Question

question 10 the graph above is a graph of what function? y = csc(x) y = cos(x) y = cot(x) y = sec(x) y = tan(x) y = sin(x)

Explanation:

Step1: Recall the properties of trigonometric functions

• \(y = \sin(x)\) and \(y=\cos(x)\) have range \([- 1,1]\). The given graph has values less than \(-1\) and greater than \(1\), so \(y = \sin(x)\) and \(y=\cos(x)\) are not correct.
• \(y=\cot(x)=\frac{\cos(x)}{\sin(x)}\), its zero - points occur when \(\cos(x) = 0\) (i.e., \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\)), and it has vertical asymptotes when \(\sin(x)=0\) (i.e., \(x = n\pi,n\in\mathbb{Z}\)).
• \(y=\tan(x)=\frac{\sin(x)}{\cos(x)}\), its zero - points occur when \(\sin(x) = 0\) (i.e., \(x=n\pi,n\in\mathbb{Z}\)) and has vertical asymptotes when \(\cos(x)=0\) (i.e., \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\)). The graph of \(y = \tan(x)\) passes through the origin \((0,0)\) when \(n = 0\) for the zero - point.
• \(y=\csc(x)=\frac{1}{\sin(x)}\), has vertical asymptotes when \(\sin(x)=0\) (i.e., \(x = n\pi,n\in\mathbb{Z}\))
• \(y=\sec(x)=\frac{1}{\cos(x)}\), has vertical asymptotes when \(\cos(x)=0\) (i.e., \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\))

Step2: Analyze the zero - points and vertical asymptotes of the given graph

The graph has zero - points at \(x=n\pi,n\in\mathbb{Z}\) (for example, \(x = 0,\pi,2\pi\) in the given interval) and vertical asymptotes at \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\) (for example, \(x=-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\) in the given interval). This matches the behavior of \(y = \tan(x)\) since \(\tan(x)=\frac{\sin(x)}{\cos(x)}\), \(\sin(x) = 0\) gives \(y = 0\) and \(\cos(x)=0\) gives vertical asymptotes.

Answer:

\(y=\tan(x)\)