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Math 20-1

Math 20-1 · Worksheets

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Ten questions of mixed difficulty, covering Inequalities and graphs, Quadratic functions, Radicals, Rational expressions, Sequences and series, Trigonometry. Print it, or work through it on screen — the answer key starts on its own page.

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Math 20-1 · maddyhelps.com

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

  1. On the graph of a function and its reciprocal, the invariant points occur where

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

  2. The sum of the infinite series 12 + 6 + 3 + 1.5 + … is

    1. a) 18
    2. b) ∞
    3. c) 21
    4. d) 24

  3. For 0° ≤ θ ≤ 360°, the solutions to tan θ = −1 are

    1. a) 45° and 315°
    2. b) 45° and 225°
    3. c) 135° only
    4. d) 135° and 315°

  4. The vertex of y = 2x² − 8x + 3 is

    1. a) (2, 5)
    2. b) (2, −5)
    3. c) (−2, 27)
    4. d) (4, 3)

  5. The value of |−7 + 3| is

    1. a) 10
    2. b) 4
    3. c) −10
    4. d) −4

  6. The solution to √(x + 7) = 4 is

    1. a) x = 16
    2. b) x = −3
    3. c) x = 9
    4. d) x = 23

  7. For √(x − 5) to be a real number, x must satisfy

    1. a) x ≥ 0
    2. b) x ≤ 5
    3. c) x ≥ 5
    4. d) x > 5

  8. The solution to x/(x − 4) = 4/(x − 4) + 3 is

    1. a) no solution
    2. b) x = 4 or x = 0
    3. c) x = 0
    4. d) x = 4

  9. The graph of y = a(x − p)² + q has no x-intercepts when

    1. a) a < 0 and q > 0
    2. b) p = 0
    3. c) a > 0 and q > 0
    4. d) a > 0 and q < 0

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

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

Answer key · Cumulative review

Math 20-1 · maddyhelps.com

  1. a) y = 1 or y = −1 — A point stays put only if it equals its own reciprocal, and the only numbers with that property are 1 and −1. Where the original function is zero, the reciprocal is undefined instead — that is an asymptote, not an invariant point.
  2. d) 24 — The ratio is ½, and |r| < 1, so the series converges to S = t₁/(1 − r) = 12/0.5 = 24. Adding forever gives a finite total here because the terms shrink fast enough.
  3. d) 135° and 315° — The reference angle is 45°, and tangent is negative where sine and cosine have opposite signs — quadrants II and IV. That gives 180° − 45° = 135° and 360° − 45° = 315°.
  4. b) (2, −5) — The axis of symmetry is x = −b/(2a) = 8/4 = 2. Substituting back: y = 2(4) − 16 + 3 = −5. So the vertex is (2, −5). Forgetting the minus sign in −b/(2a) sends you to x = −2.
  5. b) 4 — Work inside the bars first: −7 + 3 = −4. Absolute value is distance from zero, so |−4| = 4. The bars are not a minus sign and they are not brackets you can distribute over — they act on whatever the expression inside comes out to.
  6. c) x = 9 — Square both sides: x + 7 = 16, so x = 9. Check it in the original: √16 = 4 ✓. Squaring is safe here, but always check, because squaring can create roots the original equation never had.
  7. c) x ≥ 5 — A square root of a negative number is not real, so the radicand must be zero or more: x − 5 ≥ 0, giving x ≥ 5. Five itself is allowed, because √0 = 0.
  8. a) no solution — Multiplying by (x − 4) gives x = 4 + 3(x − 4), so x = 3x − 8 and x = 4. But x = 4 is the non-permissible value — it makes the original denominators zero — so it must be thrown out, and nothing is left. Finding the restriction first tells you this is coming.
  9. c) a > 0 and q > 0 — An upward parabola whose lowest point is already above the x-axis never reaches it. The same is true the other way: a < 0 with q < 0. What matters is that a and q have the same sign.
  10. c) 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.