3.2 Linear Functions, Slope & Graphs of Lines

3.2 Linear Functions, Slope & Graphs of Lines

The College Board's description of the CLEP College Algebra examination states that the test includes questions on linear and quadratic equations, inequalities, and graphs. Section 3.1 solved linear equations algebraically; this section handles the geometric half — the linear function $f(x) = mx + b$ and the straight line that represents it. Lines also underpin systems of equations (Section 3.5), the feasible regions of linear inequality systems, transformations of the parent function $f(x) = x$, and every "rate of change" word problem on the exam.

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Slope: The Constant Rate of Change

For any two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on a line, the slope is

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}, \quad x_1 \neq x_2$

What makes a function linear is that this ratio is the same for every pair of points you pick. That is the defining test: if a table of values shows equal changes in $y$ for equal changes in $x$, the relationship is linear.

Sign Trap: Subtract the coordinates in the same order top and bottom. Using $\frac{y_2 - y_1}{x_1 - x_2}$ flips the sign of the slope, and the sign-flipped line is reliably offered as a distractor.

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The Three Standard Forms of a Line

Converting is routine algebra: solve for $y$ to reach slope-intercept form, or clear fractions and move the variable terms to the left to reach standard form. From $Ax + By = C$ with $B \neq 0$, the slope is $m = -\frac{A}{B}$ — a shortcut worth memorising because it avoids a full rearrangement when a question asks only for the slope.

Intercepts of a Line

• $y$-intercept: set $x = 0$ and solve for $y$. In slope-intercept form it is simply $(0, b)$.
• $x$-intercept: set $y = 0$ and solve for $x$. For $y = mx + b$ with $m \neq 0$, it is $\left(-\frac{b}{m}, 0\right)$.

Worked Example 1: Building a Line From Two Points

Problem: Find the equation of the line through $(-2, 7)$ and $(4, -5)$ in slope-intercept form, then give both intercepts.

1. Compute the slope: $m = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2$
2. Apply point-slope form using $(-2, 7)$: $y - 7 = -2(x - (-2)) = -2(x + 2)$
3. Solve for $y$: $y - 7 = -2x - 4 \implies y = -2x + 3$
4. Check with the other point: $-2(4) + 3 = -8 + 3 = -5$ ✓.
5. Intercepts: the $y$-intercept is $(0, 3)$. Setting $y = 0$ gives $0 = -2x + 3 \implies x = \frac{3}{2}$, so the $x$-intercept is $\left(\frac{3}{2}, 0\right)$.

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Parallel and Perpendicular Lines

• Parallel: two distinct lines are parallel exactly when $m_1 = m_2$ (and their $y$-intercepts differ). Same slope, same $y$-intercept means the lines coincide, not that they are parallel.
• Perpendicular: $m_1 \cdot m_2 = -1$, so each slope is the negative reciprocal of the other. A slope of $\frac{3}{5}$ pairs with $-\frac{5}{3}$.
• The exception: a horizontal line ($m = 0$) and a vertical line (undefined slope) are perpendicular, but their slopes cannot be multiplied. Handle this pair by inspection rather than by formula.

Worked Example 2: Perpendicular Line Through a Point

Problem: Find the equation of the line perpendicular to $3x - 4y = 12$ that passes through $(6, -1)$.

1. Find the given line's slope using $m = -\frac{A}{B}$: $m = -\frac{3}{-4} = \frac{3}{4}$.
2. Take the negative reciprocal: the perpendicular slope is $-\frac{4}{3}$.
3. Apply point-slope form: $y - (-1) = -\frac{4}{3}(x - 6) \implies y + 1 = -\frac{4}{3}x + 8$
4. Solve for $y$: $y = -\frac{4}{3}x + 7$
5. Verify perpendicularity: $\frac{3}{4} \cdot \left(-\frac{4}{3}\right) = -1$ ✓.

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Horizontal and Vertical Lines

Students routinely swap these. Anchor them by remembering that $y = k$ states what $y$ always equals, which is a flat line, while $x = h$ states what $x$ always equals, which is an upright line.

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Linear Models: Interpreting $m$ and $b$ in Context

When a linear function models a real situation, both parameters carry meaning:

$f(x) = \underbrace{m}_{\text{change per unit of } x} \cdot\, x + \underbrace{b}_{\text{value when } x = 0}$

CLEP asks these as verbal-to-symbolic translation items, so practise reading the two numbers out of the sentence.

Worked Example 3: Building and Interpreting a Linear Model

Problem: A rental company charges a flat fee of 45 dollars plus 28 cents per mile driven. Write the cost $C$ as a function of miles $m$, state what each parameter means, find the cost of a 150-mile trip, and determine how many miles produce a charge of 115 dollars.

1. Identify the fixed amount: the 45-dollar deposit is charged regardless of distance, so it is the value at $m = 0$ — the intercept.
2. Identify the rate: 28 cents per mile is the change per unit of input — the slope.
3. Write the model: $C(m) = 0.28m + 45$
4. Evaluate at 150 miles: $C(150) = 0.28(150) + 45 = 42 + 45$ The cost is 87 dollars.
5. Solve backwards for $C = 115$: $115 = 0.28m + 45 \implies 70 = 0.28m \implies m = 250 \text{ miles}$

Exam Trap: "Flat fee plus a rate" always gives an intercept plus a slope. Answer choices frequently multiply the flat fee by the variable as well — $C(m) = 45m + 0.28$ or $C(m) = 45(m + 0.28)$ — so check which quantity is genuinely fixed before choosing.

Recognising a Linear Relationship in a Table

Each step of $+2$ in $x$ produces $-6$ in $y$, so the ratio $\frac{-6}{2} = -3$ is constant and the relationship is linear with $m = -3$. Back-solve for $b$ using $(1, 11)$: $11 = -3(1) + b \implies b = 14$, giving $y = -3x + 14$. A table whose $y$-differences are not constant for constant $x$-steps is not linear — check it before assuming.