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Linear Relationships Checkpoint

A free Algebra I lesson from the “Functions, Linear Relationships, and Rate of Change” unit, with a worked example and practice problems including step-by-step solutions.

This checkpoint samples all eleven lessons in Functions, Linear Relationships, and Rate of Change: functions, domain and range, slope, slope-intercept form, tables and graphs, standard form, point-slope form, parallel and perpendicular lines, direct and inverse variation, average rate of change, and linear inequalities in two variables. Questions are mixed, so start by naming what you are given (an equation, a table, two points, a described graph, or a context) and then choose a method. Watch the special cases: a horizontal line has slope 0, a vertical line has an undefined slope, and inverse variation (y = k/x) is not linear. For an inequality, decide whether the boundary is solid or dashed and which side to shade. The 25 review questions and the 25 checkpoint questions were written separately from each other and from the lesson practice and quizzes. Two review questions show a graph; checkpoint questions describe any graph in words, and many are multiple choice so that equations and graph descriptions are graded reliably. A score reports performance on this sample of proposed outcomes A1.4 and A1.5; it is not an external certification. The written tasks below, including drawing graphs, are reviewed with a rubric and do not change the score.

What you'll learn

Why it matters: Rates appear wherever one quantity changes with another: a fare with a fixed fee and a per-mile charge is a line, a melting snowpack has a negative rate, a crew that shares a fixed job is an inverse variation, and a fundraising target is a two-variable inequality.

Worked example

Problem. Find the line through (−3, 5) that is perpendicular to 2x − 3y = 6. Write it in slope-intercept form.

  1. Rewrite the given line to see its slope: −3y = −2x + 6, so y = (2/3)x − 2. Its slope is 2/3.
  2. A perpendicular slope is the opposite reciprocal: −3/2. Check: (2/3)(−3/2) = −1.
  3. Use point-slope form with (−3, 5): y − 5 = −(3/2)(x + 3), so y − 5 = −(3/2)x − 9/2.
  4. Add 5 to both sides: y = −(3/2)x + 1/2. Check the point: −(3/2)(−3) + 1/2 = 9/2 + 1/2 = 5.

Answer: y = −(3/2)x + 1/2

Lesson preview: this page shows 5 practice items. The interactive activity contains 25 practice items and 25 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. A school database pairs each student ID number with exactly one homeroom, and each homeroom has about 25 students. Which statement is correct?

Choices: Homeroom is a function of student ID, but student ID is not a function of homeroom · Student ID is a function of homeroom, but homeroom is not a function of student ID · Each quantity is a function of the other · Neither quantity is a function of the other

Show solution
  1. A relation is a function of its input when every input is paired with exactly one output.
  2. With student ID as the input, each ID has exactly one homeroom, so homeroom is a function of student ID.
  3. With homeroom as the input, one homeroom is paired with about 25 different IDs, so student ID is not a function of homeroom.

Answer: Homeroom is a function of student ID, but student ID is not a function of homeroom

2. For g(t) = 15 − 4t, which input t gives an output of −9?

Show solution
  1. Set the output equal to −9: 15 − 4t = −9.
  2. Subtract 15 from both sides to get −4t = −24, then divide by −4 to get t = 6.
  3. Check: g(6) = 15 − 4(6) = 15 − 24 = −9.

Answer: 6

3. A parking garage charges P(h) = 3h + 2 dollars for h hours, where h is a whole number from 1 to 8. What is the greatest value in the range of P?

Show solution
  1. The domain is the whole numbers 1, 2, 3, …, 8.
  2. P has a positive rate (+3 dollars per hour), so its greatest output comes from the greatest input: P(8) = 3(8) + 2 = 26.
  3. The range is {5, 8, 11, …, 26}, so its greatest value is 26 dollars.

Answer: 26

4. A bike-share dock holds up to 12 bikes. B(t) is the number of bikes docked t minutes after 8:00 a.m., for 0 ≤ t ≤ 60. Which describes the domain and range of B?

Choices: Domain: every time from 0 to 60 minutes; range: some or all of the whole numbers from 0 to 12 · Domain: the whole numbers from 0 to 12; range: every time from 0 to 60 minutes · Domain: every time from 0 to 60 minutes; range: every real number from 0 to 12 · Domain: the whole numbers from 0 to 60; range: every real number from 0 to 12

Show solution
  1. The input is the time t, which can be any value from 0 to 60 minutes, so the domain is 0 ≤ t ≤ 60.
  2. The output is a count of bikes, so it is a whole number, and the dock holds at most 12.
  3. So every output is one of 0, 1, 2, …, 12; a value such as 4.5 bikes is impossible, which rules out every real number from 0 to 12.

Answer: Domain: every time from 0 to 60 minutes; range: some or all of the whole numbers from 0 to 12

5. Which pair of points lies on a line whose slope is undefined?

Choices: (3, −1) and (3, 6) · (−1, 3) and (6, 3) · (0, 0) and (3, 3) · (1, 3) and (2, 6)

Show solution
  1. Slope is (change in y)/(change in x).
  2. For (3, −1) and (3, 6), the change in x is 3 − 3 = 0, so the ratio would divide by 0: the slope is undefined and the line is vertical (x = 3).
  3. The other pairs give slopes 0/7 = 0 (a horizontal line), 3/3 = 1, and 3/1 = 3.

Answer: (3, −1) and (3, 6)

Explain and apply

Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.

Graph a linear inequality and justify the shading

Graph 2x − 3y < 12 on a coordinate plane. Label both axes, mark the intercepts of the boundary line, and decide whether the boundary is solid or dashed. Use a test point to choose the side to shade. Then explain whether the point (6, 0) is a solution.

Review rubric and solution
  • Identifies the boundary 2x − 3y = 12 and plots its intercepts (6, 0) and (0, −4), or two other correct points.
  • Draws the boundary dashed because < does not include equality.
  • Tests a point not on the line, such as (0, 0), where 0 < 12 is true, and shades the side containing that point (above the line).
  • Labels both axes and the intercepts so the graph can be read.
  • Explains that (6, 0) gives 12 < 12, which is false, so a point on the dashed boundary is not a solution.
  1. The boundary is 2x − 3y = 12. Setting y = 0 gives x = 6; setting x = 0 gives y = −4. Plot (6, 0) and (0, −4) and connect them with a dashed line, because < excludes equality.
  2. Test (0, 0): 2(0) − 3(0) = 0, and 0 < 12 is true, so shade the side containing the origin.
  3. Solving for y gives y > (2/3)x − 4 (dividing by −3 reverses the symbol), which confirms that the shaded region is above the line.
  4. (6, 0) gives 2(6) − 3(0) = 12, and 12 < 12 is false. It lies on the dashed boundary, so it is not a solution.

Represent one linear relationship four ways

A streaming service charges a one-time $15 setup fee plus $9 per month. (a) Write the total cost C after m months in slope-intercept form, in point-slope form using the total after 4 months, and in standard form. (b) Make a table for m = 0 to 4 and graph it with labeled axes and units. (c) Interpret the slope and the C-intercept with units. (d) State a reasonable domain and range for the first year, and explain whether the graph should be drawn as separate points or as a continuous line.

Review rubric and solution
  • Writes C = 9m + 15, C − 51 = 9(m − 4), and 9m − C = −15 (or another correct standard form), and shows the forms are equivalent.
  • Makes a correct table (15, 24, 33, 42, 51) and graphs it with labeled axes and units.
  • Interprets 9 as dollars per month and 15 as the setup fee charged at m = 0.
  • States the domain as the whole numbers 0 to 12 and the range as 15, 24, …, 123 dollars.
  • Explains that the cost is charged in whole months, so separate points (or a line drawn with that restriction stated) fit the context.
  1. Slope-intercept: C = 9m + 15. After 4 months C = 51, so point-slope: C − 51 = 9(m − 4). Standard form: 9m − C = −15. Distributing the point-slope form gives C = 9m − 36 + 51 = 9m + 15.
  2. Table: m = 0, 1, 2, 3, 4 gives C = 15, 24, 33, 42, 51. Put months on the horizontal axis and dollars on the vertical axis.
  3. The slope 9 means the total rises $9 for each additional month; the C-intercept 15 is the setup fee paid before any months of service.
  4. For the first year, the domain is m = 0, 1, …, 12 and the range is 15, 24, …, 123 dollars. Because charges occur per whole month, the graph is a set of separate points that lie on the line C = 9m + 15.

Compare direct variation, inverse variation, and linear change

Table A pairs x = 1, 2, 3, 6 with y = 4, 8, 12, 24. Table B pairs x = 1, 2, 3, 6 with y = 24, 12, 8, 4. (a) Decide whether y varies directly or inversely with x in each table, and write each equation. (b) Find the rate of change from x = 1 to x = 2 and from x = 3 to x = 6 in each table. (c) Use those rates to explain which relationship is linear. (d) Sketch both graphs for x > 0 and describe how each one behaves.

Review rubric and solution
  • Identifies Table A as direct variation, y = 4x, because y/x = 4 for every pair.
  • Identifies Table B as inverse variation, y = 24/x, because xy = 24 for every pair.
  • Computes the rates: Table A gives 4 and 4; Table B gives −12 and −4/3.
  • Explains that Table A has a constant rate and is linear, while Table B's rate changes, so inverse variation is not linear.
  • Sketches a line through the origin for A and a decreasing curve for B that approaches, but never reaches, the axes.
  1. Table A: y/x = 4 for every pair, so y = 4x (direct variation). Table B: xy = 24 for every pair, so y = 24/x (inverse variation).
  2. Table A rates: (8 − 4)/(2 − 1) = 4 and (24 − 12)/(6 − 3) = 4. Table B rates: (12 − 24)/(2 − 1) = −12 and (4 − 8)/(6 − 3) = −4/3.
  3. Table A's rate is constant, so it is linear: a line through the origin with slope 4. Table B's rate changes, so it is not linear.
  4. The graph of y = 4x is a straight line rising through the origin. The graph of y = 24/x for x > 0 is a curve that falls quickly and then levels off, staying above the x-axis and never touching either axis.

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