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Math 30-1 · Worksheets

Transformations · Set A

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

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Transformations · Set A

Math 30-1 · maddyhelps.com

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

  1. h(x) = (2x − 1)². Which pair of functions gives h(x) = f(g(x))?

    1. a) f(x) = 2x − 1, g(x) = x²
    2. b) f(x) = 2x, g(x) = x² − 1
    3. c) f(x) = x², g(x) = 2x − 1
    4. d) f(x) = x − 1, g(x) = 4x²

  2. If f(x) = x² − 4 and g(x) = x − 2, what is (f/g)(x)?

    1. a) x − 2, x ≠ 2
    2. b) x + 2, with no restriction
    3. c) x + 2, x ≠ 2
    4. d) x² − x − 2

  3. What is the inverse of f(x) = 2x − 6?

    1. a) f⁻¹(x) = 1/(2x − 6)
    2. b) f⁻¹(x) = (x − 6)/2
    3. c) f⁻¹(x) = (x + 6)/2
    4. d) f⁻¹(x) = 2x + 6

  4. The graph of y = −f(x) is a reflection of y = f(x) in:

    1. a) the x-axis
    2. b) the line y = x
    3. c) the origin
    4. d) the y-axis

  5. If f(x) = x + 1 and g(x) = 2x, what is f(g(3))?

    1. a) 6
    2. b) 8
    3. c) 9
    4. d) 7

  6. The graph of y = f(x) is translated 3 units to the right. What is the new equation?

    1. a) y = f(x) − 3
    2. b) y = f(x + 3)
    3. c) y = f(x − 3)
    4. d) y = f(x) + 3

  7. The point (2, 5) is on the graph of y = f(x). Which point is on the graph of its inverse?

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

  8. The point (4, −6) is on y = f(x). Which point is on y = 3f(x − 1) + 2?

    1. a) (5, −16)
    2. b) (5, −12)
    3. c) (3, −16)
    4. d) (13, −16)

  9. If f(x) = x² and g(x) = x + 3, what is g(f(x))?

    1. a) x² + 3
    2. b) (x + 3)²
    3. c) x³ + 3x²
    4. d) x² + x + 3

  10. How could the domain of f(x) = x² be restricted so that its inverse is also a function?

    1. a) x ≥ 1
    2. b) x ≥ 0
    3. c) no restriction is needed
    4. d) y ≥ 0

Answer key · Transformations · Set A

Math 30-1 · maddyhelps.com

  1. c) f(x) = x², g(x) = 2x − 1 — The inside step is 2x − 1, so that is g. The outside step squares the result, so that is f. The second option builds 2x² − 1 instead.
  2. c) x + 2, x ≠ 2 — (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2. But you can never divide by g(x) = 0, so x ≠ 2 stays even after cancelling.
  3. c) f⁻¹(x) = (x + 6)/2 — Swap x and y: x = 2y − 6. Solve for y: y = (x + 6)/2. The inverse undoes the steps in reverse order — add 6, then divide by 2.
  4. a) the x-axis — The negative is outside, so every y-value changes sign and the graph flips over the x-axis. y = f(−x) would flip it over the y-axis.
  5. d) 7 — Work from the inside out: g(3) = 6, then f(6) = 7. Doing f first gives g(f(3)) = 8, which is a different composition.
  6. c) y = f(x − 3) — Horizontal shifts act on x and go the opposite way to the sign: x − 3 moves the graph right. Adding 3 outside the function would move it up.
  7. c) (5, 2) — An inverse swaps x and y, so (2, 5) becomes (5, 2).
  8. a) (5, −16) — x: shift right 1, so 4 + 1 = 5. y: stretch first, then shift — 3 × (−6) + 2 = −16. Shifting before stretching gives the wrong −12.
  9. a) x² + 3 — f goes in first: g(x²) = x² + 3. (x + 3)² is the other order, f(g(x)) — composition is not usually the same both ways.
  10. b) x ≥ 0 — The inverse of the whole parabola fails the vertical line test. Keeping only one half, such as x ≥ 0, makes each output come from one input. y ≥ 0 restricts the range, not the domain.