## Trigonometry (11th Edition) Clone

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