Study guide
3.1 Slope, Intercepts, and Graphing Lines
ACCUPLACER Advanced Algebra and Functions Exam
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1 Introduction & Exam Overview
1.1 Current Exam Facts 1.2 Scoring, Skills Insight Bands, and Placement 1.3 How to Study for AAF
2 Linear Equations
2.1 Writing Linear Equations in One and Two Variables 2.2 Solving Linear Equations 2.3 Linear Inequalities 2.4 Systems of Two Linear Equations
3 Linear Applications and Graphs
3.1 Slope, Intercepts, and Graphing Lines 3.2 Parallel and Perpendicular Lines 3.3 Modeling with Linear Functions 3.4 Graphing Linear Inequalities and Systems
4 Factoring
4.1 GCF, Grouping, and Special Products 4.2 Factoring Quadratic Trinomials 4.3 Factoring Cubics and Higher Polynomials
5 Quadratics
5.1 Quadratic Forms, Graphs, Vertex, and Intercepts 5.2 Solving Quadratics by Factoring 5.3 Quadratic Formula and Completing the Square 5.4 Quadratic Inequalities and Quadratic-Linear Systems
6 Functions
6.1 Function Notation, Evaluation, Domain, and Range 6.2 Graphs of Functions, Vertical Line Test, Maxima and Minima 6.3 Function Operations, Composition, and Inverses 6.4 Transformations and Interpreting Functions in Context
7 Radical and Rational Equations
7.1 Simplifying Radical Expressions 7.2 Solving Radical Equations and Extraneous Roots 7.3 Rational Expressions and Equations 7.4 Graphs, Domain, and Range of Radical and Rational Functions
8 Polynomial Equations
8.1 Polynomial Operations and Structure 8.2 Solving Polynomial Equations 8.3 Graphing Polynomials, Roots, and End Behavior
9 Exponential and Logarithmic Equations
9.1 Exponential Functions, Growth, and Decay 9.2 Graphs of Exponential Functions 9.3 Logarithms and Logarithmic Properties 9.4 Solving Exponential and Logarithmic Equations
10 Geometry Concepts for Algebra 1
10.1 Area, Perimeter, and Prism Volume 10.2 Distance Formula and Pythagorean Theorem 10.3 Dilations, Rotations, Translations, and Reflections
11 Geometry Concepts for Algebra 2
11.1 Volume of Cylinders, Cones, and Spheres 11.2 Intersecting Lines and Angle Relationships 11.3 Triangle Similarity and Congruence 11.4 Circles in the Coordinate Plane
12 Trigonometry
12.1 Right-Triangle Trigonometry and Special Triangles 12.2 Unit Circle, Radians, and Arc Length 12.3 Trigonometric Graphs and Identities 12.4 Solving Trig Equations; Law of Sines and Cosines
3.1 Slope, Intercepts, and Graphing Lines
Key Takeaways Explain This Topic AI
The line through (−2, 5) and (4, −1) has slope (−1 − 5)/(4 − (−2)) = −6/6 = −1 and equation y = −x + 3, with intercepts (0, 3) and (3, 0).
Slope m = (y2 − y1)/(x2 − x1) is both rise/run and the rate of change in y per one-unit increase in x.
Horizontal lines have slope 0 and equation y = k; vertical lines have undefined slope and equation x = h and cannot be written as y = mx + b.
Graph y = mx + b by plotting the y-intercept (0, b) and stepping with rise/run; graph Ax + By = C by plotting both intercepts and connecting them.
College Board Skills Insight band 200–236 on AAF already evaluates a linear function in context; the slope and intercept work in this section is the machinery behind that evaluation.
Last updated: August 2026
3.1 Slope, Intercepts, and Graphing Lines
Quick Answer: Slope is rise over run , m = ( y 2 − y 1 ) / ( x 2 − x 1 ) m = (y_2 - y_1)/(x_2 - x_1) m = ( y 2 − y 1 ) / ( x 2 − x 1 ) . The y-intercept is where the line meets the y-axis ( x = 0 x = 0 x = 0 ); the x-intercept is where it meets the x-axis ( y = 0 y = 0 y = 0 ). Graph y = m x + b y = mx + b y = m x + b by plotting b b b and stepping with m m m , or plot both intercepts and connect them. Horizontal lines have slope 0 0 0 (equation y = k y = k y = k ); vertical lines have undefined slope (equation x = h x = h 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 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) is m = rise run = y 2 − y 1 x 2 − x 1 . m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}. m = run rise = x 2 − x 1 y 2 − y 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 m 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 x 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 -3 − 3 means y y y drops 3 units every time x x x increases by 1; a slope of 4 / 5 4/5 4/5 means y y 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 = − 1 − 5 4 − ( − 2 ) = − 6 6 = − 1. m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1. m = 4 − ( − 2 ) − 1 − 5 = 6 − 6 = − 1.
The line falls 1 unit for every 1 unit it runs to the right : a negative slope of − 1 -1 − 1 . To write the equation, plug one point into point-slope form y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) : y − 5 = − 1 ( x − ( − 2 ) ) = − ( x + 2 ) . y - 5 = -1(x - (-2)) = -(x + 2). y − 5 = − 1 ( x − ( − 2 )) = − ( x + 2 ) . y − 5 = − x − 2 ⇒ y = − x + 3. y - 5 = -x - 2 \Rightarrow y = -x + 3. y − 5 = − x − 2 ⇒ y = − x + 3.
Check with the unused point: at x = 4 x = 4 x = 4 , y = − 4 + 3 = − 1 y = -4 + 3 = -1 y = − 4 + 3 = − 1 , which matches ( 4 , − 1 ) (4, -1) ( 4 , − 1 ) . In slope-intercept form the slope is − 1 -1 − 1 and the y-intercept is 3 3 3 , so the graph crosses the y-axis at ( 0 , 3 ) (0, 3) ( 0 , 3 ) .
The x-intercept occurs when y = 0 y = 0 y = 0 : 0 = − x + 3 ⇒ x = 3. 0 = -x + 3 \Rightarrow x = 3. 0 = − x + 3 ⇒ x = 3.
The graph crosses the x-axis at ( 3 , 0 ) (3, 0) ( 3 , 0 ) . Two intercepts plus the known slope fully determine the picture: start at ( 0 , 3 ) (0, 3) ( 0 , 3 ) and step down 1, right 1 (or up 1, left 1) through ( 3 , 0 ) (3, 0) ( 3 , 0 ) , ( − 2 , 5 ) (-2, 5) ( − 2 , 5 ) , and ( 4 , − 1 ) (4, -1) ( 4 , − 1 ) .
If you start from the other point instead — y − ( − 1 ) = − 1 ( x − 4 ) y - (-1) = -1(x - 4) y − ( − 1 ) = − 1 ( x − 4 ) — you get y + 1 = − x + 4 y + 1 = -x + 4 y + 1 = − x + 4 , then y = − x + 3 y = -x + 3 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 b b values, the slope arithmetic is wrong.
Four Slope Cases
- Case Slope m m m Equation shape Graph
- Positive m > 0 m > 0 m > 0 y = m x + b y = mx + b y = m x + b with m m m positive Rises left to right
- Negative m < 0 m < 0 m < 0 y = m x + b y = mx + b y = m x + b with m m m negative Falls left to right
- Zero (horizontal) m = 0 m = 0 m = 0 y = k y = k y = k Horizontal through ( 0 , k ) (0, k) ( 0 , k )
- Undefined (vertical) none x = h x = h x = h Vertical through ( h , 0 ) (h, 0) ( h , 0 )
A horizontal line has equal y-coordinates at every point, so the rise is 0 0 0 and m = 0 m = 0 m = 0 . Example: through ( 1 , 4 ) (1, 4) ( 1 , 4 ) and ( 7 , 4 ) (7, 4) ( 7 , 4 ) , m = 0 / 6 = 0 m = 0/6 = 0 m = 0/6 = 0 , equation y = 4 y = 4 y = 4 . A vertical line has equal x-coordinates, so the run is 0 0 0 and you would divide by zero: slope is undefined , equation x = h x = h x = h . Example: through ( 2 , − 3 ) (2, -3) ( 2 , − 3 ) and ( 2 , 8 ) (2, 8) ( 2 , 8 ) , equation x = 2 x = 2 x = 2 . Never treat undefined as a number you then plug into y = m x + b y = mx + b y = m x + b . Vertical lines are not functions of x x 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 m m ; a line that is flat has m = 0 m = 0 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:
Method What you plot Best when
Slope-intercept y = m x + b y = mx + b y = m x + b Plot ( 0 , b ) (0, b) ( 0 , b ) , then use m m m as a rise/run step Equation is already solved for y y y
Intercepts Plot ( a , 0 ) (a, 0) ( a , 0 ) and ( 0 , b ) (0, b) ( 0 , b ) , then connect Equation is in standard form A x + B y = C Ax + By = C A x + B y = C
Point-slope Plot the given point, then step with m m m You have a point and a slope, not yet rewritten
To graph 2 x + 3 y = 12 2x + 3y = 12 2 x + 3 y = 12 from intercepts: set y = 0 y = 0 y = 0 to get x = 6 x = 6 x = 6 (x-intercept ( 6 , 0 ) (6, 0) ( 6 , 0 ) ); set x = 0 x = 0 x = 0 to get y = 4 y = 4 y = 4 (y-intercept ( 0 , 4 ) (0, 4) ( 0 , 4 ) ). Draw the unique line through those two points. Solving for y y y first gives y = − 2 3 x + 4 y = -\frac{2}{3}x + 4 y = − 3 2 x + 4 , the same line: slope − 2 / 3 -2/3 − 2/3 , y-intercept 4 4 4 . From ( 0 , 4 ) (0, 4) ( 0 , 4 ) step down 2, right 3 (or up 2, left 3) to land on lattice points such as ( 3 , 2 ) (3, 2) ( 3 , 2 ) and ( 6 , 0 ) (6, 0) ( 6 , 0 ) .
A fractional slope is a ready-made step. For m = 3 / 4 m = 3/4 m = 3/4 , rise 3 and run 4. For m = − 5 / 2 m = -5/2 m = − 5/2 , rise − 5 -5 − 5 and run 2 (down 5, right 2), or rise 5 and run − 2 -2 − 2 (up 5, left 2). Reducing the fraction first keeps the steps on grid points. If you leave m = − 10 / 4 m = -10/4 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 m 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 m = -4/2 = -2 m = − 4/2 = − 2 . Then read b b 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 ) (0, 8) ( 0 , 8 ) , ( 2 , 5 ) (2, 5) ( 2 , 5 ) , ( 4 , 2 ) (4, 2) ( 4 , 2 ) , ( 6 , − 1 ) (6, -1) ( 6 , − 1 ) . Consecutive x-steps of 2 drop y y y by 3, so m = − 3 / 2 m = -3/2 m = − 3/2 . Because ( 0 , 8 ) (0, 8) ( 0 , 8 ) is in the table, b = 8 b = 8 b = 8 and the equation is y = − 3 2 x + 8 y = -\frac{3}{2}x + 8 y = − 2 3 x + 8 . Confirm with ( 4 , 2 ) (4, 2) ( 4 , 2 ) : − ( 3 / 2 ) ( 4 ) + 8 = − 6 + 8 = 2 -(3/2)(4) + 8 = -6 + 8 = 2 − ( 3/2 ) ( 4 ) + 8 = − 6 + 8 = 2 .
If the table skipped x = 0 x = 0 x = 0 — say ( 1 , 10 ) (1, 10) ( 1 , 10 ) and ( 4 , 1 ) (4, 1) ( 4 , 1 ) — then m = ( 1 − 10 ) / ( 4 − 1 ) = − 9 / 3 = − 3 m = (1 - 10)/(4 - 1) = -9/3 = -3 m = ( 1 − 10 ) / ( 4 − 1 ) = − 9/3 = − 3 , and y − 10 = − 3 ( x − 1 ) y - 10 = -3(x - 1) y − 10 = − 3 ( x − 1 ) gives y = − 3 x + 13 y = -3x + 13 y = − 3 x + 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 ) = m x + b f(x) = mx + b f ( x ) = m x + 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 ) = − 2 x + 7 f(x) = -2x + 7 f ( x ) = − 2 x + 7 , then f ( 3 ) = 1 f(3) = 1 f ( 3 ) = 1 and f ( 0 ) = 7 f(0) = 7 f ( 0 ) = 7 , the y-intercept. The graph of y = f ( x ) y = f(x) 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 f(2) = 5 f ( 2 ) = 5 and f ( 6 ) = 13 f(6) = 13 f ( 6 ) = 13 , the implied slope is ( 13 − 5 ) / ( 6 − 2 ) = 8 / 4 = 2 (13 - 5)/(6 - 2) = 8/4 = 2 ( 13 − 5 ) / ( 6 − 2 ) = 8/4 = 2 , so f ( x ) = 2 x + b f(x) = 2x + b f ( x ) = 2 x + b ; then 5 = 2 ( 2 ) + b 5 = 2(2) + b 5 = 2 ( 2 ) + b gives b = 1 b = 1 b = 1 and f ( x ) = 2 x + 1 f(x) = 2x + 1 f ( x ) = 2 x + 1 .
After you can compute m m m , name both intercepts, and graph y = m x + b y = mx + b y = m x + b , y = k y = k y = k , and x = h x = h 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.