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.
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
\( f \) 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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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)
▶️ Answer/Explanation
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.
▶️ Answer/Explanation
The correct option is (A).
The transformation \(y = f(x + 1) – 2\) involves two shifts applied to the parent function \(f(x)\):
- Horizontal Shift: \(f(x + 1)\) shifts the graph 1 unit left.
- Vertical Shift: \(- 2\) shifts the graph 2 units down.
By applying the rule \((x, y) \to (x – 1, y – 2)\) to key points:
| Original Point | Transformed Point |
|---|---|
| \((-3, 2)\) (Start) | \((-4, 0)\) |
| \((-1, 0)\) (V-minimum) | \((-2, -2)\) |
| \((2, -2)\) (Semicircle bottom) | \((1, -4)\) |
Graph (A) is the only option that matches these coordinates and the overall shape.
▶️ Answer/Explanation
The correct answer is (A).
Step 1 (Statement I): The definition of an odd function is \( k(-x) = -k(x) \). For \( x = 2 \), we have \( k(-2) = -k(2) \). Since it is given that \( k(2) = 0 \), it follows that \( k(-2) = -0 = 0 \). Thus, Statement I is true.
Step 2 (Statement II): Using the odd function property for \( x = 3 \), we have \( k(-3) = -k(3) \). Since \( k(3) = -7 \), substitution gives \( k(-3) = -(-7) = 7 \). The statement claims \( k(-3) = -7 \), which is incorrect. Thus, Statement II is false.
Step 3 (Statement III): We are given \( k(x) \leq 0 \) for \( x \geq 2 \). Due to symmetry, for \( x \leq -2 \), \( k(x) \) must be \( \geq 0 \). However, a function being non-negative does not imply it approaches infinity; it could approach a constant limit. Thus, Statement III is not necessarily true.
▶️ Answer/Explanation
The function transformation $h(x) = f(x – 1) + 3$ indicates a horizontal shift and a vertical shift.
The term $(x – 1)$ shifts the graph 1 unit to the right, so each $x$-coordinate increases by $1$.
The term $+ 3$ shifts the graph 3 units up, so each $f(x)$ value increases by $3$.
Taking the first point $(-1, 0)$, the new $x$ is $-1 + 1 = 0$ and the new $y$ is $0 + 3 = 3$.
Taking the second point $(0, 5)$, the new $x$ is $0 + 1 = 1$ and the new $y$ is $5 + 3 = 8$.
Taking the third point $(1, 4)$, the new $x$ is $1 + 1 = 2$ and the new $y$ is $4 + 3 = 7$.
Taking the fourth point $(2, 9)$, the new $x$ is $2 + 1 = 3$ and the new $y$ is $9 + 3 = 12$.
Comparing these calculated points $(0, 3), (1, 8), (2, 7), (3, 12)$ to the options, Table (C) is the correct choice.
▶️ Answer/Explanation
The point $(3, -2)$ on $y = h(x)$ implies that $h(3) = -2$.
The transformation $y = 2h(x) – 1$ affects the $y$-coordinates of the function.
To find the new point, substitute $x = 3$ into the transformed equation.
$y = 2h(3) – 1$
Substitute $-2$ for $h(3)$: $y = 2(-2) – 1$.
Simplify the expression: $y = -4 – 1 = -5$.
Therefore, the point $(3, -5)$ must be on the graph.
The correct option is (B).
▶️ Answer/Explanation
The function $g(x) = \log_{10}(x^3)$ can be rewritten using the logarithm power rule.
The power rule states that $\log_b(a^n) = n \cdot \log_b a$.
Applying this to $g(x)$, we get $g(x) = 3 \cdot \log_{10} x$.
Since $f(x) = \log_{10} x$, we can express $g$ as $g(x) = 3 \cdot f(x)$.
A transformation of the form $y = a \cdot f(x)$ represents a vertical dilation.
Because $a = 3$, the graph of $f$ is stretched vertically by a factor of $3$.
Therefore, the correct choice is (A).
▶️ Answer/Explanation
An odd function satisfies the property $f(-x) = -f(x)$.
To find $a$, we use the table entry $x = -2$ where $f(-2) = 10$.
Since $f(2) = -f(-2)$, it follows that $a = -10$.
To find $b$, we look for $f(b) = -7$ and notice $f(-8) = -7$.
Using the odd property $f(-x) = -f(x)$, we see $f(8) = -f(-8) = -(-7) = 7$.
From the table, $f(4) = -f(-4) = -7$, so $b$ must be $4$.
Therefore, $a + b = -10 + 4 = -6$.
The correct option is (A).
▶️ Answer/Explanation
The correct answer is (C).
A horizontal dilation by a factor of $3$ replaces $x$ with $\frac{x}{3}$, giving $f\left(\frac{x}{3}\right)$.
A vertical dilation by a factor of $5$ multiplies the entire function by $5$, giving $5f\left(\frac{x}{3}\right)$.
A vertical translation by $-7$ units subtracts $7$ from the function, giving $5f\left(\frac{x}{3}\right) – 7$.
Combining these steps in the specified order results in the equation $g(x) = 5f\left(\frac{x}{3}\right) – 7$.
▶️ Answer/Explanation
Compare the values at opposite inputs: \( g(4) = -7 \) and \( g(-4) = 7 \).
Similarly, check \( g(2) = -3 \) and \( g(-2) = 3 \).
In both cases, \( g(-x) = -g(x) \), which is the definition of an odd function.
Since \( g \) is an odd function, the relationship \( g(-1) = -g(1) \) must hold true.
From the table, we identify that \( g(1) = -6 \).
Substituting this value, we find \( g(-1) = -(-6) = 6 \).
Therefore, \( g \) is an odd function and \( g(-1) = 6 \), which matches option (B).
▶️ Answer/Explanation
The original function is $f(x) = x^2 + 2$.
The transformed function is $g(x) = \frac{x^2}{4} + 2$, which can be rewritten as $g(x) = (\frac{x}{2})^2 + 2$.
Comparing the two, we see that $g(x) = f(\frac{x}{2})$.
A transformation of the form $f(\frac{x}{k})$ represents a horizontal dilation by a factor of $k$.
In this case, $k = 2$, which means the graph is stretched horizontally.
Therefore, the transformation is a horizontal dilation by a factor of $2$.
The correct option is (C).
▶️ Answer/Explanation
The domain of $g$ is found by setting the argument $x + 3$ within the domain of $f$: $-2 \le x + 3 \le 2$.
Subtracting $3$ from all sides gives the new domain: $-5 \le x \le -1$, or $[-5, -1]$.
For the range, the original outputs $f(x)$ are in the interval $[1, 5]$.
Multiplying the range by $-2$ scales and reflects it: $[-10, -2]$.
Adding $4$ to this interval shifts it to $[-10 + 4, -2 + 4] = [-6, 2]$.
Therefore, the domain is $[-5, -1]$ and the range is $[-6, 2]$.
The correct option is (A).
▶️ Answer/Explanation
Observe $f(x)$ peaks at $f(3) = 9$ within the first period $x \in [0, 6]$.
Observe $g(x)$ peaks at $g(1) = 2$ and $g(2) = 2$ within its first period.
The maximum value of $f$ is $9$ and the maximum value of $g$ is $2$, implying a vertical dilation by $\frac{2}{9}$.
However, looking at the pattern, $f$ returns to $0$ at $x=6$, while $g$ returns to $0$ at $x=3$.
This indicates the period of $g$ is half the period of $f$, meaning a horizontal dilation by a factor of $\frac{1}{2}$.
Comparing $f(1)=1$ to $g(x)$ values, a vertical dilation by factor $\frac{1}{3}$ relates $f(3)=9$ to $g$ peaks if looking at specific points.
By analyzing the transformation $g(x) = \frac{1}{3} f(2x)$, we see $g(1) = \frac{1}{3}f(2) = \frac{4}{3}$ (approx 2 in discrete table) and $g(1.5) = \frac{9}{3}=3$.
Based on the provided options, (B) is the closest fit for the observed period change and amplitude reduction.
▶️ Answer/Explanation
To transform the graph, apply a horizontal translation left by \(\frac{1}{2}\) unit followed by a horizontal compression by a factor of \(\frac{2}{3}\).
Reasoning: Shifting left by \(\frac{1}{2}\) changes \(x \to x + \frac{1}{2}\), giving \(f(2(x+0.5)+1) = f(2x+2)\). Compressing by \(\frac{2}{3}\) changes \(x \to \frac{3}{2}x\), resulting in \(f(2(\frac{3}{2}x)+2) = f(3x+2)\).
First, apply a horizontal translation left by 1 unit to get \(y = f(x+1)\). Then, apply a horizontal compression by a factor of \(\frac{1}{2}\). This replaces \(x\) with \(2x\), resulting in \(y = f(2x+1)\).
First, apply a horizontal compression by a factor of \(\frac{1}{2}\) to get \(y = f(2x)\). Then, apply a horizontal translation left by \(\frac{1}{2}\) unit. This replaces \(x\) with \(x+\frac{1}{2}\), resulting in \(y = f(2(x+0.5)) = f(2x+1)\).