## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Review Exercises - Page 645: 70

#### Answer

The amplitude of the function is 2 and period is $4\pi$

#### Work Step by Step

The given equation is in the form of $y=A\cos Bx$. Here, $A=-2\text{ and }B=\frac{1}{2}$ So, the amplitude is: \begin{align} & \text{Amplitude=}\left| A \right| \\ & =\left| 2 \right| \\ & =2 \end{align} The period is: \begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{\left( \frac{1}{2} \right)} \\ & =4\pi \end{align} $\text{Phase shift=}\frac{C}{B}=0$ The quarter period is as follows: \begin{align} & \text{Quarter-period}=\frac{4\pi }{4} \\ & =\pi \end{align} Now, add quarter periods starting from $x=\text{0}$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=\text{0}$ And the x-value for the second key point is: \begin{align} & x=0+\pi \\ & =\pi \end{align} And the x-value for the third key point is: \begin{align} & x=\pi +\pi \\ & =2\pi \end{align} And the x-value for the fourth key point is: \begin{align} & x=2\pi +\pi \\ & =3\pi \end{align} And the x-value for the fifth key point is: \begin{align} & x=3\pi +\pi \\ & =4\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.