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: 68

Answer

The amplitude of the function is 3 and period is $\frac{\pi }{2}$

Work Step by Step

The given equation is in the form of $y=A\sin Bx$. Here, $A=3\text{ and }B=4$ So, the amplitude is: $\begin{align} & \text{Amplitude}=\left| A \right| \\ & =\left| 3 \right| \\ & =3 \end{align}$ The period is given below: $\begin{align} & \text{Period }=\frac{2\pi }{B} \\ & =\frac{2\pi }{4} \\ & =\frac{\pi }{2} \end{align}$ $\text{Phase shift=}\frac{C}{B}=0$ And the quarter period is as follows: $\begin{align} & \text{Quarter-period}=\frac{\left( \frac{\pi }{2} \right)}{4} \\ & =\frac{\pi }{8} \end{align}$ Now, add quarter periods starting from $x=0$ 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: $\begin{align} & x=0+\frac{\pi }{8} \\ & =\frac{\pi }{8} \end{align}$ And the x-value for the third key point is: $\begin{align} & x=\frac{\pi }{8}+\frac{\pi }{8} \\ & =\frac{\pi }{4} \end{align}$ And the x-value for the fourth key point is: $\begin{align} & x=\frac{\pi }{4}+\frac{\pi }{8} \\ & =\frac{3\pi }{8} \end{align}$ And the x-value for the fifth key point is: $\begin{align} & x=\frac{3\pi }{8}+\frac{\pi }{8} \\ & =\frac{\pi }{2} \end{align}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.