AP Precalculus Quiz: Transformations Of Functions

What this quiz covers

This quiz focuses on Transformations Of Functions, 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. Domain: all real numbers except $x=6$. Range: all real numbers except $y=-4$.
2. Domain: all real numbers except $x=-2$. Range: all real numbers except $y=-4$. (correct answer)
3. Domain: all real numbers except $x=-2$. Range: all real numbers except $y=6$.
4. Domain: all real numbers except $x=6$. Range: all real numbers except $y=6$.

Explanation: The transformation $r(x+4)$ shifts the graph of $r(x)$ four units to the left. This shifts the vertical asymptote and the excluded value in the domain from $x=2$ to $x=2-4=-2$. The transformation $-5$ shifts the graph five units down. This shifts the horizontal asymptote and the excluded value in the range from $y=1$ to $y=1-5=-4$.

Question 2

1. $x = 2.5$ (correct answer)
2. $x = 10$
3. $x = 7$
4. $x = 2$

Explanation: The vertical asymptote of $h(x)$ occurs where its input is 5. For the function $k(x)$, the input to $h$ is $2x$. Therefore, the vertical asymptote of $k(x)$ will occur when $2x = 5$. Solving for $x$ gives $x = 5/2 = 2.5$. The vertical shift of $-3$ does not affect the location of the vertical asymptote.

Question 3

1. As $x \to \infty$, $g(x) \to \infty$, and as $x \to -\infty$, $g(x) \to -\infty$. (correct answer)
2. As $x \to \infty$, $g(x) \to -\infty$, and as $x \to -\infty$, $g(x) \to \infty$.
3. As $x \to \infty$, $g(x) \to \infty$, and as $x \to -\infty$, $g(x) \to \infty$.
4. As $x \to \infty$, $g(x) \to -\infty$, and as $x \to -\infty$, $g(x) \to -\infty$.

Explanation: A polynomial $p(x)$ of odd degree with a positive leading coefficient has the end behavior: as $x \to \infty$, $p(x) \to \infty$, and as $x \to -\infty$, $p(x) \to -\infty$. The function $g(x) = -p(-x)$ is a reflection of $p(x)$ across the y-axis, followed by a reflection across the x-axis. To find the end behavior of $g(x)$, we consider the limits. As $x \to \infty$, the input to $p$ is $-x$, which approaches $-\infty$. So, $p(-x) \to -\infty$. Then $g(x) = -p(-x) \to -(-\infty) = \infty$. As $x \to -\infty$, the input to $p$ is $-x$, which approaches $\infty$. So, $p(-x) \to \infty$. Then $g(x) = -p(-x) \to -(\infty) = -\infty$. Thus, the end behavior of $g(x)$ is the same as $p(x)$

Question 4

1. A translation 3 units to the left and 5 units down.
2. A translation 3 units to the right and 5 units up. (correct answer)
3. A translation 3 units to the left and 5 units up.
4. A translation 3 units to the right and 5 units down.

Explanation: The function $g(x) = f(x-3) + 5$. The term $(x-3)$ indicates a horizontal translation of 3 units to the right. The term $+5$ outside the function indicates a vertical translation of 5 units up.

Question 5

1. $N(x) = C(2x) + 50$
2. $N(x) = 2C(x) + 50$
3. $N(x) = C(\frac{1}{2}x) + 50$ (correct answer)
4. $N(x) = \frac{1}{2}C(x) + 50$

Explanation: A horizontal stretch by a factor of 2 corresponds to replacing $x$ with $(\frac{1}{2}x)$ in the function definition. This transformation is applied first. A fixed fee of _50 added to the total cost corresponds to a vertical shift up by 50 units, which means adding 50 to the entire function. Therefore, the new cost function is $N(x) = C(\frac{1}{2}x) + 50$.

Question 6

1. A horizontal compression by a factor of $1/3$ followed by a horizontal shift of 2 units to the right. (correct answer)
2. A horizontal compression by a factor of $1/3$ followed by a horizontal shift of 6 units to the right.
3. A horizontal stretch by a factor of 3 followed by a horizontal shift of 2 units to the right.
4. A horizontal shift of 6 units to the right followed by a horizontal compression by a factor of $1/3$.

Explanation: To correctly identify the transformations, the expression inside the function must be factored: $3x-6 = 3(x-2)$. The factor of 3 corresponds to a horizontal compression by a factor of $1/3$. The term $(x-2)$ corresponds to a horizontal shift of 2 units to the right. The standard order is to apply stretches/compressions/reflections first, then shifts.

Question 7

1. $g(x) = (x-2)^3 - 4(x-2) + 6$
2. $g(x) = (x+2)^3 - 4(x+2) - 4$
3. $g(x) = (x+2)^3 - 4(x+2) + 6$ (correct answer)
4. $g(x) = (x-2)^3 - 4(x-2) - 4$

Explanation: To shift the graph of $f(x)$ two units to the left, we replace $x$ with $(x+2)$. To shift the graph five units up, we add 5 to the function. Thus, $g(x) = f(x+2) + 5 = ((x+2)^3 - 4(x+2) + 1) + 5 = (x+2)^3 - 4(x+2) + 6$.

Question 8

1. $q(x) = -3x^2 + 18x$ (correct answer)
2. $q(x) = 3x^2 - 18x$
3. $q(x) = -\frac{1}{3}x^2 + 2x$
4. $q(x) = (-3x)^2 - 6(-3x)$

Explanation: A vertical stretch by a factor of 3 means multiplying the function by 3, resulting in $3p(x) = 3(x^2 - 6x) = 3x^2 - 18x$. A reflection across the x-axis means multiplying the result by -1. Therefore, $q(x) = -1(3p(x)) = -(3x^2 - 18x) = -3x^2 + 18x$.

Question 9

1. A horizontal compression by a factor of $\frac{1}{2}$ and a reflection across the x-axis.
2. A horizontal stretch by a factor of 2 and a reflection across the y-axis.
3. A horizontal compression by a factor of $\frac{1}{2}$ and a reflection across the y-axis. (correct answer)
4. A horizontal stretch by a factor of 2 and a reflection across the x-axis.

Explanation: The transformation is of the form $g(x) = f(bx)$ where $b = -2$. The negative sign indicates a reflection across the y-axis. The magnitude $|b| = 2$ indicates a horizontal compression by a factor of $\frac{1}{|b|} = \frac{1}{2}$.

Question 10

1. $y = 1$ (correct answer)
2. $y = 5$
3. $y = -2$
4. $y = -5$

Explanation: The transformation $-f(x)$ reflects the graph of $f(x)$ across the x-axis. This transformation reflects the horizontal asymptote from $y=2$ to $y=-2$. The transformation $+3$ then shifts the graph, including the asymptote, up by 3 units. The new horizontal asymptote is at $y = -2 + 3 = 1$.

Question 11

1. A horizontal compression by a factor of $1/2$, a horizontal shift 4 units to the left, and a vertical shift 1 unit down. (correct answer)
2. A horizontal compression by a factor of $1/2$, a horizontal shift 8 units to the left, and a vertical shift 1 unit down.
3. A horizontal stretch by a factor of 2, a horizontal shift 4 units to the left, and a vertical shift 1 unit down.
4. A horizontal compression by a factor of $1/2$, a horizontal shift 4 units to the right, and a vertical shift 1 unit down.

Explanation: To identify the horizontal transformations correctly, the expression inside the function must be factored: $g(x) = (2(x+4))^4 - 1$. This shows a horizontal compression by a factor of $1/2$ (from the factor of 2) and a horizontal shift 4 units to the left (from the $(x+4)$ term). The $-1$ outside the function represents a vertical shift 1 unit down.

Question 12

1. The average rate of change of $g(x)$ is $c$ units more than the average rate of change of $f(x)$.
2. The average rate of change of $g(x)$ is $c$ units less than the average rate of change of $f(x)$.
3. The average rate of change of $g(x)$ is $c$ times the average rate of change of $f(x)$.
4. The average rate of change of $g(x)$ is the same as the average rate of change of $f(x)$. (correct answer)

Explanation: The average rate of change of $g(x)$ over $[a, b]$ is $\frac{g(b)-g(a)}{b-a}$. Substituting $g(x) = f(x)+c$, this becomes $\frac{(f(b)+c) - (f(a)+c)}{b-a} = \frac{f(b)-f(a)}{b-a}$, which is the average rate of change of $f(x)$ over $[a, b]$. A vertical shift does not affect the average rate of change.

Question 13

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

Explanation: A reflection across the y-axis means replacing $x$ with $-x$. A horizontal stretch by a factor of 3 means replacing $x$ with $(\frac{1}{3}x)$. Combining these gives replacing $x$ with $(-\frac{1}{3}x)$. A vertical shift up by 1 unit means adding 1 to the function. So, $g(x) = f(-\frac{1}{3}x) + 1 = \frac{1}{-\frac{1}{3}x} + 1$.

Question 14

1. $(4, -4)$ (correct answer)
2. $(2, -4)$
3. $(4, -1)$
4. $(2, -1)$

Explanation: The transformation $y = 2f(x-1)$ involves a horizontal shift and a vertical stretch. The horizontal transformation is $x \to x-1$. To find the new x-coordinate, we set the new input equal to the old input: $x_{new}-1 = 3$, so $x_{new} = 4$. The vertical transformation is $y \to 2y$. The new y-coordinate is $2 \times (-2) = -4$. Thus, the point $(3, -2)$ on the graph of $f(x)$ corresponds to the point $(4, -4)$ on the graph of $y = 2f(x-1)$

Question 15

1. A reflection across the x-axis, followed by a shift 4 units to the left. (correct answer)
2. A reflection across the y-axis, followed by a shift 4 units to the left.
3. A reflection across the x-axis, followed by a shift 4 units to the right.
4. A reflection across the y-axis, followed by a shift 4 units to the right.

Explanation: The term $(x+4)$ inside the function indicates a horizontal shift of 4 units to the left. The negative sign outside the function indicates a reflection across the x-axis. The order of these two transformations does not matter in this case. Note that for an odd function like $f(x)=x^3$, a reflection across the y-axis, $f(-x)=(-x)^3=-x^3$, is equivalent to a reflection across the x-axis, $-f(x) = -x^3$. However, the shift is unambiguously to the left.

Question 16

1. $g(x) = f(x - 1)$
2. $g(x) = f(x) + 1$
3. $g(x) = f(2x)$
4. $g(x) = 2f(x)$ (correct answer)

Explanation: A real zero of a function occurs at an x-value where the function's output is 0. If $c$ is a zero of $f(x)$, then $f(c) = 0$. For the transformation $g(x) = 2f(x)$, the new output at $x=c$ is $g(c) = 2f(c) = 2(0) = 0$. Therefore, $c$ is also a zero of $g(x)$. Horizontal shifts and compressions change the x-values of the zeros. A vertical shift changes the output values, so if $f(c)=0$, then $f(c)+1 = 1$, meaning $c$ is no longer a zero.

Question 17

1. $(2, -4)$
2. $(-4, 4)$
3. $(-4, -4)$ (correct answer)
4. $(2, 4)$

Explanation: The point $(-1, 4)$ on the graph of $f$ means $f(-1) = 4$. The transformation involves a horizontal shift and a vertical reflection. The horizontal shift is 3 units to the left. The new x-coordinate satisfies $x_{new} + 3 = -1$, so $x_{new} = -4$. The vertical transformation is a reflection across the x-axis. The new y-coordinate is $-f(-1) = -4$. Therefore, the point $(-4, -4)$ must be on the graph of $g$.

Question 18

1. The vertical asymptote shifts from $x=0$ to $x=1$, and the horizontal asymptote shifts from $y=0$ to $y=3$. (correct answer)
2. The vertical asymptote shifts from $x=0$ to $x=-1$, and the horizontal asymptote shifts from $y=0$ to $y=3$.
3. The vertical asymptote shifts from $x=0$ to $x=1$, and the horizontal asymptote is stretched from $y=0$ to $y=2$.
4. The vertical asymptote is stretched from $x=0$ to $x=2$, and the horizontal asymptote is shifted from $y=0$ to $y=3$.

Explanation: The parent function $f(x) = 1/x$ has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=0$. The function $g(x)$ can be written as $g(x) = 2f(x-1) + 3$. The horizontal shift of 1 unit right changes the vertical asymptote to $x=1$. The vertical stretch by a factor of 2 does not move the horizontal asymptote from $y=0$, but the vertical shift of 3 units up moves it to $y=3$.