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: 78

Answer

The amplitude of the function is $3$, period is $6$, and phase shift is $9$.

Work Step by Step

The given equation is in the form of $y=A\sin \left( Bx-C \right)$. Here, $A=-3,\text{ }B=\frac{\pi }{3}\text{ and }C=3\pi $ So, the amplitude is: $\begin{align} & \text{Amplitude=}\left| A \right| \\ & =\left| -3 \right| \\ & =3 \end{align}$ The period is given below: $\begin{align} & \text{Period = }\frac{2\pi }{B} \\ & =\frac{2\pi }{\left( \frac{\pi }{3} \right)} \\ & =6 \end{align}$ The phase shift is: $\begin{align} & \text{Phase shift = }\frac{C}{B} \\ & =\frac{3\pi }{\left( \frac{\pi }{3} \right)} \\ & =9 \end{align}$ And the quarter period is as follows: $\begin{align} & \text{Quarter-period}=\frac{6}{4} \\ & =\frac{3}{2} \end{align}$ Now, add quarter periods starting from $x=9$ to generate x-values for the key points. The x-value for the first key point is as follows: $x=9$ And the x-value for the second key point is: $\begin{align} & x=9+\frac{3}{2} \\ & =\frac{21}{2} \end{align}$ And the x-value for the third key point is: $\begin{align} & x=\frac{21}{2}+\frac{3}{2} \\ & =12 \end{align}$ And the x-value for the fourth key point is: $\begin{align} & x=12+\frac{3}{2} \\ & =\frac{27}{2} \end{align}$ And the x-value for the fifth key point is: $\begin{align} & x=\frac{27}{2}+\frac{3}{2} \\ & =15 \end{align}$
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