## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 4 - Graphs of the Circular Functions - Section 4.4 Graphs of the Secant and Cosecant Functions - 4.4 Exercises - Page 179: 10

#### Answer

$A$

#### Work Step by Step

The cosecant function is undefined when $x=..., -\pi, 0, \pi, 2\pi, ...$. Thus, its graph has the vertical asymptotes $x=..., -\pi, 0, \pi, 2\pi, ...$. RECALL: The graph of the function $y=\csc{(x-d)}$ involves a $|d|$phase (horizontal) shift of the parent function $y=\csc{x}$. The shift is to the right when $d\gt0$ and to the left when $d\lt0$. The given function has $d=-\frac{\pi}{2}$ so it involves a phase shift of $\frac{\pi}{2}$ to the left. Thus, its vertical asymptotes are $x=..., -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, ...$ The only possible graph among the choices are the ones in Options A and C. Note that when $x=0$ , the value of $y=\csc{(x+\frac{\pi}{2})}$ is $1$. Only the graph in Option $A$ satisfies this. Thus, the graph of $y=\csc{(x+\frac{\pi}{2})}$ must be the one in Option $A$.

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