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Math 20-1

Math 20-1 · Worksheets

Inequalities and graphs

Ten questions of mixed difficulty, covering Inequalities and graphs. Print it, or work through it on screen — the answer key starts on its own page.

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Inequalities and graphs

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. The graph of y = |x − 2| is

    1. a) a parabola with vertex (2, 0)
    2. b) a V with its vertex at (2, 0)
    3. c) a V with its vertex at (0, 2)
    4. d) a V with its vertex at (−2, 0)

  2. The graph of y < x² − 4 is drawn with

    1. a) a dashed parabola, shaded below
    2. b) a solid parabola, shaded below
    3. c) a dashed parabola, shaded above
    4. d) a solid parabola, shaded above

  3. Testing (0, 0) in y ≥ x² − 4 shows that

    1. a) the origin is a solution, so shade the region containing it
    2. b) the inequality has no solutions
    3. c) the origin is not a solution
    4. d) the boundary should be dashed

  4. The point (1, 5) is a solution to

    1. a) y > x² + 3
    2. b) y < x² + 3
    3. c) y < x + 3
    4. d) y ≤ x²

  5. The solution to x² − x − 6 ≤ 0 is

    1. a) −3 ≤ x ≤ 2
    2. b) −2 ≤ x ≤ 3
    3. c) x ≤ 3
    4. d) x ≤ −2 or x ≥ 3

  6. The graph of y = 1/(x² − 4) has

    1. a) a vertical asymptote at x = 0
    2. b) vertical asymptotes at x = ±2 and a local minimum at (0, −¼)
    3. c) no vertical asymptotes
    4. d) vertical asymptotes at x = ±2 and a local maximum at (0, −¼)

  7. The vertical asymptote of y = 1/(x − 3) is

    1. a) y = 0
    2. b) x = −3
    3. c) y = 3
    4. d) x = 3

  8. Compared with y = 2x − 6, the graph of y = |2x − 6|

    1. a) is the whole line reflected in the x-axis
    2. b) is identical everywhere
    3. c) is identical for x ≥ 3 and reflected in the x-axis for x < 3
    4. d) is the whole line shifted up

  9. The solution to x² > 16 is

    1. a) x > 4
    2. b) x > ±4
    3. c) x < −4 or x > 4
    4. d) −4 < x < 4

  10. The solution to |x − 4| < 3 is

    1. a) x < 7
    2. b) −3 < x < 3
    3. c) 1 < x < 7
    4. d) x < 1 or x > 7

Answer key · Inequalities and graphs

Math 20-1 · maddyhelps.com

  1. b) a V with its vertex at (2, 0) — The graph of y = x − 2 crosses the x-axis at x = 2; absolute value folds everything below the axis upward, creating a corner exactly there. Like vertex form, the minus sign inside means a shift to the right.
  2. a) a dashed parabola, shaded below — A strict inequality excludes the boundary, so the curve is dashed. Less than means the region below the parabola. Using a solid curve claims the points on it are solutions, and they are not.
  3. a) the origin is a solution, so shade the region containing it — Substituting gives 0 ≥ −4, which is true, so the origin lies in the solution region and you shade that side. One test point settles the whole region, as long as it is not on the boundary itself.
  4. a) y > x² + 3 — Substitute and check: x² + 3 = 4, and 5 > 4 is true. The other options fail — for instance 5 < 4 is false. Testing is quicker and safer than reasoning about the picture.
  5. b) −2 ≤ x ≤ 3 — Factor to (x − 3)(x + 2) ≤ 0, so the roots are −2 and 3. The parabola opens upward, so it is at or below zero between the roots. Picture the graph rather than trying to divide by a bracket, which is not allowed when its sign is unknown.
  6. d) vertical asymptotes at x = ±2 and a local maximum at (0, −¼) — The denominator is zero at x = 2 and x = −2, so those are the asymptotes. At x = 0 the original is −4, so the reciprocal is −¼ — and because the original has a minimum there, the reciprocal has a maximum. Reciprocals turn minima into maxima whenever the value is negative.
  7. d) x = 3 — A reciprocal blows up where its denominator is zero, which is x = 3. That is a vertical line, so it is written as an equation in x. The horizontal asymptote here is y = 0, which is a different thing.
  8. c) is identical for x ≥ 3 and reflected in the x-axis for x < 3 — Absolute value leaves non-negative outputs alone and flips negative ones. The line is negative to the left of its x-intercept x = 3, so only that part is reflected, producing a corner at (3, 0).
  9. c) x < −4 or x > 4 — Rewrite as x² − 16 > 0, so (x − 4)(x + 4) > 0, with roots ±4. An upward parabola is above zero outside its roots, so the solution is two separate pieces. Taking a square root and writing x > 4 quietly loses every negative solution.
  10. c) 1 < x < 7 — Read it as distance: x is less than 3 away from 4, which is everything between 1 and 7. Algebraically, −3 < x − 4 < 3, then add 4 throughout. A less-than absolute value always gives one interval; greater-than gives two pieces.