Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.4 Graphs of the Sine and Cosine Functions - 6.4 Assess Your Understanding - Page 408: 21

Answer

amplitude: $\frac{5}{3}$ period: $3$

Work Step by Step

If a function has form $y=A\sin(\omega x)$, then there $|A|$ stands for the amplitude. First we identify $\omega$ and $A$ from the form of $y=A\sin{(\omega x)}$. Hence, $A=\left|\frac{5}{3}\right|=\frac{5}{3}$, $\omega=\frac{-2\pi}{3}.$ To get the period $T$ we can use the formula $T=\left|\frac{2\pi}{\omega}\right|$ to obtain $T=\left|\dfrac{2\pi}{\frac{-2\pi}{3}}\right|=3$
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