Question

question 11
the graph above is a graph of what function?
$y = \cot(x)$
$y = \sin(x)$
$y = \tan(x)$
$y = \cos(x)$
$y = \csc(x)$
$y = \sec(x)$

Explanation:

Step1: Analyze the properties of each trigonometric function

• \(y = \cot(x)=\frac{\cos(x)}{\sin(x)}\), its graph has vertical asymptotes at \(x = n\pi\), \(n\in\mathbb{Z}\) and is a decreasing function in each of its intervals \((n\pi,(n + 1)\pi)\).
• \(y=\sin(x)\) has a range of \([- 1,1]\) and is a periodic function that oscillates between \(-1\) and \(1\).
• \(y = \tan(x)=\frac{\sin(x)}{\cos(x)}\), its graph has vertical asymptotes at \(x=\frac{\pi}{2}+n\pi\), \(n\in\mathbb{Z}\) and is an increasing function in each of its intervals \((-\frac{\pi}{2}+n\pi,\frac{\pi}{2}+n\pi)\).
• \(y=\cos(x)\) has a range of \([-1,1]\) and is a periodic function that oscillates between \(-1\) and \(1\).
• \(y=\csc(x)=\frac{1}{\sin(x)}\), its graph has vertical asymptotes at \(x = n\pi\), \(n\in\mathbb{Z}\).
• \(y=\sec(x)=\frac{1}{\cos(x)}\), its graph has vertical asymptotes at \(x=\frac{\pi}{2}+n\pi\), \(n\in\mathbb{Z}\).

Step2: Check the range and asymptotes of the given graph

The given graph has vertical asymptotes (where the function is undefined) and the range of the function is \(y\leq - 1\) or \(y\geq1\).
We know that \(y = \sec(x)=\frac{1}{\cos(x)}\), when \(\cos(x)=0\), \(x=\frac{\pi}{2}+n\pi\) (\(n\in\mathbb{Z}\)) (vertical asymptotes). And since \(-1\leq\cos(x)\leq1\), then \(y=\sec(x)\) has \(y\leq - 1\) or \(y\geq1\)

Answer:

\(y = \sec(x)\)