## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 4 - Graphs of the Circular Functions - Section 4.2 Translations of the Graphs of the Sine and Cosine Functions - 4.2 Exercises - Page 162: 44

#### Answer

Refer to the graph below. #### Work Step by Step

RECALL: The graph of $y=a \cdot \sin{[b(x-d)]}$ has: amplitude = $|a|$ period = $\frac{2\pi}{b}$ phase shift = $|d|$, to the left when $d\lt0$, to the right when $d\gt0$ The given function has: $a=-\frac{1}{2}$ $b=4$ $d=-\frac{\pi}{2}$ Thus, the given function has: amplitude = $|-\frac{1}{2}|=\frac{1}{2}$ period = $\frac{2\pi}{4} = \frac{\pi}{2}$ phase shift = $|-\frac{\pi}{2}|=\frac{\pi}{2}$ to the left Therefore, the graph of the given function has the following properties/characteristics: amplitude = $\frac{1}{2}$ so the y-values range from $-\frac{1}{2}$ to $\frac{1}{2}$ phase shift = $\frac{\pi}{2}$ units to the left one period interval = $[-\frac{\pi}{2}, 0]$ Refer to the graph in the answer part above.

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