Find the phase shift of y = -2 sin (4x -) 2, 2n units to the left B) units to the left C) - units to the right D) 4m units dow. TE 8.
Find the phase shift of y = -2 sin (4x -) 2, 2n units to the left B) units to the left C) - units to the right D) 4m units dow. TE 8.
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
Related questions
Question
![### Finding the Phase Shift of a Trigonometric Function
**Problem Statement:**
6. Find the phase shift of \( y = -2 \sin \left( 4x - \frac{\pi}{2} \right) \)
**Options:**
- A) \( 2\pi \) units to the left
- B) \( \frac{\pi}{2} \) units to the left
- C) \( \frac{\pi}{8} \) units to the right
- D) \( 4\pi \) units down
**Solution:**
The phase shift for a function of the form \( y = a \sin(bx - c) \) can be determined using the formula:
\[ \text{Phase Shift} = \frac{c}{b} \]
Given the function:
\[ y = -2 \sin \left( 4x - \frac{\pi}{2} \right) \]
In this case:
- \( c = \frac{\pi}{2} \)
- \( b = 4 \)
So, the phase shift is:
\[ \frac{c}{b} = \frac{\frac{\pi}{2}}{4} = \frac{\pi}{8} \]
Since the phase shift is positive, it is to the right.
**Correct Answer:**
- C) \( \frac{\pi}{8} \) units to the right](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F357de21d-4dd5-43e5-ad45-c8ac77f9a858%2F9515fa5e-67d1-415b-ad0b-6c461edd0a3f%2Fydb3dqw.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Phase Shift of a Trigonometric Function
**Problem Statement:**
6. Find the phase shift of \( y = -2 \sin \left( 4x - \frac{\pi}{2} \right) \)
**Options:**
- A) \( 2\pi \) units to the left
- B) \( \frac{\pi}{2} \) units to the left
- C) \( \frac{\pi}{8} \) units to the right
- D) \( 4\pi \) units down
**Solution:**
The phase shift for a function of the form \( y = a \sin(bx - c) \) can be determined using the formula:
\[ \text{Phase Shift} = \frac{c}{b} \]
Given the function:
\[ y = -2 \sin \left( 4x - \frac{\pi}{2} \right) \]
In this case:
- \( c = \frac{\pi}{2} \)
- \( b = 4 \)
So, the phase shift is:
\[ \frac{c}{b} = \frac{\frac{\pi}{2}}{4} = \frac{\pi}{8} \]
Since the phase shift is positive, it is to the right.
**Correct Answer:**
- C) \( \frac{\pi}{8} \) units to the right
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