## 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: 45

#### 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 $d\gt0$ Write the given function in the form $y=a \cdot \sin{[b(x-d)]}$ by factoring out $2$ inside the sine function to obtain: $y=-4\sin{[2(x-\frac{\pi}{2})]}$ The given function has: $a=-4$ $b=2$ $d=\frac{\pi}{2}$ Thus, the given function has: amplitude = $|-4|=4$ period = $\frac{2\pi}{2} = \pi$ phase shift = $|\frac{\pi}{2}|=\frac{\pi}{2}$ to the right Therefore, the graph of the given function has the following properties/characteristics: amplitude = $4$ so the y-values range from $-4$ to $4$ phase shift = $\frac{\pi}{2}$ units to the right one period interval = $[\frac{\pi}{2}, \frac{3\pi}{2}]$ Refer to the graph in the answer part above.

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