Question

the graph above is a graph of what function?
( y=sec (x) )
( y=\tan (x) )
( y=sin (x) )
( y=cot (x) )
( y=cos (x) )
( y=csc (x) )

Explanation:

• \(y = \sin(x)\) and \(y=\cos(x)\) are continuous functions (their graphs have no vertical asymptotes), so we can rule out these two options.
• \(y = \tan(x)\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\) (e.g., \(x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\cdots\)). The period of \(y = \tan(x)\) is \(\pi\). The graph of \(y=\tan(x)\) passes through the origin \((0,0)\) and has a shape that is different from the given graph (it is an increasing function between its vertical asymptotes).
• \(y=\sec(x)=\frac{1}{\cos(x)}\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\).
• \(y=\cot(x)=\frac{\cos(x)}{\sin(x)}\) has vertical asymptotes at \(x = n\pi,n\in\mathbb{Z}\) (e.g., \(x = 0,\pm\pi,\pm2\pi,\cdots\)). The period of \(y=\cot(x)\) is \(\pi\). The function \(y = \cot(x)\) is a decreasing function between its vertical asymptotes.
• \(y=\csc(x)=\frac{1}{\sin(x)}\) has vertical asymptotes at \(x = n\pi,n\in\mathbb{Z}\) (e.g., \(x=0,\pm\pi,\pm2\pi,\cdots\)).

Answer:

\(y = \csc(x)\)