Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 645: 76

Answer

The amplitude of the function is $\frac{3}{2}$, period is $\pi $, and phase shift is $-\frac{\pi }{8}$

Work Step by Step

Rewrite the given equation as $y=\frac{3}{2}\cos \left( 2x-\left( -\frac{\pi }{4} \right) \right)$. Now, the given equation is in the form of $y=A\cos \left( Bx-C \right)$. Here, $A=\frac{3}{2},\text{ }B=2\text{ and }C=-\frac{\pi }{4}$ So, the amplitude is: $\begin{align} & \text{Amplitude=}\left| A \right| \\ & =\left| \frac{3}{2} \right| \\ & =\frac{3}{2} \end{align}$ The period is given below: $\begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{2} \\ & =\pi \end{align}$ And the phase shift is: $\begin{align} & \text{Phase shift = }\frac{C}{B} \\ & =\frac{\left( \frac{-\pi }{4} \right)}{2} \\ & =-\frac{\pi }{8} \end{align}$ And the quarter period is as follows: $\text{Quarter-period}=\frac{\pi }{4}$ Now, add quarter periods starting from $x=-\frac{\pi }{8}$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=-\frac{\pi }{8}$ And the x-value for the second key point is: $\begin{align} & x=-\frac{\pi }{8}+\frac{\pi }{4} \\ & =\frac{\pi }{8} \end{align}$ And the x-value for the third key point is: $\begin{align} & x=\frac{\pi }{8}+\frac{\pi }{4} \\ & =\frac{3\pi }{8} \end{align}$ And the x-value for the fourth key point is: $\begin{align} & x=\frac{3\pi }{8}+\frac{\pi }{4} \\ & =\frac{5\pi }{8} \end{align}$ And the x-value for the fifth key point is: $\begin{align} & x=\frac{5\pi }{8}+\frac{\pi }{4} \\ & =\frac{7\pi }{8} \end{align}$
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