Algebra 1 Graphing Linear Equations Test Solutions and Scoring (PDF)

Algebra 1 math test solutions: Graphing Linear Equations – Maths PDF


This graphing linear test answers for Algebra 1 covers graphing linear equations (Common Core A-REI.D.10, F-IF.C.7a, A-CED.A.2), with a free PDF to print.

Graphing linear test answers (Algebra 1) – first page of the printable PDF
Use this key to grade the graphing linear equations test. Each solution shows the full work, and the scoring note explains how the points are awarded. The test totals 20 points. Give partial credit for correct steps, even when the final answer is wrong.

Algebra 1 · Common Core State Standards: A-REI.D.10, F-IF.C.7a, A-CED.A.2

Graphing linear test answers: worked solutions

Question 1 – Slope and intercept [2 pts]

a) Slope \(-4\), y-intercept \((0, 7)\).

b) Slope \(\frac{3}{5}\), y-intercept \((0, -2)\).

Scoring: 1 point for each part (both values correct).

Question 2 – Graph a line [2 pts]

xy-6-5-4-3-2-1123456-6-5-4-3-2-1123456(0, -1)(4, 2)(-4, -4)
Graph of the line.

Plot the y-intercept \((0, -1)\). The slope is \(\frac{3}{4}\): up 3, right 4 to \((4, 2)\). Draw the line.

Scoring: 1 point for the correct y-intercept; 1 point for the correct slope and line.

Question 3 – Find intercepts [2 pts]

Let \(y = 0\): \(5x = 20\), so \(x = 4\).

Let \(x = 0\): \(-2y = 20\), so \(y = -10\).

(4, 0) and (0, -10)

Scoring: 1 point for each intercept.

Question 4 – Rewrite an equation [2 pts]

Subtract \(4x\): \(2y = -4x + 10\). Divide by 2.

\(y = -2x + 5\)

Scoring: 1 point for isolating the y-term; 1 point for dividing every term.

Question 5 – Point-slope form [2 pts]

Point-slope: \(y – 5 = -3(x – 2)\).

Distribute: \(y – 5 = -3x + 6\), so \(y = -3x + 11\).

\(y – 5 = -3(x – 2)\); \(y = -3x + 11\)

Scoring: 1 point for each form.

Question 6 – Equation from a graph [2 pts]

xy-112345-112345678(0, 6)(2, 3)
Slope \(-\frac{3}{2}\), intercept 6.

The y-intercept is 6. From \((0, 6)\) to \((2, 3)\), the level changes by -3 in 2 hours.

\(y = -\frac{3}{2}x + 6\)

Scoring: 1 point for the slope; 1 point for the equation.

Question 7 – Data plan [3 pts]

xy-2-112345678910-22468101214(0, 12)(8, 0)(4, 6)
The point (4, 6) is also on the line.

a) Slope \(\frac{0 – 12}{8 – 0} = -1.5\), so \(y = -1.5x + 12\).

b) The x-intercept \((8, 0)\) means all the data is used after 8 days.

c) \(y = -1.5(4) + 12 = 6\) GB.

\(y = -1.5x + 12\); 8 days; 6 GB

Scoring: 1 point per part.

Question 8 – Points on a line [3 pts]

a) \(2(6) – 3(0) = 12\) true. \(2(0) – 3(4) = -12\) false. \(2(-3) – 3(-6) = -6 + 18 = 12\) true. \(2(5) – 3(1) = 7\) false.

b) \(18 – 3y = 12\), so \(-3y = -6\) and \(y = 2\).

\((6, 0)\) and \((-3, -6)\); \(y = 2\)

Scoring: part a, 1 point for testing all points and 1 point for the correct two; part b, 1 point.

Question 9 – Horizontal and vertical [2 pts]

Every point on the horizontal line has y-coordinate -3. Every point on the vertical line has x-coordinate 4.

\(y = -3\) and \(x = 4\)

Scoring: 1 point for each equation.

Total score

Total

20 points

Key facts behind this graphing linear test answers

Key facts
  • Every point on the line is a solution of the equation, and every solution is on the line.
  • In y = mx + b, m is the slope and b is the y-intercept.
  • Find intercepts by setting x = 0 for the y-intercept and y = 0 for the x-intercept.
  • Point-slope form y – y₁ = m(x – x₁) uses a known point and the slope.
  • y = k is horizontal with slope 0; x = h is vertical with undefined slope.

How to use this graphing linear test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

All Algebra 1 chapters: Algebra 1 math.

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