AP Precalculus -1.12 Transformations of Functions- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.12 Transformations of Functions- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.12 Transformations of Functions- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The function $ f $ is given by $ f(x) = 3x^2 + 2x + 1 $. The graph of which of the following functions is the image of the graph of $ f $ after a vertical dilation of the graph of $ f $ by a factor of 2?

(A) $ m(x) = 12x^2 + 4x + 1 $, because this is a multiplicative transformation of $ f $ that results from multiplying each input value $ x $ by 2.
(B) $ k(x) = 6x^2 + 4x + 2 $, because this is a multiplicative transformation of $ f $ that results from multiplying $ f(x) $ by 2.
(C) $ p(x) = 3(x + 2)^2 + 2(x + 2) + 1 $, because this is an additive transformation of $ f $ that results from adding 2 to each input value $ x $.
(D) $ n(x) = 3x^2 + 2x + 3 $, because this is an additive transformation of $ f $ that results from adding 2 to $ f(x) $.

▶️ Answer/Explanation

Detailed solution

Vertical dilation by factor 2 means: multiply $ f(x) $ by 2.
$ 2f(x) = 2(3x^2 + 2x + 1) = 6x^2 + 4x + 2 $.
This matches $ k(x) $ in option (B).
✅ Answer: (B)

Question

$x$

$-8$

$-4$

$-2$

$-1$

$0$

$3$

$f(x)$

$87$

$55$

$5$

$-4$

$-7$

$20$

The table gives values for a polynomial function $f$ at selected values of $x$. Let $g(x) = af(bx) + c$, where $a, b,$ and $c$ are positive constants. In the $xy$-plane, the graph of $g$ is constructed by applying three transformations to the graph of $f$ in this order: a horizontal dilation by a factor of 2, a vertical dilation by a factor of 3, and a vertical translation by 5 units. What is the value of $g(-4)$?

(A) 266
(B) 170
(C) 28
(D) 20

▶️ Answer/Explanation

Detailed solution

Horizontal dilation by factor 2 ⇒ input scaled by $ \frac{1}{2} $ (since dilation by factor $ k $ means $ x $ replaced by $ x/k $).
So $ b = \frac{1}{2} $.
Vertical dilation by factor 3 ⇒ multiply output by 3 ⇒ $ a = 3 $.
Vertical translation up 5 ⇒ add 5 ⇒ $ c = 5 $.
Thus $ g(x) = 3f\left(\frac{1}{2}x\right) + 5 $.
To find $ g(-4) $:
First compute $ \frac{1}{2}(-4) = -2 $.
From table, $ f(-2) = 5 $.
Then $ g(-4) = 3 \cdot 5 + 5 = 15 + 5 = 20 $.
✅ Answer: (D)

Question

The graph of $ y = f(x) $, consisting of four line segments and a semicircle, is shown for $-3 \leq x \leq 3$.

Which of the following is the transformed graph for $ y = f(x + 1) – 2 $?

▶️ Answer/Explanation

Detailed solution

The transformation $ y = f(x + 1) – 2 $ represents:
– Horizontal shift: $ x + 1 $ ⇒ shift left by 1 unit.
– Vertical shift: $ -2 $ ⇒ shift down by 2 units.
Thus the entire original graph (line segments and semicircle) moves 1 unit left and 2 units down.
The only choice showing this translated shape is option (A).
✅ Answer: (A)

Question

The polynomial function $ f $ is given by $ f(x) = ax^4 + bx^3 + cx^2 + dx + k $, where $ a \neq 0 $ and $ b, c, d, $ and $ k $ are constants. Which of the following statements about $ f $ is true?

(A) $ f $ has both a global maximum and a global minimum.
(B) $ f $ has either a global maximum or a global minimum, but not both.
(C) $ f $ has neither a global maximum nor a global minimum.
(D) The nature of a global maximum or a global minimum for $ f $ cannot be determined without more information about $ b, c, d, $ and $ k $.

▶️ Answer/Explanation

Detailed solution

$ f $ is a quartic polynomial (degree 4, even).
For even-degree polynomials with leading coefficient $ a \neq 0 $:
– If $ a > 0 $, as $ x \to \pm\infty $, $ f(x) \to +\infty $ ⇒ global minimum exists, no global maximum.
– If $ a < 0 $, as $ x \to \pm\infty $, $ f(x) \to -\infty $ ⇒ global maximum exists, no global minimum.
Thus $ f $ has either a global max or a global min, but not both.
✅ Answer: (B)

Question

During a month in a certain town, the temperature increases and decreases over the course of a day. A graph of the average temperatures during that month for any given day is shown. The data in the graph can be modeled by the function $ T $, where $ T(h) $ gives the temperature, in degrees Fahrenheit (°F), at time $ h $ hours after midnight. At a certain point in that month, there is an adjustment to clocks for daylight saving time, at which point clocks are adjusted 1 hour forward. For example, 6 a.m. instantly becomes 7 a.m. The function $ D $ models the same data as function $ T $, after the shift to daylight saving time. If $ h $ still represents the number of hours after midnight, which of the following defines $ D(h) $ in terms of $ T $?

(A) $ D(h) = T(h + 1) $
(B) $ D(h) = T(h – 1) $
(C) $ D(h) = T(h) + 1 $
(D) $ D(h) = T(h) – 1 $

▶️ Answer/Explanation

Detailed solution

After daylight saving time starts, clocks move forward 1 hour.
This means the temperature at a given clock time $ h $ (hours after midnight) used to occur at time $ h – 1 $ before the shift.
Thus $ D(h) = T(h – 1) $.
Example: At new time 7 a.m. ($ h = 7 $), the temperature is what it used to be at old time 6 a.m. ($ h = 6 $), so $ D(7) = T(6) $.
✅ Answer: (B)

Question

The functions $ f $ and $ g $ are defined for all real numbers such that $ g(x) = f(2(x – 4)) $. Which of the following sequences of transformations maps the graph of $ f $ to the graph of $ g $ in the same $ xy $-plane?

(A) A horizontal dilation of the graph of $ f $ by a factor of 2, followed by a horizontal translation of the graph of $ f $ by $-8$ units
(B) A horizontal dilation of the graph of $ f $ by a factor of 2, followed by a horizontal translation of the graph of $ f $ by 8 units
(C) A horizontal dilation of the graph of $ f $ by a factor of $\frac{1}{2}$, followed by a horizontal translation of the graph of $ f $ by $-4$ units
(D) A horizontal dilation of the graph of $ f $ by a factor of $\frac{1}{2}$, followed by a horizontal translation of the graph of $ f $ by 4 units

▶️ Answer/Explanation

Detailed solution

Inside $ f $: argument is $ 2(x – 4) $.
Factor inside: $ 2(x – 4) = 2x – 8 $.
Transformation form: $ f(bx + c) $ with $ b = 2, c = -8 $.
Order: horizontal dilation by factor $ \frac{1}{|b|} = \frac{1}{2} $ (compression), then horizontal shift by $ -\frac{c}{b} = -\frac{-8}{2} = 4 $ units right.
So sequence: horizontal dilation factor $ \frac{1}{2} $ → horizontal translation right 4 units.
✅ Answer: (D)

Question

The function $ f $ is given by $ f(x) = x^2 + 2 $. The function $ g $ is the result of a transformation of $ f $ and is given by $ g(x) = \frac{x^2}{4} + 2 $. Which of the following describes the transformation of the graph of $ f $ whose image is the graph of $ g $?

(A) A vertical dilation by a factor of $\frac{1}{4}$
(B) A horizontal dilation by a factor of $\frac{1}{2}$
(C) A horizontal dilation by a factor of 2
(D) A horizontal dilation by a factor of 4

▶️ Answer/Explanation

Detailed solution

Compare: $ f(x) = x^2 + 2 $, $ g(x) = \frac{x^2}{4} + 2 = \left(\frac{x}{2}\right)^2 + 2 $.
Thus $ g(x) = f\left(\frac{x}{2}\right) $.
Inside $ f $: input multiplied by $ \frac{1}{2} $ ⇒ horizontal dilation by factor $ \frac{1}{1/2} = 2 $ (stretch).
✅ Answer: (C)

Question

The function $ f $ has domain $[-2, 2]$ and range $[1, 5]$. The function $ g $ is given by $ g(x) = -2f(x + 3) + 4 $. What are the domain and range of $ g $?

(A) domain: $[-5, -1]$, range: $[-6, 2]$
(B) domain: $[-5, -1]$, range: $[-2, 6]$
(C) domain: $[1, 5]$, range: $[-2, 6]$
(D) domain: $[1, 5]$, range: $[-6, 2]$

▶️ Answer/Explanation

Detailed solution

Domain transformation: $ x+3 $ inside $ f $ means shift left 3.
Domain of $ f $: $ -2 \leq x \leq 2 $.
For $ g $, require $ -2 \leq x+3 \leq 2 $ ⇒ $ -5 \leq x \leq -1 $.
Domain of $ g $: $[-5, -1]$.
Range transformation: $ -2f(x+3) + 4 $.
Range of $ f $: $ 1 \leq f \leq 5 $.
Multiply by $-2$ (reflect over $x$-axis and stretch by 2):
Max $ f = 5 $ → $ -2 \cdot 5 = -10 $ (becomes min after reflection).
Min $ f = 1 $ → $ -2 \cdot 1 = -2 $ (becomes max after reflection).
Then add 4: $ -10 + 4 = -6 $, $ -2 + 4 = 2 $.
So range of $ g $: $[-6, 2]$.
✅ Answer: (A)

Question

$x$

$f(x)$

$g(x)$

The table gives values of the functions $f$ and $g$ for selected values of $x$. The pattern of the values of $f$ and $g$ continue, repeating every interval of width 6, for $0 \leq x \leq 48$. The graph of the function $g$ is the result of a sequence of dilations of the graph of the function $f$. Which of the following could describe those dilations?

(A) A horizontal dilation by a factor of $\frac{1}{3}$ and a vertical dilation by a factor of $\frac{1}{2}$
(B) A horizontal dilation by a factor of $\frac{1}{2}$ and a vertical dilation by a factor of $\frac{1}{3}$
(C) A horizontal dilation by a factor of $\frac{1}{2}$ and a vertical dilation by a factor of $\frac{1}{2}$
(D) A horizontal dilation by a factor of 3 and a vertical dilation by a factor of 2

▶️ Answer/Explanation

Detailed solution

From the table, $ f $ has period 6: pattern $ 0, 1, 4, 9, 4, 1 $ repeats.
$ g $ has period 6: pattern $ 0, 2, 2, 0, 2, 2 $ repeats.
We check if $ g(x) = a f(bx) $ fits.
Try $ b = 2 $ (horizontal dilation factor $ 1/b = 1/2 $): Compare values:
For $ x = 0 $: $ g(0) = 0 $, $ f(2\cdot0) = f(0) = 0 $.
$ x = 1 $: $ g(1) = 2 $, $ f(2) = 4 $. If $ a = 1/2 $, $ \frac12 f(2) = 2 $ ✔.
$ x = 2 $: $ g(2) = 2 $, $ f(4) = 4 $, $ \frac12 f(4) = 2 $ ✔.
$ x = 3 $: $ g(3) = 0 $, $ f(6) = 0 $, $ \frac12 \cdot 0 = 0 $ ✔.
Continues to match.
Thus $ g(x) = \frac12 f(2x) $, meaning:
– Horizontal dilation by factor $ 1/2 $ (since $ 2x $ compresses horizontally).
– Vertical dilation by factor $ 1/2 $ (multiply by 1/2).
✅ Answer: (C)

Question

The function $ f $ is not explicitly given. In the $ xy $-plane, the graph of the function $ g $ is the result of a sequence of transformations to the graph of $ f $. The graph of $ g $ is the result of dilating the graph of $ f $ vertically by a factor of 2, then horizontally by a factor of 3, then translating the result up by 7 units, and then left by 11 units. Which of the following defines $ g $ in terms of $ f $?

(A) $ g(x) = \frac{1}{3} f(2(x – 7)) – 11 $
(B) $ g(x) = \frac{1}{2} f(3(x – 11)) – 7 $
(C) $ g(x) = 2f\left(\frac{x+11}{3}\right) + 7 $
(D) $ g(x) = 3f\left(\frac{x+7}{2}\right) + 11 $

▶️ Answer/Explanation

Detailed solution

Transformations in order:
1. Vertical dilation by 2 ⇒ multiply $ f $ by 2.
2. Horizontal dilation by 3 ⇒ multiply input by $ \frac{1}{3} $ (since dilation factor $ k $ means $ x $ replaced by $ x/k $).
3. Translate up 7 ⇒ add 7 outside.
4. Translate left 11 ⇒ replace $ x $ by $ x + 11 $.
Combine: Start with $ f(x) $.
After step 1: $ 2f(x) $.
After step 2: $ 2f\left(\frac{x}{3}\right) $ (horizontal dilation by factor 3 ⇒ input divided by 3).
After step 3: $ 2f\left(\frac{x}{3}\right) + 7 $.
After step 4: replace $ x $ with $ x+11 $ ⇒ $ 2f\left(\frac{x+11}{3}\right) + 7 $.
✅ Answer: (C)

Question

The function $ f $ is given by $ f(x) = x^4 – 3x^2 + 2 $. In the $ xy $-plane, the graph of the function $ g $ is a vertical translation of the graph of the function $ f $ downward by 3 units. Which of the following defines $ g $?

(A) $ g(x) = x^4 – 3x^2 – 1 $
(B) $ g(x) = x^4 – 3x^2 + 5 $
(C) $ g(x) = (x – 3)^4 – 3(x – 3)^2 + 2 $
(D) $ g(x) = (x + 3)^4 – 3(x + 3)^2 + 2 $

▶️ Answer/Explanation

Detailed solution

Vertical translation downward by 3 units ⇒ subtract 3 from $ f(x) $:
$ g(x) = f(x) – 3 = (x^4 – 3x^2 + 2) – 3 = x^4 – 3x^2 – 1 $.
✅ Answer: (A)

Question

The graph of $ y = f(x) $, consisting of four line segments and a semicircle, is shown for $-3 \leq x \leq 3$. Which of the following is the transformed graph for $ y = f(x+1) – 2 $?

▶️ Answer/Explanation

Detailed solution

The transformation $ y = f(x+1) – 2 $ consists of:
• $ f(x+1) $: horizontal shift left by 1 unit.
• $ -2 $ outside: vertical shift down by 2 units.
Thus, every point on the original graph moves 1 unit left and 2 units down.
From the original domain $[-3,3]$, the new domain becomes $[-4,2]$.
Among the choices, only one graph matches this left-and-down translation of the original shape.

✅ Answer: (A)

Question

▶️ Answer/Explanation

Detailed solution

A vertical dilation by a factor of 2 means multiplying the output ($ f(x) $) by 2.
So the transformed function is $ 2f(x) = 2(3x^2 + 2x + 1) = 6x^2 + 4x + 2 $.
This matches option (B) and the given reason: multiplying $ f(x) $ by 2.
✅ Answer: (B)

Question

▶️ Answer/Explanation

Detailed solution

Given $ g(x) = f(2(x – 4)) $.
This is in the form $ f(b(x – h)) $ with $ b = 2 $, $ h = 4 $.

Transformations (in order applied to $ x $):
1. Horizontal dilation by factor $ \frac{1}{b} = \frac{1}{2} $ (compression).
2. Horizontal translation right by $ h = 4 $ units.

Thus: horizontal dilation by factor $ \frac{1}{2} $, then horizontal translation right 4 units.
✅ Answer: (D)

Question

A function $g(x) = \frac{x-5}{x+2}$ has been transformed from the parent function $f(x) = \frac{1}{x}$. Which of the following correctly describes the transformations from the parent function $f(x)$ to $g(x)$?

a. Vertical Dilation of $7$, Up $1$, Right $2$, Flip over $y$-axis.
b. Vertical Dilation of $7$, Up $1$, Right $2$, Flip over $x$-axis.
c. Vertical Dilation of $7$, Up $1$, Left $2$, Flip over $y$-axis.
d. Vertical Dilation of $7$, Up $1$, Left $2$, Flip over $x$-axis.

▶️ Answer/Explanation

Detailed solution

First, use polynomial long division to rewrite $g(x) = \frac{x-5}{x+2}$ as $1 – \frac{7}{x+2}$.
This can be expressed in transformation form as $g(x) = -7 \left( \frac{1}{x+2} \right) + 1$.
The term $(x+2)$ in the denominator indicates a horizontal shift Left $2$ units.
The multiplier of $7$ indicates a Vertical Dilation (stretch) of $7$.
The negative sign in front of the $7$ represents a Flip (reflection) over the $x$-axis.
The $+1$ at the end indicates a vertical shift Up $1$ unit.
Comparing these steps to the given options, the correct choice is d.

AP Precalculus -1.12 Transformations of Functions- MCQ Exam Style Questions - Effective Fall 2023