3.1 Slope, Intercepts, and Graphing Lines

3.1 Slope, Intercepts, and Graphing Lines

Quick Answer: Slope is rise over run, $m = (y_2 - y_1)/(x_2 - x_1)$. The y-intercept is where the line meets the y-axis ($x = 0$); the x-intercept is where it meets the x-axis ($y = 0$). Graph $y = mx + b$ by plotting $b$ and stepping with $m$, or plot both intercepts and connect them. Horizontal lines have slope $0$ (equation $y = k$); vertical lines have undefined slope (equation $x = h$).

College Board's Next-Generation ACCUPLACER Advanced Algebra and Functions (AAF) test weights linear applications and graphs at 10–15% of the exam — typically 2–3 computer-adaptive items. Skills Insight scores in the 200–236 band already expect you to evaluate a linear function in context, and calculating slope and writing the equation from a graph or a table is the machinery under that evaluation. This section builds that geometric toolkit so parallel lines, linear models, and inequality graphs later in the chapter rest on a line you can actually draw. FREE practice for the whole AAF blueprint is at /practice/accuplacer-advanced-algebra.

Slope as Rise over Run

The slope of a nonvertical line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is

$$m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}.$$

Rise is the vertical change; run is the horizontal change. Keep the two points in the same order in the numerator and the denominator. Swapping both signs leaves $m$ unchanged; swapping only one sign flips the slope and is a standard AAF distractor.

Slope is also a rate of change: the number of y-units gained or lost for each 1-unit increase in $x$. On application items that rate is what the stem is asking you to interpret — dollars per credit hour, miles per gallon, degrees per minute. A slope of $-3$ means $y$ drops 3 units every time $x$ increases by 1; a slope of $4/5$ means $y$ rises 4 units for every 5 units of run.

Worked Example: Line through (−2, 5) and (4, −1)

Use the two given points.

$$m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1.$$

The line falls 1 unit for every 1 unit it runs to the right: a negative slope of $-1$. To write the equation, plug one point into point-slope form $y - y_1 = m(x - x_1)$:

$$y - 5 = -1(x - (-2)) = -(x + 2).$$ $$y - 5 = -x - 2 \Rightarrow y = -x + 3.$$

Check with the unused point: at $x = 4$, $y = -4 + 3 = -1$, which matches $(4, -1)$. In slope-intercept form the slope is $-1$ and the y-intercept is $3$, so the graph crosses the y-axis at $(0, 3)$.

The x-intercept occurs when $y = 0$:

$$0 = -x + 3 \Rightarrow x = 3.$$

The graph crosses the x-axis at $(3, 0)$. Two intercepts plus the known slope fully determine the picture: start at $(0, 3)$ and step down 1, right 1 (or up 1, left 1) through $(3, 0)$, $(-2, 5)$, and $(4, -1)$.

If you start from the other point instead — $y - (-1) = -1(x - 4)$ — you get $y + 1 = -x + 4$, then $y = -x + 3$, the same line. That is the built-in check: any point on the line must produce the same slope-intercept equation. If two different points produce two different $b$ values, the slope arithmetic is wrong.

Four Slope Cases

A horizontal line has equal y-coordinates at every point, so the rise is $0$ and $m = 0$. Example: through $(1, 4)$ and $(7, 4)$, $m = 0/6 = 0$, equation $y = 4$. A vertical line has equal x-coordinates, so the run is $0$ and you would divide by zero: slope is undefined, equation $x = h$. Example: through $(2, -3)$ and $(2, 8)$, equation $x = 2$. Never treat undefined as a number you then plug into $y = mx + b$. Vertical lines are not functions of $x$ and cannot be written in slope-intercept form.

Sign of slope is a left-to-right statement, not an up-versus-down statement. A line that looks steep downward still has a negative $m$; a line that is flat has $m = 0$, not no slope. Save undefined exclusively for vertical lines. Mixing those two phrases — calling a horizontal line undefined, or calling a vertical line slope 0 — is one of the highest-yield traps on this 10–15% content area.

Graphing from Slope-Intercept and from Intercepts

Three practical methods cover every AAF graphing item in this cluster:

To graph $2x + 3y = 12$ from intercepts: set $y = 0$ to get $x = 6$ (x-intercept $(6, 0)$); set $x = 0$ to get $y = 4$ (y-intercept $(0, 4)$). Draw the unique line through those two points. Solving for $y$ first gives $y = -\frac{2}{3}x + 4$, the same line: slope $-2/3$, y-intercept $4$. From $(0, 4)$ step down 2, right 3 (or up 2, left 3) to land on lattice points such as $(3, 2)$ and $(6, 0)$.

A fractional slope is a ready-made step. For $m = 3/4$, rise 3 and run 4. For $m = -5/2$, rise $-5$ and run 2 (down 5, right 2), or rise 5 and run $-2$ (up 5, left 2). Reducing the fraction first keeps the steps on grid points. If you leave $m = -10/4$ unreduced, you still have the right line, but you are more likely to miscount boxes on a CAT grid.

Reading Slope from a Graph or a Table

AAF graph items often show a graph or a table of ordered pairs and ask either for $m$ or for the equation. From a graph, pick two lattice points the line actually hits — not a point that merely looks close — and compute rise/run. Count the boxes: if the line goes down 4 and right 2, $m = -4/2 = -2$. Then read $b$ from the y-axis crossing, or use point-slope if the y-intercept is off the visible window.

From a table, pick two clean rows. Suppose the table lists $(0, 8)$, $(2, 5)$, $(4, 2)$, $(6, -1)$. Consecutive x-steps of 2 drop $y$ by 3, so $m = -3/2$. Because $(0, 8)$ is in the table, $b = 8$ and the equation is $y = -\frac{3}{2}x + 8$. Confirm with $(4, 2)$: $-(3/2)(4) + 8 = -6 + 8 = 2$.

If the table skipped $x = 0$ — say $(1, 10)$ and $(4, 1)$ — then $m = (1 - 10)/(4 - 1) = -9/3 = -3$, and $y - 10 = -3(x - 1)$ gives $y = -3x + 13$. Always test a third row when one exists; a table that is not linear will fail that check, and AAF will occasionally mix a nonlinear distractor table into a linear-looking stem.

Elementary Linear Functions

An elementary linear function is $f(x) = mx + b$. On AAF, elementary means you evaluate, graph, and interpret it — composition, inverses, and transformations sit in the separate Functions content area. Evaluating is substitution: if $f(x) = -2x + 7$, then $f(3) = 1$ and $f(0) = 7$, the y-intercept. The graph of $y = f(x)$ is exactly the line you just learned to draw.

When two points determine a unique nonvertical line, they determine a unique linear function. Vertical lines are excluded from function language because they fail the vertical-line test. If an item gives $f(2) = 5$ and $f(6) = 13$, the implied slope is $(13 - 5)/(6 - 2) = 8/4 = 2$, so $f(x) = 2x + b$; then $5 = 2(2) + b$ gives $b = 1$ and $f(x) = 2x + 1$.

After you can compute $m$, name both intercepts, and graph $y = mx + b$, $y = k$, and $x = h$, you have the geometric toolkit for parallel and perpendicular lines, linear models, and inequality graphs in the rest of this chapter. Drill the arithmetic until rise/run is automatic.