Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 4 - Review Exercises - Page 183: 13

Answer

The amplitude is $\frac{1}{2}$ The period is $\pi$ The vertical translation is 0 The phase shift is $\frac{\pi}{8}$

Work Step by Step

$y = A~csc(Bx+C)+D$ $y = \frac{1}{2}csc(2x-\frac{\pi}{4})$ $A = \frac{1}{2}$ $B = 2$ $C = -\frac{\pi}{4}$ $D = 0$ $\vert A \vert$ is the amplitude. The amplitude is $\frac{1}{2}$ $\frac{2\pi}{B}$ is the period. The period is $\pi$ $D$ is the vertical translation. The vertical translation is 0 $\frac{-C}{B}$ is the phase shift. The phase shift is $\frac{\pi}{8}$
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