Unit 3 · Checkpoint quiz

Unit 3 checkpoint: Lines & graphs

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next. The last two questions review earlier units.

Find the x-intercept of the line \(2x - 5y = 10\). Give it as a point.

On the x-axis, \(y = 0\). Substitute: \(2x - 5(0) = 10\).

\(2x = 10\), so \(x = 5\).

Answer: \((5, 0)\).

Which point lies on the line \(y = -2x + 3\)?

  1. \((-1, 1)\)
  2. \((2, 1)\)
  3. \((1, 1)\)
  4. \((0, -3)\)

Put each \(x\) into \(-2x + 3\) and compare with the \(y\).

a: \(-2(-1) + 3 = 5\), not 1. b: \(-2(2) + 3 = -1\), not 1. d: \(-2(0) + 3 = 3\), not \(-3\).

c: \(-2(1) + 3 = 1\). It matches. Answer: c.

Find the slope of the line through \((-3, 2)\) and \((1, -4)\).

\(m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-4 - 2}{1 - (-3)} = \dfrac{-6}{4}\).

Simplify: \(m = -\dfrac{3}{2}\).

What is the slope of the line \(x = 4\)?

  1. It is \(0\)
  2. It is undefined
  3. It is \(4\)
  4. It is \(\tfrac{1}{4}\)

\(x = 4\) is a vertical line: every point has \(x = 4\), for example \((4, 0)\) and \((4, 1)\).

The run is \(4 - 4 = 0\), and you cannot divide by 0: \(\dfrac{1 - 0}{0}\).

Answer: b, the slope is undefined. (Zero slope belongs to horizontal lines.)

Rewrite \(4x - 3y = 12\) in slope-intercept form \(y = mx + b\).

Subtract \(4x\) from both sides: \(-3y = -4x + 12\).

Divide every term by \(-3\): \(y = \dfrac{-4x}{-3} + \dfrac{12}{-3}\).

Answer: \(y = \tfrac{4}{3}x - 4\).

Write the equation of the line in the graph, in the form \(y = mx + b\).

A line through the two marked points.

The line crosses the y-axis at \((0, -2)\), so \(b = -2\).

From \((0, -2)\) to \((3, 0)\): rise 2, run 3, so \(m = \dfrac{2}{3}\).

Answer: \(y = \tfrac{2}{3}x - 2\).

Write the equation of the line through \((-2, 3)\) with slope 4. Give it in the form \(y = mx + b\).

Point-slope form: \(y - 3 = 4(x - (-2)) = 4(x + 2)\).

Distribute: \(y - 3 = 4x + 8\).

Add 3: \(y = 4x + 11\).

Write the equation of the line through \((1, 5)\) and \((3, -1)\). Give it in the form \(y = mx + b\).

Slope: \(m = \dfrac{-1 - 5}{3 - 1} = \dfrac{-6}{2} = -3\).

Point-slope with \((1, 5)\): \(y - 5 = -3(x - 1) = -3x + 3\).

Add 5: \(y = -3x + 8\). (Check with \((3, -1)\): \(-9 + 8 = -1\).)

Write the equation of the line through \((3, 2)\) that is perpendicular to \(y = \tfrac{3}{4}x - 1\). Give it in the form \(y = mx + b\).

The given slope is \(\tfrac{3}{4}\). The perpendicular slope is the opposite reciprocal: \(-\tfrac{4}{3}\).

Point-slope: \(y - 2 = -\tfrac{4}{3}(x - 3) = -\tfrac{4}{3}x + 4\).

Add 2: \(y = -\tfrac{4}{3}x + 6\).

To graph \(y \ge -3x + 2\), what kind of boundary line do you draw, and which side do you shade?

  1. Solid line, shade above it
  2. Dashed line, shade above it
  3. Solid line, shade below it
  4. Dashed line, shade below it

The sign \(\ge\) includes "equal to", so points on the line count: draw it solid.

\(y\) is greater than or equal to the line, so shade above. (Test \((0, 0)\): \(0 \ge 2\) is false, and \((0, 0)\) is below the line, so shade the other side.)

Answer: a.

Solve \(5 - 3x \le 14\). Give the answer in interval notation.

Subtract 5: \(-3x \le 9\).

Divide by \(-3\) and flip the sign: \(x \ge -3\).

Answer: \([-3, \infty)\).

Simplify \(-(x - 3) + 2(x + 4)\).

Distribute: \(-(x - 3) = -x + 3\) and \(2(x + 4) = 2x + 8\).

Combine like terms: \(-x + 2x = x\) and \(3 + 8 = 11\).

Answer: \(x + 11\).

Want to look something up first? Lines and systems cheat sheet · Glossary

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