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 4.4 Graphs of the Secant and Cosecant Functions - 4.4 Exercises - Page 180: 30


$y=1 - \csc{(\frac{1}{2}x)}$

Work Step by Step

The graph looks like a reflection about the x-axis of the graph of the basic cosecant function so its tentative equation is $y=-\csc{bx}+d$ RECALL: The period of $y=-\csc{(bx)}+d$ is $\frac{2\pi}{b}$. The period of the given graph is $4\pi$. Thus, $4\pi=\frac{2\pi}{b} \\4\pi(b) = 2\pi \\\frac{4\pi(b)}{2\pi}= \frac{2\pi}{4\pi} \\b=\frac{1}{2}$ This means that the tentative equation of the given graph is $y=-\csc{(\frac{x}{2})}+d$ Notice, that instead of having vertices whose y-coordinates are either $-1$ or $1$, the vertices have the y-coordinates $0$ and $2$. This means that the given graph involves a 1-unit upward shift of the parent function $y=\csc{x}$. Therefore, the equation of the function whose graph is given is $y=-\csc{(\frac{1}{2}x)} +1$ or $y=1 - \csc{(\frac{1}{2}x)}$.
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