Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Graphs of the Circular Functions - Section 4.2 Translations of the Graphs of the Sine and Cosine Functions - 4.2 Exercises - Page 162: 44

Answer

Refer to the graph below.
1523854328

Work Step by Step

RECALL: The graph of $y=a \cdot \sin{[b(x-d)]}$ has: amplitude = $|a|$ period = $\frac{2\pi}{b}$ phase shift = $|d|$, to the left when $d\lt0$, to the right when $d\gt0$ The given function has: $a=-\frac{1}{2}$ $b=4$ $d=-\frac{\pi}{2}$ Thus, the given function has: amplitude = $|-\frac{1}{2}|=\frac{1}{2}$ period = $\frac{2\pi}{4} = \frac{\pi}{2}$ phase shift = $|-\frac{\pi}{2}|=\frac{\pi}{2}$ to the left Therefore, the graph of the given function has the following properties/characteristics: amplitude = $\frac{1}{2}$ so the y-values range from $-\frac{1}{2}$ to $\frac{1}{2}$ phase shift = $\frac{\pi}{2}$ units to the left one period interval = $[-\frac{\pi}{2}, 0]$ Refer to the graph in the answer part above.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.