Trigonometry (10th Edition)

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

Chapter 4 - Review Exercises - Page 183: 12

Answer

The amplitude is 1 The period is $2\pi$ The vertical translation is 0 The phase shift is $\frac{3\pi}{4}$

Work Step by Step

$y = A~sin(Bx+C)+D$ $y = -sin(x-\frac{3\pi}{4})$ $A = -1$ $B = 1$ $C = -\frac{3\pi}{4}$ $D = 0$ $\vert A \vert$ is the amplitude. The amplitude is 1 $\frac{2\pi}{B}$ is the period. The period is $2\pi$ $D$ is the vertical translation. The vertical translation is 0 $\frac{-C}{B}$ is the phase shift. The phase shift is $\frac{3\pi}{4}$
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