AP Precalculus Quiz: Sinusoidal Function Transformations

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.

Question 1

1. $12$
2. $8$
3. $6$ (correct answer)
4. $4$

Explanation: The amplitude of a sinusoidal function is half the difference between its maximum and minimum values. Amplitude = $\frac{\text{Maximum Value} - \text{Minimum Value}}{2}$. For this function, the amplitude is $\frac{10 - (-2)}{2} = \frac{12}{2} = 6$.

Question 2

1. $\pi$ units to the left
2. $\frac{\pi}{4}$ units to the left (correct answer)
3. $\pi$ units to the right
4. $\frac{\pi}{4}$ units to the right

Explanation: To find the phase shift, we must factor the coefficient of $x$ from the argument of the sine function. $4x+\pi = 4(x+\frac{\pi}{4})$. The function becomes $f(x) = -2\sin(4(x+\frac{\pi}{4}))+1$. This is of the form $y = a\sin(b(x-c))+d$ where $c = -\frac{\pi}{4}$. This represents a phase shift of $\frac{\pi}{4}$ units to the left.

Question 3

1. $y = -\cos(x+\frac{\pi}{6}) + 4$ (correct answer)
2. $y = \cos(-(x+\frac{\pi}{6})) + 4$
3. $y = -\cos(x-\frac{\pi}{6}) + 4$
4. $y = -\cos(x+\frac{\pi}{6}) - 4$

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

Question 4

1. The graph of $y = \sin(-3x)$ is a reflection of the graph of $y=\sin(3x)$ over the y-axis. (correct answer)
2. The graph of $y = \sin(-3x)$ is identical to the graph of $y=\sin(3x)$ because sine is an even function.
3. The graph of $y = \sin(-3x)$ is a horizontal shift of the graph of $y=\sin(3x)$ by $3\pi$ units.
4. The graph of $y = \sin(-3x)$ is a vertical shift of the graph of $y=\sin(3x)$ by 3 units.

Explanation: A transformation of the form $y = f(-x)$ is a reflection of the graph of $y=f(x)$ over the y-axis. Therefore, the graph of $y = \sin(-3x)$ is a reflection of the graph of $y=\sin(3x)$ over the y-axis.

Question 5

1. A horizontal shift of $\frac{\pi}{3}$ units to the left and a vertical shift of 5 units down. (correct answer)
2. A horizontal shift of $\frac{\pi}{3}$ units to the right and a vertical shift of 5 units down.
3. A horizontal shift of $\frac{\pi}{3}$ units to the left and a vertical shift of 5 units up.
4. A horizontal shift of $\frac{\pi}{3}$ units to the right and a vertical shift of 5 units up.

Explanation: In the general form $y = a\sin(b(x-c))+d$, $c$ represents the horizontal shift and $d$ represents the vertical shift. In $g(x) = \sin(x + \frac{\pi}{3}) - 5$, the argument can be written as $(x - (-\frac{\pi}{3}))$, so $c = -\frac{\pi}{3}$, which corresponds to a shift to the left by $\frac{\pi}{3}$ units. The value of $d$ is $-5$, which corresponds to a vertical shift of 5 units down.

Question 6

1. The amplitude is 3 and the midline is $y=-1$. (correct answer)
2. The period is $\pi$ and the amplitude is $-3$.
3. The phase shift is $\frac{\pi}{2}$ units to the right and the period is $4\pi$.
4. The maximum value is 3 and the minimum value is -3.

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

Question 7

1. $y = 4\sin(2(x - \frac{\pi}{2})) - 1$ (correct answer)
2. $y = 4\sin(\frac{1}{2}(x + \frac{\pi}{2})) - 1$
3. $y = -1\sin(2(x - \frac{\pi}{2})) + 4$
4. $y = 4\sin(\pi(x + \frac{\pi}{2})) - 1$

Explanation: For a sinusoidal function $y = a\sin(b(x-c))+d$: The amplitude $|a|=4$. The midline $y=d$ is $y=-1$. The phase shift $c$ is $\frac{\pi}{2}$ to the right. The period is $\frac{2\pi}{|b|} = \pi$, which implies $|b|=2$. Combining these parameters gives the equation $y = 4\sin(2(x - \frac{\pi}{2})) - 1$.

Question 8

1. $\frac{\pi}{3}$
2. $6$ (correct answer)
3. $3$
4. $2\pi$

Explanation: The period of a sinusoidal function of the form $y = a\cos(b(x-c))+d$ is given by the formula $P = \frac{2\pi}{|b|}$. In the function $g(x) = -4\cos(\frac{\pi}{3}x + \pi) - 2$, the value of $b$ is $\frac{\pi}{3}$. Therefore, the period is $P = \frac{2\pi}{\pi/3} = 2\pi \cdot \frac{3}{\pi} = 6$.

Question 9

1. $5$
2. $2$
3. $3$ (correct answer)
4. $8$

Explanation: The amplitude of a sinusoidal function of the form $f(t) = d + a\sin(b(t-c))$ is the absolute value of $a$. In the function $f(t) = 5 - 3\sin(4t + 1)$, the value of $a$ is $-3$. The amplitude is $|-3| = 3$.

Question 10

1. $y = 2$
2. $y = 1$
3. $y = 9$
4. $y = 7$ (correct answer)

Explanation: The midline of a sinusoidal function of the form $h(x) = a\sin(b(x-c))+d$ is the horizontal line $y = d$. For the function $h(x) = 2\sin(\pi(x-1)) + 7$, the value of $d$ is 7. Thus, the equation of the midline is $y = 7$.

Question 11

1. $\frac{\pi}{4}$ units to the right (correct answer)
2. $\frac{\pi}{2}$ units to the right
3. $\frac{\pi}{4}$ units to the left
4. $\frac{\pi}{2}$ units to the left

Explanation: To determine the phase shift, the function must be written in the form $y = a\cos(b(x-c))+d$. We factor out the $b$ value from the argument: $2x - \frac{\pi}{2} = 2(x - \frac{\pi}{4})$. The equation becomes $y = 3\cos(2(x - \frac{\pi}{4}))$. Here, $c = \frac{\pi}{4}$, which represents a phase shift of $\frac{\pi}{4}$ units to the right.

Question 12

1. A horizontal compression by a factor of $\frac{1}{3}$
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 $\frac{1}{3}$

Explanation: A transformation of the form $y = f(bx)$ results in a horizontal dilation of the graph of $y=f(x)$. If $|b| < 1$, it is a horizontal stretch. If $|b| > 1$, it is a horizontal compression. In $y = \cos(\frac{x}{3})$, $b=\frac{1}{3}$. Since $|\frac{1}{3}| < 1$, the transformation is a horizontal stretch by a factor of $1/b = 1/(1/3) = 3$.

Question 13

1. $A$, which controls the amplitude.
2. $B$, which controls the horizontal dilation. (correct answer)
3. $C$, which contributes to the phase shift.
4. $D$, which controls the vertical shift.

Explanation: The period of the function $g(x) = A\cos(Bx+C)+D$ is given by $P = \frac{2\pi}{|B|}$. Therefore, changing the value of $B$ alters the period by causing a horizontal stretch or compression.

Question 14

1. The parameter $a$, which represents the amplitude.
2. The parameter $b$, which affects the period.
3. The parameter $c$, which represents the phase shift.
4. The parameter $d$, which represents the vertical shift. (correct answer)

Explanation: The parameter $d$ in the general form $y = a\sin(b(x-c))+d$ represents the vertical shift of the graph, which determines the midline $y=d$. A shift down by 4 units changes the value of $d$.

Question 15

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 $g(x) = -5\cos(x)$ is of the form $y=af(x)$. 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 16

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 = $d + |a| = 2 + 5 = 7$. Minimum Value = $d - |a| = 2 - 5 = -3$.

Question 17

1. $\cos(x - \frac{\pi}{2})$ (correct answer)
2. $\cos(x + \frac{\pi}{2})$
3. $\cos(x - \pi)$
4. $\cos(x) + 1$

Explanation: The graph of $y=\cos(x)$ shifted to the right by $\frac{\pi}{2}$ units results in the graph of $y=\sin(x)$. This is a known trigonometric identity: $\sin(x) = \cos(x - \frac{\pi}{2})$.

Question 18

1. $y = \cos(3x)$
2. $y = 3\cos(x)$
3. $y = \cos(\frac{1}{3}x)$ (correct answer)
4. $y = \cos(x) + 6\pi$

Explanation: The period of $y = a\cos(bx)+d$ is $P = \frac{2\pi}{|b|}$. We are given that the new period is $6\pi$. So, $6\pi = \frac{2\pi}{|b|}$. Solving for $|b|$ gives $|b| = \frac{2\pi}{6\pi} = \frac{1}{3}$. The function is $y = \cos(\frac{1}{3}x)$.

Question 19

1. A horizontal compression by a factor of $\frac{1}{2}$ and a horizontal shift of $\frac{\pi}{2}$ units to the right. (correct answer)
2. A horizontal compression by a factor of $\frac{1}{2}$ and a horizontal shift of $\pi$ units to the left.
3. A horizontal stretch by a factor of 2 and a horizontal shift of $\frac{\pi}{2}$ units to the right.
4. A horizontal stretch by a factor of 2 and a horizontal shift of $\pi$ units to the left.

Explanation: The transformed function is $g(x) = \sin(2x-\pi)$. To identify the transformations, we factor out the coefficient of $x$: $g(x) = \sin(2(x-\frac{\pi}{2}))$. This form shows a horizontal compression by a factor of $\frac{1}{2}$ (due to the 2 multiplying $x$) and a horizontal shift of $\frac{\pi}{2}$ units to the right (due to the $(x-\frac{\pi}{2})$).

Question 20

1. $g(x) = 10\sin(4x) + 3$ (correct answer)
2. $g(x) = 10\sin(x) + 3$
3. $g(x) = 5\sin(4x) + 3$
4. $g(x) = 10\sin(2x) + 6$

Explanation: For $f(x) = 5\sin(2x) + 3$, the amplitude is 5, the period is $\frac{2\pi}{2} = \pi$, and the midline is $y=3$. For $g(x)$, the amplitude is twice that of $f(x)$, so $a=2 \cdot 5 = 10$. The period is half that of $f(x)$, so $P = \frac{\pi}{2}$. The period formula $P=\frac{2\pi}{b}$ gives $\frac{\pi}{2} = \frac{2\pi}{b}$, so $b=4$. The midline is the same, so $d=3$. Thus, $g(x) = 10\sin(4x) + 3$.