Precalculus (10th Edition)

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

Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Cumulative Review - Page 637: 12

Answer

Amplitude: $4$, period: $2$.

Work Step by Step

If a function has form $y=A\cos(\alpha x)$, then there $|A|$ stands for the amplitude and $T =\frac{2\pi}{\omega}$ stands for the period. Hence here $|A|=|-4|=4$ and $\omega=\pi$, hence $T=\frac{2\pi}{\omega}=\frac{2\pi}{\pi}=2.$
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