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Lesson 3 — Linear (Affine) Functions

Linear function f(x) = ax + b. Slope as a CONSTANT rate of change — conceptual bridge to the derivative.

Used in: 1.º ano EM

f(x) = ax + b

Linear (affine) function: the coefficient a is the rate of change (how much y changes when x increases by 1 unit). The b is the value of y when x = 0. When b = 0, it is a linear function. When a = 0, it is a constant function.

Choose your door

Rigorous notation, full derivation, hypotheses

Definition and properties

$a$ : slope (angular coefficient, inclination)
$b$ : y-intercept (linear coefficient)
Graph: a straight line. $a > 0$ : increasing. $a < 0$ : decreasing. $a = 0$ : constant.

\frac{f(x_2) - f(x_1)}{x_2 - x_1} = a

(1)

what this means · Rate of change between two points on the line. For an affine function, this value is CONSTANT — it does not depend on which pair of points you choose. It is this constancy that characterizes the affine function among all others.

"The slope of a line that passes through two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = (y_2 - y_1)/(x_2 - x_1)$ ." — OpenStax College Algebra 2e, §2.2

Zero of the function and intercepts

$f(x) = 0 \iff x = -b/a$ (when $a \neq 0$ ). The pair $(0, b)$ is the vertical intercept. The pair $(-b/a, 0)$ is the zero (or horizontal intercept).

Uniqueness theorem: two points determine a line

Proof (sketch). Existence: define $a$ by the formula above and $b = y_1 - a x_1$ . One verifies $f(x_1) = y_1$ by construction, and $f(x_2) = a(x_2 - x_1) + y_1 = (y_2 - y_1) + y_1 = y_2$ . Uniqueness: if $g(x) = a' x + b'$ also satisfies $g(x_i) = y_i$ , then $a' = (y_2 - y_1)/(x_2 - x_1) = a$ and $b' = y_1 - a' x_1 = b$ . ∎

Composition and operations

Let $f(x) = a_1 x + b_1$ and $g(x) = a_2 x + b_2$ . Then:

Sum: $(f + g)(x) = (a_1 + a_2) x + (b_1 + b_2)$ — affine, with slopes added.
Composition: $(f \circ g)(x) = a_1 (a_2 x + b_2) + b_1 = a_1 a_2 x + (a_1 b_2 + b_1)$ — affine, with slopes multiplied.
Inverse (if $a_1 \neq 0$ ): $f^{-1}(y) = (y - b_1)/a_1$ — also affine, with slope $1/a_1$ .

The set of invertible affine functions ( $a \neq 0$ ) under composition forms a group — the structure $(\mathbb{R}^* \times \mathbb{R}, \circ)$ . This observation will be used in linear algebra (Lesson 31+) and in affine geometry.

Family of parallel lines

Family of lines with the same slope a = 1 and different intercepts b. Vertical translation: changing b only shifts the line up or down, without rotating it.

Family of concurrent lines

Family with the same intercept (0, 1) and different slopes — all crossing at that point. Rotation: changing a rotates the line around the intercept.

Worked examples

Five examples with increasing difficulty — from direct evaluation of a given line to modeling a break-even point for transport plans. Each example cites its source: the original problem always comes from an open-licensed book.

Example— 1· Example 1 — Identify coefficients and zero (application)

Problem: For $f(x) = -\dfrac{2}{3} x + 4$ , determine: slope $a$ , y-intercept $b$ , zero of the function, and the value of $f(6)$ .

Strategy: Direct comparison with the slope-intercept form $y = ax + b$ . The zero is $x_0 = -b/a$ ; $f(6)$ is straightforward numerical evaluation.

Solution:

Coefficients. By comparison: $a = -2/3$ , $b = 4$ .
Zero. $f(x) = 0 \iff -\tfrac{2}{3}x + 4 = 0 \iff x = 6$ .
$f(6)$ . $f(6) = -\tfrac{2}{3}(6) + 4 = -4 + 4 = 0$ . Consistent with item 2 — $x = 6$ is the zero.

Verification: Since $a < 0$ , the function is decreasing. At $x = 0$ , $f(0) = 4$ (vertical intercept). At $x = 6$ , it crosses the $x$ -axis. ✓

Source. OpenStax — College Algebra 2e §2.2 — CC-BY 4.0 license.

Example— 2· Example 2 — Line through two points (intermediate)

Problem: Determine the equation of the affine function whose graph passes through $(2, -1)$ and $(5, 8)$ .

Strategy: Apply the formula $a = (y_2 - y_1)/(x_2 - x_1)$ for the slope, then use one of the points to find $b$ .

Solution:

Slope. $a = \dfrac{8 - (-1)}{5 - 2} = \dfrac{9}{3} = 3$ .
Intercept. Using the point $(2, -1)$ : $-1 = 3 \cdot 2 + b \Rightarrow b = -7$ .
Equation. $f(x) = 3x - 7$ .

Verification: $f(2) = 6 - 7 = -1$ ✓ and $f(5) = 15 - 7 = 8$ ✓.

Source. OpenStax — College Algebra 2e §2.2 — CC-BY 4.0 license.

Example— 3· Example 3 — Parallel and perpendicular lines (intermediate)

Problem: Find the equation of the line parallel to $y = -2x + 7$ passing through $(3, 1)$ , and the equation of the line perpendicular to that same line passing through the same point.

Strategy: Parallel lines have the same slope ( $a' = a$ ). Perpendicular lines have slopes whose product is $-1$ ( $a' = -1/a$ ). In both cases, find $b'$ using the given point.

Solution:

Parallel line. Slope $a' = -2$ . Using $(3, 1)$ : $1 = -2(3) + b' \Rightarrow b' = 7$ . Equation: $y = -2x + 7$ . (Note — it coincides with the original line. This happens because $(3, 1)$ lies on the original line.)
Verification. $-2(3) + 7 = 1$ . ✓ Indeed, $(3, 1)$ belongs to the original line; the only parallel line is itself.
Perpendicular line. Slope $a' = -1/(-2) = 1/2$ . Using $(3, 1)$ : $1 = (1/2)(3) + b' \Rightarrow b' = -1/2$ . Equation: $y = \tfrac{1}{2}x - \tfrac{1}{2}$ .

Verification: At $x = 3$ : $\tfrac{1}{2}(3) - \tfrac{1}{2} = 1$ . ✓ Product of slopes: $(-2) \cdot (1/2) = -1$ . ✓

Source. Stitz–Zeager — Precalculus §2.1 — CC-BY-NC-SA license.

Example— 4· Example 4 — Composition of affine functions is affine (advanced)

Problem: Let $f(x) = 2x + 1$ and $g(x) = -3x + 5$ . Compute $f \circ g$ , $g \circ f$ and show that both are affine. Determine the resulting coefficients in each case.

Strategy: Substitute one function into the other and simplify to slope-intercept form. The composition of two affine functions is always affine — a key property that allows building groups of transformations.

Solution:

$f \circ g$ . $f(g(x)) = f(-3x + 5) = 2(-3x + 5) + 1 = -6x + 10 + 1 = -6x + 11$ . Coefficients: $a = -6$ , $b = 11$ . Note that the resulting $a$ is the product $2 \cdot (-3) = -6$ .
$g \circ f$ . $g(f(x)) = g(2x + 1) = -3(2x + 1) + 5 = -6x - 3 + 5 = -6x + 2$ . Coefficients: $a = -6$ , $b = 2$ .
Observation. Both have the same slope $-6$ (multiplication of the individual slopes — commutative) but different intercepts (not commutative in the constant term).

Verification: At $x = 0$ : $(f \circ g)(0) = f(5) = 11$ ✓; $(g \circ f)(0) = g(1) = 2$ ✓.

Source. OpenStax — College Algebra 2e §5.1 — CC-BY 4.0 license.

Example— 5· Example 5 — Real-world modeling: comparing taxi plans (real modeling)

Problem: Two taxi companies compete in the same city. Company X: base fare $6.00 +$ 2.80/km. Company Y: base fare $4.00 +$ 3.50/km. Above what distance is Company X cheaper?

Strategy: Model each cost as an affine function, set them equal to find the break-even point, and analyze the comparison for longer and shorter distances.

Solution:

Models. $C_X(d) = 6 + 2{,}80\,d$ and $C_Y(d) = 4 + 3{,}50\,d$ .
Break-even point. Set equal: $6 + 2{,}80\,d = 4 + 3{,}50\,d \Rightarrow 2 = 0{,}70\,d \Rightarrow d = 2/0{,}70 \approx 2{,}86$ km.
Comparison. For $d < 2{,}86$ km, Company Y is cheaper (smaller base fare wins). For $d > 2{,}86$ km, Company X is cheaper (lower per-km rate wins).

Verification: At $d = 5$ km: $C_X = 6 + 14 = 20{,}00$ ; $C_Y = 4 + 17{,}50 = 21{,}50$ . ✓ X is cheaper. At $d = 1$ km: $C_X = 8{,}80$ , $C_Y = 7{,}50$ . ✓ Y is cheaper.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 1 — free license.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 20Understanding 10Modeling 12Challenge 3

Ex. 3.1Modeling
Terry is hiking down a steep hill. Her elevation, $E(t)$ , in feet after $t$ seconds is $E(t) = 3000 - 70t$ . What does this affine model tell us about her initial elevation and how it changes over time?
Select the correct option
Initial elevation of 3,000 feet, decreasing 70 feet per second.Initial elevation of 70 feet, increasing 3,000 feet per second.Constant elevation of 3,000 feet.Initial elevation of 3,000 feet, increasing 70 feet per second.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 1
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In $E(t) = 3000 - 70t$ , the initial value $E(0) = 3000$ is the starting elevation (feet), and $-70$ is the rate of change: the elevation drops 70 feet each second.
Show step-by-step (with the why)
1. Compare with $f(t) = at + b$ : here $a = -70$ and $b = 3000$ .
2. The y-intercept $b = 3000$ is the value at $t = 0$ : the initial elevation.
3. The slope $a = -70$ indicates a drop of 70 feet per second.
4. Conclusion: starts at 3,000 feet and descends 70 ft/s.
Ex. 3.2ModelingAnswer key
Jessica is walking home from a friend's house. After 2 minutes she is 1.4 miles from home; 12 minutes after leaving, she is 0.9 miles away. What is her speed in miles per hour?
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3 miles per hour.0.5 miles per hour.1.4 miles per hour.0.05 miles per hour.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 2
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The rate of change is $a = (0{,}9 - 1{,}4)/(12 - 2) = -0{,}5/10 = -0{,}05$ miles per minute. Converting: $0{,}05 \times 60 = 3$ miles per hour.
Show step-by-step (with the why)
1. Compute the change in distance: $0{,}9 - 1{,}4 = -0{,}5$ mi in 10 minutes.
2. Rate per minute: $0{,}5/10 = 0{,}05$ mi/min.
3. Convert to mi/h: $0{,}05 \times 60 = 3$ mi/h.
Ex. 3.3ModelingAnswer key
A boat is 100 miles from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after $t$ hours.
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$D(t) = 100 - 10t$ , where $t$ is the time in hours. $D(t) = 10t$ . $D(t) = 100 + 10t$ . $D(t) = 100t - 10$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 3
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The boat starts 100 miles away and approaches at 10 mi/h, so $D(t) = 100 - 10t$ . The slope $a = -10$ indicates the distance decreases by 10 miles per hour.
Ex. 3.4Understanding
Determine whether the function $f(x) = 4x + 3$ is increasing or decreasing.
Select the correct option
Increasing, because the slope is positive.Decreasing, because the y-intercept is positive.Constant.Decreasing, because the slope is negative.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 14
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In $f(x) = 4x + 3$ , the slope $a = 4 > 0$ , therefore the function is increasing.
Ex. 3.5Understanding
Determine whether the function $a(x) = 5 - 2x$ is increasing or decreasing.
Select the correct option
Decreasing, because the slope is negative.Increasing, because the constant term is positive.Constant.Increasing, because the slope is positive.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 16
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In $a(x) = 5 - 2x$ , the slope $a = -2 < 0$ , therefore the function is decreasing.
Ex. 3.6Understanding
Determine whether the function $b(x) = 8 - 3x$ is increasing or decreasing.
Select the correct option
Decreasing, because the slope is negative.Increasing, because the y-intercept is positive.Constant.Increasing, because the slope is positive.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 17
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In $b(x) = 8 - 3x$ , the slope $a = -3 < 0$ , therefore the function is decreasing.
Ex. 3.7UnderstandingAnswer key
Determine whether the function $j(x) = \dfrac{1}{2}x - 3$ is increasing or decreasing.
Select the correct option
Increasing, because the slope is positive.Decreasing, because there is a negative term.Constant.Decreasing, because the y-intercept is negative.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 20
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In $j(x) = \frac{1}{2}x - 3$ , the slope $a = \frac{1}{2} > 0$ , therefore the function is increasing.
Ex. 3.8Application
Calculate the slope of the line passing through the points $(2, 4)$ and $(4, 10)$ .
Select the correct option
$3$ $2$ $6$ $1$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 24
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The slope is $m = \frac{10 - 4}{4 - 2} = \frac{6}{2} = 3$ .
Ex. 3.9ApplicationAnswer key
Calculate the slope of the line passing through the points $(1, 5)$ and $(4, 11)$ .
Select the correct option
$2$ $3$ $\dfrac{1}{2}$ $6$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 25
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The slope is $m = \frac{11 - 5}{4 - 1} = \frac{6}{3} = 2$ .
Ex. 3.10Application
Calculate the slope of the line passing through the points $(-1, 4)$ and $(5, 2)$ .
Select the correct option
$-\dfrac{1}{3}$ $\dfrac{1}{3}$ $3$ $-3$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 26
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The slope is $m = \frac{2 - 4}{5 - (-1)} = \frac{-2}{6} = -\frac{1}{3}$ .
Ex. 3.11Application
Calculate the slope of the line passing through the points $(8, -2)$ and $(4, 6)$ .
Select the correct option
$-2$ $2$ $\dfrac{1}{2}$ $-\dfrac{1}{2}$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 27
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The slope is $m = \frac{6 - (-2)}{4 - 8} = \frac{8}{-4} = -2$ .
Ex. 3.12ApplicationAnswer key
Calculate the slope of the line passing through the points $(6, 11)$ and $(-4, 3)$ .
Select the correct option
$\dfrac{4}{5}$ $-\dfrac{4}{5}$ $\dfrac{5}{4}$ $2$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 28
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The slope is $m = \frac{3 - 11}{-4 - 6} = \frac{-8}{-10} = \frac{4}{5}$ .
Show step-by-step (with the why)
1. Compute $\Delta y = 3 - 11 = -8$ .
2. Compute $\Delta x = -4 - 6 = -10$ .
3. Divide: $m = -8 / -10 = 4/5$ .
Ex. 3.13ApplicationAnswer key
Find the equation of the affine function satisfying $f(-5) = -4$ and $f(5) = 2$ .
Select the correct option
$f(x) = \dfrac{3}{5}x - 1$ $f(x) = \dfrac{3}{5}x + 1$ $f(x) = -\dfrac{3}{5}x - 1$ $f(x) = 5x + 2$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 29
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The slope is $m = \frac{2-(-4)}{5-(-5)} = \frac{6}{10} = \frac{3}{5}$ . Using $f(-5) = -4$ : $b = -4 - \frac{3}{5}(-5) = -4 + 3 = -1$ . Therefore $f(x) = \frac{3}{5}x - 1$ .
Ex. 3.14Application
Find the equation of the affine function satisfying $f(-1) = 4$ and $f(5) = 1$ .
Select the correct option
$f(x) = -\dfrac{1}{2}x + \dfrac{7}{2}$ $f(x) = \dfrac{1}{2}x + \dfrac{7}{2}$ $f(x) = -\dfrac{1}{2}x - \dfrac{7}{2}$ $f(x) = 2x + 7$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 30
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The slope is $m = \frac{1-4}{5-(-1)} = \frac{-3}{6} = -\frac{1}{2}$ . Using $(-1, 4)$ : $b = 4 - (-\frac{1}{2})(-1) = 4 - \frac{1}{2} = \frac{7}{2}$ . Therefore $f(x) = -\frac{1}{2}x + \frac{7}{2}$ .
Ex. 3.15Application
Find the equation of the affine function whose graph passes through the points $(2, 4)$ and $(4, 10)$ .
Select the correct option
$f(x) = 3x - 2$ $f(x) = 3x + 2$ $f(x) = -3x - 2$ $f(x) = \dfrac{1}{3}x - 2$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 31
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The slope is $m = \frac{10 - 4}{4 - 2} = 3$ . Using $(2, 4)$ : $b = 4 - 3(2) = -2$ . Therefore $f(x) = 3x - 2$ .
Ex. 3.16Application
Find the equation of the affine function whose zero is $x = -2$ (x-intercept: $(-2, 0)$ ) and whose y-intercept is $(0, -3)$ .
Select the correct option
$f(x) = -\dfrac{3}{2}x - 3$ $f(x) = \dfrac{3}{2}x - 3$ $f(x) = -\dfrac{3}{2}x + 3$ $f(x) = -\dfrac{2}{3}x - 3$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 35
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The zero is $(-2, 0)$ and the y-intercept is $(0, -3)$ . The slope is $m = \frac{-3 - 0}{0 - (-2)} = -\frac{3}{2}$ . Therefore $f(x) = -\frac{3}{2}x - 3$ .
Show step-by-step (with the why)
1. Identify the two points: $(-2, 0)$ and $(0, -3)$ .
2. Compute the slope: $m = (-3 - 0)/(0 - (-2)) = -3/2$ .
3. The y-intercept is $b = -3$ (given directly).
4. Write: $f(x) = -\frac{3}{2}x - 3$ .
Ex. 3.17Application
Find the equation of the affine function whose zero is $x = -5$ (x-intercept: $(-5, 0)$ ) and whose y-intercept is $(0, 4)$ .
Select the correct option
$f(x) = \dfrac{4}{5}x + 4$ $f(x) = -\dfrac{4}{5}x + 4$ $f(x) = \dfrac{4}{5}x - 4$ $f(x) = \dfrac{5}{4}x + 4$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 36
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The zero is $(-5, 0)$ and the y-intercept is $(0, 4)$ . The slope is $m = \frac{4 - 0}{0 - (-5)} = \frac{4}{5}$ . Therefore $f(x) = \frac{4}{5}x + 4$ .
Ex. 3.18Understanding
Are the lines $4x - 7y = 10$ and $7x + 4y = 1$ parallel, perpendicular, or neither?
Select the correct option
Perpendicular, because the product of the slopes is $-1$ .Parallel, because they have the same slope.Neither parallel nor perpendicular.Coincident.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 37
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Converting: $4x - 7y = 10 \Rightarrow y = \frac{4}{7}x - \frac{10}{7}$ and $7x + 4y = 1 \Rightarrow y = -\frac{7}{4}x + \frac{1}{4}$ . The product of the slopes is $\frac{4}{7} \cdot (-\frac{7}{4}) = -1$ , so they are perpendicular.
Show step-by-step (with the why)
1. Isolate $y$ in the first equation: $m_1 = 4/7$ .
2. Isolate $y$ in the second equation: $m_2 = -7/4$ .
3. Compute $m_1 \cdot m_2 = (4/7)(-7/4) = -1$ .
4. Conclude: perpendicular.
Ex. 3.19Understanding
Are the lines $3y + x = 12$ and $-y = 8x + 1$ parallel, perpendicular, or neither?
Select the correct option
Neither parallel nor perpendicular.Parallel.Perpendicular.Coincident.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 38
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Converting: $3y + x = 12 \Rightarrow y = -\frac{1}{3}x + 4$ , slope $-1/3$ . And $-y = 8x + 1 \Rightarrow y = -8x - 1$ , slope $-8$ . The slopes are different and the product $(-1/3)(-8) = 8/3 \neq -1$ , so they are neither parallel nor perpendicular.
Ex. 3.20UnderstandingAnswer key
Are the lines $3y + 4x = 12$ and $-6y = 8x + 1$ parallel, perpendicular, or neither?
Select the correct option
Parallel, because both have slope $-\dfrac{4}{3}$ .Perpendicular.Neither parallel nor perpendicular.Coincident.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 39
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Converting: $3y + 4x = 12 \Rightarrow y = -\frac{4}{3}x + 4$ and $-6y = 8x + 1 \Rightarrow y = -\frac{4}{3}x - \frac{1}{6}$ . Both have slope $-4/3$ , so they are parallel.
Ex. 3.21Understanding
Are the lines $6x - 9y = 10$ and $3x + 2y = 1$ parallel, perpendicular, or neither?
Select the correct option
Perpendicular.Parallel.Neither parallel nor perpendicular.Coincident.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 40
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Converting: $6x - 9y = 10 \Rightarrow y = \frac{2}{3}x - \frac{10}{9}$ , slope $2/3$ . And $3x + 2y = 1 \Rightarrow y = -\frac{3}{2}x + \frac{1}{2}$ , slope $-3/2$ . Product: $(2/3)(-3/2) = -1$ , so they are perpendicular.
Ex. 3.22Application
Find the zero (x-intercept) and the y-intercept of the function $f(x) = -x + 2$ .
Select the correct option
Zero at $x = 2$ ; y-intercept at $y = 2$ .Zero at $x = -2$ ; y-intercept at $y = -2$ .Zero at $x = 2$ ; y-intercept at $y = -2$ .Zero at $x = -2$ ; y-intercept at $y = 2$ .
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 41
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For $f(x) = -x + 2$ : the zero is $f(x)=0 \Rightarrow x = 2$ ; the y-intercept is $f(0) = 2$ .
Ex. 3.23Application
Find the zero (x-intercept) and the y-intercept of the function $g(x) = 2x + 4$ .
Select the correct option
Zero at $x = -2$ ; y-intercept at $y = 4$ .Zero at $x = 2$ ; y-intercept at $y = -4$ .Zero at $x = 4$ ; y-intercept at $y = -2$ .Zero at $x = -4$ ; y-intercept at $y = 2$ .
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 42
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For $g(x) = 2x + 4$ : the zero is $2x + 4 = 0 \Rightarrow x = -2$ ; the y-intercept is $g(0) = 4$ .
Ex. 3.24Application
Find the zero (x-intercept) and the y-intercept of the function $h(x) = 3x - 5$ .
Select the correct option
Zero at $x = \dfrac{5}{3}$ ; y-intercept at $y = -5$ .Zero at $x = -5$ ; y-intercept at $y = \dfrac{5}{3}$ .Zero at $x = 5$ ; y-intercept at $y = -\dfrac{5}{3}$ .Zero at $x = -\dfrac{5}{3}$ ; y-intercept at $y = 5$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 43
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For $h(x) = 3x - 5$ : the zero is $3x - 5 = 0 \Rightarrow x = 5/3$ ; the y-intercept is $h(0) = -5$ .
Ex. 3.25Application
Write the equation of the line parallel to $f(x) = -5x - 3$ that passes through $(2, -12)$ .
Select the correct option
$f(x) = -5x - 2$ $f(x) = -5x + 2$ $f(x) = 5x - 2$ $f(x) = \dfrac{1}{5}x - 2$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 52
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A line parallel to $f(x) = -5x - 3$ has slope $-5$ . Using $(2, -12)$ : $b = -12 - (-5)(2) = -12 + 10 = -2$ . Therefore $f(x) = -5x - 2$ .
Ex. 3.26ApplicationAnswer key
Write the equation of the line parallel to $g(x) = 3x - 1$ that passes through $(4, 9)$ .
Select the correct option
$g(x) = 3x - 3$ $g(x) = 3x + 3$ $g(x) = -3x - 3$ $g(x) = \dfrac{1}{3}x - 3$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 53
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A line parallel to $g(x) = 3x - 1$ has slope $3$ . Using $(4, 9)$ : $b = 9 - 3(4) = 9 - 12 = -3$ . Therefore $g(x) = 3x - 3$ .
Ex. 3.27Application
Write the equation of the line perpendicular to $h(t) = -2t + 4$ that passes through $(-4, -1)$ .
Select the correct option
$h(t) = \dfrac{1}{2}t + 1$ $h(t) = -\dfrac{1}{2}t + 1$ $h(t) = 2t + 1$ $h(t) = \dfrac{1}{2}t - 1$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 54
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A line perpendicular to $h(t) = -2t + 4$ has slope $m = -1/(-2) = 1/2$ . Using $(-4, -1)$ : $b = -1 - (1/2)(-4) = -1 + 2 = 1$ . Therefore $h(t) = \frac{1}{2}t + 1$ .
Show step-by-step (with the why)
1. Slope of $h(t) = -2t + 4$ is $-2$ .
2. Perpendicular slope: $m = -1/(-2) = 1/2$ .
3. Using $(-4, -1)$ : $b = -1 - (1/2)(-4) = 1$ .
4. Result: $h(t) = \frac{1}{2}t + 1$ .
Ex. 3.28ApplicationAnswer key
Write the equation of the line perpendicular to $p(t) = 3t + 4$ that passes through $(3, 1)$ .
Select the correct option
$p(t) = -\dfrac{1}{3}t + 2$ $p(t) = 3t + 2$ $p(t) = -\dfrac{1}{3}t - 2$ $p(t) = \dfrac{1}{3}t + 2$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 55
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A line perpendicular to $p(t) = 3t + 4$ has slope $m = -1/3$ . Using $(3, 1)$ : $b = 1 - (-1/3)(3) = 1 + 1 = 2$ . Therefore $p(t) = -\frac{1}{3}t + 2$ .
Ex. 3.29Understanding
Can the table below represent a linear function? If so, find the equation.
$x$ 0 5 10 15
$f(x)$ $-5$ $20$ $45$ $70$
Select the correct option
Yes, it is linear: $f(x) = 5x - 5$ .Yes, it is linear: $f(x) = 5x + 5$ .No, it is not linear, because the change in $x$ is not constant.Yes, it is linear: $f(x) = 25x$ .
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 91
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Checking the differences: $\Delta x = 5$ and $\Delta f = 25$ in all intervals, so $m = 25/5 = 5$ . Using $(0,-5)$ : $f(x) = 5x - 5$ .
Ex. 3.30UnderstandingAnswer key
Can the table below represent a linear function? If so, find the equation.
$x$ 2 4 6 8
$f(x)$ $-4$ $16$ $36$ $56$
Select the correct option
Yes, it is linear: $f(x) = 10x - 24$ .Yes, it is linear: $f(x) = 10x + 24$ .Not linear.Yes, it is linear: $f(x) = 5x - 24$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 95
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Checking: $\Delta x = 2$ and $\Delta f = 20$ in all intervals, so $m = 20/2 = 10$ . Using $(2, -4)$ : $b = -4 - 10(2) = -24$ . Therefore $f(x) = 10x - 24$ .
Ex. 3.31ChallengeAnswer key
If $f$ is a linear function with $f(0{,}1) = 11{,}5$ and $f(0{,}4) = -5{,}9$ , find the equation of $f$ .
Select the correct option
$f(x) = -58x + 17{,}26$ $f(x) = 58x - 17{,}26$ $f(x) = -58x - 17{,}26$ $f(x) = 58x + 17{,}26$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 97
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The slope is $m = \frac{-5{,}9 - 11{,}5}{0{,}4 - 0{,}1} = \frac{-17{,}4}{0{,}3} = -58$ . Using $(0{,}1,\; 11{,}5)$ : $b = 11{,}5 - (-58)(0{,}1) = 11{,}5 + 5{,}8 = 17{,}3$ . Therefore $f(x) = -58x + 17{,}3$ .
Ex. 3.32Modeling
At noon, a barista has $20 in the tip jar. If they receive an average of$ 0.50 per customer, what is the affine function that models the total amount in the jar after serving $n$ customers?
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$T(n) = 20 + 0{,}50\,n$ (in currency units, with $n$ customers served). $T(n) = 0{,}50\,n$ . $T(n) = 20n + 0{,}50$ . $T(n) = 20 - 0{,}50\,n$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 114
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The initial amount is 20 (y-intercept) and it grows by 0.50 per customer (slope). Therefore $T(n) = 20 + 0{,}50n$ .
Ex. 3.33Modeling
A gym membership with 2 personal training sessions costs $125, and with 5 sessions costs$ 260. What is the cost per individual session?
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$45 per session.$125 per session.$65 per session.$30 per session.
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 115
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The price difference for 3 extra sessions is $260 - 125 = 135$ . Cost per session: $135 / 3 = 45$ .
Show step-by-step (with the why)
1. Cost difference: $260 - 125 = 135$ .
2. Session difference: $5 - 2 = 3$ sessions.
3. Cost per session: $135 / 3 = 45$ .
Ex. 3.34Modeling
A city's population was growing linearly. In 2003 there were 45,000 inhabitants, with a growth of 1,700 people per year. Write the function $P(t)$ that models the population.
Select the correct option
$P(t) = 45\ 000 + 1\ 700\,t$ , where $t$ is years after 2003. $P(t) = 1\ 700\,t$ . $P(t) = 45\ 000 - 1\ 700\,t$ . $P(t) = 45\ 000 + 1\ 700$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 120
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The population starts at 45,000 in 2003 and grows by 1,700 per year. Therefore $P(t) = 45000 + 1700t$ , where $t = 0$ corresponds to 2003.
Ex. 3.35Modeling
When the temperature is 0 °C, Fahrenheit is 32. When Celsius is 100, Fahrenheit is 212. Express Fahrenheit as an affine function of Celsius.
Select the correct option
$F = 1{,}8C + 32$ . $F = C + 32$ . $F = 1{,}8C - 32$ . $F = \dfrac{C}{1{,}8} + 32$ .
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 122
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Two points: $(C, F) = (0, 32)$ and $(100, 212)$ . The slope is $m = (212-32)/(100-0) = 180/100 = 1{,}8$ . With $b = 32$ : $F = 1{,}8C + 32$ .
Show step-by-step (with the why)
1. Identify two known points: $(0, 32)$ and $(100, 212)$ .
2. Compute the slope: $m = (212-32)/100 = 1{,}8$ .
3. The y-intercept is $b = 32$ (when $C = 0$ ).
4. Write: $F = 1{,}8C + 32$ .
Ex. 3.36Modeling
A city's population was decreasing at a constant rate. In 2010 it was 5,900; in 2012 it fell to 4,700. Write the affine function $P(t)$ modeling the population, with $t = 0$ in 2010.
Select the correct option
$P(t) = 5\ 900 - 600\,t$ , where $t$ is years after 2010. $P(t) = 5\ 900 + 600\,t$ . $P(t) = 600\,t$ . $P(t) = 5\ 900 - 1\ 200\,t$ .
Select an option first
Solve online OpenStax College Algebra 2e · §4.2 · ex. 9
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Rate of change: $(4700 - 5900)/(2012 - 2010) = -1200/2 = -600$ inhabitants/year. With $P(0) = 5900$ : $P(t) = 5900 - 600t$ .
Ex. 3.37Modeling
In 2004, a school's population was 1,001. In 2008 it grew to 1,697. Assuming linear growth, what was the annual growth rate?
Select the correct option
174 people per year.696 people per year.2,784 people per year.1,001 people per year.
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Solve online OpenStax College Algebra 2e · §4.2 · ex. 45
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Growth rate: $(1697 - 1001)/(2008 - 2004) = 696/4 = 174$ people/year.
Ex. 3.38Challenge
You are choosing between two prepaid cell phone plans. Plan A charges $0.26/minute. Plan B charges a flat fee of$ 19.95 plus $0.11/minute. After how many minutes does Plan B become cheaper?
Select the correct option
After 133 minutes.After 100 minutes.After 200 minutes.After 66 minutes.
Select an option first
Solve online OpenStax College Algebra 2e · §4.2 · ex. 53
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Plan 1: $C_1(m) = 0{,}26m$ . Plan 2: $C_2(m) = 19{,}95 + 0{,}11m$ . Setting equal: $0{,}26m = 19{,}95 + 0{,}11m \Rightarrow 0{,}15m = 19{,}95 \Rightarrow m = 133$ minutes.
Show step-by-step (with the why)
1. Model: $C_1 = 0{,}26m$ and $C_2 = 19{,}95 + 0{,}11m$ .
2. Set equal to find the break-even: $0{,}26m = 19{,}95 + 0{,}11m$ .
3. Isolate $m$ : $0{,}15m = 19{,}95$ .
4. Solve: $m = 19{,}95 / 0{,}15 = 133$ minutes.
Ex. 3.39Challenge
At a new sales job, you have two salary options:
- Option A: base salary of $17,000/year + 12% commission on sales.
- Option B: base salary of $20,000/year + 5% commission on sales.
Above what sales value does Option A earn more?
Select the correct option
Above approximately $42,857 in sales.Above $17,000 in sales.Above $50,000 in sales.Above $12,000 in sales.
Select an option first
Solve online OpenStax College Algebra 2e · §4.2 · ex. 55
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Option A: $17000 + 0{,}12s$ ; Option B: $20000 + 0{,}05s$ . Setting equal: $17000 + 0{,}12s = 20000 + 0{,}05s \Rightarrow 0{,}07s = 3000 \Rightarrow s = 3000/0{,}07 \approx 42857$ . Above approximately 42,857 in sales, Option A is more profitable.
Ex. 3.40Modeling
Company A washes windows for $5 per window. Company B charges a flat fee of$ 40 plus $3 per window. For how many windows does Company B become cheaper?
Select the correct option
Above 20 windows.Above 5 windows.Above 40 windows.Above 10 windows.
Select an option first
Solve online OpenStax College Algebra 2e · §4.2 · ex. 54
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Company 1: $C_1 = 5w$ ; Company 2: $C_2 = 40 + 3w$ . Setting equal: $5w = 40 + 3w \Rightarrow 2w = 40 \Rightarrow w = 20$ windows. Above 20 windows, Company 2 is cheaper.
Ex. 3.41Application
Determine whether each function is increasing or decreasing: $g(x) = 5x + 6$ and $a(x) = 5 - 2x$ .
Select the correct option
$g(x) = 5x + 6$ is increasing; $a(x) = 5 - 2x$ is decreasing.Both are increasing.Both are decreasing. $g(x)$ is decreasing; $a(x)$ is increasing.
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 15
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In $g(x) = 5x + 6$ the slope is $5 > 0$ , so it is increasing. In $a(x) = 5 - 2x$ the slope is $-2 < 0$ , so it is decreasing.
Ex. 3.42Application
Find the equation of the affine function whose graph passes through the points $(1, 5)$ and $(4, 11)$ .
Select the correct option
$f(x) = 2x + 3$ $f(x) = 2x - 3$ $f(x) = -2x + 3$ $f(x) = \dfrac{1}{2}x + 3$
Select an option first
Solve online OpenStax College Algebra 2e · §4.1 · ex. 32
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The slope is $m = (11 - 5)/(4 - 1) = 6/3 = 2$ . With $(1, 5)$ : $b = 5 - 2(1) = 3$ . Therefore $f(x) = 2x + 3$ .
Show step-by-step (with the why)
1. Compute $m = (11-5)/(4-1) = 2$ .
2. Use $(1,5)$ : $b = 5 - 2 \cdot 1 = 3$ .
3. Write: $f(x) = 2x + 3$ .
4. Verify: $f(4) = 8 + 3 = 11$ . Correct.
Ex. 3.43Application
Find the equation of the affine function whose graph passes through the points $(-1, 4)$ and $(5, 2)$ .
Select the correct option
$f(x) = -\dfrac{1}{3}x + \dfrac{11}{3}$ $f(x) = \dfrac{1}{3}x + \dfrac{11}{3}$ $f(x) = -3x + \dfrac{11}{3}$ $f(x) = -\dfrac{1}{3}x - \dfrac{11}{3}$
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Solve online OpenStax College Algebra 2e · §4.1 · ex. 33
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The slope is $m = (2-4)/(5-(-1)) = -2/6 = -1/3$ . With $(-1,4)$ : $b = 4 - (-1/3)(-1) = 4 - 1/3 = 11/3$ . Therefore $f(x) = -\frac{1}{3}x + \frac{11}{3}$ .
Ex. 3.44Modeling
A newborn weighs 7.5 pounds. In the first year, they gain half a pound per month. Write the affine function $W(m)$ that models the weight after $m$ months.
Select the correct option
$W(m) = 7{,}5 + 0{,}5\,m$ pounds, with $m$ the number of months. $W(m) = 0{,}5\,m$ pounds. $W(m) = 7{,}5 - 0{,}5\,m$ pounds. $W(m) = 7{,}5 + m$ pounds.
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Solve online OpenStax College Algebra 2e · §4.2 · ex. 19
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The baby is born weighing 7.5 lb (y-intercept) and gains 0.5 lb/month (slope). Therefore $W(m) = 7{,}5 + 0{,}5m$ .
Ex. 3.45Modeling
Using the model $W(m) = 7{,}5 + 0{,}5m$ from the previous exercise, after how many months did the baby weigh 10.0 pounds?
Select the correct option
After 5 months.After 3 months.After 7 months.After 10 months.
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Solve online OpenStax College Algebra 2e · §4.2 · ex. 23
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Using $W(m) = 7{,}5 + 0{,}5m$ , solve $10{,}0 = 7{,}5 + 0{,}5m \Rightarrow 0{,}5m = 2{,}5 \Rightarrow m = 5$ months.

$x$	0	5	10	15
$f(x)$	$-5$	$20$	$45$	$70$

$x$	2	4	6	8
$f(x)$	$-4$	$16$	$36$	$56$

Sources

Only books that directly fed the text and exercises. Full catalog at /livros.

OpenStax College Algebra 2e — Jay Abramson et al. · 2022 · EN · CC-BY 4.0 · §2.1–2.4, §5.1–5.3. Primary source for blocks A, B, C.
Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · free · ch. 1. Primary source for block E (modeling).
Active Calculus — Matt Boelkins · 2024, ed. 2.0 · EN · CC-BY-NC-SA · §1.2–1.3 (rate of change as motivation for the derivative). Source for Door 25.
Stitz–Zeager Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §2.1, §10.6.
Hammack — Book of Proof — Richard Hammack · 2018 · EN · free · §12. Source for exercise 3.15 (composition of affine functions).