AP Precalculus -3.6 Sinusoidal Transformations- MCQ Exam Style Questions - Effective Fall 2023
AP Precalculus -3.6 Sinusoidal Transformations- MCQ Exam Style Questions – Effective Fall 2023
AP Precalculus -3.6 Sinusoidal Transformations- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
From the graphs (described in the original question):
– \( g(x) \) appears horizontally wider than \( f(x) = \tan^{-1}x \), indicating a horizontal stretch (dilation) by a factor of 2, which corresponds to replacing \( x \) with \( \frac{x}{2} \).
– \( g(x) \) is also shifted upward by 1 unit compared to the stretched curve.
Thus, \( g(x) = \tan^{-1}\left(\frac{x}{2}\right) + 1 \).
✅ Answer: (C)
▶️ Answer/Explanation
\( g(\theta) = \cos(2\theta) \) has period \( \pi \).
Minima of \( g \) occur when \( \cos(2\theta) = -1 \), i.e., \( 2\theta = \pi + 2\pi k \) → \( \theta = \frac{\pi}{2} + \pi k \).
Two consecutive minima: at \( \theta = -\frac{\pi}{2} \) and \( \theta = \frac{\pi}{2} \).
A phase shift of \( -\frac{\pi}{3} \) means \( h(\theta) = g\left(\theta + \frac{\pi}{3}\right) \) (shift left by \( \frac{\pi}{3} \)).
Thus minima of \( h \) occur at \( \theta + \frac{\pi}{3} = -\frac{\pi}{2} \) → \( \theta = -\frac{5\pi}{6} \), and \( \theta + \frac{\pi}{3} = \frac{\pi}{2} \) → \( \theta = \frac{\pi}{6} \).
✅ Answer: (A)
▶️ Answer/Explanation
From the graph:
– Max = 2, Min = -4 → Amplitude = \( \frac{2 – (-4)}{2} = 3 \)
– Midline: \( y = \frac{2 + (-4)}{2} = -1 \)
– Period: distance between consecutive peaks (or troughs) = \( \pi \) → \( b = \frac{2\pi}{\pi} = 2 \)
– Shape matches a sine curve shifted right by \( \frac{\pi}{4} \) (since sine normally starts at midline going up at 0, here that occurs at \( x = \frac{\pi}{4} \))
Thus: \( f(x) = 3\sin(2(x – \frac{\pi}{4})) – 1 \).
✅ Answer: (C)
▶️ Answer/Explanation
\( g(x) = \frac{1}{f(x)} = \frac{1}{3\sec x} = \frac{1}{3} \cos x \).
Thus \( g \) is a cosine function with amplitude \( \frac13 \) and period \( 2\pi \).
However, \( f(x) = 3\sec x \) is undefined where \( \cos x = 0 \) (\( x = \frac{\pi}{2} + k\pi \)), so \( g \) will have holes at those \( x \)-values even though \( \cos x = 0 \) there, because \( \frac{1}{3\sec x} \) would involve division by infinity.
Graph (A) matches \( y = \frac13\cos x \) with holes at the vertical asymptotes of \( \sec x \).
✅ Answer: (A)
▶️ Answer/Explanation
Horizontal translation of \( f(x) = \sin x \) by \( h \) units to the left is given by \( f(x + h) \).
Here \( h = \frac{\pi}{6} \), so \( g(x) = \sin\left(x + \frac{\pi}{6}\right) \).
✅ Answer: (B)
▶️ Answer/Explanation
Amplitude \( A = \frac{7 – (-1)}{2} = 4 \).
Midline \( d = \frac{7 + (-1)}{2} = 3 \).
Since the graph has maxima at \( x = 0 \) and \( x = 2\pi \) and no horizontal shift, a cosine function (which has a maximum at \( x = 0 \)) fits best.
Thus \( f(x) = 4\cos x + 3 \) or equivalently \( 3 + 4\cos x \).
✅ Answer: (A)
▶️ Answer/Explanation
1. Apply Horizontal Translation Rule:
To translate a function \(f(x)\) horizontally by \(h\) units to the right, we replace every \(x\) in the function definition with \((x – h)\).
2. Substitute the Value:
Here, the shift is \(h = \frac{\pi}{2}\) units right.
Therefore, \(g(x) = f(x – \frac{\pi}{2})\).
3. Write the Expression:
\(g(x) = 2\cos(\pi(x – \frac{\pi}{2})) + 3\).
✅ Answer: (B)
▶️ Answer/Explanation
From the graph:
• Midline: \( y = -1 \) → vertical shift \( d = -1 \).
• Amplitude: max \(=2\), min \(=-4\), amplitude \(= \frac{2 – (-4)}{2} = 3\) → \( a = 3 \).
• Period: distance between consecutive maxima \( \frac{3\pi}{2} – \frac{\pi}{2} = \pi \) → \( \frac{2\pi}{b} = \pi \) ⇒ \( b = 2 \).
• Shape: starts increasing through midline at \( x = \frac{\pi}{4} \) (like sine shifted right \( \frac{\pi}{4} \)).
A sine function with phase shift \( \frac{\pi}{4} \) to the right:
\[ f(x) = 3\sin\left(2\left(x – \frac{\pi}{4}\right)\right) – 1 \]
This matches option (C).
✅ Answer: (C)