Given: a value u in the domain of y = f(x).
• Result: you can immediately plot the point (u, f(u)) on the graph.
• This follows directly from the definition: every x in the domain has exactly one corresponding y = f(x).

❓ How to test whether a given point is on the graph

• Given: a point (u, v).
• Test: the point is on the graph only if v = f(u) is true.
• Don't confuse: just because u is in the domain does not mean any arbitrary v will work; you must check that v equals f(u).

🧮 Example: parabola function

The excerpt gives s = h(t) = (15/8)(t - 4)² - 10 on domain 0 ≤ t ≤ 10:

• The graph is a portion of a parabola.
• Data points from an earlier seagull example all lie on this parabola (verified by the procedure).
• The point (0, 0) is not on the graph, because h(0) = 20/6 ≠ 0.
• Example: To check if (4, -10) is on the graph, compute h(4) = (15/8)(4 - 4)² - 10 = -10, which is true, so (4, -10) is on the graph.

🧪 The vertical line test

🧪 Why the test works

• Since (x, f(x)) is the only point on the graph with first coordinate equal to x, a vertical line passing through x on the x-axis (with x in the domain) crosses the graph once and only once.
• This pictorial property is "very revealing" and gives a decisive way to test if a curve is the graph of a function.

📏 The test procedure

Vertical line test: Draw a curve in the xy-plane and specify a set D of x-values. If every vertical line through a value in D intersects the curve exactly once, then the curve is the graph of some function on domain D. If any single vertical line through some value in D intersects the curve more than once, then the curve is not the graph of a function on domain D.

• "Exactly once" means the curve passes the test.
• "More than once" (or zero times, implicitly) means the curve fails the test.

📐 Example: straight lines

• Non-vertical line: By the vertical line test, any straight line m that is not vertical is the graph of a function.
• Vertical line: If the line m is vertical, then m is not the graph of a function (because a vertical line through that x-value would intersect the curve infinitely many times, not exactly once).
• Don't confuse: a vertical line in the plane vs. the vertical lines used in the test—the test uses imaginary vertical lines to check the curve, while the curve itself may or may not be vertical.

🔵 Example: circle

The excerpt mentions the equation x² + y² = 1, whose graph is the unit circle:

• A vertical line through x = 0 intersects the circle at (0, 1) and (0, -1)—two points.
• Therefore, the circle fails the vertical line test and is not the graph of a function.

🧭 Overview

🧠 One-sentence thesis

The vertical line test provides a decisive graphical method to determine whether a curve represents a function by checking if every vertical line through the domain intersects the curve exactly once.

📌 Key points (3–5)

• What the test checks: whether each x-value in the domain corresponds to exactly one y-value on the curve.
• How to apply it: draw vertical lines through x-values in the domain; if any line crosses the curve more than once, it's not a function.
• Why it works: by definition, a function assigns only one output to each input, so the graph can have only one point per x-coordinate.
• Common confusion: vertical lines themselves fail the test (they are not graphs of functions), but non-vertical lines always pass.
• Practical use: domain constraints in physical problems may restrict which portion of a curve we test.

📐 The fundamental principle

📐 Why functions have unique y-values

Since (x, f(x)) is the only point on the graph with first coordinate equal to x, a vertical line passing through x on the x-axis (with x in the domain) crosses the graph of y = f(x) once and only once.

• A function assigns exactly one output to each input.
• On a graph, this means: for any x in the domain, there is exactly one point (x, y) on the curve.
• A vertical line at x = some value will hit all points with that x-coordinate.
• If the curve is a function graph, the vertical line can only hit one point.

🔍 The test procedure

The vertical line test: Draw a curve in the xy-plane and specify a set D of x-values. Suppose every vertical line through a value in D intersects the curve exactly once. Then the curve is the graph of some function on the domain D. If we can find a single vertical line through some value in D that intersects the curve more than once, then the curve is not the graph of a function on the domain D.

How to use it:

• Draw vertical lines through various x-values in the specified domain D.
• Count how many times each line crosses the curve.
• Passes test (is a function): every vertical line crosses exactly once.
• Fails test (not a function): at least one vertical line crosses more than once.

🧪 Examples and applications

🧪 Straight lines

Type of line	Passes test?	Reason
Non-vertical line	Yes	Any vertical line crosses it exactly once; it is the graph of a function
Vertical line	No	A vertical line intersects itself infinitely many times; it is not the graph of a function

Example: Any non-vertical line m in the plane passes the vertical line test and is the graph of a function. A vertical line m fails the test.

⭕ The unit circle

• Equation: x² + y² = 1
• Domain specified: −1 ≤ x ≤ 1
• Test result: Fails

Why it fails:

• The vertical line passing through the point (1/2, 0) will intersect the unit circle twice.
• By the vertical line test, the unit circle is not the graph of a function on the domain −1 ≤ x ≤ 1.
• Don't confuse: the circle is a valid curve, but it does not represent a function because some x-values correspond to two y-values.

🎯 The parabola example

• Function: s = h(t) = 15/8 (t − 4)² − 10
• Domain: 0 ≤ t ≤ 10
• Graph: a portion of a parabola (opens upward or downward)

Verification:

• The point (0, 0) is NOT on the graph, since h(0) = 20 ≠ 0.
• This shows that you must check actual function values, not just assume points lie on the graph.

🔒 Domain constraints in physical problems

🔒 Why constrain the domain

In physical problems, it might be natural to constrain (meaning to "limit" or "restrict") the domain.

• Real-world contexts often make only part of a mathematical graph physically meaningful.
• Example: time cannot be negative, heights cannot be negative, etc.

⚾ Ball height example

• Function: s = h(t) = −16t² + 4
• Physical meaning: height s (in feet) of a ball above the ground after t seconds.
• Graph: a parabola in the ts-plane.

Natural constraint:

• Only consider the portion of the graph in the first quadrant (where both t ≥ 0 and s ≥ 0).
• Why? Negative time doesn't make sense, and negative height (below ground) is not physically relevant for a ball in the air.
• Domain restriction: 0 ≤ t ≤ 1/2 (the ball hits the ground at t = 1/2).

How to specify:

• Write the constraint notationally: 0 ≤ t ≤ 1/2.
• This restricts which portion of the parabola we consider when applying the vertical line test or interpreting the function.

🧭 Overview

🧠 One-sentence thesis

Linear functions produce straight-line graphs and are defined by the rule f(x) = mx + b, where m is the slope and b is the y-intercept, making them useful for modeling constant-rate relationships like distance over time.

📌 Key points (3–5)

• Definition: A linear function has the form f(x) = mx + b on a specified domain, producing a line with slope m and y-intercept b.
• Real-world modeling: Linear functions naturally describe situations with constant rates (e.g., distance = rate × time).
• Domain matters: The domain should reflect realistic constraints (e.g., 0 ≤ t ≤ 2 for a two-hour drive).
• Multiple perspectives: The same situation can be modeled by different linear functions depending on what you're measuring (e.g., distance from start vs. distance from destination).

📐 What is a linear function

📐 The rule and its graph

Linear function: a function of the form f(x) = mx + b on some specified domain, whose graph is a line of slope m and y-intercept b.

• This builds on the equation in x and y from Chapter 4.
• The graph must be a straight line.
• m controls the steepness (slope).
• b controls where the line crosses the y-axis (y-intercept).

🔢 The form f(x) = mx + b

• m: the slope—how much y changes for each unit change in x.
• b: the y-intercept—the value of f(x) when x = 0.
• The domain specifies which x-values are allowed or meaningful.

🚗 Modeling with linear functions

🚗 Distance from a starting point

Example: Driving 65 mph from mile marker 0 (Kansas state line) to mile marker 130 (Salina) along I-35.

• Rate: 65 mph.
• Time: t hours.
• Distance traveled: Using "distance = rate × time," the distance after t hours is 65t.
• Function rule: d(t) = 65t (mile marker after t hours).
• Domain: 0 ≤ t ≤ 2, because it takes 2 hours to reach Salina (130 miles ÷ 65 mph = 2 hours).

🎯 Distance from a destination

Example: Same drive, but measuring distance from Salina instead of from the start.

• Setup: Define s(t) as your distance from Salina after t hours.
• Logic:
  • Mile marker of Salina = 130.
  • Your mile marker at t hours = d(t) = 65t.
  • Distance from Salina = 130 − 65t.
• Function rule: s(t) = 130 − 65t.
• Domain: Again 0 ≤ t ≤ 2.

🔄 Two functions, same situation

• Both d(t) and s(t) are linear functions describing the same drive.
• d(t) = 65t increases from 0 to 130 (positive slope).
• s(t) = 130 − 65t decreases from 130 to 0 (negative slope).
• Don't confuse: The same real-world scenario can yield different linear functions depending on what quantity you choose to measure.

💰 Application: Profit analysis

💰 Gross income and expense functions

Example: A software company with sales price $15 per unit and expense $100(1 + √x) to produce and market x units.

• Gross income function: g(x) = 15x (dollars from selling x units).
  • This is a linear function: slope = 15, y-intercept = 0.
• Expense function: e(x) = 100(1 + √x) (dollars to produce x units).
  • This is not a linear function (it involves √x).

📊 Profit, loss, and break-even

Condition	Meaning	Mathematical test
Profit	Gross income exceeds expenses	g(x) > e(x)
Loss	Expenses exceed gross income	e(x) > g(x)
Break-even	Zero profit and zero loss	e(x) = g(x)

📈 Visual interpretation of graphs

• Plot both g(x) and e(x) in the same coordinate system with domain 0 ≤ x ≤ 100.
• For any sales figure x, you can relate three things:
  • x on the horizontal axis.
  • A point on the graph of g(x) or e(x).
  • y on the vertical axis (the function value).
• Example at x = 20:
  • Point P = (20, g(20)) = (20, 300) on the gross income graph.
  • Point Q = (20, e(20)) = (20, 547) on the expense graph.
  • The vertical distance between P and Q represents the loss: e(20) − g(20) = 547 − 300 = 247 dollars.

🔍 Reading profit and loss from the graph

• Where e(x) is above g(x): Loss occurs; loss amount = e(x) − g(x) (length of vertical segment).
• Where the graphs cross (point B): Break-even point; e(x) = g(x), so profit = 0.
• Where g(x) is above e(x): Profit occurs; profit amount = g(x) − e(x) (length of vertical segment).

⚠️ Solving for break-even: watch for extraneous solutions

To find the break-even point B, solve g(x) = e(x):

• 15x = 100(1 + √x)
• 15x − 100 = 100√x
• Squaring both sides: 225x² − 3000x + 10000 = 10000x
• 225x² − 13000x + 10000 = 0
• Quadratic formula gives two answers: x = 0.78 or x = 57.

Problem: Which solution is correct?

• Only x = 57 is the true break-even point.
• x = 0.78 is an extraneous solution introduced by squaring both sides.
• Check: Plugging x = 0.78 into the original equation 15(0.78) ≠ 100(1 + √0.78).
• Lesson: Whenever you square both sides of an equation, you risk adding extraneous solutions; always check your answers in the original equation.

🧭 Overview

🧠 One-sentence thesis

Profit analysis uses graphical comparison of income and expense functions to identify break-even points where profit transitions from loss to gain, requiring careful algebraic solving and verification of solutions.

📌 Key points (3–5)

• Visual profit analysis: plotting income and expense graphs together reveals where loss occurs (expense above income), where break-even happens (graphs cross), and where profit occurs (income above expense).
• Measuring profit vs. loss: the vertical distance between graphs determines the amount—loss is e(x) − g(x) when expense is higher; profit is g(x) − e(x) when income is higher.
• Break-even point: found by solving g(x) = e(x), where income equals expense and profit is zero.
• Common confusion: squaring both sides of an equation can introduce extraneous solutions that don't satisfy the original equation—always check final answers by substituting back.
• Why it matters: identifies the sales volume needed to cover costs and start generating profit.

📊 Reading the profit-loss graph

📊 Three regions on the graph

The excerpt describes plotting both expense e(x) and income g(x) functions in the same coordinate system, with x representing sold units and y representing dollars.

Three distinct regions emerge:

Region	Graph relationship	Financial meaning	Distance formula
Left of B	Expense above income	Loss is realized	Loss = e(x) − g(x)
Point B	Graphs cross	Break-even (zero profit, zero loss)	Distance = 0
Right of B	Income above expense	Profit is realized	Profit = g(x) − e(x)

📏 Measuring the amount

• The length of the vertical line segment between the two graphs at any x-value shows the exact dollar amount of profit or loss.
• Don't confuse: the distance formula switches depending on which graph is higher—always subtract the lower value from the higher value to get a positive amount.

Example: If at 50 units sold the expense graph shows $600 and the income graph shows $500, the loss is $600 − $500 = $100 (the length of the segment between them).

🎯 Finding the break-even point

🎯 Setting up the equation

Break-even point B: the point where expense and income agree, so there is zero profit (and zero loss).

The analysis is complete once we "pin down" point B by solving g(x) = e(x).

The excerpt's example:

• Income function: g(x) = 15x
• Expense function: e(x) = 100(1 + √x)
• Break-even equation: 15x = 100(1 + √x)

🔧 Algebraic steps

The excerpt shows the solving process:

1. 15x = 100(1 + √x)
2. 15x − 100 = 100√x
3. 225x² − 3000x + 10000 = 10000x (both sides squared)
4. 225x² − 13000x + 10000 = 0
5. Apply quadratic formula → two answers: x = 0.78 or x = 57

⚠️ Which solution is correct?

The excerpt argues that only x = 57 gives the break-even point.

Why x = 0.78 fails:

• Plugging x = 0.78 into the original equation: 15(0.78) ≠ 100(1 + √0.78)
• The equation does not hold true

The final break-even point:

• B = (57, g(57)) = (57, 855) = (57, e(57))
• At 57 units sold, both income and expense equal $855

⚠️ Extraneous solutions warning

⚠️ Why extra solutions appear

Whenever we square both sides of an equation, we run the risk of adding extraneous solutions.

• The excerpt identifies the step from line 2 to line 3 (squaring both sides) as the source of the problem.
• Squaring can introduce solutions that satisfy the squared equation but not the original equation.

✅ How to avoid mistakes

The excerpt provides clear guidance:

After solving any equation:

1. Look back at your steps
2. Ask whether you may have added (or lost) solutions
3. Be wary when squaring or taking the square root of both sides
4. Always check your final answer in the original equation

Example: The check 15(0.78) ≠ 100(1 + √0.78) immediately reveals that x = 0.78 is extraneous, even though it satisfies the squared equation.

Don't confuse: a solution to the transformed equation is not automatically a solution to the original equation—verification is mandatory, not optional.

🧭 Overview

🧠 One-sentence thesis

This exercise set reinforces graphical analysis techniques by asking students to sketch graphs from verbal descriptions, interpret function behavior from graphs, work with multipart and absolute-value functions, and apply function concepts to real-world scenarios involving distance, cost optimization, and geometric constraints.

📌 Key points (3–5)

• Graph sketching from scenarios: translate real-world motion (walking, jumping, dribbling) and geometric constraints (cables, tunnels) into function graphs.
• Reading graphs: identify intercepts, increasing/decreasing intervals, domain/range, and whether a curve represents a function (vertical line test).
• Multipart functions: handle functions defined by different rules on different intervals; graph them by piecing together separate cases.
• Common confusion: distinguishing between "distance from A" vs. "distance from B" changes the graph; also, a curve may look like a function but fail the vertical line test.
• Algebraic mechanics: practice function evaluation (substituting symbols like ♥ or expressions like x + 3) and solving equations involving radicals and fractions.

🚶 Motion and distance graphs

🚶 Dave's walk scenarios (Problem 5.3)

• Each sub-problem describes Dave walking from Padelford Hall to Gould Hall (or back) with different speeds, pauses, or direction changes.
• The task is to sketch "distance from Padelford" vs. time.
• Key features to capture:
  • Constant speed → straight-line segment (slope = speed).
  • Pause → horizontal segment (distance stays constant).
  • Speed change → change in slope.
  • Turning back → distance decreases (downward slope).
• Example: (d) Dave walks halfway, stops, then returns → graph rises to the midpoint, flattens for 1 minute, then falls back to zero.
• Part (e) adds Angela walking in the opposite direction at twice Dave's speed; both start at the same time, so you plot two lines with different slopes.
• Part (f) asks how graphs change if you track "distance from Gould Hall" instead of "distance from Padelford" → the roles of starting and ending points swap; graphs flip or shift accordingly.

🧘 The monk problem (Problem 5.4)

A monk walks up a mountain from 5 AM to 11 AM one day, then walks down the same path from 5 AM to 11 AM the next day.

• (a) Was there necessarily a time when the monk was at the same place on both days?
  • Yes. Imagine overlaying both trips on the same graph (time vs. position). The upward journey and downward journey are continuous curves that must cross at least once (by the Intermediate Value Theorem, though the excerpt does not name it).
  • Example: if you graph both trips on the same axes, one curve rises and one falls; they must intersect.
• (b) Faster return (arrives at 9 AM): the downward curve is steeper but still crosses the upward curve at some point.
• (c) Later start (10 AM to 3 PM): the time windows differ, but the same logic applies—there is still a crossing point during the overlapping time interval.
• Don't confuse: the monk's speed or arrival time does not matter; the key is that both trips cover the same path continuously.

📊 Sketching and interpreting graphs

📊 Reasonable graphs from descriptions (Problem 5.5)

Each part asks you to sketch a function, specify domain and range, and describe largest/smallest values.

Scenario	Key features	Domain example	Range example
(a) Height vs. age	Increases rapidly in childhood, levels off in adulthood	0 ≤ age ≤ 100 years	~0.5 to ~2 meters
(b) Head height on pogo stick	Periodic up-and-down motion	0 ≤ t ≤ 5 seconds	~1 to ~2 meters
(c) Postage vs. letter weight	Step function (jumps at weight thresholds)	0 < weight ≤ some max	Discrete postage rates
(d) Toe height while biking	Periodic (pedal rotation)	0 ≤ t ≤ 10 seconds	~0 to ~0.5 meters
(e) Height after diving	Starts high, drops through zero (water surface), reaches minimum underwater, then rises	0 ≤ t ≤ a few seconds	Negative (underwater) to positive (board height)

• State assumptions: e.g., "assume constant pedaling speed" or "ignore air resistance."
• Largest/smallest values: e.g., maximum head height when pogo stick is fully extended; minimum when compressed.

🔍 Analyzing a given graph (Problem 5.6)

Given the graph of f(x) = 3x² − 3x − 2:

• (a) Intercepts:
  • y-intercept: set x = 0 → f(0) = −2 → point (0, −2).
  • x-intercepts: solve 3x² − 3x − 2 = 0 (use quadratic formula or factoring).
• (b) Points with y = 5: solve 3x² − 3x − 2 = 5 → 3x² − 3x − 7 = 0 → find exact x-values.
• (c) Points with y = −3: solve 3x² − 3x − 2 = −3 → 3x² − 3x + 1 = 0 → find exact x-values.
• (d) Check if given points are on the graph: substitute each x into f(x) and see if it equals the given y.
• (e) Find coordinates for x = √(1 + √2): compute f(√(1 + √2)) exactly.

Procedure reference: the excerpt mentions "procedure 5.3.1 on page 63" (not shown here), which likely outlines steps for finding intercepts and evaluating functions.

💰 Optimization and cost functions

💰 Island cable problem (Problem 5.7)

• Setup: island 1 km offshore; power station 4 km down the coast; cable costs $50,000/km underwater, $30,000/km overland.
• (a) Why the path is two straight segments:
  • Straight lines minimize distance (and cost) between two points.
  • Any curved path would be longer.
  • The cable should reach shore somewhere between the power station and the point directly opposite the island (otherwise you're going the wrong direction).
• (b) Cost function f(x):
  • Let x = distance from power station to where cable reaches shore.
  • Overland distance = x km → cost = 30,000x.
  • Underwater distance = √(1² + (4 − x)²) km (Pythagorean theorem) → cost = 50,000√(1 + (4 − x)²).
  • Total: f(x) = 30,000x + 50,000√(1 + (4 − x)²).
• (c) Table of values: compute f(0), f(0.5), f(1), …, f(4) and estimate the minimum cost by inspection.

Don't confuse: the variable x is measured from the power station, not from the island's nearest point.

🔧 Function mechanics and algebra

🔧 Evaluating functions (Problem 5.8)

For each function, compute f(0), f(−2), f(x + 3), f(♥), f(♥ + △).

• (a) f(x) = ½(x − 3):
  • f(0) = ½(0 − 3) = −3/2.
  • f(−2) = ½(−2 − 3) = −5/2.
  • f(x + 3) = ½((x + 3) − 3) = ½x.
  • f(♥) = ½(♥ − 3).
  • f(♥ + △) = ½((♥ + △) − 3) = ½(♥ + △ − 3).
• (b) f(x) = 2x² − 6x: substitute each input into the rule.
• (c) f(x) = 4π²: this is a constant function → f(anything) = 4π².

Key idea: treat ♥ and △ as variables; substitute them just like you would substitute x or any number.

🔧 Solving equations (Problem 5.10)

Each part asks for an exact answer (no decimals).

• (a) Rational equation: x/(x + 3) + 5/(x − 7) = 30/(x² − 4x − 21).
  • Factor denominator: x² − 4x − 21 = (x + 3)(x − 7).
  • Multiply through by (x + 3)(x − 7) to clear fractions.
  • Solve the resulting linear or quadratic equation.
• (b) Radical equation: √(5x − 4) = x/2 + 2.
  • Square both sides (watch for extraneous solutions).
  • Solve the resulting quadratic.
• (c) Nested radicals: √x + √(x − 20) = 10.
  • Isolate one radical, square, then isolate the remaining radical and square again.
• (d) Sum of radicals: √(2t − 1) + √(3t + 3) = 5.
  • Similar strategy: isolate, square, repeat.

Don't confuse: squaring both sides can introduce extraneous solutions; always check your answers in the original equation.

🧩 Curves and the vertical line test

🧩 Which curves are functions? (Problem 5.9)

• Vertical line test: a curve is the graph of a function if and only if every vertical line intersects it at most once.
• For each curve in Figure 5.14 (not shown here), determine:
  • Does any vertical line hit the curve more than once?
  • If yes, describe what goes wrong (e.g., "a vertical line at x = 2 hits the curve twice").
  • Suggest a "fix": cut the curve into pieces so each piece passes the vertical line test.
• Example fix: a full circle fails the test; split it into upper and lower semicircles, each of which is a function.

🧩 Absolute value function (Problem 6.1)

The absolute value function is defined by: |x| = { x if 0 ≤ x; −x if x < 0 }

• This is a multipart function with two cases.
• Graph: V-shaped, with vertex at the origin; right branch has slope +1, left branch has slope −1.
• Key feature: the function is always non-negative; |x| ≥ 0 for all x.
• Evaluating: |3| = 3, |−5| = 5, |0| = 0.

Don't confuse: |x| is not the same as x; for negative x, |x| = −x (which is positive).

🧭 Overview

🧠 One-sentence thesis

Graphical analysis allows us to visually extract key information about a function—including its domain, range, sign, intercepts, and where it increases or decreases—by interpreting geometric features of the graph.

📌 Key points (3–5)

• What the graph represents: the set of all points (x, y) in the plane where y = f(x), linking input x-values to output y-values through the function rule.
• Domain and range visualization: the domain projects onto the x-axis, and the range is found by projecting all corresponding graph points onto the y-axis.
• Sign and intercepts: function values control vertical position—positive values place points above the x-axis, negative below; intercepts mark where the graph crosses the axes.
• Increasing vs decreasing: "walking right" along the graph, uphill portions correspond to increasing function values, downhill to decreasing; peaks and valleys are local extrema.
• Common confusion: a circle equation (x − h)² + (y − k)² = r² is not a function because solving for y yields y = k ± √[r² − (x − h)²], which gives two y-values for most x-values (the ± sign splits into two separate functions).

📐 Reading domain and range from the graph

📐 What the graph encodes

The graph of a function f(x) is the set of all points P = (x, y) in the plane such that y = f(x).

• The graph is not arbitrary; every point on it satisfies the function rule.
• A point (x, y) lies on the graph exactly when y = f(x).

🔍 Visualizing domain and range

Domain: the subset of the x-axis representing allowed x-values.
Range: the subset of the y-axis representing output y-values.

Procedure (Important Procedure 6.1.1):

1. Identify all x-values in the domain on the x-axis.
2. For each domain x-value, go up (or down) to the graph.
3. From each graph point, project horizontally to the y-axis.
4. The collection of all y-values you hit is the range.

Example from the excerpt:

• Domain = [3, 5] on the graph of f(x) = −2x + 3 → Range = [−7, −3].
• Domain = [−3, 1] on the same function → Range = [1, 9].

The arrows in the excerpt's diagrams show: x-value → up/down to graph → over to y-axis.

🎯 Interpreting points and sign

🎯 Vertical position and function value

Important Fact 6.1.2: The function value f(x) controls the height of the point P(x) = (x, f(x)) above the x-axis; if f(x) is negative, P(x) is below the x-axis.

• Positive f(x): the point is above the x-axis by f(x) units.
• Negative f(x): the point is below the x-axis by |f(x)| units.
• Zero f(x): the point is on the x-axis.

📊 Sign plot

A sign plot tracks where the function is positive, negative, or zero along the domain (a labeled number line).

• Divide the domain into intervals.
• Label each interval "positive" or "negative" based on whether the graph is above or below the x-axis.
• The excerpt includes a "shadow" of the graph above the number line to show how the labels are derived, but in practice only the labeled number line is needed.

🔄 Dynamic interpretation

• As x moves from left to right (e.g., from 1 to 5), the corresponding points on the graph move along the curve.
• The function values f(x) change accordingly: they may rise, fall, or stay constant.
• The excerpt's Figure 6.4 shows: x-values move → points on graph move from P to Q → f(x) values trace out a path.

🔗 Intercepts and crossings

🔗 y-intercept

The y-intercept is the point where the graph crosses the y-axis, which occurs at (0, f(0)).

• There can be at most one y-intercept (because x = 0 is a single value).
• The y-intercept value is simply f(0).

🔗 x-intercepts (roots or zeros)

The x-intercepts are points where the graph crosses the x-axis, of the form (x₀, f(x₀)) where f(x₀) = 0.
The values x₀ are called roots or zeros of the function.

• There can be several x-intercepts, one, or none.
• To find them, solve f(x) = 0 for x.

🔗 Crossing horizontal and vertical lines

Vertical line x = h:

• The graph crosses x = h at the point (h, f(h)).

Horizontal line y = k:

• To find where the graph crosses y = k, solve the equation k = f(x) for x.
• If the solutions are x₁, x₂, x₃, x₄, then the intersection points are (x₁, k), (x₂, k), (x₃, k), (x₄, k).

Example from the excerpt (semicircle):

• The upper semicircle of radius 2 centered at (2, 1) crosses y = k twice if and only if 1 ≤ k < 3.
• It crosses y = k once if and only if k = 3.
• It does not intersect y = 1/2.
• It crosses the vertical line x = h if and only if 0 ≤ h ≤ 4.

📈 Increasing, decreasing, and local extrema

📈 Uphill and downhill

If you "walk to the right" along the graph, the function values are increasing if you walk uphill, and decreasing if you walk downhill.

• Increasing: as x moves from left to right, f(x) gets larger.
• Decreasing: as x moves from left to right, f(x) gets smaller.
• Graphically, this corresponds to the slope of the curve: upward slope = increasing, downward slope = decreasing.

🏔️ Local extrema

Local maxima and local minima (collectively, local extrema) are the "peaks" and "valleys" where the graph changes from uphill to downhill or vice versa.

• Local maximum: a peak where the graph switches from increasing to decreasing.
• Local minimum: a valley where the graph switches from decreasing to increasing.
• The excerpt notes that finding local extrema is a major topic in mathematics, with applications ranging from business (optimizing profit) to biology (understanding molecular shapes).
• Tools from calculus are typically needed for precise methods.

🪂 Example: Hang glider elevation

Scenario: A hang glider launches from a 500 ft cliff. The graph shows elevation above the gliderport over time (in minutes).

Questions and graphical answers:

1. When is the pilot climbing and descending?
   • Climbing (increasing): 0 ≤ t ≤ 2 and 7 ≤ t ≤ 9.
   • Descending (decreasing): 3 ≤ t ≤ 5 and 9 ≤ t ≤ 10.
2. When is the pilot at gliderport elevation (elevation = 0)?
   • Find where the graph crosses the horizontal axis: t = 0, 4, 8, 10.
3. How much time flying level?
   • Horizontal line segments: 2 ≤ t ≤ 3 and 5 ≤ t ≤ 7.
   • Total: 3 minutes.

Don't confuse: "level flight" means the graph is a horizontal segment (constant elevation), not just crossing zero elevation.

⭕ Circles and the function issue

⭕ Circle equation

The equation (x − h)² + (y − k)² = r² describes a circle of radius r centered at (h, k).

• This is the standard form from earlier chapters.

⚠️ Why a circle is not a function

• If we solve for y, we get (y − k)² = r² − (x − h)², then y − k = ±√[r² − (x − h)²].
• This yields two equations:
  • y = k + √[r² − (x − h)²]
  • y = k − √[r² − (x − h)²]
• The ± sign means for most x-values, there are two corresponding y-values (one above the center, one below).
• A function requires exactly one y for each x in the domain, so the full circle is not a function.

🌗 Semicircles as functions

• Each of the two equations above is a function:
  • f(x) = k + √[r² − (x − h)²] (upper semicircle)
  • f(x) = k − √[r² − (x − h)²] (lower semicircle)
• By splitting the circle into top and bottom halves, each half passes the vertical line test and defines a function.

Don't confuse: The full circle equation is a valid relation and has a graph, but it is not the graph of a function because it fails the vertical line test (a vertical line can intersect the circle at two points).

🧭 Overview

🧠 One-sentence thesis

A circle equation can be split into two separate semicircle functions—upper and lower—which allows us to apply function-based graphical analysis to circular shapes.

📌 Key points (3–5)

• Circle equation is not a function: the standard circle equation (x − h)² + (y − k)² = r² fails the vertical line test.
• How to split into functions: solving for y yields a ± sign, which means two separate equations—one for the upper semicircle, one for the lower.
• Upper vs lower semicircle: the function with +√ graphs the upper half; the function with −√ graphs the lower half.
• Common confusion: don't treat the ± as a single expression; it represents two distinct functions that together form the full circle.
• Why it matters: splitting circles into semicircle functions lets us solve intersection problems (e.g., finding where a horizontal line crosses a circle).

🔄 From circle equation to two functions

🔄 Why the circle equation is not a function

Circle equation: (x − h)² + (y − k)² = r² describes a circle of radius r centered at (h, k).

• This equation does not define a function because a single x-value can correspond to two y-values (top and bottom of the circle).
• The excerpt warns: "It is possible to manipulate this equation and become confused."

➗ Splitting the equation

1. Rearrange: (y − k)² = r² − (x − h)²
2. Take the square root of both sides: y = k ± √(r² − (x − h)²)
3. The ± sign means two separate equations:
   • y = k + √(r² − (x − h)²)
   • y = k − √(r² − (x − h)²)

• Each equation defines its own function.
• Don't confuse: the ± is not a single output; it signals that you must write two functions.

🌗 Upper and lower semicircle functions

🔼 Upper semicircle function

f(x) = k + √(r² − (x − h)²)

• The plus sign in front of the square root gives the upper half of the circle.
• Graphically: this function traces the top arc from the leftmost point to the rightmost point of the circle.

🔽 Lower semicircle function

g(x) = k − √(r² − (x − h)²)

• The minus sign in front of the square root gives the lower half of the circle.
• Graphically: this function traces the bottom arc.

🖼️ Visual relationship

Part of circle	Function	Graph
Upper semicircle	f(x) = k + √(r² − (x − h)²)	Top half of the circle
Lower semicircle	g(x) = k − √(r² − (x − h)²)	Bottom half of the circle

• Together, the graphs of f(x) and g(x) reconstruct the full circle.

🛠️ Solving intersection problems with semicircles

🛠️ The space station tunnel example

Scenario: A circular tunnel with radius 15 feet; two horizontal decks (A at y = 6, B at y = −8); need to find where a 6-foot-tall person can walk safely.

Setup:

• Circle equation: x² + y² = 225 (centered at origin, radius 15).
• Upper semicircle: f(x) = √(225 − x²)
• Lower semicircle: g(x) = −√(225 − x²)

🔍 Finding intersection points

For Deck A (upper semicircle, y = 6):

• Solve the system: y = √(225 − x²) and y = 6.
• Substitute y = 6: 6 = √(225 − x²) → x² = 225 − 36 = 189 → x = ±√189 ≈ ±13.75.
• Points: P = (−13.75, 6) and Q = (13.75, 6).
• Safe zone width: 13.75 feet to the right and left of center.

For Deck B (lower semicircle, y = −8):

• Solve the system: y = −√(225 − x²) and y = −8.
• Substitute y = −8: −8 = −√(225 − x²) → x² = 225 − 64 = 161 → x ≈ ±12.69.
• Points: R = (−12.69, −8) and S = (12.69, −8).
• Safe zone width: 12.69 feet to the right and left of center.

🎯 Key technique

• Intersecting a horizontal line with a circle = intersecting that line with the appropriate semicircle function.
• Upper semicircle for positive y-values; lower semicircle for negative y-values.
• Example: To find where y = 6 crosses the circle, use the upper semicircle function f(x) and solve f(x) = 6.

🧭 Overview

🧠 One-sentence thesis

Multipart functions allow us to describe curves made of separate pieces by defining different rules for different portions of the domain, and we can translate between graphs and symbolic rules by analyzing each piece individually.

📌 Key points (3–5)

• What multipart functions are: functions whose rule consists of multiple cases, each applying to a different part of the domain.
• How to work with them: "home in" on the appropriate case for the x-value you're evaluating, then apply only that piece of the rule.
• Graph-to-rule and rule-to-graph: you can start with either a piecewise graph or a multi-case formula and construct the other.
• Common confusion: open vs closed circles matter—they translate to strict inequalities (< or >) versus weak inequalities (≤ or ≥) and determine whether endpoint values belong to that piece.
• Why they're useful: real-world phenomena (like a bouncing basketball) often behave differently in different intervals, requiring separate descriptions for each phase.

🧩 What multipart functions are

🧩 Definition and structure

Multipart function: a function whose rule is broken into multiple cases, each case corresponding to a separate piece of the graph.

• The excerpt's first example has five pieces: four line segments and one point.
• Each piece has its own formula and its own domain restriction (e.g., "if 0 ≤ x < 1" or "if x = 4").
• The symbolic notation lists all cases in a single large brace, showing which rule applies when.

🔍 Verifying it's a function

• Even though the graph is made of separate pieces, it still passes the vertical line test: any vertical line in the domain intersects the curve exactly once.
• This ensures that for each x-value, there is exactly one y-value, so the curve defines a function.

🎯 How to evaluate multipart functions

🎯 The "home in" method

• To compute f(x) for a specific x-value, first identify which case covers that x.
• Then apply only that piece of the rule.
• Example from the excerpt: to find f(3.56), notice that 3.56 falls in the interval 3 ≤ x < 4, where the rule says f(x) = 1, so f(3.56) = 1.

⚠️ Open vs closed circles

• Closed circle (filled dot): the endpoint is included; use ≤ or ≥.
• Open circle (hollow dot): the endpoint is excluded; use < or >.
• The excerpt emphasizes that "care with open and closed circles is really needed" to ensure the curve defines a function—otherwise two pieces might both claim the same x-value.
• Don't confuse: an open circle at x = 2 on one piece and a closed circle at x = 2 on the next piece means the function "jumps" but is still well-defined at x = 2.

🖼️ Translating between graphs and rules

🖼️ From graph to rule

• Look at how many separate pieces make up the curve.
• For each piece, determine:
  • Its shape (horizontal line, semicircle, etc.).
  • Its domain (which x-values it covers).
  • Whether endpoints are included (closed) or excluded (open).
• Write one case in the rule for each piece.
• Example: the excerpt's Figure 6.11 shows five pieces, so the rule has five cases.

📐 From rule to graph (Example 6.3.1)

The excerpt walks through sketching g(x) = {1 if x ≤ −1; 1 + √(1 − x²) if −1 ≤ x ≤ 1; 1 if x ≥ 1}.

Step-by-step approach:

1. First case (x ≤ −1): graph y = 1 on the domain x ≤ −1 → horizontal line to the left of and including (−1, 1).
2. Third case (x ≥ 1): graph y = 1 on the domain x ≥ 1 → horizontal line to the right of and including (1, 1).
3. Middle case (−1 ≤ x ≤ 1): graph y = 1 + √(1 − x²) → the upper semicircle of a circle with radius 1 centered at (0, 1).
4. Paste the pieces together to get the complete graph of g(x).

• The excerpt notes that each piece is "lassoed" or highlighted separately, then combined.

🏀 Real-world example: bouncing basketball

🏀 Modeling height over time (Example 6.3.2)

• The function s = h(t) tracks the height of a basketball's center above the floor after t seconds.
• The graph (Figure 6.13) shows alternating decreasing and increasing portions: three decreasing (ball falling) and two increasing (ball bouncing up).
• This is naturally a multipart function because the ball's motion changes direction at each bounce.

🤔 Why the graph doesn't touch the t-axis

• The excerpt asks, "Why doesn't the graph touch the t-axis?"
• The ball's center is always some distance above the floor (the radius of the ball), so the height never reaches zero.
• This illustrates that domain restrictions and physical constraints shape the pieces of a multipart function.

📋 Summary table: key distinctions

Aspect	What to remember
Structure	Multiple cases, each with its own formula and domain
Evaluation	Find which case applies, then use only that piece
Endpoints	Open circle (< or >) vs closed circle (≤ or ≥) determines inclusion
Graph ↔ rule	Count pieces in the graph; write one case per piece; or sketch each case from the rule and combine
Real-world use	Different behaviors in different intervals (e.g., bouncing, switching modes)

🧭 Overview

🧠 One-sentence thesis

These exercises apply multipart function definitions, absolute value, and geometric reasoning to model real-world scenarios ranging from pizza cuts to drainage ditches and moving objects.

📌 Key points (3–5)

• Absolute value as a multipart function: the absolute value function is defined piecewise (x if x ≥ 0, −x if x < 0) and appears in many equations.
• Graphing and solving: problems require both sketching multipart functions and solving equations involving absolute values or piecewise rules.
• Real-world modeling: distance, area, volume, and motion problems are expressed as multipart functions of time or position.
• Common confusion: multipart functions change their formula at boundary points—always check which piece applies for a given input value.
• Geometric applications: problems involve triangular pizza slices, baseball diamonds, drainage ditches with circular cross-sections, and parabolic graphs built from line segments and semicircles.

📐 Absolute value fundamentals

📐 Definition and basic properties

Absolute value function: |x| = { x if 0 ≤ x; −x if x < 0 }

• The absolute value measures distance from zero, always non-negative.
• The graph is V-shaped, symmetric about the y-axis.
• Example calculations requested: |0|, |2|, |−3|.

🔍 Solving absolute value equations

• Problem 6.1(b): Solve |x| = 4, |x| = 0, |x| = −1.
  • For |x| = 4, both x = 4 and x = −4 work (two solutions).
  • For |x| = 0, only x = 0 works.
  • For |x| = −1, no solution exists (absolute value cannot be negative).
• Don't confuse: an equation like |x| = a has two solutions if a > 0, one solution if a = 0, and no solution if a < 0.

📊 Intersection and area problems

• Problem 6.1(c): Sketch y = (1/2)x + 2 and y = |x| in the same system, find intersections, then compute the area bounded by the two graphs.
• This combines graphing, solving systems, and geometric area calculation.

🧩 Graphing and converting to multipart form

🧩 From formula to absolute value graph

• Problem 6.2: Given f(x), graph both f(x) and g(x) = |f(x)|, then write the multipart rule for g(x).
  • Example parts: f(x) = −0.5x − 1, f(x) = 2x − 5, f(x) = x + 3.
  • The absolute value "flips" any negative portion of f(x) above the x-axis.
  • The multipart rule for g(x) will have two cases: where f(x) ≥ 0 (keep f(x)) and where f(x) < 0 (use −f(x)).

🔢 Solving equations with multipart functions

• Problem 6.3: Solve for x in equations involving absolute value or piecewise-defined functions.
  • (a) g(x) = 17 where g(x) = |3x + 5|.
  • (b) f(x) = 1.5 where f(x) = { 2x if x < 3; 4 − x if x ≥ 3 }.
  • (c) h(x) = −1 where h(x) = { −8 − 4x if x ≤ −2; 1 + (1/3)x if x > −2 }.
• Strategy: check each piece of the domain separately, then verify the solution satisfies the domain condition for that piece.

🧮 Combined absolute value expressions

• Problem 6.4: Solve equations involving sums of absolute values.
  • (a) f(x) = x + |2x − 1| = 8.
  • (b) g(a) = 3a − 3 + |a + 5| = 2a + 8.
  • (c) h(x) = |x| − 3x + 4, solve h(x − 1) = x − 2 (composition with shift).
• These require breaking the absolute value into cases, solving in each case, and checking validity.

🏗️ Geometric area and volume modeling

🏗️ Area as a function of position

• Problem 6.5: Express the area of a shaded region as a function of x, given dimensions in centimeters (3, x, 6, 5 shown).
• The area formula will depend on how x changes the shape; likely a multipart function if the shape changes character at certain x values.

🍕 Triangular pizza problem

• Problem 6.6: A triangular pizza (base 30 inches, height 20 inches) is cut vertically at position x.
  • (a) Find y (the cut length) as a multipart function of x for 0 ≤ x ≤ 30, sketch the graph, calculate the range.
  • (b) Find the area of Alice's portion as a multipart function of x.
  • (c) If Alice wants half the total area, where should she cut?
• Why multipart? The triangle's geometry changes depending on whether the cut is in the left or right half (the triangle is symmetric or the cut crosses the midpoint).

📦 Cardboard box volume and surface area

• Problem 6.10: A delivery box is made by removing shaded squares of side x from a 20 in × 50 in cardboard sheet and folding.
  • (a) Find polynomial function v(x) for volume; determine its degree.
  • (b) Find polynomial function a(x) for exposed surface area; determine its degree and find dimensions when area = 600 sq. in.
• Key idea: cutting squares of side x and folding creates a box with height x; length and width depend on the original dimensions minus 2x.

🚗 Motion and distance modeling

🚗 Cars traveling between cities

• Problem 6.7: Bellevue and Spokane are 280 miles apart; time t measured from noon.
  • (a) Joan drives Bellevue → Spokane, departs 11 am, arrives 3:30 pm at constant speed. Find j(t) (distance from Bellevue), sketch, give domain and range.
  • (b) Steve drives Spokane → Bellevue at 70 mph, departs noon. Find s(t) (distance from Bellevue), sketch, domain, range.
  • (c) Find d(t), the distance between Joan and Steve at time t.
• Modeling steps: determine speed from distance/time, write linear function for position, then compute |j(t) − s(t)| for separation.

🏃 Arthur's run with direction changes

• Problem 6.8: Arthur runs east 250 ft at 10 ft/sec, then north 400 ft at 12 ft/sec, then west 90 ft at 9 ft/sec.
• Express straight-line distance from start as a function of t (seconds since start).
• Multipart structure: three time intervals, each with a different leg; use Pythagorean theorem to compute distance from origin at each stage.

⚾ Baseball diamond distance

• Problem 6.9: Square diamond, 90 ft sides; Edgar runs counterclockwise at 18 ft/sec after hitting a home run.
• Express distance d(t) from home plate as a function of time t.
• Hint: multipart function—distance formula changes as Edgar rounds each base (four legs of the square).

🌊 Drainage ditch cross-section

🌊 Circular geometry and water level

• Problem 6.11: Vertical cross-section of a ditch consists of circles of radius 10 ft; centers lie on a horizontal line 10 ft above the bottom; water rises at 2 inches/minute.
  • (a) When will the ditch be completely full?
  • (b) Find a multipart function modeling the vertical cross-section.
  • (c) Width of filled portion after 1 hour 18 minutes?
  • (d) When will the filled portion be 42 ft, 50 ft, 73 ft wide?

📏 Width as a function of depth

• The width depends on how many circles are intersected by the water level and where the level cuts each circle.
• Multipart reasoning: as water rises, the geometry changes—first one circle is partially filled, then multiple circles contribute to the width.
• Use circle equations and the relationship between depth and chord length.

📈 Piecewise graph analysis

📈 Line segments and semicircles

• Problem 6.12: Graph of y = g(x) on −6 ≤ x ≤ 6 consists of line segments and semicircles (radius 2) connecting (−6,0), (−4,4), (0,4), (4,4), (6,0).
  • (a) What is the range of g?
  • (b) Where is g increasing? Where decreasing?
  • (c) Find the multipart formula for y = g(x).
  • (d) Range on restricted domain −5 ≤ x ≤ 0?
  • (e) Range on restricted domain 0 ≤ x ≤ 4?

🔍 Breaking down the graph

• Identify pieces: line segments between certain points, semicircles between others.
• Formulas: line segments use point-slope form; semicircles use the equation of a circle shifted and restricted to upper or lower half.
• Range: determined by the highest and lowest y-values on the graph (or restricted domain).

🧮 Algebra review problems

🧮 Simplification and systems

• Problem 6.13(a): Simplify (1 / (1 + 1/a)) − (a / (a + 1)).

  • Combine fractions, find common denominators, reduce.

• Problem 6.13(b): Solve the system:

  • a + b − c = 5
  • 2a − 3b + c = 4
  • a + b + c = −1
  • Use substitution or elimination to find a, b, c.

• Purpose: these algebra skills support solving multipart function equations and modeling problems.

🧭 Overview

🧠 One-sentence thesis

Any standard parabola can be obtained by applying horizontal shifts, vertical shifts, reflections, and vertical dilations to the basic parabola y = x squared, resulting in the vertex form y = a(x − h) squared + k.

📌 Key points (3–5)

• Standard parabolas: Only parabolas that pass the vertical line test (are graphs of functions) qualify; they open either upward or downward and have a vertex and axis of symmetry.
• Vertex form: Every standard parabola has the equation y = a(x − h) squared + k, where (h, k) is the vertex, x = h is the axis of symmetry, and a determines opening direction and width.
• Three geometric maneuvers: Shifting (horizontal and vertical), reflection (across the x-axis), and vertical dilation (stretching or compressing) transform y = x squared into any standard parabola.
• Common confusion: The sign of h in (x − h) squared—positive h shifts right, negative h shifts left; writing x + 2 means h = −2, so the shift is 2 units left, not right.
• Connection to quadratic functions: Vertex form y = a(x − h) squared + k can be algebraically expanded into the standard quadratic form y = ax squared + bx + c.

📐 What makes a parabola "standard"

📐 The vertical line test requirement

A parabola that is the graph of a function is called a standard parabola.

• Not all parabolas are graphs of functions; some fail the vertical line test (e.g., parabolas opening sideways).
• The excerpt shows four parabolas labeled I, II, III, IV; only I and IV pass the vertical line test.
• Standard parabolas are the only ones we can describe with a function equation y = f(x).

🎯 Three basic features

Every standard parabola has:

• Opening direction: either upward or downward.
• Vertex: the highest point (if opening downward) or lowest point (if opening upward).
• Axis of symmetry: a vertical line about which the parabola is symmetric.

🔄 Building parabolas through geometric maneuvers

🔄 Starting point: y = x squared

• The excerpt begins with the graph of y = x squared (Figure 7.3), a parabola opening upward with vertex at (0, 0).
• The strategy: show that every other standard parabola can be obtained from this one graph by applying specific geometric operations.
• As we perform these operations, we track how the function equation changes.

➡️ Horizontal shifts

Shifting right (positive h):

• The graph of y = (x − h) squared is the parabola y = x squared shifted h units to the right.
• Example: y = (x − 2) squared has its lowest point at (2, 0); y = (x − 4) squared has its lowest point at (4, 0).

Shifting left (negative h):

• If h is negative, shifting h units to the right is the same as shifting |h| units left.
• Example: y = (x − (−2)) squared = (x + 2) squared has its lowest point at (−2, 0).
• Don't confuse: In y = (x + 2) squared, the vertex is at x = −2, not x = 2; the form is (x − h) squared with h = −2.

⬆️ Vertical shifts

• The graph of y = x squared + k is the parabola y = x squared shifted k units vertically upward.
• Positive k shifts upward; negative k shifts downward.
• Example: y = x squared + 4 moves the vertex to (0, 4); y = x squared − 10 moves the vertex to (0, −10).

Combined shifts:

• Combining horizontal and vertical shifts gives y = (x − h) squared + k.
• Example: The four cases y = (x ± 2) squared ± 4 produce four parabolas with vertices at different locations.

🪞 Reflection across the x-axis

• Reflecting any curve y = p(x) across the x-axis produces the graph of y = −p(x).
• Example: Starting from y = (x ± 2) squared ± 4, reflecting gives y = −(x ± 2) squared ± 4.
• This flips the parabola upside down: an upward-opening parabola becomes downward-opening.

📏 Vertical dilation

If a is a positive number, the graph of y = a·x squared is usually called a vertical dilation of the graph of y = x squared.

Two cases:

• If a > 1: vertically expanded (stretched) graph—parabola becomes narrower.
• If 0 < a < 1: vertically compressed graph—parabola becomes wider.
• Example: y = 2·x squared is narrower (upper dashed plot); y = (1/2)·x squared is wider (lower dashed plot).

🎯 Vertex form and its meaning

🎯 The vertex form equation

y = a(x − h) squared + k, for some constants a, h, and k and a ≠ 0.

• This is called the vertex form of a quadratic function.
• Every standard parabola can be written this way.

📍 Reading information from vertex form

Parameter	What it tells you
(h, k)	The vertex of the parabola
x = h	The axis of symmetry (vertical line)
a > 0	Parabola opens upward
a < 0	Parabola opens downward
|a| > 1	Parabola is narrower (vertically stretched)
0 < |a| < 1	Parabola is wider (vertically compressed)

• A great deal of qualitative information can simply be "read off" from vertex form.
• Example: y = −3(x − 1) squared + 2 has vertex (1, 2), axis of symmetry x = 1, opens downward (a = −3 < 0), and is narrower than y = x squared (|a| = 3 > 1).

🔄 Converting vertex form to standard form

• Vertex form can be algebraically expanded into y = ax squared + bx + c.
• Example: Starting with y = 2(x − 1) squared + 3 (vertex form with a = 2, h = 1, k = 3):
  • Expand: 2(x squared − 2x + 1) + 3 = 2x squared − 4x + 5.
  • Standard form: y = 2x squared − 4x + 5 (with a = 2, b = −4, c = 5).

🛠️ Applying the maneuvers in order

🛠️ The correct sequence

The excerpt demonstrates a procedure that works:

1. Horizontal shift
2. Vertical dilate
3. Reflect
4. Vertical shift

• The order matters; the full explanation involves function compositions (covered later in the course).
• Example: To obtain y = −3(x − 1) squared + 2 from y = x squared:
  • Horizontal shift by h = 1 → y = (x − 1) squared (vertex at (1, 0)).
  • Vertical dilate by a = 3 → y = 3(x − 1) squared (narrower, vertex still (1, 0)).
  • Reflect across x-axis → y = −3(x − 1) squared (opens downward, vertex (1, 0)).
  • Vertical shift by k = 2 → y = −3(x − 1) squared + 2 (vertex now (1, 2)).

🧩 Why this approach works

• The excerpt uses a geometric and visual approach: start with one specific example (y = x squared), then show every other standard parabola can be obtained via specific geometric maneuvers.
• By tracking how the function equation changes with each maneuver, we derive the vertex form.
• This is a concrete application of more general transformation tools developed in the following section.

🔗 Connection to quadratic functions

🔗 Quadratic functions defined

A function of the form y = ax squared + bx + c (for constants a, b, c with a ≠ 0) is called a quadratic function.

• Quadratic functions play a central role throughout the course.
• The excerpt's two-step task:
  1. Show every standard parabola arises as the graph of y = a(x − h) squared + k (vertex form).
  2. Show any quadratic function can be written in vertex form (using a technique called completing the square).

🔄 From standard form to vertex form

• The second step is more involved and uses completing the square.
• Example: Given y = −3x squared + 6x − 1 (standard form), the vertex form is y = −3(x − 1) squared + 2.
• The reason behind this equality is the completing-the-square technique (covered later).
• In practice, vertex form is almost always preferred because it reveals the parabola's key features directly.

🧭 Overview

🧠 One-sentence thesis

Completing the square is an algebraic procedure that rewrites any quadratic function into vertex form by matching coefficients of like powers of x, enabling us to describe the graph through a sequence of geometric transformations from y = x².

📌 Key points (3–5)

• What completing the square does: transforms a standard quadratic (e.g., y = −3x² + 6x − 1) into vertex form y = a(x − h)² + k by solving for the constants a, h, and k.
• How the method works: expand the vertex form, match coefficients of x², x, and the constant term on both sides, then solve the resulting system of three equations.
• Order of geometric operations: horizontal shift → vertical dilation → reflection → vertical shift (this specific order works; the full explanation involves function composition).
• Common confusion: the order of transformations matters—applying them in the wrong sequence will not produce the correct graph.
• Why it matters: once in vertex form, you can immediately read off the vertex (h, k), determine the axis of symmetry x = h, and describe all geometric maneuvers needed to obtain the graph from y = x².

🔧 The algebraic procedure

🔧 Setting up the equation

Completing the square: a procedure for rewriting a given quadratic function in vertex form by matching coefficients.

• Start with a quadratic in standard form, e.g., y = −3x² + 6x − 1.
• Write it equal to the vertex form: −3x² + 6x − 1 = a(x − h)² + k.
• The goal is to find the unknown constants a, h, and k.

🔧 Expanding and matching coefficients

• Expand the right-hand side: a(x − h)² + k = a(x² − 2xh + h²) + k = ax² − 2axh + ah² + k.
• Rewrite both sides to group coefficients of like powers:
  • Left side: (−3)x² + (6)x + (−1)
  • Right side: (a)x² + (−2ah)x + (ah² + k)
• For the equation to hold, coefficients of x², x, and the constant term must match on both sides.

🔧 Solving the system

The matching gives three equations:

1. Coefficient of x²: −3 = a
2. Coefficient of x: 6 = −2ah
3. Constant term: −1 = ah² + k

Solution steps:

• From equation 1: a = −3 (immediate).
• Substitute a = −3 into equation 2: 6 = −2(−3)h = 6h, so h = 1.
• Substitute a = −3 and h = 1 into equation 3: −1 = (−3)(1²) + k = −3 + k, so k = 2.
• Result: −3x² + 6x − 1 = −3(x − 1)² + 2.

Example: For y = −4x² + 5x + 2, the same procedure yields a = −4, h = 5/8 = 0.625, k = 57/16 = 3.562, so y = −4(x − 0.625)² + 3.562.

🎨 From vertex form to geometric transformations

🎨 The standard order of operations

Once you have vertex form y = a(x − h)² + k, apply transformations in this sequence:

1. Horizontal shift by h units → y = (x − h)²
2. Vertical dilation by factor |a| → y = |a|(x − h)²
3. Reflection across the x-axis (if a < 0) → y = a(x − h)²
4. Vertical shift by k units → y = a(x − h)² + k

Don't confuse: Applying these in a different order will not work. The excerpt notes that the full explanation involves function compositions (covered later in Chapter 8).

🎨 Example walkthrough

For y = −3(x − 1)² + 2 (from the first example):

• Start with y = x² (dashed reference curve).
• Horizontal shift by h = 1 → y = (x − 1)² (fat upward parabola, vertex at (1, 0)).
• Vertical dilation by a = 3 → y = 3(x − 1)² (skinny upward parabola, vertex at (1, 0)).
• Reflection → y = −3(x − 1)² (downward parabola, vertex at (1, 0)).
• Vertical shift by k = 2 → y = −3(x − 1)² + 2 (downward parabola, vertex at (1, 2)).

🎨 Reading the vertex and symmetry

From vertex form y = a(x − h)² + k:

• Vertex: (h, k)
• Axis of symmetry: the vertical line x = h
• Direction: parabola opens upward if a > 0, downward if a < 0

Example: For y = −4(x − 0.625)² + 3.562, the vertex is (0.625, 3.562), the axis of symmetry is x = 0.625, and the parabola opens downward.

🧮 Applied example: the drainage canal

🧮 Setting up the coordinate system

The excerpt describes a parabolic drainage canal:

• 10 feet deep, 20 feet wide at the top.
• Impose an xy-coordinate system so the parabola is symmetric about the y-axis and the vertex is at the origin.
• Vertex form: y = ax² (since (h, k) = (0, 0) and the parabola opens upward, a > 0).

🧮 Using given dimensions

• The canal is 20 feet wide at the top and 10 feet deep.
• This means the points (10, 10) and (−10, 10) lie on the parabola (10 feet from the centerline horizontally, 10 feet up vertically).
• Substitute (10, 10) into y = ax²: 10 = a(10²) = 100a, so a = 0.1.
• The parabola is y = 0.1x².

Question: If the water depth is 5 feet, how wide is the water surface?

• At depth 5 feet, y = 5.
• Solve 5 = 0.1x² → x² = 50 → x ≈ ±7.07 feet.
• The water surface width is about 2 × 7.07 ≈ 14.14 feet.

(Note: The excerpt cuts off before completing this calculation, but the setup is clear.)

🧭 Overview

🧠 One-sentence thesis

The vertex of a parabola represents the maximum or minimum value of a quadratic function, and its coordinates can be computed directly from the standard form coefficients to solve optimization problems.

📌 Key points (3–5)

• What the vertex represents: the highest or lowest point on the parabola, depending on whether it opens downward or upward.
• How to find the vertex: use the formula x = −b/(2a) for the first coordinate, then compute f(−b/(2a)) for the second coordinate.
• Maximum vs minimum: if a > 0, the vertex is the minimum (lowest point); if a < 0, the vertex is the maximum (highest point).
• Common confusion: the vertex form makes the vertex obvious, but you can always find it from standard form ax² + bx + c using the formula.
• Why it matters: the second coordinate of the vertex gives the extreme value of the function, which is often the key step in solving real-world optimization problems.

📍 What the vertex tells you

📍 The vertex as an extreme point

The vertex corresponds to either the "highest point" or "lowest point" on the graph of a parabola.

• The vertex is not just any point—it is the point where the function reaches its maximum or minimum value.
• Which extreme depends on the sign of the leading coefficient a:
  • If a > 0, the parabola opens upward → the vertex is the lowest point (minimum).
  • If a < 0, the parabola opens downward → the vertex is the highest point (maximum).

🎯 Coordinates of the vertex

The vertex has two coordinates that each carry meaning:

Coordinate	What it represents	How to interpret
First coordinate (x-value)	The value of x where the extreme occurs	This is the input that produces the max/min output
Second coordinate (y-value)	The maximum or minimum value of the function	This is the actual extreme value f achieves

• Example: If the vertex is (2.5, 10), then the function reaches its extreme value of 10 when x = 2.5.

🧮 Finding the vertex from standard form

🧮 The vertex formula

Important Fact 7.3.1: For a quadratic function f(x) = ax² + bx + c, the vertex has coordinates P = (−b/(2a), f(−b/(2a))).

• You do not need vertex form to find the vertex; you can compute it directly from the coefficients a, b, and c.
• The first coordinate is always x = −b/(2a).
• The second coordinate is found by plugging that x-value back into the function: y = f(−b/(2a)).

🔄 Connection to completing the square

• The formula for the vertex coordinates comes from completing the square.
• The excerpt emphasizes: "it is always possible to obtain this formula by simply completing the square."
• Don't confuse: you can use either method (the formula or completing the square), but the formula is faster when you only need the vertex.

🛠️ Using the vertex in problem solving

🛠️ Detecting maximum or minimum values

• The second coordinate of the vertex detects the extreme value of f(x).
• The excerpt states: "this is often a key step in problem solving."
• Example: If you need to find the maximum height a ball reaches, find the vertex of the height function; the second coordinate is the maximum height.

📐 Example walkthrough: sketching a parabola

The excerpt provides Example 7.3.2: y = f(x) = −2x² + 11x − 4.

Step-by-step:

• Identify a = −2, b = 11, c = −4.
• Compute the first coordinate of the vertex: h = −b/(2a) = 11/4.
• Compute the second coordinate: k = f(11/4) = 89/8.
• Rewrite in vertex form: f(x) = −2(x − 11/4)² + 89/8.
• Interpret: the parabola opens downward (a < 0), so the vertex (11/4, 89/8) is the highest point, and the axis of symmetry is x = 11/4.

⚠️ Common pitfall

• Don't confuse the vertex with the y-intercept or x-intercepts.
• The vertex is about the extreme value, not where the graph crosses the axes.
• The axis of symmetry passes through the vertex at x = −b/(2a), not through the y-intercept.

🧭 Overview

🧠 One-sentence thesis

Quadratic functions model real-world motion and change by describing parabolic paths, and choosing the right function—whether tracking time or position—is critical to answering the correct question.

📌 Key points (3–5)

• Two modeling approaches: time-based functions y(t) tell when events happen; position-based functions y = f(x) tell where events happen, but not both.
• Vertex form reveals extrema: converting to vertex form a(x − h)² + k identifies maximum or minimum values and when/where they occur.
• Multiple heights in context: in problems like the balloon example, "height above lake level" and "height above ground level" require different functions even for the same object.
• Common confusion: the graph of a time-based function y(t) is NOT the physical path of the object; it shows height over time, not the trajectory through space.
• Three non-collinear points determine a parabola: just as two points determine a line, three points with different x-coordinates uniquely determine a quadratic function.

🎯 Two ways to model motion

⏱️ Time as the variable: y(t)

A function y(t) describes the height of an object t seconds after an event (e.g., a ball is kicked).

• What it tells you: when something happens (e.g., when the ball hits the ground by solving 0 = y(t)).
• What it cannot tell you: where horizontally the ball lands.
• Key property: y(t) is a quadratic function in t.
• Example: A kicked ball's height is y(t) = −16t² + 48t + 50. Solving y(t) = 0 gives the time of landing, not the horizontal distance.

📍 Position as the variable: y = f(x)

A function y = f(x) describes the exact path of an object, where x is horizontal position and y is height.

• What it tells you: where something happens (e.g., where the ball hits the ground by solving 0 = f(x)).
• What it cannot tell you: when in time the ball reaches that position.
• Key property: the graph of y = f(x) is the physical trajectory.
• Don't confuse: the graph of y(t) in a t-y coordinate system is not the path; it shows height changing over time, not the shape of the flight.

🔍 Analyzing quadratic models with vertex form

🔝 Finding maximum or minimum values

• Convert the quadratic to vertex form: y = a(x − h)² + k using the formulas h = −b/(2a) and k = f(h).
• The vertex (h, k) is the highest point if a < 0 (parabola opens downward) or the lowest point if a > 0 (parabola opens upward).
• Example: For y(t) = −16t² + 48t + 50, vertex form is y(t) = −16(t − 3/2)² + 86. The maximum height is 86 feet at t = 3/2 seconds.

🎯 Interpreting the vertex in context

• In a time-based model y(t), the vertex tells you the extreme value and the time it occurs.
• In a position-based model y = f(x), the vertex tells you the extreme value and the horizontal position where it occurs.
• Example: In the balloon problem, f(x) = −(2/2500)x² + (4/5)x has vertex (500, 200), meaning the balloon reaches 200 feet above lake level at horizontal position x = 500 feet.

🎈 Distinguishing multiple height functions

🏔️ Height above lake vs. height above ground

• In the hot air balloon example, the ground rises at a constant slope, so "height above lake level" and "height above ground level" are different quantities.
• Height above lake level: y = f(x) = −(2/2500)x² + (4/5)x (the balloon's path).
• Height of ground above lake level: y = (1/10)x (the rising ground).
• Height above ground level: g(x) = f(x) − (1/10)x = −(2/2500)x² + (7/10)x, which is a new quadratic function.

🧮 Why separate functions matter

• To find the maximum height above the lake, analyze f(x) → vertex at (500, 200).
• To find the maximum height above the ground, analyze g(x) → vertex at (437.5, 153.12).
• To find where the balloon is 50 feet above the ground, solve g(x) = 50, not f(x) = 50.
• Don't confuse: asking "how high above the ground?" requires g(x), not f(x), because the ground itself is rising.

🔢 Determining a quadratic from three points

📐 The three-point rule

Three distinct non-collinear points with different x-coordinates uniquely determine a standard parabola y = ax² + bx + c.

• Just as two points determine a line, three points determine a parabola.
• The points must not lie on a common line (non-collinear) and must have different x-coordinates.
• To find a, b, and c, plug each point into y = ax² + bx + c and solve the resulting system of three equations.

🏠 Example: house value model

• Given: house worth 50K at x = 0 years, 80K at x = 10 years, 200K at x = 20 years.
• Three points: P = (0, 50), Q = (10, 80), R = (20, 200) (units in thousands of dollars, K).
• Set up the system:
  • c = 50 (from the first point)
  • 100a + 10b + c = 80
  • 400a + 20b + c = 200
• Solve: substitute c = 50 into the other two equations, then solve for a and b.
• Result: v(x) = (9/20)x² − (3/2)x + 50.
• To find value at x = 26 years: v(26) = 315.2K = $315,200.
• To find when value reaches 1000K: solve 1000 = v(x) using the quadratic formula → x ≈ 47.64 years (ignore the negative solution).

⚠️ Units and interpretation

• Using K (thousands of dollars) instead of raw dollars avoids "drowning in zeros."
• Always interpret solutions in context: negative time values have no physical meaning, so discard them.
• Example: solving 0 = y(t) may give t = 3.818 sec and t = −0.818 sec; only the positive solution is meaningful (the ball hits the ground after 3.818 seconds).

🧩 Solving systems to find intersections

🎯 Landing points and ground intersections

• To find where an object lands, solve the system where the path function equals the ground function.
• Example: balloon path y = f(x) and ground y = (1/10)x. Solve f(x) = (1/10)x:
  • Substitute: (1/10)x = −(2/2500)x² + (4/5)x
  • Rearrange: x² − 875x = 0 → x(x − 875) = 0
  • Solutions: x = 0 (takeoff) and x = 875 (landing).
• Don't divide out x prematurely, or you lose the x = 0 solution.

📏 Finding specific heights

• To find where the balloon is 50 feet above the ground, solve g(x) = 50.
• This gives two x-values (one on the way up, one on the way down).
• Example: solving −(2/2500)x² + (7/10)x = 50 yields x ≈ 78.46 and x ≈ 796.54.
• Convert back to original coordinates by finding the corresponding y-values in the lake-level system: (78.46, 57.85) and (796.54, 129.6).

🧭 Overview

🧠 One-sentence thesis

A quadratic model can be completely determined either by three distinct non-collinear points or by the vertex and one other point on the graph.

📌 Key points (3–5)

• Two sufficient conditions: three distinct non-collinear points OR the vertex plus one other point will uniquely determine a quadratic model.
• Why the vertex method works: the vertex form f(x) = a(x − h)² + k contains the vertex (h, k), so one additional point provides enough information to solve for the remaining unknown coefficient a.
• Common confusion: don't confuse "three points" with "any three points"—the three points must be non-collinear (not on the same straight line).
• Connection to earlier material: the three-point approach is the same as the system-of-equations method described in Fact 7.4.3.

🔢 Two ways to determine a quadratic model

🔢 Method 1: Three distinct non-collinear points

A quadratic model is completely determined by three distinct non-collinear points.

• This approach is the same as the method in Fact 7.4.3 (referenced in the excerpt).
• Why three points: a quadratic function has the form f(x) = ax² + bx + c, which contains three unknown coefficients (a, b, c).
• Each point gives one equation, so three points yield three equations that can be solved as a system.
• Non-collinear requirement: the three points must not lie on the same straight line; otherwise, they would define a linear (not quadratic) relationship.

🎯 Method 2: The vertex and one other point

A quadratic model is completely determined by the vertex and one other point on the graph.

• This method uses the vertex form of a quadratic function:
  • f(x) = a(x − h)² + k
  • where (h, k) is the vertex.
• How it works:
  • If the vertex (h, k) is given, then h and k are known.
  • The only remaining unknown is the coefficient a.
  • One additional point (x₀, y₀) on the graph provides the equation: y₀ = a(x₀ − h)² + k.
  • This single equation can be solved for a using algebra.
• Example: if the vertex is (2, 5) and the graph passes through (4, 13), plug in: 13 = a(4 − 2)² + 5, which simplifies to 13 = 4a + 5, so a = 2.

🧩 Understanding the vertex form

🧩 What the vertex form reveals

• Every quadratic function can be rewritten as f(x) = a(x − h)² + k.
• The vertex (h, k) is the highest or lowest point on the parabola.
• The coefficient a controls the "width" and direction (upward if a > 0, downward if a < 0).

🔍 Why one point is enough

• Once h and k are known, the vertex form has only one unknown: a.
• A single point (x₀, y₀) provides one equation with one unknown, which is always solvable.
• Don't confuse: this is different from the three-point method, which uses the standard form f(x) = ax² + bx + c and requires three equations to solve for three unknowns.

📊 Comparison of the two methods

Method	What you need	Unknowns to solve	Form used
Three points	Three distinct non-collinear points	a, b, c (three unknowns)	Standard form: f(x) = ax² + bx + c
Vertex + one point	Vertex (h, k) and one other point	a (one unknown)	Vertex form: f(x) = a(x − h)² + k

• Both methods yield a complete quadratic model.
• The vertex method is often simpler algebraically because it reduces the problem to solving for a single unknown.
• The three-point method is more general and does not require knowing the vertex in advance.

🧭 Overview

🧠 One-sentence thesis

A quadratic function forms a parabola with a vertex that represents either the maximum or minimum point, and can be fully determined by three non-collinear points or by the vertex plus one additional point.

📌 Key points (3–5)

• Standard form: quadratic functions have the form f(x) = ax² + bx + c where a ≠ 0.
• Vertex location: the vertex is at (h, k) where h = -b/(2a) and k = f(h), representing the parabola's highest or lowest point.
• Direction matters: if a > 0 the parabola opens upward (vertex is minimum); if a < 0 it opens downward (vertex is maximum).
• Two equivalent forms: every quadratic can be written as f(x) = ax² + bx + c or as f(x) = a(x - h)² + k (vertex form).
• What's needed to build the model: three distinct non-collinear points, OR the vertex plus one other point.

📐 Core definition and structure

📐 What is a quadratic function

Quadratic function: a function of the form f(x) = ax² + bx + c where a ≠ 0.

• The coefficient a cannot be zero (otherwise it becomes linear, not quadratic).
• Contains three parameters: a, b, and c.
• The graph is always a parabola.

🔄 Vertex form

Vertex form: f(x) = a(x - h)² + k where (h, k) is the vertex.

• Every quadratic function can be rewritten in this form.
• This form makes the vertex immediately visible.
• The parameter a is the same in both standard and vertex forms.

📊 The parabola and its vertex

📍 Finding the vertex

The vertex location is given by:

• Horizontal coordinate: h = -b/(2a)
• Vertical coordinate: k = f(h) (plug h back into the function)

⬆️⬇️ Direction and extrema

Value of a	Parabola direction	Vertex type
a > 0	Opens upward	Minimum (lowest point)
a < 0	Opens downward	Maximum (highest point)

• The parabola is symmetric about the vertical line through the vertex.
• This vertical line is the axis of symmetry.

🎯 Why the vertex matters

• The vertex represents the extreme value (either maximum or minimum) of the function.
• Knowing whether the parabola opens up or down tells you whether you're finding a minimum or maximum.
• Example: if modeling profit over time with a downward-opening parabola, the vertex shows when profit is maximized.

🔨 Building a quadratic model

🔨 Two approaches to determine a quadratic

Important Facts 7.5.1 states a quadratic model is completely determined by:

1. Three distinct non-collinear points, OR
2. The vertex and one other point on the graph

🧮 Method 1: Three points

• Given three points (x₁, y₁), (x₂, y₂), (x₃, y₃), plug each into f(x) = ax² + bx + c.
• This creates a system of three equations with three unknowns (a, b, c).
• Solve the system using algebra to find the coefficients.
• Don't confuse: the three points must be non-collinear (not on the same line), otherwise they don't uniquely determine a parabola.

🎯 Method 2: Vertex plus one point

• If given vertex (h, k) and another point (x₀, y₀):
  • Start with vertex form: f(x) = a(x - h)² + k
  • Plug in the additional point: y₀ = a(x₀ - h)² + k
  • Solve for a (the only unknown)
• This approach is often simpler because you only need to solve for one variable.

🧭 Overview

🧠 One-sentence thesis

These exercises apply quadratic function techniques—vertex form, finding functions from points, and optimization—to a range of algebraic and real-world problems.

📌 Key points (3–5)

• Two main approaches to finding a quadratic: use three non-collinear points, or use the vertex plus one other point on the graph.
• Vertex form is central: converting to vertex form reveals the vertex, axis of symmetry, and whether the parabola opens upward or downward.
• Optimization problems: many exercises ask for maximum or minimum values (e.g., profit, area, distance) by exploiting the vertex.
• Common confusion: the vertex may lie inside or outside a given interval; maximum/minimum on an interval can occur at the vertex or at an endpoint.
• Applications span algebra and modeling: from stock prices and balloon trajectories to fencing enclosures and particle motion.

🔧 Finding quadratic functions

🔧 Two methods to determine a quadratic

The excerpt describes two distinct approaches:

1. Three non-collinear points: plug each point into f(x) = ax² + bx + c and solve the resulting system for a, b, c.
2. Vertex plus one other point: use the vertex form f(x) = a(x − h)² + k, where (h, k) is the vertex; plug in the other point (x₀, y₀) to get y₀ = a(x₀ − h)² + k, then solve for a.

• Problem 7.2 asks you to find quadratic functions through given points using these ideas.
• Example: given (0,0), (1,1), and (3,−1), set up three equations in a, b, c and solve.

🔄 Vertex form and conversion

Every quadratic function can be expressed in the form f(x) = a(x − h)² + k, where (h, k) is the vertex of the function's graph.

• Problem 7.1 requires converting standard form (ax² + bx + c) to vertex form, then identifying the vertex and axis of symmetry.
• The vertex coordinates are h = −b/(2a) and k = f(h).
• Don't confuse: the vertex is the turning point; the axis of symmetry is the vertical line x = h.

📊 Optimization on intervals

📊 Maximum and minimum values

• Problem 7.3 asks for max and min of f(x) = x² − 3x + 4 on different intervals (e.g., −3 ≤ x ≤ 5 and 2 ≤ x ≤ 7).
• Key insight: the vertex gives the global extremum (minimum if a > 0, maximum if a < 0), but on a restricted interval the extremum may occur at an endpoint.
• Example: if the vertex lies outside the interval, check the function values at both endpoints.

📊 Vertex on the x-axis

• Problem 7.4: if the vertex lies on the x-axis, then k = 0, so the quadratic touches the x-axis at exactly one point.
• This means the discriminant is zero (the quadratic has one repeated root).
• Example: for f(x) = x² + dx + 3d, set the vertex's y-coordinate to zero and solve for d.

🌍 Real-world modeling problems

💰 Stock price and profit

• Problem 7.5: model buzz.com stock price with a quadratic function given three data points (day 0: $10, day 20: $20, day 40: $25).
• The problem assumes a quadratic model while the price is nonzero, then asks for the cost of 1000 shares at day 30 and when to sell to maximize profit.
• Strategy: find the quadratic using three points, then locate the vertex to find the maximum price (and hence the optimal selling time).

🎈 Balloon trajectory

• Problem 7.7: a hot air balloon follows the path y = f(x) = −(4/2500)x² + (4/5)x above a plateau; the ground slopes downward at 1 vertical foot per 5 horizontal feet.
• Questions include maximum height above plateau, maximum height above ground, landing point, and where the balloon is 50 feet above ground.
• Key: the balloon's height above ground is the balloon's y-coordinate minus the ground's y-coordinate (a linear function of x).

🌳 Apple orchard yield

• Problem 7.9: Sylvia has 100 trees yielding 140 apples each; for every 10 additional trees, yield per tree drops by 4 apples.
• Goal: maximize total production (number of trees × apples per tree).
• Let the number of additional trees be a variable; total production becomes a quadratic function; find its vertex.

🎪 Ticket pricing

• Problem 7.10: attendance is a linear function of ticket price; at $5, 1200 attend; at $7, 970 attend.
• Revenue = (price) × (attendance); since attendance is linear in price, revenue is quadratic in price.
• Maximize revenue by finding the vertex of the revenue function.

📐 Geometry and fencing problems

🪟 Norman window

• Problem 7.11: a Norman window is a rectangle topped by a semicircle; perimeter is fixed at 24 feet.
• Express the area as a function of one dimension (e.g., the width), then maximize.
• The perimeter constraint gives a relationship between width and height; substitute to get area as a quadratic function.

🚧 Rectangular enclosures with partitions

• Problem 7.12: Jun has 300 meters of fencing to make a rectangle split into two parts by a fence parallel to two sides.

• Total fencing used: perimeter plus one internal partition.

• Express area as a function of one dimension, then find the maximum.

• Problem 7.13: budget of $6000; fencing costs $20/meter; two internal partitions cost $15/meter each.

• Set up the cost constraint, express area as a quadratic, and maximize.

⭕ Circle and square from wire

• Problem 7.14: a 60-inch wire is cut into two pieces; one forms a circle, the other a square.
• Total area = area of circle + area of square, both expressed in terms of the cut point.
• Minimize (or maximize) this total area by finding the vertex of the resulting quadratic.

🚶 Motion and distance problems

🚶 Two particles moving

• Problem 7.15: particles A and B move along straight lines at constant speed; positions are given parametrically in terms of time t.
• Find the equations of the lines, sketch the paths, and determine when the distance between A and B is minimal.
• Distance as a function of t is a quadratic (after squaring); minimize by finding the vertex.

🚶 Sven and Rudyard

• Problem 7.16: Sven walks south at 5 ft/sec from 120 feet north of an intersection; Rudyard walks east at 4 ft/sec from 150 feet west.
• Express the distance between them as a function of time t (using the Pythagorean theorem); this is a quadratic in t.
• Minimize to find when they are closest and the minimum distance.

🚶 Tina and Michael

• Problem 7.17: Tina and Michael run in straight lines at 10 ft/sec along indicated paths.
• Parametrize each person's motion, compute the distance between them as a function of time, and minimize.
• Same strategy: distance squared is quadratic in t; find the vertex.

🔢 Algebraic and parameter problems

🔢 Solving for variables in quadratic equations

• Problem 7.18: given αx² + 2α²x + 1 = 0, solve for x in terms of α, then solve for α in terms of x.
• This reverses the usual role of parameter and variable.

🔢 Exactly one solution

• Problem 7.19: for equations like αx² + x + 1 = 0, find values of α so that the equation has exactly one solution.
• A quadratic has exactly one solution when its discriminant is zero.
• Example: for αx² + x + 1 = 0, set 1² − 4(α)(1) = 0 and solve for α.

🔢 Solving quadratics in different variables

• Problem 7.20: solve s = 2(t − 1)² + 1 for t, and solve y = x² + 2x + 3 for x.
• Rearrange into standard quadratic form and apply the quadratic formula or complete the square.

🔗 Intersections and graphs

🔗 Points on graphs

• Problem 7.8(a): does (1,2) lie on y = 3x² − 2? Check by substituting x = 1 and seeing if y = 2.

🔗 Intersections with lines

• Problem 7.8(b)–(c): find where a line (with a parameter) intersects a parabola.
• Set the two expressions equal and solve the resulting quadratic.
• Example: where does y = 1 + 2b intersect y = x² + bx + b? Set 1 + 2b = x² + bx + b and solve for x.

🔗 Intersections of two parabolas

• Problem 7.8(d): where does y = −2x² + 3x + 10 intersect y = x² + x − 10?
• Set the two quadratics equal: −2x² + 3x + 10 = x² + x − 10, rearrange, and solve.

🔍 Absolute value and multipart functions

🔍 Graphing |f(x)|

• Problem 7.6: sketch y = x² − 2x − 3, then sketch y = |x² − 2x − 3|.
• The absolute value reflects the negative portions of the graph above the x-axis.
• Label x- and y-intercepts for both graphs; write the multipart rule for the absolute value function.
• Don't confuse: the vertex of the original parabola may lie below the x-axis; after taking absolute value, that portion flips upward.

🧭 Overview

🧠 One-sentence thesis

Function composition is a mechanical tool that builds a new function by replacing every occurrence of the independent variable in one function with the formula of another function, allowing us to break complicated functions into simpler parts or model situations where one quantity depends on another that itself depends on a third variable.

📌 Key points (3–5)

• What composition does: takes two functions f and g and produces a new function f(g(t)) by substituting the entire expression for g(t) wherever the variable appears in f.
• The mechanical procedure: replace every occurrence of "x" in f(x) by the expression "g(t)" to obtain f(g(t)).
• Why it matters: composition lets us model chained dependencies (e.g., area depends on radius, radius depends on time) and decompose complicated functions into simpler building blocks.
• Common confusion: the order matters—f(g(t)) means "apply g first, then f," not the other way around; the innermost function is evaluated first.
• Practical use: you can either combine simpler functions into a complex one, or break a complex function into simpler parts for easier analysis.

🔗 What composition means

🔗 The basic idea

Function composition: Given two functions y = f(x) and x = g(u), the composition y = f(g(u)) is a new function that expresses y directly in terms of u by replacing every occurrence of x in f(x) with the expression g(u).

• The excerpt describes this as a "very common occurrence": you have a function y = f(x), and the variable x itself depends on a third variable u through x = g(u).
• Given u, the rule g(u) produces a value of x; from this x, the rule f(x) produces a y value.
• So y becomes a function of the new independent variable u.
• The schematic flow: u → (apply g) → x → (apply f) → y.

🧩 Why composition is useful

The excerpt gives two main reasons:

1. Modeling chained dependencies: When one quantity depends on another, which in turn depends on a third, composition captures the full relationship.
2. Simplifying complex functions: A complicated function can be written as a composition of simpler parts, which helps when studying the original function.

Example from the excerpt: The function y = square root of (x squared + 1) can be written as f(g(x)), where f(z) = square root of z and g(x) = x squared + 1. Each of f and g is "simpler" than the original.

Don't confuse: Composition is not just "plugging in a number"—it's plugging in an entire function formula.

🛠️ The mechanical procedure

🛠️ How to compute f(g(t))

The excerpt provides a clear procedure:

Important Procedure: To obtain the formula for f(g(t)), replace every occurrence of "x" in f(x) by the expression "g(t)."

• This is described as "fairly mechanical, though perhaps somewhat unnatural."
• It is a substitution process: wherever you see the independent variable in the outer function, substitute the entire inner function's formula.

📝 Worked examples from the excerpt

Given f(x) and g(t)	Composition f(g(t))	Notes
f(x) = 2, g(t) = 2t	f(g(t)) = f(2t) = 2	Constant function stays constant
f(x) = 3x - 7, g(t) = 4	f(g(t)) = f(4) = 3·4 - 7 = 5	g is constant, so result is a number
f(x) = x² + 1, g(t) = 2t - 1	f(g(t)) = (2t - 1)² + 1 = 4t² - 4t + 2	Replace x with (2t - 1) everywhere
f(x) = 2 + square root of (1 + (x - 3)²), g(t) = 2t² - 1	f(g(t)) = 2 + square root of (1 + (2t² - 1 - 3)²) = 2 + square root of (4t⁴ - 16t² + 17)	Complex substitution
f(x) = x², g(t) = t + ♥	f(g(t)) = (t + ♥)² = t² + 2t♥ + ♥²	Works even with symbolic constants

Key insight: The substitution is purely mechanical—you don't need to understand what the functions "mean," just follow the replacement rule.

⚠️ Order matters

• The excerpt emphasizes that f(g(t)) means "apply g first, then f."
• The innermost function (g) is evaluated first, producing an intermediate result, which is then fed into the outer function (f).
• Example: If you have y = f(x) = x² and x = g(t) = 2t - 1, then f(g(t)) = (2t - 1)², not 2(t²) - 1.

Don't confuse: f(g(t)) is generally different from g(f(t))—the order of composition matters.

🌱 Real-world example: the ripple problem

🌱 The setup

The excerpt presents a botany experiment (light flashes and oxygen) as motivation, but the detailed worked example is about ripples in a pond:

• A pebble is tossed into a pond, creating circular ripples.
• The radius of the leading ripple increases at a constant rate of 2.3 feet per second.
• Questions: What is the area after 6 seconds? When does the area reach 300 square feet?

🌱 Building the composition

The solution uses composition to relate area and time:

1. Area as a function of radius: A = A(r) = π r²
   (This is the standard formula for the area of a circle.)

2. Radius as a function of time: r = r(t) = 2.3t feet
   (The radius grows at 2.3 feet per second, so after t seconds, r = 2.3t.)

3. Compose to get area as a function of time:
   Replace "r" in A(r) = π r² with "r(t) = 2.3t":
   A = π (2.3t)² = 5.29πt²

The new function a(t) = 5.29πt² gives the precise relationship between area and time.

🌱 Answering the questions

• After 6 seconds: a(6) = 5.29π(6)² ≈ 598.3 square feet.
• When area is 300 square feet: Solve 300 = 5.29πt².
  t² = 300 / (5.29π)
  t = ± square root of (300 / (5.29π)) ≈ ± 4.25
  Only the positive solution t = 4.25 seconds makes sense (time cannot be negative).

Key takeaway: Composition lets you model chained dependencies—area depends on radius, radius depends on time, so composition gives area as a function of time directly.

🔄 Decomposing functions

🔄 Breaking complex functions into simpler parts

The excerpt notes that composition can work in reverse: you can take a complicated function and express it as a composition of simpler functions.

Example from the excerpt:

• Complex function: y = h(x) = square root of (x² + 1)
• Decomposition: Write h as f(g(x)), where:
  • y = f(z) = square root of z (simpler: just a square root)
  • z = g(x) = x² + 1 (simpler: just a quadratic)

Each of f and g is "simpler" than the original h, which can help when studying h.

🔄 Why decompose?

• Analysis: Simpler functions are easier to understand, differentiate, integrate, or graph.
• Insight: Decomposition reveals the "building blocks" of a function.

Don't confuse: There is often more than one way to decompose a function—the choice depends on what you want to emphasize or simplify.

🧮 What comes next

🧮 Domains and ranges

The excerpt explicitly defers one important topic:

• The first step (this subsection) is "fairly mechanical"—combining formulas.
• The next step is "of varying complexity"—analyzing how the domains and ranges of f and g affect those of the composition.
• This will be covered in the next subsection (not included in this excerpt).

🧮 Other function-building tools

The excerpt mentions that composition is one of five tools for building new functions from known ones:

1. Composition (this section)
2. Reflection
3. Shifting
4. Dilation
5. Arithmetic

The other four will be discussed in the following two sections (not included in this excerpt).

Context: The chapter is titled "Composition," and the excerpt is from section 8.1, "The Formula for a Composition." The botany experiment with light flashes and oxygen output (mentioned at the start) will be revisited in Example 8.2.4 to show how composition solves that modeling problem.

🧭 Overview

🧠 One-sentence thesis

When composing two functions f(g(x)), the domain and range must be carefully matched so that the output values of g(x) lie within the allowed input values of f(x), and both the formula and domain conditions must be transformed according to the composition procedure.

📌 Key points (3–5)

• Range-domain matching requirement: the range values of g(x) must lie within the domain values of f(x) for the composition f(g(x)) to make sense.
• Domain of the composition: the domain values for f(g(x)) come from the domain values of g(x), but may need to be restricted to ensure g(x) outputs fall within f's domain.
• How to handle domain conditions: replace every occurrence of "x" in f's domain condition by the expression "g(x)" and solve the resulting inequality.
• Common confusion: don't forget that domain restrictions on f(x) must be translated into restrictions on the input to g(x)—the composition inherits constraints from both functions.
• Multipart functions: when composing with piecewise-defined functions, apply the composition procedure separately to each part, transforming both the formula and the domain condition for that part.

📦 The composition package: rule, domain, and range

📦 What a function consists of

A function is a "package" consisting of a rule, a domain of allowed input values, and a range of output values.

• When you compose functions, you cannot just combine the formulas—you must also track how domains and ranges interact.
• The composition f(g(x)) is itself a new function with its own rule, domain, and range.

🔗 The matching requirement

• Key constraint: the range values for g(x) must lie within the domain values for f(x).
• Why: the output of g becomes the input to f, so f must be able to accept those values.
• If g's range doesn't naturally fit into f's domain, you must modify g's domain to restrict its range.
• Example: if f(z) = √z (domain z ≥ 0) and g(x) = x + 1, then g(x) must output non-negative values, so the domain of the composition is x ≥ -1.

🛠️ Procedure for finding domain and range

🛠️ The refined composition procedure

Procedure 8.2.1: To obtain the formula for f(g(x)), replace every occurrence of "x" in f(x) by the expression "g(x)." In addition, if there is a condition on the domain of f that involves x, then replace every occurrence of "x" in that condition by the expression "g(x)."

• This is a two-step process:
  1. Substitute g(x) into the formula for f(x).
  2. Substitute g(x) into any domain condition for f(x) and solve the resulting inequality.

📐 Example: restricted domain on f

• Start with f(x) = x² on the domain -1 ≤ x ≤ 1.
• Let g(x) = x - 1.
• Step 1 (formula): f(g(x)) = f(x - 1) = (x - 1)² = x² - 2x + 1.
• Step 2 (domain): the condition -1 ≤ x ≤ 1 becomes -1 ≤ g(x) ≤ 1, i.e., -1 ≤ x - 1 ≤ 1, which simplifies to 0 ≤ x ≤ 2.
• Result: f(g(x)) = x² - 2x + 1 on the domain 0 ≤ x ≤ 2.

🔍 Finding the largest possible domain

• Sometimes you want the largest domain for which the composition makes sense.
• Strategy:
  1. Identify the domain of f(z) (e.g., z ≥ 0 for f(z) = √z).
  2. Find which x-values make g(x) fall into that domain.
  3. Solve the inequality g(x) ≥ 0 (or whatever the condition is).
• Example: for f(z) = √z and g(x) = x + 1, the largest domain is x ≥ -1 because that ensures x + 1 ≥ 0.

🧩 Composing with multipart functions

🧩 The challenge

• When f(t) is defined piecewise (different formulas on different intervals), you must apply the composition procedure to each part separately.
• Each part has its own formula and its own domain condition; both must be transformed.

🌱 Example: the botany experiment

• Original function f(t) models oxygen output with a flash at t = 1:
  • f(t) = 1 if t ≤ 1
  • f(t) = (2/3)t² - (8/3)t + 3 if 1 ≤ t ≤ 3
  • f(t) = 1 if 3 ≤ t
• To shift the flash to t = 5, compose with g(t) = t - 4.
• First part: f(t) = 1 when t ≤ 1 becomes f(g(t)) = 1 when t - 4 ≤ 1, i.e., t ≤ 5.
• Second part: f(t) = (2/3)t² - (8/3)t + 3 when 1 ≤ t ≤ 3 becomes f(g(t)) = (2/3)(t - 4)² - (8/3)(t - 4) + 3 when 1 ≤ t - 4 ≤ 3, i.e., 5 ≤ t ≤ 7.
• Third part: f(t) = 1 when 3 ≤ t becomes f(g(t)) = 1 when 3 ≤ t - 4, i.e., 7 ≤ t.
• The result is a new multipart function with the "parabolic dip" shifted from [1, 3] to [5, 7].

⚠️ Don't confuse

• Don't apply the composition only to the formula and forget the domain conditions—each piece's domain must also be transformed.
• Don't mix up which variable you're solving for: after substituting g(t) into the domain condition, solve for the original input variable (t in this case).

📊 Visual interpretation

📊 The domain-range flow

Step	What happens	Variable
Input to g	Start with x-values in the domain of g	x
Output of g	g(x) produces z-values (range of g)	z = g(x)
Input to f	z-values must lie in the domain of f	z
Output of f	f(z) produces y-values (range of f(g(x)))	y = f(z)

• The domain of f(g(x)) is the set of x-values.
• The range of f(g(x)) is the set of y-values.
• The "middle" variable (z or whatever g outputs) must satisfy f's domain requirements.

🎨 Graphical approach

• To find the largest domain for f(g(x)):
  1. Graph z = g(x) in the xz-plane.
  2. Mark the desired range (the domain of f) on the vertical z-axis.
  3. Determine which x-values lead to points on the graph with z-coordinates in that zone.
• Example: for f(z) = √z and g(x) = x + 1, mark z ≥ 0 on the vertical axis; the graph z = x + 1 enters this zone when x ≥ -1.

🧭 Overview

🧠 One-sentence thesis

These exercises apply composition of functions to multipart functions, domain/range transformations, and practical scenarios, building fluency in decomposing complex functions and understanding how compositions affect domains and ranges.

📌 Key points (3–5)

• Composing with multipart functions: absolute value and piecewise functions require case-by-case analysis when composed.
• Decomposition skill: breaking down complex functions into simpler compositions (multiple valid answers exist).
• Domain and range under composition: transformations like f(B(x - C)) shift and scale the domain; Af(x) + D scales and shifts the range.
• Common confusion: composition order matters—f(g(x)) is different from g(f(x)); also, domain restrictions propagate through compositions.
• Fixed points: values where f(x) = x remain unchanged under repeated composition.

🧩 Working with multipart and absolute value functions

🧩 Composing with absolute value (Problem 8.1)

The exercise asks you to compose functions involving the absolute value function h(t) = |t| with linear functions f(t) = t - 1 and g(t) = -t - 1.

• Because absolute value creates a "split" at t = 0, you must write separate cases for positive and negative inputs.
• Example approach: For h(f(t)), substitute f(t) into |t|, then determine where f(t) ≥ 0 and where f(t) < 0 to write the multipart rule.
• The order matters: h(f(t)) will look different from f(h(t)).

🔄 Nested compositions (Problem 8.1c)

Problem 8.1(c) asks for h(h(t) - 1), a composition where the inner function is itself a composition minus 1.

• Work from the inside out: first find h(t), then subtract 1, then apply h again.
• Each absolute value introduces potential case splits.

🔨 Decomposing complex functions

🔨 Breaking functions into simpler pieces (Problem 8.2)

Decomposition: writing a complex function as f(g(x)) where f and g are simpler functions.

The exercise provides six functions to decompose:

• (a) y = (x - 11)⁵: Think "something to the fifth power"—outer function f(u) = u⁵, inner function g(x) = x - 11.
• (c) polynomial in (x - 3): The entire expression is a polynomial in the quantity (x - 3), so let g(x) = x - 3 and f(u) be the polynomial in u.
• (e) nested square roots: Work outward from the innermost operation.

Key insight: There is more than one correct answer—you can choose different "split points" in the composition.

📐 Domain and range transformations

📐 How composition affects domain (Problem 8.8a,b)

Given: f(x) has domain 1 ≤ x ≤ 6 and range -3 ≤ y ≤ 5.

For f(2(x - 3)):

• The inner function is g(x) = 2(x - 3).
• You need g(x) to produce values in [1, 6] (the original domain of f).
• Solve: 1 ≤ 2(x - 3) ≤ 6 to find the new domain for x.
• The range stays the same: -3 ≤ y ≤ 5 (composition with the inner function doesn't change what f outputs).

📏 How outer transformations affect range (Problem 8.8c,d)

For 2f(x) - 3:

• The domain stays the same: 1 ≤ x ≤ 6 (you're not changing the input to f).
• The range transforms: multiply the original range by 2, then subtract 3.
• Original range [-3, 5] becomes [2(-3) - 3, 2(5) - 3] = [-9, 7].

🎯 Engineering specific domains and ranges (Problem 8.8e,f)

• (e) Find B and C so that f(B(x - C)) has domain 8 ≤ x ≤ 9.
• (f) Find A and D so that Af(x) + D has range 0 ≤ y ≤ 1.
• These require solving systems of inequalities to match the desired intervals.

🔁 Special composition patterns

🔁 Fixed points (Problem 8.4)

Fixed point: a value x where f(x) = x; applying f does not change it.

For f(x) = (1/2)x + 3:

• The point x = 6 is fixed because f(6) = 6.
• Why it stays fixed under repeated composition: if f(6) = 6, then f(f(6)) = f(6) = 6, and so on.
• Linear functions have at most one fixed point; quadratic functions can have at most two.

🔄 Self-composition (Problem 8.3, 8.6)

• Problem 8.3: If f(x) = ax + b, then f(f(x)) is also linear—verify by substitution.
• Problem 8.6: Compute f(g(x)), f(f(x)), and g(f(x)) for various function pairs.
• Don't confuse: f(g(x)) ≠ g(f(x)) in general; order matters.

🚗 Applied composition problems

🚗 Unit conversions (Problem 8.5)

A car's speed is given as C(m) where m is minutes.

• (a) Convert seconds to minutes: f(s) gives minutes from seconds; C(f(s)) gives speed as a function of seconds.
• (b) Convert hours to minutes: g(h) gives minutes from hours; C(g(h)) gives speed as a function of hours.
• (c) Convert mph to ft/sec: v(s) converts speed units; v(C(m)) gives speed in ft/sec as a function of minutes.

Pattern: Composition chains unit conversions—the inner function converts the input unit, the outer function uses that converted value.

🧪 Domain restrictions from context (Problem 8.7)

Given y = f(z) = √(4 - z²) and z = g(x) = 2x + 3:

• Compute f(g(x)) by substitution.
• Find the largest domain: the square root requires 4 - (2x + 3)² ≥ 0.
• Solve this inequality to find valid x-values.
• Don't forget: domain restrictions from the inner function (g) and outer function (f) both apply.

🔬 Difference quotient preparation (Problem 8.9)

🔬 Simplifying [f(x + h) - f(x)] / h

This expression appears in calculus (the definition of derivative).

The exercise asks you to:

1. Substitute x + h into the function.
2. Subtract f(x).
3. Divide by h.
4. Simplify algebraically so h is no longer in the denominator.
5. Then set h = 0 in the simplified form.

Example approach for f(x) = 1/(x - 1):

• f(x + h) = 1/((x + h) - 1)
• The difference involves combining fractions with different denominators.
• Factor out h from the numerator to cancel with the denominator.

🧭 Overview

🧠 One-sentence thesis

Inverse functions reverse the input-output roles of an original function, but a valid inverse exists only when each output corresponds to exactly one input.

📌 Key points (3–5)

• What an inverse does: takes the output of a function and returns the input that produced it—reversing the "left to right" process into "right to left."
• When a single inverse exists: linear functions with non-zero slope always have a unique reverse process that qualifies as a function.
• When multiple inverses arise: functions like y = (x − 1)² + 1 can map different inputs to the same output, so the reverse process may produce multiple answers and violate the definition of a function.
• Common confusion: not every reverse process is a function—if solving f(x) = y for x yields more than one answer, the reverse process must be split into separate functions.
• Why it matters: understanding inverse functions helps predict inputs from outputs (e.g., finding the time when a chemical reaction reaches a certain fraction).

🔄 What inverse functions do

🔄 Reversing input and output roles

• A function is a process: input x → output f(x).
• So far, we work "left to right"—plug in x, get f(x).
• An inverse function works "right to left": given an output y, find the input x such that f(x) = y.

Inverse function: a reverse process that takes an output value y and returns the input x required so that f(x) = y.

🧪 Motivation from experimental science

• Example: measuring the fraction of chemical reactants remaining over time.
• Original function: input time t → output fraction y = f(t).
• Inverse question: given a desired fraction, at what time does it occur?
• The inverse function swaps roles: input fraction → output time.

🔢 When a unique inverse exists

🔢 Linear functions always work

• For y = f(x) = mx + b (where m ≠ 0), you can always solve for x in terms of y:
  • y = mx + b
  • y − b = mx
  • x = (1/m)(y − b)
• This reverse process is a single formula, so it defines a function.

📐 Example: f(x) = 3x − 1

• Forward: x = −1, −1/2, 0, 1/2, 1, 2 → outputs −4, −5/2, −1, 1/2, 2, 5.
• Reverse process: x = (1/3)(y + 1).
• Given y = 11, compute x = (1/3)(11 + 1) = 4; check: f(4) = 3·4 − 1 = 11. ✓

📐 Example: f(x) = −0.8x + 2

• Here m = −0.8, b = 2.
• Reverse process: x = −1.25(y − 2).
• Given y = 11, compute x = −1.25(11 − 2) = −11.25; check: f(−11.25) = 11. ✓

⚠️ When the reverse process is not a function

⚠️ Multiple inputs can produce the same output

• Consider y = f(x) = (x − 1)² + 1.
• Forward process: x = −1, −1/2, 0, 1/2, 1, 2 → outputs 5, 13/4, 2, 5/4, 1, 2.
• Notice: both x = −1 and x = 0 produce different outputs, but some outputs repeat (e.g., y = 2 comes from both x = 0 and x = 2).

⚠️ Solving for x yields two answers

• Given y = 3, solve (x − 1)² + 1 = 3:
  • (x − 1)² = 2
  • x − 1 = ±√2
  • x = 1 + √2 or x = 1 − √2
• The reverse process has two outputs for a single input y = 3.
• This violates the definition of a function (one input → one output).

🔀 Solution: split into two separate reverse functions

• Create two new reverse processes, each with a unique output:
  • Reverse process 1: x = 1 + √(y − 1) (takes the positive square root)
  • Reverse process 2: x = 1 − √(y − 1) (takes the negative square root)
• Each of these is a function because each has exactly one output for each input.

🧩 Don't confuse: reverse process vs. inverse function

• A reverse process is the algebraic procedure of solving f(x) = y for x.
• An inverse function is a reverse process that satisfies the definition of a function (one output per input).
• If the reverse process produces multiple answers, it is not an inverse function—you must restrict or split it.

🔍 Connection to solving equations

🔍 Why some equations have one solution, others have two

The excerpt opens with three equations to solve for x:

• (x + 2) = 64 → add −2 to both sides → x = 62 (one solution)
• (x + 2)³ = 64 → take cube root of both sides → x + 2 = 4 → x = 2 (one solution)
• (x + 2)² = 64 → take square root of both sides → x + 2 = ±8 → x = −10 or x = 6 (two solutions)

Operation	Number of solutions	Why
Add/subtract	One	Reverse operation (subtract/add) is unique
Cube	One	Cube root is unique for any real number
Square	Two	Square root has both positive and negative results

• The square function maps both positive and negative inputs to the same output, so its reverse process is not a function unless restricted.
• This pattern will reappear with inverse circular (trigonometric) functions in later chapters.

🧭 Overview

🧠 One-sentence thesis

The graphical approach reveals that finding inverse processes means solving for input values that produce a given output, and the number of such inputs is determined by counting how many times a horizontal line intersects the function's graph.

📌 Key points (3–5)

• What the reverse process asks: given an output value, find which input(s) produce it—like working backward from answer to question.
• Graphical method: draw a horizontal line y = c and see where it crosses the graph of y = f(x); the x-coordinates of those crossings are the inputs that give output c.
• How many solutions exist: count the intersection points between y = c and y = f(x) to know how many x values produce the same output.
• Common confusion: a reverse process is only a function if each output corresponds to exactly one input; if a horizontal line crosses the graph multiple times, the reverse process is not a single function.
• Why linear functions are special: their graphs intersect any horizontal line exactly once, so they always have a unique reverse process that is itself a function.

🔄 The reverse-process question

🔄 What "reverse" means

• The excerpt frames the reverse process as asking: "What x value can be run through the process so you end up with the number [some given output]?"
• It compares this to the game show Jeopardy: you know the answer (the output), you want to find the question (the input).
• Example: if y = 3x − 1 and you want output 11, solve 11 = 3x − 1 to find x = 4.

🧩 Why some reverse processes fail to be functions

• The excerpt's third example shows f(x) = (x − 1)² + 1.
• When you try to reverse it for y = 3, you get two possible x values: x = 1 + √2 and x = 1 − √2.
• A reverse process with two outputs violates the definition of a function (one input → one output).
• The solution: split into two separate reverse processes, each giving a unique output.

📊 Graphical method for finding inputs

📊 The horizontal-line technique

Important Fact 9.2.1: Given a number c, the x values such that f(x) = c can be found by finding the x-coordinates of the intersection points of the graphs of y = f(x) and y = c.

• Instead of solving equations symbolically, draw the horizontal line y = c on the same axes as y = f(x).
• Where the line crosses the graph, read off the x-coordinates.
• Example from the excerpt: for a function graphed in Figure 9.7, the line y = 3 crosses at (−5, 3), (−1, 3), and (9, 3), so f(−5) = f(−1) = f(9) = 3.

🔢 Counting how many solutions

Important Fact 9.2.3: For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c and y = f(x) intersect.

• The number of crossings tells you how many inputs produce the same output.
• If a horizontal line crosses the graph once, there is exactly one input for that output.
• If it crosses twice, there are two inputs; if it never crosses, no input produces that output.

🧪 Example: y = x²

The excerpt works through f(x) = x² with three horizontal lines:

Output c	Equation to solve	Solutions (x values)	Intersection count
6	x² = 6	x = ±√6 ≈ ±2.449	2
3	x² = 3	x = ±√3 ≈ ±1.732	2
1	x² = 1	x = ±1	2

• Every horizontal line y = c (for c > 0) crosses the parabola twice, so the reverse process always has two outputs.
• Don't confuse: the function f(x) = x² is perfectly valid (one input → one output), but its reverse process is not a function because one output corresponds to two inputs.

🔍 Special cases: linear and constant functions

🔍 Linear functions (f(x) = mx + b, m ≠ 0)

• The graph is a non-horizontal line.
• Any horizontal line y = c intersects it exactly once.
• Therefore, the equation c = f(x) always has a unique solution.
• This means the reverse process for a linear function is always a function itself.

🔍 Constant functions (f(x) = d)

• The graph is a horizontal line at height d.
• If c = d, the horizontal line y = c coincides with the graph, so every x in the domain is a solution (infinitely many intersections).
• If c ≠ d, the lines are parallel and never intersect, so no x produces output c.
• Example from the excerpt: if f(x) = 1 and c = 1, every real number is an input that gives output 1; if c = 2, no input gives output 2.

🧭 Overview

🧠 One-sentence thesis

A function has an inverse function only when it is one-to-one, meaning each output corresponds to at most one input, and the inverse reverses the original function's input-output relationship.

📌 Key points (3–5)

• One-to-one requirement: A function must be one-to-one (pass the horizontal line test) to have an inverse function; otherwise, the "reverse process" produces multiple answers and is not a function.
• What an inverse does: The inverse function f⁻¹(y) takes an output y and returns the unique input x such that f(x) = y; it reverses the original function's process.
• Domain and range swap: The domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹.
• Common confusion: Not every function has an inverse function; non-one-to-one functions (like y = x²) can be split into one-to-one pieces by restricting the domain.
• Graphical relationship: The graph of f⁻¹ is obtained by reflecting the graph of f across the line y = x (or equivalently, swapping x and y coordinates).

🔍 Finding inputs from outputs graphically

🔍 The horizontal line method

Important Fact 9.2.1: Given a number c, the x values such that f(x) = c can be found by finding the x-coordinates of the intersection points of the graphs of y = f(x) and y = c.

• To solve f(x) = c graphically, draw the horizontal line y = c and see where it crosses the graph of y = f(x).
• Each intersection point (x, c) gives an x value that produces output c.
• Example: For f(x) = x², the line y = 3 intersects the parabola at two points, so there are two x values (x ≈ ±1.732) such that f(x) = 3.

📊 Counting solutions

Important Fact 9.2.3: For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c and y = f(x) intersect.

• The number of intersections tells you how many inputs produce the same output.
• Linear functions (f(x) = mx + b, m ≠ 0) always intersect a horizontal line exactly once → unique solution.
• Constant functions (f(x) = d) either intersect every horizontal line y = d at infinitely many points (every x works) or never intersect y = c when c ≠ d.

🎯 One-to-one functions

🎯 Definition and importance

One-to-one function: A function where, given any number c, there is at most one input x value in the domain so that f(x) = c.

• "At most one" means either exactly one or zero solutions for each output value.
• Linear functions (degree 1 polynomials) are always one-to-one.
• f(x) = x² is not one-to-one on all real numbers because f(2) = f(−2) = 4; two different inputs produce the same output.
• Why it matters: Only one-to-one functions have inverse functions; the "reverse process" must give a unique answer.

🧪 The horizontal line test

Important Fact 9.2.5 (Horizontal Line Test): On a given domain of x-values, if the graph of some function f(x) has the property that every horizontal line crosses the graph at most only once, then the function is one-to-one on this domain.

• Visual check: If any horizontal line intersects the graph more than once, the function is not one-to-one.
• Example: f(x) = x³ passes the horizontal line test on all real numbers → it is one-to-one.
• Don't confuse: A function can be made one-to-one by restricting its domain (see below).

🔄 Inverse functions

🔄 Definition and notation

Important Fact 9.3.1: Suppose a function f(x) is one-to-one on a domain of x values. Then define a NEW FUNCTION by the rule f⁻¹(y) = the x value so that f(x) = y. The domain of y values for the function f⁻¹(y) is equal to the range of the function f(x).

• Read f⁻¹(y) as "f inverse of y."
• The inverse function reverses the process: input y → output the unique x such that f(x) = y.
• CAUTION: You do NOT automatically get an inverse function from functions that are not one-to-one!
• Both the domain and the rule of f determine whether the inverse is a function.

🔁 The reversal property

Important Fact 9.3.2: For every a value in the domain of f(x), we have f⁻¹(f(a)) = a.

• The inverse "undoes" the original function.
• If you apply f to a, then apply f⁻¹ to the result, you get back a.
• Example: If f(a) produces some output, then f⁻¹ takes that output and returns a.
• Think of f⁻¹ as a "black box" running f backwards.

📍 Domain and range swap

Original function f	Inverse function f⁻¹
Domain of f	Range of f⁻¹
Range of f	Domain of f⁻¹

• The roles of input and output are reversed.
• Example: If f has domain [0, ∞) and range [0, ∞), then f⁻¹ has domain [0, ∞) and range [0, ∞).

📐 Graphing inverse functions

📐 Coordinate swap

• A typical point on the graph of y = f(x) looks like (x, f(x)).
• A point on the graph of x = f⁻¹(y) looks like (f(x), x) — the x and y coordinates are reversed.
• Graphically, this means the graph of f⁻¹ is the graph of f with the positive x-axis and positive y-axis interchanged.

🔄 Reflection across y = x

• The graph of f and the graph of f⁻¹ are mirror images of each other across the line y = x.
• Example: For f(x) = x³, the inverse is f⁻¹(y) = ³√y. The graph of f⁻¹ is obtained by reflecting the graph of f across the line y = x.
• Practical method: Rotate the graph 90° clockwise, then flip across the horizontal axis (as shown in Figure 9.14).

🚫 Non-one-to-one functions and domain restriction

🚫 Why inverses fail for non-one-to-one functions

• If f is not one-to-one, attempting to define an inverse leads to contradictions.
• Example: Try to invert f(x) = x² using f⁻¹(y) = √y. Start with x = −7:
  • f(−7) = 49, so f⁻¹(f(−7)) = f⁻¹(49) = √49 = 7.
  • But Fact 9.3.2 says f⁻¹(f(−7)) must equal −7.
  • Contradiction: 7 ≠ −7, so √y is not the inverse of x² on all real numbers.

✂️ Splitting the domain

• You can make a non-one-to-one function one-to-one by restricting its domain.
• Example: f(x) = x² on all real numbers is not one-to-one. Split it into two cases:
  1. Non-negative x: f(x) = x² for x ≥ 0 → inverse is f⁻¹(y) = √y (domain y ≥ 0).
  2. Non-positive x: f(x) = x² for x ≤ 0 → inverse is f⁻¹(y) = −√y (domain y ≥ 0).
• Each piece is one-to-one and has its own inverse function.
• This splitting explains why equations like x² = 5 have two solutions: x = √5 (from the right side) and x = −√5 (from the left side).

📝 Summary checklist

📝 Key definitions and tests

Concept	Definition / Test
Inverse functions	Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in their domains.
One-to-one	Every equation f(x) = k has at most one solution. If any k gives more than one solution, f is not one-to-one.
Horizontal line test	A function is one-to-one if every horizontal line intersects the graph at most once.
Existence of inverse	A function has an inverse if and only if the function is one-to-one.
Domain/range swap	Domain of f = range of f⁻¹; range of f = domain of f⁻¹.
Graph relationship	The graph of f and f⁻¹ are mirror images across the line y = x.

✅ Quick checks

• To verify f and g are inverses, check both compositions: f(g(x)) = x and g(f(x)) = x.
• To test one-to-one: Use the horizontal line test or check if f(x₁) = f(x₂) implies x₁ = x₂.
• Don't confuse: A function can have multiple "reverse processes" (like ±√y for x²), but only one-to-one functions have a single inverse function.

🧭 Overview

🧠 One-sentence thesis

When a function is not one-to-one, it cannot have a single inverse function, but restricting its domain to make it one-to-one allows us to define separate inverse functions for different parts of the original graph.

📌 Key points (3–5)

• Why non one-to-one functions fail: attempting to invert a function like y = x² produces contradictions because multiple inputs map to the same output.
• The domain restriction solution: by limiting the domain to only non-negative or only non-positive x-values, we can split a non one-to-one function into one-to-one pieces.
• Multiple inverses from splitting: each restricted piece has its own inverse function (e.g., √y for x ≥ 0 and -√y for x ≤ 0).
• Common confusion: the equation x² = 5 has two solutions (√5 and -√5) precisely because the left and right sides of the parabola have separate inverse functions.
• Graphical interpretation: restricting the domain means erasing part of the graph to ensure every horizontal line intersects at most once.

⚠️ The problem with non one-to-one functions

⚠️ Why direct inversion fails

The excerpt demonstrates the failure by attempting to invert y = x² using f⁻¹(y) = √y:

• Start with the input -7
• Apply f: f(-7) = 49
• Apply the proposed inverse: f⁻¹(49) = √49 = 7
• But Fact 9.3.2 requires f⁻¹(f(-7)) = -7
• Contradiction: we get 7 = -7, which is impossible

The key insight: "If you didn't have to worry about negative numbers, things would be all right."

🔄 Why the contradiction occurs

• The function y = x² maps both x = 7 and x = -7 to the same output y = 49
• An inverse function must give a unique answer
• When f is not one-to-one, f⁻¹ cannot decide which of the multiple inputs to return
• Example: Should f⁻¹(49) return 7 or -7? Both are valid pre-images under f(x) = x²

✂️ Splitting the domain

✂️ Restricting to non-negative x-values

The excerpt shows how to make y = x² invertible by limiting the domain:

• Restriction: only allow x ≥ 0 (non-negative x-values)
• Graphical effect: erase the graph to the left of the y-axis
• Result: the restricted function is now one-to-one
• Inverse function: f⁻¹(y) = √y

Original function	Restricted domain	Inverse function
y = x²	x ≥ 0	f⁻¹(y) = √y

The excerpt emphasizes: "This is now a one-to-one function!"

✂️ Restricting to non-positive x-values

Alternatively, we can restrict to the other side:

• Restriction: only allow x ≤ 0 (non-positive x-values)
• Graphical effect: erase the graph to the right of the y-axis
• Result: this piece is also one-to-one
• Inverse function: f⁻¹(y) = -√y

Original function	Restricted domain	Inverse function
y = x²	x ≤ 0	f⁻¹(y) = -√y

🔪 The splitting concept

"We have split the graph of y = x² into two parts, each of which is the graph of a one-to-one function."

• The original parabola is divided at the y-axis
• Each half is one-to-one when considered separately
• Each half has its own distinct inverse function
• Don't confuse: these are not two branches of one inverse; they are two separate inverse functions for two separate restricted functions

🎯 Why equations have multiple solutions

🎯 Connection to solving equations

The excerpt explains why x² = 5 has two solutions:

• Solution 1: x = √5 (comes from the right side of the parabola, where x ≥ 0)
• Solution 2: x = -√5 (comes from the left side of the parabola, where x ≤ 0)

The excerpt states: "It is precisely this splitting into two cases that leads us to multiple solutions."

🎯 Separate inverse functions create separate solutions

• The right side of y = x² has inverse f⁻¹(y) = √y
• The left side of y = x² has inverse f⁻¹(y) = -√y
• When solving x² = 5, we apply both inverse functions to y = 5
• Each inverse function produces one solution
• Example: For any positive number k, the equation x² = k has two solutions because we have "separate inverse functions for the left and right side of the graph"

🔍 Don't confuse

• The two solutions are not arbitrary; they come from the geometric fact that a horizontal line intersects the parabola twice
• Each intersection point corresponds to one of the two restricted one-to-one functions
• Without restricting the domain, we cannot define a single inverse function

📋 Summary of key facts

📋 Core requirements for inverses

The excerpt's summary section provides essential definitions:

"Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g."

"A function f is one-to-one if every equation f(x) = k has at most one solution."

📋 The one-to-one criterion

• Horizontal line test: a function is one-to-one if every horizontal line intersects the graph at most once
• Inverse existence: a function has an inverse if and only if the function is one-to-one
• If f(x) = k has more than one solution for some value k, then f is not one-to-one

📋 Domain and range relationships

When a function has an inverse:

• The domain of f equals the range of f⁻¹
• The range of f equals the domain of f⁻¹
• The graphs of f and f⁻¹ are mirror images across the line y = x

🧭 Overview

🧠 One-sentence thesis

A function has an inverse if and only if it is one-to-one, meaning each output corresponds to exactly one input, and the inverse function reverses the original function's input-output relationship.

📌 Key points (3–5)

• Definition of inverse functions: two functions are inverses if composing them in either order returns the original input.
• One-to-one requirement: a function must be one-to-one (each horizontal line crosses the graph at most once) to have an inverse.
• Domain and range swap: the domain of a function becomes the range of its inverse, and vice versa.
• Common confusion: non-one-to-one functions like y = x² don't have inverses unless you restrict the domain to make them one-to-one.
• Graphical relationship: a function and its inverse are mirror images across the line y = x.

🔄 What makes two functions inverses

🔄 The composition test

Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g.

• Both compositions must equal x—not just one direction.
• This means applying one function then the other "undoes" the operation and returns you to where you started.
• Example: if f takes an input, squares it, and adds 3, then f⁻¹ must subtract 3 and take the square root to reverse the process.

🎯 The one-to-one requirement

🎯 What one-to-one means

A function f is one-to-one if every equation f(x) = k has at most one solution.

• If any equation f(x) = k has more than one solution, the function is not one-to-one.
• In other words: different inputs must produce different outputs.
• Don't confuse: "at most one" means zero or one solution is allowed, but two or more solutions disqualifies the function.

📏 Horizontal line test

A function is one-to-one if every horizontal line intersects the function's graph at most once.

• This is the graphical version of the one-to-one definition.
• If any horizontal line crosses the graph twice (or more), the function fails the test.
• Example: the graph of y = x² fails because a horizontal line at y = 4 crosses at both x = 2 and x = -2.

🔗 Connection to inverses

• A function has an inverse if and only if the function is one-to-one.
• Every one-to-one function has an inverse.
• This is a two-way relationship: one-to-one ↔ has an inverse.

🔀 Domain and range relationships

🔀 The swap property

Original function	Inverse function
Domain of f	Range of f⁻¹
Range of f	Domain of f⁻¹

• The domain and range completely switch roles.
• This makes sense: if f maps inputs A to outputs B, then f⁻¹ maps inputs B back to outputs A.

📐 Graphical properties

📐 Mirror image across y = x

• The graph of a function and its inverse are mirror images of each other across the line y = x.
• This happens because the inverse swaps the x and y coordinates: the point (a, b) on f becomes (b, a) on f⁻¹.
• Reflecting across y = x is equivalent to swapping the x-axis and y-axis.

⚠️ When functions aren't one-to-one

⚠️ The y = x² problem

The excerpt illustrates why y = x² doesn't have an inverse over all real numbers:

• Starting with x = -7: if we try f⁻¹(y) = √y, then f⁻¹(f(-7)) = f⁻¹(49) = 7.
• But the inverse function formula requires f⁻¹(f(-7)) = -7.
• This produces the contradiction 7 = -7, proving √y cannot be the inverse.
• The root cause: the equation x² = 49 has two solutions (7 and -7), so the function is not one-to-one.

✂️ Restricting the domain

To create an inverse for y = x², split the graph into one-to-one pieces:

Domain restriction	Inverse function	Why it works
x ≥ 0 (non-negative)	f⁻¹(y) = √y	Each y-value corresponds to only one non-negative x
x ≤ 0 (non-positive)	f⁻¹(y) = -√y	Each y-value corresponds to only one non-positive x

• Each restricted piece passes the horizontal line test.
• This splitting explains why equations like x² = 5 have two solutions: one from each piece (√5 and -√5).
• Don't confuse: the original full function y = x² still has no inverse; only the restricted versions do.

🧭 Overview

🧠 One-sentence thesis

These exercises apply the theory of inverse functions to concrete problems involving domain restrictions, graphical analysis, and real-world scenarios where reversing a functional relationship is necessary.

📌 Key points (3–5)

• Core skill: finding inverse functions algebraically by swapping variables and solving for the new output, then determining correct domains and ranges.
• Domain restriction: functions like y = x² are not one-to-one on all real numbers, so they must be restricted (e.g., x ≥ 0 or x ≤ 0) to have inverses.
• Verification requirement: always check that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x to confirm the inverse relationship.
• Common confusion: the same function can yield different inverse formulas depending on which domain restriction you choose (left side vs. right side of a parabola).
• Applied contexts: inverse functions model "reversing" a process—finding time from a measurement, or horizontal position from height.

🔧 Finding inverse functions algebraically

🔧 Basic procedure

The excerpt provides multiple problems (9.1, 9.2, 9.7) that ask you to find inverse functions using a standard method:

1. Start with y = f(x).
2. Swap x and y.
3. Solve for y to get y = f⁻¹(x).
4. Specify the domain and range of f⁻¹.

Example (Problem 9.2a): For f(x) = (1/5)x + 8, you would write x = (1/5)y + 8, then solve for y to find the inverse rule.

✅ Verification step

Problem 9.7 explicitly requires testing each proposed inverse with two compositions:

• f(f⁻¹(x)) ?= x
• f⁻¹(f(x)) ?= x

Both must hold for all x in the appropriate domains. This confirms that the two functions truly undo each other.

📐 Domain and range swap

The excerpt reminds us (from the summary section):

The domain of a function is the range of its inverse, and the range of a function is the domain of its inverse.

When you find f⁻¹, you must specify its domain (which was the range of f) and compute its range (which was the domain of f).

🪓 Restricting domains for non-one-to-one functions

🪓 Why restriction is needed

The excerpt explains that y = x² is not one-to-one over all real numbers because equations like x² = 5 have two solutions (√5 and −√5). To create an inverse, you must split the graph into parts that pass the horizontal line test.

🪓 Two ways to restrict y = x²

The excerpt describes two cases:

• Non-negative x-values: Take f(x) = x² for x ≥ 0. Then f⁻¹(y) = √y (the positive square root).
• Non-positive x-values: Take f(x) = x² for x ≤ 0. Then f⁻¹(y) = −√y (the negative square root).

Each restriction yields a different inverse function. The graph of y = x² is split at the y-axis; one inverse comes from the right side, the other from the left side.

🪓 Parabola vertex problems

Problem 9.3 applies this idea to a general quadratic y = 2x² − 3x − 1:

• Part (a): Sketch the graph and find the vertex P = (a, b).
• Part (b): Explain why the function has no inverse on all real numbers (it fails the horizontal line test).
• Parts (c) and (d): Restrict the domain to {x ≥ a} or {x ≤ a} (the right or left side of the vertex) and find the corresponding inverse formulas and their domains/ranges.

Don't confuse: the same parabola gives two different inverse functions depending on which side you keep.

📊 Graphical and conceptual problems

📊 One-to-one identification

Problem 9.4 shows four graphs (A, B, C, D) and asks:

• Which are one-to-one?
• If not one-to-one, section the graph into parts that are one-to-one.

This tests the horizontal line test: a function is one-to-one if every horizontal line intersects its graph at most once.

📊 Self-inverse functions

Problem 9.5 asks you to show that f(x) = a + 1/(x − a) is its own inverse for every value of a. This means f(f(x)) = x. Such functions are symmetric across the line y = x.

📊 Mirror-image property

The summary states:

The graph of a function and its inverse are mirror images of each other across the line y = x.

When sketching f and f⁻¹ (as in Problems 9.1 and 9.7), the two graphs should reflect across the diagonal line y = x.

🌍 Applied inverse function problems

🌍 Rocket height above sloping ground (Problem 9.6)

Clovis launches a rocket from a cliff edge. The cliff slopes downward 4 feet vertically for every 1 foot horizontally. The rocket's path is y = −2x² + 120x.

• Part (a): Find h = f(x), the height of the rocket above the sloping ground (not above the horizontal x-axis).
• Part (b): Find the maximum height above the sloping ground and the corresponding x-coordinate.
• Part (c): While the rocket is going up, Clovis measures height h. Find x = g(h), the inverse function relating x-coordinate to h.
• Part (d): Does this inverse function still work when the rocket is going down? Explain.

The key issue: the rocket traces a parabola, so the same height h occurs at two different x-values (once going up, once going down). The inverse function is only valid on the restricted domain where the rocket is ascending (one-to-one).

🌍 Water trough with semicircular cross-section (Problem 9.8)

A trough has a semicircular cross-section with radius 5 feet. Water depth increases at 2 inches per hour.

• Part (a): Find w = f(t), the width of the water surface as a function of time t (in hours). Specify domain and range.
• Part (b): After how many hours will the surface width be 6 feet?
• Part (c): Find t = f⁻¹(w), the inverse function relating time to surface width. Specify domain and range.

This problem requires geometric reasoning (relating depth to width via the semicircle) and then inverting the relationship to answer "when" questions from "width" measurements.

🌍 Biochemical protein extract (Problem 9.9)

A function φ(t) monitors the amount (nanograms) of extract A remaining at time t (nanoseconds). You are given:

• φ is invertible.
• Specific values: φ(0) = 6, φ(1) = 5, φ(2) = 3, φ(3) = 1, φ(4) = 0.5, φ(10) = 0.

Questions:

• (a) At what time are there 3 nanograms remaining? (Use the inverse: find t such that φ(t) = 3, so t = φ⁻¹(3) = 2.)
• (b) What is φ⁻¹(0.5) and what does it mean? (Answer: 4 nanoseconds; it tells you when the amount is 0.5 nanograms.)
• (c) True or False: There is exactly one time when the amount is 4 nanograms. (True, because φ is invertible, so each output corresponds to exactly one input.)
• (d) Calculate φ(φ⁻¹(1)). (Answer: 1, by the definition of inverse.)
• (e) Calculate φ⁻¹(φ(6)). (This is asking for φ⁻¹ evaluated at φ(6); the problem does not give φ(6), so you cannot compute a numeric answer unless you interpret "6" as an input, in which case the answer is 6.)
• (f) What is the domain and range of φ? (Domain: [0, 10] or similar based on the data; range: [0, 6].)

Don't confuse: φ⁻¹(0.5) asks "what input gives output 0.5?" while φ(0.5) would ask "what is the output when input is 0.5?"

🔍 Summary of key definitions and tests

🔍 Inverse function definition

From the summary:

Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g.

🔍 One-to-one definition

From the summary:

A function f is one-to-one if every equation f(x) = k has at most one solution.

Equivalently, every horizontal line intersects the graph at most once.

🔍 Existence of inverses

From the summary:

A function has an inverse if the function is one-to-one, and every one-to-one function has an inverse.

If a function is not one-to-one, you must restrict its domain to a one-to-one piece before finding an inverse.

🧭 Overview

🧠 One-sentence thesis

Exponential functions interchange the roles of base and exponent from monomials, creating a fundamentally different class of functions that requires understanding non-integer powers and exhibits distinctive growth behavior.

📌 Key points (3–5)

• The key switch: exponential functions have the form y = b^x (variable in the exponent), whereas monomials have the form y = x^b (variable in the base).
• Why non-integer exponents matter: real-world problems (like population after 6.37 hours) require understanding expressions like 2^6.37, which cannot be computed with simple arithmetic alone.
• Standard exponential form: A(x) = A₀ · b^x, where A₀ is the initial value (the value when x = 0), b > 0 and b ≠ 1, and A₀ ≠ 0.
• Common confusion: it is easy to initially confuse exponential functions (y = b^x) with monomials (y = x^b) because they look symbolically similar but behave very differently.
• Key graph features: exponential functions are always positive, have a horizontal asymptote, and grow (or decay) without bound in one direction.

🔄 The fundamental distinction

🔄 Exponential vs monomial functions

The excerpt emphasizes a symbolic role-reversal:

Function type	Form	What varies	What is fixed
Monomial	y = x^b	x (the base)	b (positive integer exponent)
Exponential	y = b^x	x (the power/exponent)	b (positive base)

• In monomials, the input variable is raised to a fixed power.
• In exponential functions, a fixed base is raised to a variable power.
• Don't confuse: the positions of x and b are swapped, creating entirely different behavior.

🧮 Why we need non-integer exponents

The excerpt opens with a practical problem: computing population after 6.37 hours requires evaluating N(t) = (3 × 10⁶) · 2^6.37.

• Simple arithmetic rules do not suffice to calculate 2^6.37.
• We must understand expressions like 2^6.37, 2^√5, 2^(−π), etc.
• This motivates the need to review algebra for non-integer powers.
• The excerpt notes that fully understanding b^x for all real x requires Calculus (specifically limits), but we can build reasonable understanding using rational numbers.

🧱 Building blocks: rational exponents

🧱 nth roots and fractional exponents

To make sense of b^x when x is rational, we need the concept of nth roots.

An nth root of b is a solution t to the equation t^n = b.

The excerpt defines:

The symbols: ⁿ√b = b^(1/n) = the largest real nth root of b.

• For odd n: the equation t^n = b always has exactly one real solution for any b.
• For even n and b < 0: no real solutions exist (only complex solutions like ±i = ±√(−1), which the excerpt excludes).
• For even n and b > 0: there can be two solutions, but we choose the largest (positive) one to avoid ambiguity.
• Assumption going forward: the excerpt assumes b > 0 to avoid constant even/odd distinctions.

🧱 General rational exponents

For all positive integers p and q, and any real number base b > 0: b^(p/q) = (ⁿ√b)^p = ⁿ√(b^p).

Example: Any rational number r can be written as r = p/q where p and q are integers, so using exponent rules, y = b^x defines a function whenever x is rational.

📐 Rules of exponents

The excerpt lists five fundamental rules for rational exponents r and s, with positive bases a and b:

1. Product of power rule: b^r · b^s = b^(r+s)
2. Power of power rule: (b^r)^s = b^(rs)
3. Power of product rule: (ab)^r = a^r · b^r
4. Zero exponent rule: b^0 = 1
5. Negative power rule: b^(−r) = 1/(b^r)

These rules allow calculation and manipulation:

• Example: 27^(2/3) = (³√27)² = 3² = 9
• Example: 8^(−5/3) = (³√8)^(−5) = 2^(−5) = 1/(2^5) = 1/32

🌉 Extending to all real numbers

The excerpt acknowledges a gap:

• The "sticky point" is knowing that f(x) = b^x defines a function for all real values of x, not just rationals.
• This is not easy to verify and requires the concept of a limit from Calculus.
• The excerpt asks us to accept it as a fact.
• Once accepted, the exponent rules carry through for all real powers.
• Calculators have a "y to the x key" for computing expressions like π^√2 involving non-rational powers.

📋 Standard exponential form

📋 Definition and terminology

Definition: A function of exponential type has the form A(x) = A₀ · b^x, for some b > 0, b ≠ 1, and A₀ ≠ 0.

This is called the standard exponential form.

• A₀ is the initial value: if x represents time, then A(0) = A₀ · b^0 = A₀ is the value at time x = 0.
• b is the base (must be positive and not equal to 1).
• x is the power or exponent.

📋 Converting to standard form

Sometimes equations need manipulation to reach standard form.

Example from the excerpt: Write y = 8^(3x) and y = 7·(1/2)^(2x−1) in standard exponential form.

Solution:

• For y = 8^(3x): use power of power rule: y = (8³)^x = 512^x
• For y = 7·(1/2)^(2x−1):
  • Split the exponent: y = 7·(1/2)^(2x)·(1/2)^(−1)
  • Simplify: y = 7·((1/2)²)^x·2 = 14·(1/4)^x

📊 Key features of exponential graphs

📊 The graph of y = 2^x

The excerpt uses y = 2^x as a concrete example, building a table of values:

x	2^x	Point
−2	1/4	(−2, 1/4)
−1	1/2	(−1, 1/2)
0	1	(0, 1)
1	2	(1, 2)
2	4	(2, 4)
3	8	(3, 8)

📊 Four qualitative features

The excerpt identifies four key characteristics of the graph of y = 2^x:

1. Always positive: The graph is always above the horizontal axis; function values are always positive.
2. y-intercept and increasing: The graph has y-intercept 1 and is increasing (moving upward as x increases).
3. Horizontal asymptote (left): The graph becomes closer and closer to the horizontal axis as we move left; the x-axis is a horizontal asymptote for the left-hand portion.
4. Unbounded growth (right): The graph becomes higher and higher above the horizontal axis as we move to the right (the excerpt text cuts off here, but the pattern is clear).

Don't confuse: The horizontal asymptote is approached as x → −∞ (moving left), not as x → +∞ (moving right), where the function grows without bound.

🧭 Overview

🧠 One-sentence thesis

Exponential functions of the form y = A₀bˣ exhibit fundamentally different growth or decay behavior depending on whether the base b is greater than or less than 1, and these functions model phenomena where quantities change by a constant ratio over equal intervals.

📌 Key points (3–5)

• What exponential type functions are: functions of the form A(x) = A₀bˣ where b > 0, b ≠ 1, and A₀ ≠ 0; A₀ is the initial value (the value when x = 0).
• Two fundamentally different behaviors: when b > 1, the function exhibits exponential growth (rapidly increasing); when 0 < b < 1, it exhibits exponential decay (rapidly approaching zero).
• Common confusion: the case b = 1 is excluded because it produces only a horizontal line (y = 1), not an exponential function.
• Key graphical features: all exponential graphs pass through (0, A₀), stay above or below the x-axis (never cross it), and have the x-axis as a horizontal asymptote on one side.
• Standard form matters: equations must sometimes be rewritten using exponent rules to identify A₀ and b clearly.

📐 Exponent rules foundation

📐 Five essential rules

The excerpt lists five rules that make exponential expressions work for all real numbers:

Rule	Formula	What it means
Product of powers	bʳ · bˢ = bʳ⁺ˢ	Multiply same bases → add exponents
Power of power	(bʳ)ˢ = bʳˢ	Power raised to power → multiply exponents
Power of product	(ab)ʳ = aʳbʳ	Power distributes over multiplication
Zero exponent	b⁰ = 1	Any base to the zero equals 1
Negative power	b⁻ʳ = 1/bʳ	Negative exponent → reciprocal

🔢 Why these rules matter

• They allow us to define y = bˣ for all rational numbers x (numbers of the form p/q where p and q are integers).
• They enable calculation and manipulation of expressions.
• Example from the excerpt: 27 to the power 2/3 equals (cube root of 27) squared, which equals 3 squared, which equals 9.
• Example from the excerpt: 8 to the power negative 5/3 equals (cube root of 8) to the negative 5, which equals 2 to the negative 5, which equals 1/32.

⚠️ The calculus gap

• The excerpt acknowledges that proving y = bˣ works for all real numbers (not just rationals) requires the concept of a limit from Calculus.
• For this chapter, we accept it as fact.
• Calculators have a "y to the x key" that can compute non-rational powers like π to the square root of 2.

🎯 Standard exponential form

🎯 Definition and components

Exponential type function: A(x) = A₀bˣ, for some b > 0, b ≠ 1, and A₀ ≠ 0.

• A₀ is the initial value: when x = 0, A(0) = A₀b⁰ = A₀ · 1 = A₀.
• If x represents time, A₀ is the value at time zero (the starting value).
• b is the base: determines growth (b > 1) or decay (0 < b < 1).

🔄 Converting to standard form

Sometimes equations need manipulation using exponent rules to reach standard form.

Example from the excerpt: Write y = 8 · 3ˣ in standard form.

• Use power of power rule: 8 · 3ˣ = (8³)ˣ = 512ˣ
• Standard form: y = 512ˣ (here A₀ = 1, b = 512)

Example from the excerpt: Write y = 7 · (1/2) to the power (2x - 1) in standard form.

• Split the exponent: (1/2) to the (2x - 1) = (1/2) to the 2x times (1/2) to the negative 1
• Simplify: = ((1/2)²)ˣ · 2 = (1/4)ˣ · 2
• Multiply: y = 7 · 2 · (1/4)ˣ = 14 · (1/4)ˣ
• Standard form: y = 14 · (1/4)ˣ (here A₀ = 14, b = 1/4)

📊 The case b > 1: Exponential growth

📊 Key example: y = 2ˣ

The excerpt uses y = 2ˣ as the prototype. Sample points:

x	2ˣ	Point
-2	1/4	(-2, 1/4)
-1	1/2	(-1, 1/2)
0	1	(0, 1)
1	2	(1, 2)
2	4	(2, 4)
3	8	(3, 8)

📈 Four qualitative features

When b > 1, the graph of y = bˣ has these properties:

1. Always positive: the graph is always above the horizontal axis (function values are always positive).
2. y-intercept is 1: the graph passes through (0, 1) and is increasing.
3. Left-hand asymptote: as we move left (x → negative), the graph gets closer and closer to the x-axis; the x-axis is a horizontal asymptote for the left-hand portion.
4. Unbounded on the right: as we move right (x → positive), the graph becomes higher and higher; the graph is unbounded as we move to the right.

🚀 What "exponential growth" means

• The term codifies the fact that function values grow rapidly as we move to the right along the x-axis.
• Different values of b > 1 produce similar shapes, differing only in the exact amount of "concavity."
• All such graphs pass through (0, 1).

📉 The case 0 < b < 1: Exponential decay

📉 Understanding through reflection

When 0 < b < 1:

• Notice that 1/b > 1, so y = (1/b)ˣ is a growth function (previous case).
• Using exponent rules: bˣ = (1/b)⁻ˣ
• By the reflection principle, the graph of y = (1/b)⁻ˣ is the graph of y = (1/b)ˣ reflected about the y-axis.
• Therefore, the graph of y = bˣ (when 0 < b < 1) is a mirror image of a growth graph.

🔻 Four qualitative features (mirrored)

When 0 < b < 1, the graph of y = bˣ has these properties:

1. Always positive: still always above the horizontal axis.
2. y-intercept is 1: still passes through (0, 1).
3. Right-hand asymptote: as we move right (x → positive), the graph gets closer to the x-axis; the x-axis is a horizontal asymptote for the right-hand portion.
4. Unbounded on the left: as we move left (x → negative), the graph becomes higher and higher; unbounded as we move to the left.

🍂 What "exponential decay" means

• Function values rapidly approach zero as we move to the right along the x-axis.
• Example: radioactive decay, cooling, depreciation.
• Don't confuse: decay doesn't mean the function becomes negative; it means it approaches zero from above.

🔧 The case b = 1 and the role of A₀

🔧 Why b = 1 is excluded

• When b = 1, y = 1ˣ = 1 for all x.
• This is just a horizontal line, not an exponential function.
• The excerpt says "this is not too exciting" and "we will ignore this case."

🎚️ The effect of A₀

• If A₀ > 0: the graph of y = A₀bˣ is a vertically expanded or compressed version of y = bˣ.
  • The y-intercept becomes A₀ instead of 1.
  • All four qualitative features remain, but scaled vertically.
• If A₀ < 0: additionally reflect the graph about the x-axis.
  • The graph is now always below the horizontal axis.
  • Growth becomes "downward growth" (more negative); decay becomes "upward decay" (approaching zero from below).

🎹 Application: Piano frequency range

🎹 Pure tone model

The excerpt introduces a sound application:

• A pure tone causes eardrum motion modeled by d(t) = A · sin(2πft).
• f is the frequency in Hertz (Hz), meaning periods per unit time.
• A relates to displacement and loudness.
• Humans perceive sounds from 20 Hz to 20,000 Hz.

🎼 Piano tuning rule

• A piano keyboard has white keys (A, B, C, D, E, F, G repeating) and black keys (sharps: A#, C#, etc.).
• Tuning rule: each key has a frequency 2 to the power 1/12 times the frequency of the key to its immediate left.
• This makes the ratio of adjacent keys always the same: 2 to the power 1/12 ≈ 1.0595.
• Keys 12 keys apart have a frequency ratio of exactly 2, because (2 to the power 1/12) to the power 12 = 2.
• Two keys with frequency ratio 2 are called an octave apart.

🎵 Calculating frequencies

Starting from A below middle C at 220 Hz:

• A# (one key right): 220 × 2 to the power 1/12 ≈ 233.08 Hz.
• B (two keys right): 233.08 × 2 to the power 1/12 (or equivalently, 220 × 2 to the power 2/12).
• This is an exponential function: frequency = 220 × 2 to the power (n/12), where n is the number of keys to the right.
• Example: the A one octave up (12 keys right) has frequency 220 × 2 = 440 Hz.

🧭 Overview

🧠 One-sentence thesis

Piano tuning follows an exponential rule where each key's frequency is 2^(1/12) times the previous key's frequency, creating octaves (frequency ratio of 2) every 12 keys and allowing calculation of all keyboard frequencies from a single reference pitch.

📌 Key points (3–5)

• Pure tone model: Sound waves cause eardrum motion modeled by d(t) = A sin(2πft), where f is frequency in Hz and A relates to loudness.
• Human hearing range: People typically perceive sounds from 20 Hz to 20,000 Hz.
• Piano tuning rule: Each key has frequency 2^(1/12) times the key to its immediate left, making adjacent key ratios constant.
• Octave relationship: Keys 12 positions apart have frequencies in a 2:1 ratio because (2^(1/12))^12 = 2.
• Common confusion: The exponential spacing means frequency differences grow larger as you move right on the keyboard, even though the ratio stays constant.

🎵 Sound wave fundamentals

🎵 Pure tone mathematical model

A pure tone is modeled by the function d(t) = A sin(2πft), where f is the frequency and A is related to eardrum displacement.

• d(t): displacement of the eardrum at time t
• f: frequency, measured in "periods/unit time" called Hertz (Hz)
• A: coefficient related to actual eardrum displacement, which connects to loudness
• The function describes back-and-forth motion of the eardrum caused by sound waves

👂 Human perception range

• Typical human hearing: 20 Hz to 20,000 Hz
• This range sets the context for musical instruments like the piano
• Frequencies outside this range exist but cannot be perceived by most people

🎹 Piano keyboard structure

🎹 Key layout pattern

The keyboard follows a repeating 12-key sequence:

• White keys: A, B, C, D, E, F, G (repeating left to right)
• Black keys: sharps (e.g., A# between A and B)
• Complete 12-key sequence: A, A#, B, C, C#, D, D#, E, F, F#, G, G#, then repeats
• Not all adjacent white keys have black keys between them

🎹 Reference point

• The excerpt uses "A below middle C" as the reference frequency: 220 Hz
• Middle C is shown on the keyboard diagram as a labeled position
• All other frequencies can be calculated from this single reference

📐 Exponential tuning rule

📐 Adjacent key ratio

Piano tuning rule: each key (white and black) has a frequency 2^(1/12) times the frequency of the key to its immediate left.

• The ratio between any two adjacent keys is always the same: 2^(1/12)
• This constant ratio makes the tuning system exponential, not linear
• 2^(1/12) ≈ 1.059463094...

📐 Calculating specific frequencies

Starting from A = 220 Hz:

• A# (one key right): 220 × 2^(1/12) ≈ 233.08188 Hz
• B (two keys right): 233.08188 × 2^(1/12) ≈ 246.94165 Hz
• Each successive key multiplies by the same factor

Example: To find any key n positions to the right of A220, multiply 220 × (2^(1/12))^n.

🎼 Octave relationship

Two keys are an octave apart when they are 12 keys apart and have a frequency ratio of exactly 2.

• Why 12 keys: (2^(1/12))^12 = 2^(12/12) = 2^1 = 2
• Moving 12 keys right doubles the frequency
• Moving 12 keys left halves the frequency
• Example: If A = 220 Hz, then the A one octave higher = 220 × 2 = 440 Hz

🎼 Don't confuse ratio with difference

• The ratio between adjacent keys is constant (always 2^(1/12))
• The difference in Hz between adjacent keys grows as you move right
• Example: A to A# adds ~13 Hz, but at higher frequencies the same ratio adds more Hz
• This is characteristic of exponential growth, not linear spacing

🧭 Overview

🧠 One-sentence thesis

These exercises develop computational fluency with exponential and logarithmic functions through calculator practice, algebraic manipulation, real-world modeling (population growth, compound interest, biological systems), and applications requiring inverse-function reasoning.

📌 Key points (3–5)

• Calculator skills and standard form: Practice approximating exponential expressions and rewriting equations in the form y = a·bˣ.
• Exponential modeling from data: Two data points uniquely determine an exponential function; use this to build models and check consistency.
• Compound interest mechanics: Discrete compounding uses P(t) = P₀(1 + r/n)^(nt); continuous compounding uses P(t) = P₀e^(rt); more frequent compounding increases future value but approaches a limit.
• Common confusion: Different-looking exponential formulas (e.g., base e vs. base 2) can represent the same function—always check whether predictions match.
• Real-world constraints: Exponential models can predict unrealistic outcomes (e.g., population exceeding Earth's total); context and data validation matter.

🧮 Calculator and algebraic skills

🧮 Approximating exponential expressions

• Problem 10.1 drills basic calculator use: approximate values like 3^π, 4^(2+√5), π^π, etc.
• These are not "nice" numbers; you need a calculator's exponential key.
• Why it matters: Real models use irrational bases and exponents; fluency with technology is essential.

📐 Standard exponential form

Standard exponential form: y = a·bˣ, where a is the initial value and b is the base (growth factor).

• Problem 10.2 asks you to rewrite equations like y = 3(2^(−x)) or y = 4^(−x/2) into standard form.
• How: Use exponent rules (e.g., b^(−x) = (1/b)^x, b^(x/2) = (√b)^x) to isolate the exponent as a simple multiple of x.
• Example: y = 4^(−x/2) = (4^(1/2))^(−x) = 2^(−x) = (1/2)^x, so a = 1, b = 1/2.

🧬 Exponential modeling from data

🧬 Yeast population (Problem 10.3)

• Given: y = 10⁶ e^(0.495105t) models yeast cells at time t hours.
• (a) Plug in t = 1 to find the population after one hour.
• (b) "Doubles every 1.4 hours" means y(t + 1.4) = 2·y(t). Check whether e^(0.495105·1.4) ≈ 2.
• (c) Cherie's formula y = 10⁶(2.042727)^(0.693147t) looks different but may be equivalent.
  • Key insight: Different bases can represent the same exponential function if the exponent is scaled appropriately.
  • To check: evaluate both formulas at a few t values; if they match, no need to worry.
• (d) Anja's measurements: 7.246×10⁶ cells at t = 4, 16.504×10⁶ at t = 6.
  • Compare these to your model's predictions y(4) and y(6).
  • If they differ, decide whether your model over- or underestimates.

🎹 Piano frequencies (Problem 10.4)

• The excerpt mentions "frequency of middle C" and "A above middle C" but does not give the formula.
• (a–b) You would use the exponential model for musical pitch (not detailed in this excerpt).
• (c) "Lowest note on the keyboard": the problem hints at a shortcut instead of computing every key below A220.
• (d) Bosendorfer piano extends lower; find the frequency of that lowest C.

♟️ Doubling dimes on a chessboard (Problem 10.5)

• Place 1 dime on square 1, 2 on square 2, 4 on square 3, …, doubling each time.
• (a) Square 10: 2^(10−1) = 2⁹ = 512 dimes.
• (b) Square n: 2^(n−1) dimes.
• (c) Square 64: 2^63 dimes (a huge number).
• (d) Height: 2^63 mm; convert to km.
• (e) Compare to Earth–Sun distance (≈150 million km).
• Lesson: Exponential growth becomes astronomically large very quickly.

🩸 Biological oxygen transport (Problem 10.6)

🩸 Myoglobin and hemoglobin saturation

• Myoglobin saturation: M(p) = p/(1 + p)
• Hemoglobin saturation: H(p) = p^2.8 / (26^2.8 + p^2.8)

• Both functions give the fraction (0 to 1) of the molecule saturated with oxygen at pressure p torrs.
• (a) Two graphs are shown; identify which is M(p) and which is H(p).
  • Hint: M(1) = 0.5 (half-saturated at 1 torr); use this to match the graph.
• (b) Lungs at 100 torrs: compute H(100).
• (c) Active muscle at 20 torrs: compute M(20) and H(20).
  • Myoglobin in muscle picks up oxygen; hemoglobin in blood releases it.
• (d) Oxygen transport efficiency: M(p) − H(p).
  • Compute at 20, 40, 60 torrs.
  • Sketch the difference; look for a maximum.
  • Interpretation: Efficiency is highest where the gap between M and H is largest—this is where oxygen transfer from blood to muscle is most effective.

🔍 Don't confuse

• M(p) and H(p) are both increasing functions, but they have different shapes (different exponents and denominators).
• The difference M(p) − H(p) measures how much more saturated myoglobin is than hemoglobin at a given pressure—this is the driving force for oxygen delivery.

💰 Compound interest fundamentals (Chapter 11 excerpt)

💰 Discrete compounding formula

P(t) = P₀(1 + r/n)^(nt)

• P₀ = principal (initial amount)
• r = annual decimal interest rate (e.g., 8% → 0.08)
• n = number of compounding periods per year
• t = time in years

• Periodic rate: r/n is the interest rate applied each compounding period.
• Example: $1,000 at 8% compounded monthly (n = 12):
  • After 1 year: P(1) = 1,000(1 + 0.08/12)^12 ≈ $1,083.
• Why it's exponential: The base (1 + r/n) > 1, so the function grows exponentially.

🔄 Effect of increasing compounding frequency

• The excerpt shows a table: yearly, quarterly, monthly, weekly, daily, hourly compounding for $1,000 at 8%.
• As n increases, P(1) increases but approaches a limit: ≈$1,083.29.
• Key insight: More frequent compounding helps, but there is a ceiling—the continuous compounding limit.

🌀 Continuous compounding and the number e

e ≈ 2.71828… is defined as the limit of (1 + 1/z)^z as z becomes large.

• Rewrite discrete formula: set z = n/r, so P(t) = P₀[(1 + 1/z)^z]^(rt).
• As n → ∞, (1 + 1/z)^z → e, giving the continuous compounding formula:

  P(t) = P₀e^(rt)

• This is the maximum future value for a given rate r; it is always ≥ any discrete compounding result.

📊 Comparison table

Compounding	Formula	Example (P₀=$1,000, r=0.08, t=1)
Yearly (n=1)	P₀(1+r)^t	$1,080.00
Monthly (n=12)	P₀(1+r/n)^(nt)	$1,083.00
Continuous	P₀e^(rt)	$1,083.29

🎁 Uncle Hans's bond (Example 11.1.2)

• Purchase price: $2,500; maturity value: $5,000; r = 8.75%, compounded quarterly, t = 35 years.
• Future value: P(35) = 2,500(1 + 0.0875/4)^(4·35) ≈ $51,716.42.
• Capital gain: $51,716.42 − $2,500 = $49,216.42.
• Tax (28%): $13,780.60.
• Cash after tax: $37,935.82.
• Lesson: Taxes reduce the effective gain; always account for them in financial planning.

🔢 Exponential modeling exercises (Chapter 11.3 excerpt)

📈 Minimum wage model (Problem 11.1)

• Data: 1968 → $1.60/hour; 1976 → $2.30/hour.
• Let t = years after 1960, so two points are (8, 1.60) and (16, 2.30).
• (a) Find w(t) = w₀b^t using the two-point method (divide equations to solve for b, then find w₀).
• (b) Predict w(0) (the 1960 wage).
• (c) Compare model prediction for 1996 (t = 36) with actual $5.15.

🏘️ Pinedale population (Problem 11.2)

• 1990: 860 people; 1995: 1210 people.
• (a) Build both a linear model l(x) and an exponential model p(x), where x = years after 1990.
• (b) Predict population in 2000 (x = 10) using both models; compare.

🩺 AIDS cases model (Problem 11.3)

• Given cubic model: a(t) = 155(t − 1980)/10)³ (thousands).
• The paper was a "relief" because exponential growth would be much faster.
• (Task) Pick two points from a(t), build an exponential model b(t), and compare.
• Why "relief": Cubic growth is slower than exponential in the long run; exponential models predict runaway growth.

🌉 Hyperbolic functions and hanging cables (Problem 11.4)

• cosh(x) = (e^x + e^(−x))/2
• sinh(x) = (e^x − e^(−x))/2

• (a) Sketch both functions (cosh is even, always ≥1; sinh is odd).
• (b) Verify [cosh(x)]² − [sinh(x)]² = 1, so (cosh(a), sinh(a)) lies on the unit hyperbola x² − y² = 1.
• (c) A suspension bridge cable hangs as y = a·cosh((x − h)/a) + C.
  • Given: towers 100 ft high, 400 ft apart; a = 500, h = 0; roadway at y = 0.
  • Find minimum distance from cable to road (the lowest point of the catenary).

🔁 Inverse functions and logarithms (Chapter 12 excerpt)

🔁 Solving exponential equations

• Problem: How long until $1,000 at 8% continuous compounding grows to $5,000?
• Equation: 5,000 = 1,000e^(0.08t) → 5 = e^(0.08t).
• Need: The inverse of f(t) = e^t, called the natural logarithm ln(t).
• Apply ln to both sides: ln(5) = 0.08t → t = ln(5)/0.08 ≈ 20.12 years.

📐 The natural logarithm function

ln(c) is defined as the unique solution x to the equation c = e^x, for c > 0.

• Domain: All positive real numbers (the range of e^x).
• Range: All real numbers (the domain of e^x).
• Graph: Reflection of y = e^x across the line y = x.
• Key property: ln(e^x) = x and e^(ln(y)) = y (inverse-function identities).

🔍 Don't confuse

• ln(c) is only defined for c > 0; you cannot take the logarithm of zero or a negative number.
• The horizontal line test guarantees that e^x is one-to-one, so the inverse function exists and is unique.

🧭 Overview

🧠 One-sentence thesis

The natural logarithm function is the inverse of the exponential function, enabling us to solve equations where the unknown appears in an exponent, and logarithmic scales (such as decibels) help measure quantities that vary over enormous ranges.

📌 Key points (3–5)

• Why logarithms exist: algebraic manipulation alone cannot solve equations like 5 = e^(0.08t); the inverse function of e^x, called ln(x), is needed.
• Domain and range: ln(x) is defined only for positive x; its range is all real numbers; the graph has an x-intercept at 1 and a vertical asymptote at the y-axis.
• Key properties: ln(e^x) = x and e^(ln(x)) = x; ln(b^t) = t·ln(b); ln(ab) = ln(a) + ln(b); ln(a/b) = ln(a) − ln(b).
• General logarithms: log_b(x) is the inverse of b^x; any log_b(x) can be converted to natural log via log_b(x) = ln(x)/ln(b).
• Real-world application: sound pressure level (decibels) uses a logarithmic scale because the human ear detects intensities over a huge range; β = 10·log₁₀(I/I₀).

🔄 The natural logarithm as an inverse function

🔄 Why we need an inverse

• The motivating problem: if 1,000·e^(0.08t) = 5,000, then 5 = e^(0.08t).
• Algebraic manipulation cannot isolate t when it appears in the exponent.
• Solution: apply the inverse function of e^x to both sides.
• The inverse is denoted ln(x) and called the natural logarithm.
• Example: ln(5) ≈ 1.60944, so t = ln(5)/0.08 ≈ 20.12 years.

📊 Graphical properties of y = e^x

• Every horizontal line above the x-axis crosses y = e^x exactly once (horizontal line test).
• Horizontal lines at or below the x-axis miss the graph entirely.
• Domain: all real numbers; Range: all positive numbers.
• Because the function is one-to-one, the inverse exists.

📐 Definition of ln(x)

Natural logarithm: ln(c) is the unique solution of the equation c = e^x if c > 0; ln(c) is undefined if c ≤ 0.

• The domain of ln(x) is all positive numbers (the range of e^x).
• The range of ln(x) is all real numbers (the domain of e^x).
• The graph of y = ln(x) is obtained by flipping the graph of y = e^x across the line y = x.

🖼️ Graphical features of y = ln(x)

• Domain: positive numbers only; ln(−1) makes no sense.
• x-intercept: 1 (because e^0 = 1).
• Increasing: the graph rises as x increases.
• Vertical asymptote: the y-axis; as x approaches 0 from the right, ln(x) → −∞.
• Unbounded: as x → ∞, ln(x) → ∞.

🧮 Properties and symbolic manipulation

🧮 Core inverse properties

• (a) For any real x: ln(e^x) = x.
• (b) For any positive x: e^(ln(x)) = x.
• These are the fundamental inverse relationships.

🔢 Logarithm arithmetic rules

• (c) ln(b^t) = t·ln(b), for b > 0 and any real t.
• (d) ln(ba) = ln(a) + ln(b), for all a, b > 0.
• (e) ln(b/a) = ln(b) − ln(a), for all a, b > 0.
• These mirror the exponent rules and are essential for solving equations.

🧪 Example manipulations

• ln(8³) = 3·ln(8) ≈ 6.2383.
• ln(6π) = ln(6) + ln(π) ≈ 2.9365.
• ln(3/5) = ln(3) − ln(5) ≈ −0.5108.
• ln(√x) = ln(x^(1/2)) = (1/2)·ln(x).
• ln(x² − 1) = ln((x−1)(x+1)) = ln(x−1) + ln(x+1).
• ln(x⁵/(x²+1)) = 5·ln(x) − ln(x²+1).

🧩 Solving exponential equations

Example: Solve 3^(x+1) = 12.

• Take ln of both sides: ln(3^(x+1)) = ln(12).
• Use property (c): (x+1)·ln(3) = ln(12).
• Solve: x = ln(12)/ln(3) − 1 ≈ 1.2619.

Example: If $2,000 grows to $130,000 at continuous compounding, find the rate r over 12 years.

• 130,000 = 2,000·e^(12r) → 65 = e^(12r).
• ln(65) = 12r → r = ln(65)/12 ≈ 0.3479 (34.79% annual).

Example: Same principal at 6.4% annual; when does it reach $130,000?

• 65 = e^(0.064t) → t = ln(65)/0.064 ≈ 65.22 years.

🔁 Alternate exponential form

🔁 Rewriting A₀·b^t using e

• Any exponential function A(t) = A₀·b^t can be rewritten as A(t) = A₀·e^(at).
• Key insight: b^t = (e^(ln(b)))^t = e^(t·ln(b)).
• Set a = ln(b); then A(t) = A₀·e^(at).
• Implication: you only need the e^t and ln(t) keys on a calculator.

📈 Growth vs decay from the exponent

• If a > 0, the function A(t) = A₀·e^(at) exhibits exponential growth.
• If a < 0, the function exhibits exponential decay.
• Example: A(t) = 200·(2^t) = 200·e^(0.69315t) (growth, since 0.69315 > 0).
• Example: A(t) = 4·e^(−0.2t) = 4·(0.81873)^t (decay, since −0.2 < 0).

🔢 General logarithms (base b)

🔢 Definition of log_b(x)

Logarithm base b: log_b(c) is the unique solution of the equation c = b^x if c > 0; undefined if c ≤ 0.

• For b > 0 and b ≠ 1.
• The inverse function of y = b^x.
• Domain: all positive numbers; range: all real numbers.

📊 Graphical features of y = log_b(x)

• Domain: positive numbers only.
• x-intercept: 1.
• Increasing if b > 1; decreasing if 0 < b < 1.
• Vertical asymptote: the y-axis.
• Unbounded as x → ∞.

🔢 Properties of log_b

• (a) log_b(b^x) = x for any real x.
• (b) b^(log_b(x)) = x for any positive x.
• (c) log_b(r^t) = t·log_b(r), for r > 0 and any real t.
• (d) log_b(rs) = log_b(r) + log_b(s), for all r, s > 0.
• (e) log_b(r/s) = log_b(r) − log_b(s), for all r, s > 0.

🔄 Conversion formula

Log conversion formula: log_b(x) = ln(x)/ln(b), for x > 0 and b > 0, b ≠ 1.

• Derivation: start with y = log_b(x) → b^y = x → ln(b^y) = ln(x) → y·ln(b) = ln(x) → y = ln(x)/ln(b).
• Example: log₁₀(5) = ln(5)/ln(10) ≈ 0.699.
• Example: log₀.₀₂(11) = ln(11)/ln(0.02) ≈ −0.613.
• Don't confuse: "log(x)" on a calculator usually means log₁₀(x), not ln(x).

🧩 Solving with general logs

Example: A house bought for $80,000 grows at 4.8% annual interest compounded quarterly. When will it be worth $1,000,000?

• 1,000,000 = 80,000·(1.012)^(4t) → 12.5 = (1.012)^(4t).
• Apply log₁.₀₁₂ to both sides: log₁.₀₁₂(12.5) = 4t.
• t = ln(12.5)/(4·ln(1.012)) ≈ 52.934 years.

🔊 Application: measuring sound loudness

🔊 Why a logarithmic scale

• The human ear can hear a huge range of sound intensities.
• A linear scale would be impractical.
• A logarithmic scale compresses the range into manageable numbers.

🔊 Sound pressure level (decibels)

Sound pressure level: β = 10·log₁₀(I/I₀), where I₀ is the reference intensity (threshold of hearing) and I is the measured intensity.

• Units: decibels (db).
• At I = I₀, β = 10·log₁₀(1) = 0 db (threshold of hearing).
• At the threshold of pain, β ≈ 120 db.
• Example sound levels:
  • Threshold of hearing: 0 db
  • Rustle of leaves: 10 db
  • Whisper: 20 db
  • Ordinary conversation: 65 db
  • Busy street traffic: 70 db
  • Riveter: 95 db
  • Threshold of pain: 120 db

🔊 Comparing intensities

Example: Speaker A produces 87 db; Speaker B produces 93 db. What is the ratio of intensities?

• 87 = 10·log₁₀(I₁/I₀) → log₁₀(I₁) = 8.7 + log₁₀(I₀) → I₁ = 10^(8.7)·I₀.
• Similarly, I₂ = 10^(9.3)·I₀.
• Ratio: I₂/I₁ = 10^(9.3)/10^(8.7) = 10^(0.6) ≈ 3.98.
• Speaker B is nearly 4 times as intense as Speaker A, even though the decibel difference is only 6 db.

Example: What sound pressure level is twice the intensity of 87 db?

• I₃ = 2·I₁ = 2·10^(8.7)·I₀.
• β/10 = log₁₀(I₃/I₀) = log₁₀(2·10^(8.7)) = log₁₀(2) + 8.7.
• β = 10·(log₁₀(2) + 8.7) ≈ 90 db.
• Don't confuse: doubling intensity adds about 3 db, not doubling the decibel number.

🔊 Frequency dependence

• The threshold of hearing varies with frequency.
• At 20 Hz (low rumble), the threshold is about 100 db.
• At 2000 Hz (mid-range), the threshold drops to about 0 db.
• The threshold of pain is roughly 120 db across all frequencies from 20 Hz to 20,000 Hz.
• A hearing specialist can plot your personal "envelope of hearing" to diagnose problems.