Answer
$y=\pm 3\sin(\pi x)$
Work Step by Step
Amplitude=$|A|=3$
$\implies A=\pm3$
Period $T=\frac{2\pi}{\omega}=2\implies\omega=\frac{2\pi}{2}=\pi$
Hence,
$y=A\sin(\omega x)=\pm 3\sin(\pi x)$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 5 - Trigonometric Functions - Section 5.4 Graphs of the Sine and Cosine Functions* - 5.4 Assess Your Understanding - Page 432: 61
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