Question

1. which function below has this graph? a. $y = \sec x$ b. $y = \csc x$ c. $y = \cot x$ d. none of the above

Explanation:

1. Recall the graphs of the trigonometric functions:

• The function \( y = \sec x=\frac{1}{\cos x} \) has vertical asymptotes at \( x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z} \) (where \( \cos x = 0 \)) and its graph consists of U - shaped curves (called secant curves) above and below the x - axis, with the minimum value of \( |y| = 1 \) (since \( |\sec x|=\frac{1}{|\cos x|}\geq1 \)).
• The function \( y=\csc x=\frac{1}{\sin x} \) has vertical asymptotes at \( x = n\pi,n\in\mathbb{Z} \) (where \( \sin x=0 \)) and its graph also has U - shaped curves but the asymptotes are at \( x = n\pi \).
• The function \( y = \cot x=\frac{\cos x}{\sin x} \) has vertical asymptotes at \( x=n\pi,n\in\mathbb{Z} \) and its graph is a series of curves that are decreasing (or increasing) between the asymptotes, similar to the graph of \( y=\tan x \) but with a different orientation.

1. Analyze the given graph:

• The vertical asymptotes of the given graph seem to be at \( x=(2n + 1)\frac{\pi}{2} \) (for example, around \( x=\frac{\pi}{2},x = \frac{3\pi}{2} \) etc.). The shape of the graph (U - shaped curves with \( |y|\geq1 \)) matches the graph of \( y = \sec x \). The graph of \( y=\csc x \) would have asymptotes at \( x = n\pi \) (like \( x = 0,x=\pi,x = 2\pi \) etc.), which does not match the asymptotes of the given graph. The graph of \( y=\cot x \) has a different shape (it is not U - shaped with \( |y|\geq1 \) in the same way as the secant graph) and different asymptotes.

Answer:

A. \( y=\sec x \)