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question 11 the graph above is a graph of what function? $y = \\cot(x)$…

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 trigonometric functions

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

Step2: Match the graph with the function

The given graph has vertical asymptotes at \(x=(n+\frac{1}{2})\pi\) (\(n\in\mathbb{Z}\)) (since the graph has breaks at these \(x\) - values) and the shape of the graph (the U - shaped and inverted U - shaped curves) matches the general form of \(y = \sec(x)\).

Answer:

\(y=\sec(x)\)