Question

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

Explanation:

• Recall the properties of trigonometric functions:
• The function \(y = \sin(x)\) and \(y=\cos(x)\) are continuous functions with range \([- 1,1]\). The given graph is not continuous (has vertical asymptotes), so \(y = \sin(x)\) and \(y=\cos(x)\) can be eliminated.
• The function \(y=\tan(x)=\frac{\sin(x)}{\cos(x)}\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\). The function \(y = \cot(x)=\frac{\cos(x)}{\sin(x)}\) has vertical asymptotes at \(x=n\pi,n\in\mathbb{Z}\). The function \(y=\csc(x)=\frac{1}{\sin(x)}\) has vertical asymptotes at \(x = n\pi,n\in\mathbb{Z}\) and the function \(y=\sec(x)=\frac{1}{\cos(x)}\) has vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\).
• The \(x -\)intercepts of \(y = \cot(x)=\frac{\cos(x)}{\sin(x)}\) occur when \(\cos(x)=0\) (i.e., \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\)). The \(x -\)intercepts of \(y=\csc(x)\) do not exist since \(\csc(x)=\frac{1}{\sin(x)}\) and \(\sin(x) = 0\) makes \(\csc(x)\) undefined.
• Looking at the graph, the vertical asymptotes are at \(x=-4,-1,2,5\) (which can be thought of in terms of \(x=n\pi\) if we assume a non - standard scale, and the general form of vertical asymptotes of \(y = \cot(x)\) is \(x = n\pi\) (when considering the basic form \(y=\cot(x)\) with period \(\pi\)).

Answer:

\(y = \cot(x)\)