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Question 1

The polynomial function \(p\) is given by \(p(x) = ax^n + b\), where \(a\) and \(b\) are nonzero constants and \(n\) is a positive integer. As the input values of \(p\) increase without bound, the output values of \(p\) decrease without bound. Which of the following must be true?
A. \(a\) is negative because \(\lim_{x \to \infty} p(x) = -\infty\).
B. \(a\) is positive because \(\lim_{x \to \infty} p(x) = -\infty\).
C. \(a\) is negative because \(\lim_{x \to \infty} p(x) = \infty\).
D. \(a\) is positive because \(\lim_{x \to \infty} p(x) = \infty\).
▶️ Answer/Explanation
Detailed solution

The phrase “input values of \(p\) increase without bound” corresponds to the limit as \(x \to \infty\).
The phrase “output values of \(p\) decrease without bound” corresponds to the limit equaling \(-\infty\).
Therefore, the condition is defined as \(\lim_{x \to \infty} p(x) = -\infty\).
Considering the leading term, \(\lim_{x \to \infty} (ax^n + b) = \lim_{x \to \infty} ax^n\).
Since \(n\) is a positive integer, \(x^n \to \infty\) as \(x\) increases.
For the product \(ax^n\) to approach \(-\infty\), the coefficient \(a\) must be negative.
Thus, the correct statement is that \(a\) is negative because \(\lim_{x \to \infty} p(x) = -\infty\).

Question 2

The function \(f\) is given by \(f(x) = \frac{x^2 – 9}{(x+3)(x^2 + 2x – 15)}\). Which of the following is true about the graph of \(f\) in the \(xy\)-plane?
A. The graph of \(f\) has no vertical asymptotes, and there are exactly two holes in the graph of \(f\), one at \(x = -3\) and one at \(x = 3\).
B. The graph of \(f\) has a vertical asymptote at \(x = -5\) only, and there are exactly two holes in the graph of \(f\), one at \(x = -3\) and one at \(x = 3\).
C. The graph of \(f\) has vertical asymptotes at \(x = -3\) and \(x = 3\) only, and there is exactly one hole in the graph of \(f\) at \(x = -5\).
D. The graph of \(f\) has vertical asymptotes at \(x = -5\), \(x = -3\), and \(x = 3\), and there are no holes in the graph of \(f\).
▶️ Answer/Explanation
Detailed solution
First, factor the numerator completely: \(x^2 – 9 = (x-3)(x+3)\).
Next, factor the quadratic expression in the denominator: \((x^2 + 2x – 15) = (x+5)(x-3)\).
Substitute these back into the function: \(f(x) = \frac{(x-3)(x+3)}{(x+3)(x+5)(x-3)}\).
Identify holes where factors cancel out: The terms \((x+3)\) and \((x-3)\) appear in both numerator and denominator, causing holes at \(x = -3\) and \(x = 3\).
Identify vertical asymptotes where factors remain in the denominator: The term \((x+5)\) remains, causing a vertical asymptote at \(x = -5\).
Therefore, the graph has a vertical asymptote at \(x = -5\) only, and holes at \(x = -3\) and \(x = 3\).
Correct Option: B

Question 3

The figure below shows a portion of the graph of a function \(f\) that has the \(x\)-axis as a horizontal asymptote. The function \(f\) exhibits exponential decay. Which of the following describes \(f\)?
(A) \(\lim_{x \to \infty} f(x) = -\infty\)
(B) \(\lim_{x \to \infty} f(x) = \infty\)
(C) \(\lim_{x \to -\infty} f(x) = 0\)
(D) \(\lim_{x \to -\infty} f(x) = \infty\)
▶️ Answer/Explanation
Detailed solution

The function \(f\) is described as an exponential decay function with the \(x\)-axis as a horizontal asymptote.
Looking at the graph, as \(x\) moves to the right (\(x \to \infty\)), the curve approaches the x-axis, implying \(\lim_{x \to \infty} f(x) = 0\).
Looking at the left side of the graph, as \(x\) moves to the left (\(x \to -\infty\)), the curve goes up steeply.
This upward trend indicates that the function values increase without bound as \(x\) decreases.
Mathematically, this behavior is written as \(\lim_{x \to -\infty} f(x) = \infty\).
Comparing this result with the given choices, option (D) is the correct description.

Question 4

The figure shows a portion of the graph of a function \(f\) that has the \(x\)-axis as a horizontal asymptote. The points \((1, 3)\), \((2, 9)\), and \((3, 27)\) are on the graph of \(f\). Which of the following could be the inverse function of \(f\), for \(x > 0\)?
(A) \(f^{-1}(x) = \frac{1}{3^x}\)
(B) \(f^{-1}(x) = 3^x\)
(C) \(f^{-1}(x) = 3\log_{10} x\)
(D) \(f^{-1}(x) = \log_3 x\)
▶️ Answer/Explanation
Detailed solution

The correct option is (D).

Step 1: Analyze the given points on the graph to identify the function \(f(x)\). The points are \((1, 3)\), \((2, 9)\), and \((3, 27)\).

Step 2: Observe the exponential pattern: \(3^1 = 3\), \(3^2 = 9\), and \(3^3 = 27\). This indicates that the function is \(f(x) = 3^x\). This function has the \(x\)-axis as a horizontal asymptote.

Step 3: To find the inverse function, let \(y = f(x)\), so we have \(y = 3^x\).

Step 4: Interchange the variables \(x\) and \(y\) to solve for the inverse: \(x = 3^y\).

Step 5: Convert the exponential equation into logarithmic form. By definition, \(x = 3^y\) is equivalent to \(y = \log_3 x\).

Step 6: Therefore, the inverse function is \(f^{-1}(x) = \log_3 x\).

Question 5

The function \( f \) is given by \( f(x) = \log_2 x \), and the function \( g \) is given by \( g(x) = \log_2(8x) \). Consider the graphs of both functions in the same \( xy \)-plane. Which of the following statements with reasoning is true?
(A) The graph of \( g \) is a vertical dilation of the graph of \( f \) because \( \log_2(8x) = \log_2 8 + \log_2 x \).
(B) The graph of \( g \) is a vertical dilation of the graph of \( f \) because \( \log_2(8x) = \log_2 8 \cdot \log_2 x \).
(C) The graph of \( g \) is a vertical translation of the graph of \( f \) because \( \log_2(8x) = \log_2 8 + \log_2 x \).
(D) The graph of \( g \) is a vertical translation of the graph of \( f \) because \( \log_2(8x) = \log_2 8 \cdot \log_2 x \).
▶️ Answer/Explanation
Detailed solution

The correct option is (C).

First, apply the product property of logarithms: \( \log_b(mn) = \log_b m + \log_b n \).
Using this property on \( g(x) \), we get: \( g(x) = \log_2(8x) = \log_2 8 + \log_2 x \).
Next, evaluate the constant term: since \( 2^3 = 8 \), it follows that \( \log_2 8 = 3 \).
Substituting this value back into the equation gives: \( g(x) = 3 + \log_2 x \), or \( g(x) = 3 + f(x) \).
In function transformations, adding a constant \( k \) (where \( f(x) + k \)) results in a vertical translation.
Therefore, the graph of \( g \) is a vertical translation of the graph of \( f \) shifted up by 3 units.

Question 6

A sinusoidal function model \( y = f(x) \) is constructed from the data in the table and results in \( f(x) = a\sin(b(x + c)) + d \), where \( a \), \( b \), \( c \), and \( d \) are constants. Of the following, which is the best approximation for the value of \( b \)?
(A) 0.7
(B) 1
(C) 6
(D) 9
▶️ Answer/Explanation
Detailed solution

The constant \( b \) determines the frequency and is related to the period \( P \) by the formula \( b = \frac{2\pi}{P} \).
To find the period, we identify the distance between two consecutive peaks (maximum values) in the data.
The first peak occurs at \( x = 4.5 \) with \( y = 7.02 \), and the next peak is at \( x = 13.5 \) with \( y = 7.04 \).
Calculating the difference gives the period: \( P = 13.5 – 4.5 = 9 \).
Now, we substitute \( P = 9 \) into the formula for \( b \): \( b = \frac{2\pi}{9} \).
Approximating \( \pi \approx 3.14 \), we get \( b \approx \frac{6.28}{9} \approx 0.698 \).
Rounding to the nearest option, the value is approximately \( 0.7 \).

Question 7

A circular clock face on a wall has hours marked with integers from \(1\) to \(12\) and the hour hand that rotates as time passes. The hour hand points to the number \(12\) at midnight. The height above the floor, in inches, of the tip of the hour hand at \(t\) hours after midnight is modeled by the periodic function \(y\). Values of \(y(t)\) at selected values of \(t\) are given in the table. For \(t \geq 0\), which of the following could be an expression for \(y(t)\)?
\(t\)
hours		\(0\)\(3\)\(6\)\(9\)\(12\)
\(y(t)\)
inches		\(78\)\(70\)\(62\)\(70\)\(78\)
(A) \(-8 \sin\left(\frac{\pi}{12}(t-3)\right) + 70\)
(B) \(-8 \sin\left(\frac{\pi}{6}(t-3)\right) + 70\)
(C) \(8 \cos\left(\frac{\pi}{12}(t-3)\right) + 70\)
(D) \(8 \cos\left(\frac{\pi}{6}(t-3)\right) + 70\)
▶️ Answer/Explanation
Detailed solution

The maximum height is \(78\) and the minimum is \(62\), so the vertical shift (midline) is \(\frac{78 + 62}{2} = 70\) and amplitude is \(78 – 70 = 8\).
The function repeats its values from \(t=0\) to \(t=12\), so the period is \(12\).
The coefficient \(B\) is calculated as \(\frac{2\pi}{\text{Period}} = \frac{2\pi}{12} = \frac{\pi}{6}\). This eliminates options (A) and (C).
We test the point \(t=0\) (where \(y=78\)) on the remaining options to find the correct expression.
For option (D): \(8 \cos\left(\frac{\pi}{6}(0-3)\right) + 70 = 8 \cos\left(-\frac{\pi}{2}\right) + 70 = 0 + 70 = 70\) (Incorrect).
For option (B): \(-8 \sin\left(\frac{\pi}{6}(0-3)\right) + 70 = -8 \sin\left(-\frac{\pi}{2}\right) + 70 = -8(-1) + 70 = 78\) (Correct).
Therefore, the correct expression is (B).

Question 8

The function \( h \) is given by \( h(\theta) = \tan(3\theta) + 1 \). Which of the following statements about the graph of \( h \) in the \( xy \)-plane is true?
(A) The vertical asymptotes of the graph of \( h \) occur at input values \( \theta = \frac{\pi}{6} + \frac{\pi}{6}k \), where \( k \) is an integer
(B) The vertical asymptotes of the graph of \( h \) occur at input values \( \theta = \frac{\pi}{6} + \frac{\pi}{3}k \), where \( k \) is an integer
(C) The vertical asymptotes of the graph of \( h \) occur at input values \( \theta = \frac{\pi}{3} + \frac{\pi}{6}k \), where \( k \) is an integer
(D) The vertical asymptotes of the graph of \( h \) occur at input values \( \theta = \frac{\pi}{3} + \frac{\pi}{3}k \), where \( k \) is an integer
▶️ Answer/Explanation
Detailed solution

The standard tangent function, \( y = \tan(x) \), has vertical asymptotes where the function is undefined, which is at \( x = \frac{\pi}{2} + \pi k \) for any integer \( k \).
For the function \( h(\theta) = \tan(3\theta) + 1 \), the argument is \( 3\theta \).
We set the argument equal to the asymptotic condition: \( 3\theta = \frac{\pi}{2} + \pi k \).
Solving for \( \theta \), we divide the entire equation by 3: \( \theta = \frac{1}{3}(\frac{\pi}{2} + \pi k) \).
This simplifies to \( \theta = \frac{\pi}{6} + \frac{\pi}{3}k \).
Comparing this result with the given options, it corresponds to statement (B).

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