Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 5

Answer

See below:
1571562678

Work Step by Step

Equation (1) is of the form $ y=A\sin \left( Bx-C \right)$. We can undertake the following steps to determine the graph of the equation: Step 1. Here, $ A=3,B=2$, and $ C=\pi $, such that Amplitude $=3$ $\begin{align} & \text{Period}=\frac{2\pi }{B} \\ & =\frac{2\pi }{2} \\ & =\pi \end{align}$ $\begin{align} & \text{Phase shift}=\frac{C}{B} \\ & =\frac{\pi }{2} \end{align}$ $\text{Quarter period}=\frac{\pi }{4}$ Step 2. Now we will find the $ x $ -values by adding the quarter periods. $\begin{align} & x=\frac{\pi }{2} \\ & x=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4} \\ & x=\frac{3\pi }{4}+\frac{\pi }{4}=\pi \\ & x=\pi +\frac{\pi }{4}=\frac{5\pi }{4} \end{align}$ Step 3. We will evaluate the function at each value of $ x $. At $ x=\frac{\pi }{2}$, $\begin{align} & y=3\sin \left( 2\cdot \frac{\pi }{2}-\pi \right) \\ & =3\sin \left( 0{}^\circ \right) \\ & =0 \end{align}$ Therefore, the coordinates are $\left( \frac{\pi }{2},0 \right)$. At $ x=\frac{3\pi }{4}$, $\begin{align} & y=3\sin \left( 2\cdot \frac{3\pi }{4}-\pi \right) \\ & =3\sin \left( \frac{\pi }{2} \right) \\ & =3 \end{align}$ Therefore, the coordinates are $\left( \frac{3\pi }{4},3 \right)$. At $ x=\pi $, $\begin{align} & y=3\sin \left( 2\cdot \pi -\pi \right) \\ & =3\sin \left( \pi \right) \\ & =0 \end{align}$ Therefore, the coordinates are $\left( \pi,0 \right)$. At $ x=\frac{5\pi }{4}$, $\begin{align} & y=3\sin \left( 2\cdot \frac{3\pi }{2}-\pi \right) \\ & =3\sin \left( 2\pi \right) \\ & =0 \end{align}$ Therefore, the coordinates are $\left( \frac{3\pi }{2},0 \right)$. Step 4. On connecting these points, the graph can be obtained. Step 5. Using all the steps above, we produce the final graph.
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