AP Precalculus Quiz: Sinusoidal Function Transformations
20 questions · exam conditions
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Sinusoidal Function TransformationsQuestion 1 of 20

The graph of y=cos⁡(x)y=\cos(x) is shifted π6\frac{\pi}{6} units to the left, then reflected over the x-axis, and finally shifted 4 units up. Which of the following is the equation of the transformed function?

y=−cos⁡(x+π6)+4y = -\cos(x+\frac{\pi}{6}) + 4
y=cos⁡(−(x+π6))+4y = \cos(-(x+\frac{\pi}{6})) + 4
y=−cos⁡(x−π6)+4y = -\cos(x-\frac{\pi}{6}) + 4
y=−cos⁡(x+π6)−4y = -\cos(x+\frac{\pi}{6}) - 4
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AP Precalculus Quiz

AP Precalculus Quiz: Sinusoidal Function Transformations

Practice Sinusoidal Function Transformations in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sinusoidal Function Transformations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The graph of y=cos⁡(x)y=\cos(x) is shifted π6\frac{\pi}{6} units to the left, then reflected over the x-axis, and finally shifted 4 units up. Which of the following is the equation of the transformed function?

  1. y=−cos⁡(x+π6)+4y = -\cos(x+\frac{\pi}{6}) + 4 (correct answer)
  2. y=cos⁡(−(x+π6))+4y = \cos(-(x+\frac{\pi}{6})) + 4
  3. y=−cos⁡(x−π6)+4y = -\cos(x-\frac{\pi}{6}) + 4
  4. y=−cos⁡(x+π6)−4y = -\cos(x+\frac{\pi}{6}) - 4
Explanation: Let's apply the transformations in order. 1. Shift left by π6\frac{\pi}{6} units: y=cos⁡(x+π6)y = \cos(x+\frac{\pi}{6}). 2. Reflect over the x-axis (multiply the entire function by -1): y=−cos⁡(x+π6)y = -\cos(x+\frac{\pi}{6}). 3. Shift up by 4 units (add 4 to the function): y=−cos⁡(x+π6)+4y = -\cos(x+\frac{\pi}{6}) + 4.

Question 2

What is the phase shift of the graph of the function y=3cos⁡(2x−π2)y = 3\cos(2x - \frac{\pi}{2})?

  1. π4\frac{\pi}{4} units to the right (correct answer)
  2. π2\frac{\pi}{2} units to the right
  3. π4\frac{\pi}{4} units to the left
  4. π2\frac{\pi}{2} units to the left
Explanation: To determine the phase shift, the function must be written in the form y=acos⁡(b(x−c))+dy = a\cos(b(x-c))+d. We factor out the bb value from the argument: 2x−π2=2(x−π4)2x - \frac{\pi}{2} = 2(x - \frac{\pi}{4}). The equation becomes y=3cos⁡(2(x−π4))y = 3\cos(2(x - \frac{\pi}{4})). Here, c=π4c = \frac{\pi}{4}, which represents a phase shift of π4\frac{\pi}{4} units to the right.

Question 3

The graph of y=cos⁡(x3)y = \cos(\frac{x}{3}) is obtained from the graph of y=cos⁡(x)y = \cos(x) by which of the following transformations?

  1. A horizontal compression by a factor of 13\frac{1}{3}
  2. A horizontal stretch by a factor of 3 (correct answer)
  3. A vertical stretch by a factor of 3
  4. A vertical compression by a factor of 13\frac{1}{3}
Explanation: A transformation of the form y=f(bx)y = f(bx) results in a horizontal dilation of the graph of y=f(x)y=f(x). If ∣b∣<1|b| < 1, it is a horizontal stretch. If ∣b∣>1|b| > 1, it is a horizontal compression. In y=cos⁡(x3)y = \cos(\frac{x}{3}), b=13b=\frac{1}{3}. Since ∣13∣<1|\frac{1}{3}| < 1, the transformation is a horizontal stretch by a factor of 1/b=1/(1/3)=31/b = 1/(1/3) = 3.

Question 4

In the function g(x)=Acos⁡(Bx+C)+Dg(x) = A\cos(Bx+C)+D, which parameter must be changed to alter the period of the function?

  1. AA, which controls the amplitude.
  2. BB, which controls the horizontal dilation. (correct answer)
  3. CC, which contributes to the phase shift.
  4. DD, which controls the vertical shift.
Explanation: The period of the function g(x)=Acos⁡(Bx+C)+Dg(x) = A\cos(Bx+C)+D is given by P=2π∣B∣P = \frac{2\pi}{|B|}. Therefore, changing the value of BB alters the period by causing a horizontal stretch or compression.

Question 5

The graph of a sinusoidal function is shifted down by 4 units. Which parameter in the general form y=asin⁡(b(x−c))+dy = a\sin(b(x-c))+d is directly affected by this transformation?

  1. The parameter aa, which represents the amplitude.
  2. The parameter bb, which affects the period.
  3. The parameter cc, which represents the phase shift.
  4. The parameter dd, which represents the vertical shift. (correct answer)
Explanation: The parameter dd in the general form y=asin⁡(b(x−c))+dy = a\sin(b(x-c))+d represents the vertical shift of the graph, which determines the midline y=dy=d. A shift down by 4 units changes the value of dd.

Question 6

The graph of g(x)=−5cos⁡(x)g(x) = -5\cos(x) is obtained from the graph of f(x)=cos⁡(x)f(x) = \cos(x) by what sequence of transformations?

  1. A vertical stretch by a factor of 5, followed by a reflection over the x-axis. (correct answer)
  2. A vertical stretch by a factor of 5, followed by a reflection over the y-axis.
  3. A horizontal stretch by a factor of 5, followed by a reflection over the x-axis.
  4. A vertical shift of 5 units down, followed by a reflection over the y-axis.
Explanation: The function g(x)=−5cos⁡(x)g(x) = -5\cos(x) is of the form y=af(x)y=af(x). The factor of 5 causes a vertical stretch by a factor of 5. The negative sign causes a reflection over the x-axis. The order of these two transformations does not matter.

Question 7

A sinusoidal function has an amplitude of 5 and a midline of y=2y=2. What are the maximum and minimum values of the function?

  1. Maximum value is 7, and minimum value is -3. (correct answer)
  2. Maximum value is 5, and minimum value is -5.
  3. Maximum value is 10, and minimum value is -8.
  4. Maximum value is 7, and minimum value is 2.
Explanation: The maximum value of a sinusoidal function is the midline plus the amplitude, and the minimum value is the midline minus the amplitude. Maximum Value = d+∣a∣=2+5=7d + |a| = 2 + 5 = 7. Minimum Value = d−∣a∣=2−5=−3d - |a| = 2 - 5 = -3.

Question 8

The function f(x)=sin⁡(x)f(x) = \sin(x) can also be expressed as a phase-shifted cosine function. Which of the following expressions is equivalent to f(x)f(x)?

  1. cos⁡(x−π2)\cos(x - \frac{\pi}{2}) (correct answer)
  2. cos⁡(x+π2)\cos(x + \frac{\pi}{2})
  3. cos⁡(x−π)\cos(x - \pi)
  4. cos⁡(x)+1\cos(x) + 1
Explanation: The graph of y=cos⁡(x)y=\cos(x) shifted to the right by π2\frac{\pi}{2} units results in the graph of y=sin⁡(x)y=\sin(x). This is a known trigonometric identity: sin⁡(x)=cos⁡(x−π2)\sin(x) = \cos(x - \frac{\pi}{2}).

Question 9

A transformation is applied to the function y=cos⁡(x)y=\cos(x) to change its period to 6π6\pi without altering its amplitude or midline. Which of the following functions represents this transformation?

  1. y=cos⁡(3x)y = \cos(3x)
  2. y=3cos⁡(x)y = 3\cos(x)
  3. y=cos⁡(13x)y = \cos(\frac{1}{3}x) (correct answer)
  4. y=cos⁡(x)+6πy = \cos(x) + 6\pi
Explanation: The period of y=acos⁡(bx)+dy = a\cos(bx)+d is P=2π∣b∣P = \frac{2\pi}{|b|}. We are given that the new period is 6π6\pi. So, 6π=2π∣b∣6\pi = \frac{2\pi}{|b|}. Solving for ∣b∣|b| gives ∣b∣=2π6π=13|b| = \frac{2\pi}{6\pi} = \frac{1}{3}. The function is y=cos⁡(13x)y = \cos(\frac{1}{3}x).

Question 10

A sinusoidal function has an amplitude of 4, a period of π\pi, a midline of y=−1y=-1, and a phase shift of π2\frac{\pi}{2} to the right. Which of the following equations could represent this function?

  1. y=4sin⁡(2(x−π2))−1y = 4\sin(2(x - \frac{\pi}{2})) - 1 (correct answer)
  2. y=4sin⁡(12(x+π2))−1y = 4\sin(\frac{1}{2}(x + \frac{\pi}{2})) - 1
  3. y=−1sin⁡(2(x−π2))+4y = -1\sin(2(x - \frac{\pi}{2})) + 4
  4. y=4sin⁡(π(x+π2))−1y = 4\sin(\pi(x + \frac{\pi}{2})) - 1
Explanation: For a sinusoidal function y=asin⁡(b(x−c))+dy = a\sin(b(x-c))+d: The amplitude ∣a∣=4|a|=4. The midline y=dy=d is y=−1y=-1. The phase shift cc is π2\frac{\pi}{2} to the right. The period is 2π∣b∣=π\frac{2\pi}{|b|} = \pi, which implies ∣b∣=2|b|=2. Combining these parameters gives the equation y=4sin⁡(2(x−π2))−1y = 4\sin(2(x - \frac{\pi}{2})) - 1.

Question 11

The function g(x)g(x) has twice the amplitude and half the period of the function f(x)=5sin⁡(2x)+3f(x) = 5\sin(2x) + 3. The midline of g(x)g(x) is the same as the midline of f(x)f(x). Which of the following is an equation for g(x)g(x)?

  1. g(x)=10sin⁡(4x)+3g(x) = 10\sin(4x) + 3 (correct answer)
  2. g(x)=10sin⁡(x)+3g(x) = 10\sin(x) + 3
  3. g(x)=5sin⁡(4x)+3g(x) = 5\sin(4x) + 3
  4. g(x)=10sin⁡(2x)+6g(x) = 10\sin(2x) + 6
Explanation: For f(x)=5sin⁡(2x)+3f(x) = 5\sin(2x) + 3, the amplitude is 5, the period is 2π2=π\frac{2\pi}{2} = \pi, and the midline is y=3y=3. For g(x)g(x), the amplitude is twice that of f(x)f(x), so a=2⋅5=10a=2 \cdot 5 = 10. The period is half that of f(x)f(x), so P=π2P = \frac{\pi}{2}. The period formula P=2πbP=\frac{2\pi}{b} gives π2=2πb\frac{\pi}{2} = \frac{2\pi}{b}, so b=4b=4. The midline is the same, so d=3d=3. Thus, g(x)=10sin⁡(4x)+3g(x) = 10\sin(4x) + 3.

Question 12

Which of the following statements correctly describes a characteristic of the function f(x)=−3sin⁡(12x+π2)−1f(x) = -3\sin(\frac{1}{2}x + \frac{\pi}{2}) - 1?

  1. The amplitude is 3 and the midline is y=−1y=-1. (correct answer)
  2. The period is π\pi and the amplitude is −3-3.
  3. The phase shift is π2\frac{\pi}{2} units to the right and the period is 4π4\pi.
  4. The maximum value is 3 and the minimum value is -3.
Explanation: For f(x)=−3sin⁡(12x+π2)−1f(x)=-3\sin(\frac{1}{2}x + \frac{\pi}{2}) - 1: The amplitude is ∣−3∣=3|-3|=3. The midline is y=−1y=-1. The period is 2π1/2=4π\frac{2\pi}{1/2}=4\pi. The phase shift requires factoring: 12(x+π)\frac{1}{2}(x+\pi), so the shift is π\pi units to the left. The maximum value is −1+3=2-1+3=2 and the minimum is −1−3=−4-1-3=-4. Therefore, the statement that the amplitude is 3 and the midline is y=−1y=-1 is correct.

Question 13

A sinusoidal function f(x)=asin⁡(bx)+df(x) = a\sin(bx)+d has a maximum value of 10 and a minimum value of -2. What is the amplitude of the function?

  1. 1212
  2. 88
  3. 66 (correct answer)
  4. 44
Explanation: The amplitude of a sinusoidal function is half the difference between its maximum and minimum values. Amplitude = Maximum Value−Minimum Value2\frac{\text{Maximum Value} - \text{Minimum Value}}{2}. For this function, the amplitude is 10−(−2)2=122=6\frac{10 - (-2)}{2} = \frac{12}{2} = 6.

Question 14

What is the phase shift of the function f(x)=−2sin⁡(4x+π)+1f(x) = -2\sin(4x+\pi)+1?

  1. π\pi units to the left
  2. π4\frac{\pi}{4} units to the left (correct answer)
  3. π\pi units to the right
  4. π4\frac{\pi}{4} units to the right
Explanation: To find the phase shift, we must factor the coefficient of xx from the argument of the sine function. 4x+π=4(x+π4)4x+\pi = 4(x+\frac{\pi}{4}). The function becomes f(x)=−2sin⁡(4(x+π4))+1f(x) = -2\sin(4(x+\frac{\pi}{4}))+1. This is of the form y=asin⁡(b(x−c))+dy = a\sin(b(x-c))+d where c=−π4c = -\frac{\pi}{4}. This represents a phase shift of π4\frac{\pi}{4} units to the left.

Question 15

How does the graph of y=sin⁡(−3x)y = \sin(-3x) relate to the graph of y=sin⁡(3x)y = \sin(3x)?

  1. The graph of y=sin⁡(−3x)y = \sin(-3x) is a reflection of the graph of y=sin⁡(3x)y=\sin(3x) over the y-axis. (correct answer)
  2. The graph of y=sin⁡(−3x)y = \sin(-3x) is identical to the graph of y=sin⁡(3x)y=\sin(3x) because sine is an even function.
  3. The graph of y=sin⁡(−3x)y = \sin(-3x) is a horizontal shift of the graph of y=sin⁡(3x)y=\sin(3x) by 3π3\pi units.
  4. The graph of y=sin⁡(−3x)y = \sin(-3x) is a vertical shift of the graph of y=sin⁡(3x)y=\sin(3x) by 3 units.
Explanation: A transformation of the form y=f(−x)y = f(-x) is a reflection of the graph of y=f(x)y=f(x) over the y-axis. Therefore, the graph of y=sin⁡(−3x)y = \sin(-3x) is a reflection of the graph of y=sin⁡(3x)y=\sin(3x) over the y-axis.

Question 16

Which of the following describes the transformations applied to the graph of y=sin⁡(x)y = \sin(x) to obtain the graph of g(x)=sin⁡(x+π3)−5g(x) = \sin(x + \frac{\pi}{3}) - 5?

  1. A horizontal shift of π3\frac{\pi}{3} units to the left and a vertical shift of 5 units down. (correct answer)
  2. A horizontal shift of π3\frac{\pi}{3} units to the right and a vertical shift of 5 units down.
  3. A horizontal shift of π3\frac{\pi}{3} units to the left and a vertical shift of 5 units up.
  4. A horizontal shift of π3\frac{\pi}{3} units to the right and a vertical shift of 5 units up.
Explanation: In the general form y=asin⁡(b(x−c))+dy = a\sin(b(x-c))+d, cc represents the horizontal shift and dd represents the vertical shift. In g(x)=sin⁡(x+π3)−5g(x) = \sin(x + \frac{\pi}{3}) - 5, the argument can be written as (x−(−π3))(x - (-\frac{\pi}{3})), so c=−π3c = -\frac{\pi}{3}, which corresponds to a shift to the left by π3\frac{\pi}{3} units. The value of dd is −5-5, which corresponds to a vertical shift of 5 units down.

Question 17

What is the period of the function g(x)=−4cos⁡(π3x+π)−2g(x) = -4\cos(\frac{\pi}{3}x + \pi) - 2?

  1. π3\frac{\pi}{3}
  2. 66 (correct answer)
  3. 33
  4. 2π2\pi
Explanation: The period of a sinusoidal function of the form y=acos⁡(b(x−c))+dy = a\cos(b(x-c))+d is given by the formula P=2π∣b∣P = \frac{2\pi}{|b|}. In the function g(x)=−4cos⁡(π3x+π)−2g(x) = -4\cos(\frac{\pi}{3}x + \pi) - 2, the value of bb is π3\frac{\pi}{3}. Therefore, the period is P=2ππ/3=2π⋅3π=6P = \frac{2\pi}{\pi/3} = 2\pi \cdot \frac{3}{\pi} = 6.

Question 18

What is the amplitude of the function f(t)=5−3sin⁡(4t+1)f(t) = 5 - 3\sin(4t + 1)?

  1. 55
  2. 22
  3. 33 (correct answer)
  4. 88
Explanation: The amplitude of a sinusoidal function of the form f(t)=d+asin⁡(b(t−c))f(t) = d + a\sin(b(t-c)) is the absolute value of aa. In the function f(t)=5−3sin⁡(4t+1)f(t) = 5 - 3\sin(4t + 1), the value of aa is −3-3. The amplitude is ∣−3∣=3|-3| = 3.

Question 19

Which of the following is the equation of the midline for the function h(x)=2sin⁡(π(x−1))+7h(x) = 2\sin(\pi(x-1)) + 7?

  1. y=2y = 2
  2. y=1y = 1
  3. y=9y = 9
  4. y=7y = 7 (correct answer)
Explanation: The midline of a sinusoidal function of the form h(x)=asin⁡(b(x−c))+dh(x) = a\sin(b(x-c))+d is the horizontal line y=dy = d. For the function h(x)=2sin⁡(π(x−1))+7h(x) = 2\sin(\pi(x-1)) + 7, the value of dd is 7. Thus, the equation of the midline is y=7y = 7.

Question 20

The function f(x)=sin⁡(x)f(x)=\sin(x) is transformed to g(x)=f(2x−π)g(x)=f(2x-\pi). Which of the following correctly describes the transformations?

  1. A horizontal compression by a factor of 12\frac{1}{2} and a horizontal shift of π2\frac{\pi}{2} units to the right. (correct answer)
  2. A horizontal compression by a factor of 12\frac{1}{2} and a horizontal shift of π\pi units to the left.
  3. A horizontal stretch by a factor of 2 and a horizontal shift of π2\frac{\pi}{2} units to the right.
  4. A horizontal stretch by a factor of 2 and a horizontal shift of π\pi units to the left.
Explanation: The transformed function is g(x)=sin⁡(2x−π)g(x) = \sin(2x-\pi). To identify the transformations, we factor out the coefficient of xx: g(x)=sin⁡(2(x−π2))g(x) = \sin(2(x-\frac{\pi}{2})) . This form shows a horizontal compression by a factor of 12\frac{1}{2} (due to the 2 multiplying xx) and a horizontal shift of π2\frac{\pi}{2} units to the right (due to the (x−π2)(x-\frac{\pi}{2})).