## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Review Exercises - Page 645: 75

#### Answer

The amplitude of the function is $3$, period is $2\pi$ and phase shift is $-\pi$

#### Work Step by Step

The given equation is in the form of $y=A\cos \left( Bx-C \right)$. Here, $A=-3,\text{ }B=1\text{ and }C=-\pi$ So, the amplitude is: \begin{align} & \text{Amplitude}=\left| A \right| \\ & =\left| -3 \right| \\ & =3 \end{align} The period is given below: \begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{1} \\ & =2\pi \end{align} The phase shift is: \begin{align} & \text{Phase shift = }\frac{C}{B} \\ & =\frac{-\pi }{1} \\ & =-\pi \end{align} And the quarter period is as follows: \begin{align} & \text{Quarter-period}=\frac{2\pi }{4} \\ & =\frac{\pi }{2} \end{align} Now, add quarter periods starting from $x=-\pi$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=-\pi$ And the x-value for the second key point is: \begin{align} & x=-\pi +\frac{\pi }{2} \\ & =-\frac{\pi }{2} \end{align} And the x-value for the third key point is: \begin{align} & x=-\frac{\pi }{2}+\frac{\pi }{2} \\ & =0 \end{align} And the x-value for the fourth key point is: \begin{align} & x=0+\frac{\pi }{2} \\ & =\frac{\pi }{2} \end{align} And the x-value for the fifth key point is: \begin{align} & x=\frac{\pi }{2}+\frac{\pi }{2} \\ & =\pi \end{align} After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.