Question
Describe how the graph of the function f(x) = \sin \left(x - (\pi)/(6)\right) differs from the graph of \sin x.
A. It shifts the graph of \sin x to the left by (\pi)/(6) units.
B. It shifts the graph of \sin x down by (\pi)/(6) units.
C. It shifts the graph of \sin x up by (\pi)/(6) units.
D. It shifts the graph of \sin x to the right by (\pi)/(6) units.
A. It shifts the graph of \sin x to the left by (\pi)/(6) units.
B. It shifts the graph of \sin x down by (\pi)/(6) units.
C. It shifts the graph of \sin x up by (\pi)/(6) units.
D. It shifts the graph of \sin x to the right by (\pi)/(6) units.
Answers
Answer
Are you sure
Answered by
GPT-4o mini
To analyze how the graph of the function \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) differs from the graph of \( \sin x \), we should look at the transformation applied to \( x \).
The function \( \sin(x - c) \) represents a horizontal shift of the sine function. Specifically:
- If \( c \) is positive, the graph shifts to the right by \( c \) units.
- If \( c \) is negative, the graph shifts to the left by \( |c| \) units.
In your case, you have \( c = \frac{\pi}{6} \) which is positive. Therefore, the graph of \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) is shifted to the right by \( \frac{\pi}{6} \) units.
Thus, the correct option is:
**D. It shifts the graph of \(\sin x\) to the right by \(\frac{\pi}{6}\) units.**
The function \( \sin(x - c) \) represents a horizontal shift of the sine function. Specifically:
- If \( c \) is positive, the graph shifts to the right by \( c \) units.
- If \( c \) is negative, the graph shifts to the left by \( |c| \) units.
In your case, you have \( c = \frac{\pi}{6} \) which is positive. Therefore, the graph of \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) is shifted to the right by \( \frac{\pi}{6} \) units.
Thus, the correct option is:
**D. It shifts the graph of \(\sin x\) to the right by \(\frac{\pi}{6}\) units.**
Answered by
GPT-4o mini
Yes, I am sure. Let's clarify again how the transformation works:
The function \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) can be analyzed by considering the argument of the sine function, which is \( x - \frac{\pi}{6} \).
In general, for a sine function of the form \( \sin(x - c) \):
- If \( c > 0 \), the graph shifts **to the right** by \( c \) units.
- If \( c < 0 \), the graph shifts **to the left** by \( -c \) units.
In your function, \( c = \frac{\pi}{6} \), which is positive, meaning the graph of \( \sin x \) will shift **to the right** by \( \frac{\pi}{6} \) units.
Therefore, the correct conclusion is that the graph of \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) is indeed shifted to the right by \( \frac{\pi}{6} \) units.
So, the answer is:
**D. It shifts the graph of \(\sin x\) to the right by \(\frac{\pi}{6}\) units.**
The function \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) can be analyzed by considering the argument of the sine function, which is \( x - \frac{\pi}{6} \).
In general, for a sine function of the form \( \sin(x - c) \):
- If \( c > 0 \), the graph shifts **to the right** by \( c \) units.
- If \( c < 0 \), the graph shifts **to the left** by \( -c \) units.
In your function, \( c = \frac{\pi}{6} \), which is positive, meaning the graph of \( \sin x \) will shift **to the right** by \( \frac{\pi}{6} \) units.
Therefore, the correct conclusion is that the graph of \( f(x) = \sin \left(x - \frac{\pi}{6}\right) \) is indeed shifted to the right by \( \frac{\pi}{6} \) units.
So, the answer is:
**D. It shifts the graph of \(\sin x\) to the right by \(\frac{\pi}{6}\) units.**
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