## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 4 - Quiz (Sections 4.1-4.2) - Page 164: 1

#### Answer

The amplitude is $4$, the period is $\pi$, thevertical translation is $3$ units up since $c$ is more than zero and the phase shift is $\frac{\pi}{4}$ units to the left since $d$ is less than zero.

#### Work Step by Step

We first write the equation in the form $y=c+a \sin [b(x-d)]$. Therefore, $y=3-4\sin(2x+\frac{\pi}{2})$ becomes $y=3-4\sin [2(x+\frac{\pi}{4})]$. Comparing the two equations, $a=-4,b=2,c=3$ and $d=-\frac{\pi}{4}$. The amplitude is $|a|=|-4|=4.$ The period is $\frac{2\pi}{b}=\frac{2\pi}{2}=\pi$. The vertical translation is $c=3$. The phase shift is $|d|=|-\frac{\pi}{4}|=\frac{\pi}{4}$ Therefore, the amplitude is $4$, the period is $\pi$, thevertical translation is $3$ units up since $c$ is more than zero and the phase shift is $\frac{\pi}{4}$ units to the left since $d$ is less than zero.

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