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: 6

Answer

See below:
1571562710

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=-4$ and $ B=\pi $ such that $\begin{align} & \text{Amplitude}=\left| -4 \right| \\ & =4 \end{align}$ $\begin{align} & \text{Period}=\frac{2\pi }{B} \\ & =\frac{2\pi }{\pi } \\ & =2 \end{align}$ $\begin{align} & \text{Quarter period}=\frac{2}{4} \\ & =\frac{1}{2} \end{align}$ Step 2. Now we will find the $ x $ -values by adding the quarter periods. The cycle starts at $ x=0$, such that $\begin{align} & x=0 \\ & x=0+\frac{1}{2}=\frac{1}{2} \\ & x=\frac{1}{2}+\frac{1}{2}=1 \\ & x=1+\frac{1}{4}=\frac{3}{2} \end{align}$ Step 3. Now, we will evaluate the function at each value of $ x $. At $ x=0$, $\begin{align} & y=-4\cos \left( \pi x \right) \\ & =-4\cos \left( {{0}^{\circ }} \right) \\ & =-4 \end{align}$ Therefore, the coordinates are $\left( 0,-4 \right)$. At $ x=\frac{1}{2}$, $\begin{align} & y=-4\cos \left( \pi \cdot \frac{1}{2} \right) \\ & =-4\cos \left( \frac{\pi }{2} \right) \\ & =0 \end{align}$ Therefore, the coordinates are $\left( \frac{1}{2},0 \right)$. At $ x=1$, $\begin{align} & y=-4\cos \left( \pi \cdot 1 \right) \\ & =-4\cos \left( \pi \right) \\ & =4 \end{align}$ Therefore, the coordinates are $\left( 1,4 \right)$. At $ x=\frac{3}{2}$, $\begin{align} & y=-4\cos \left( \pi \cdot \frac{3}{2} \right) \\ & =-4\cos \left( \frac{3\pi }{2} \right) \\ & =0 \end{align}$ Therefore, the coordinates are $\left( \frac{3}{2},0 \right)$. Step 4. On connecting these points, the graph can be obtained.
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