Question

question 12 (1 point) ✓ saved
determine an equation for the function shown.

a) ( y = \tan x )
b) ( y = sec x - 1 )
c) ( y = sec x )
d) ( y = csc x - 1 )

Explanation:

Step1: Recall the properties of trigonometric functions

• The function \(y = \tan x\) has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) and passes through the origin \((0,0)\) with a range of \((-\infty,\infty)\). Its graph is a set of S - shaped curves. So, option a) is incorrect.
• The function \(y=\sec x=\frac{1}{\cos x}\) has vertical asymptotes at \(x = \frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) and its range is \(y\leq - 1\) or \(y\geq1\). The standard \(y = \sec x\) has a value of \(y = 1\) when \(x = 2k\pi,k\in\mathbb{Z}\).
• The function \(y=\csc x=\frac{1}{\sin x}\) has vertical asymptotes at \(x=k\pi,k\in\mathbb{Z}\).
• For the function \(y = \sec x-1\), we know that the range of \(y=\sec x\) is \(y\leq - 1\) or \(y\geq1\). Then the range of \(y=\sec x - 1\) is \(y\leq-2\) or \(y\geq0\). When \(x = 0\), \(y=\sec(0)-1=1 - 1=0\).

Step2: Analyze the graph

• The given graph has vertical asymptotes at \(x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\) (same as \(y = \sec x\) and \(y=\tan x\) but different from \(y=\csc x\)).
• When \(x = 0\), the function value is \(y = 0\). For \(y=\sec x\), when \(x = 0\), \(y=\sec(0)=1\). For \(y=\sec x-1\), when \(x = 0\), \(y=\sec(0)-1=1 - 1=0\)

Answer:

B. \(y = \sec x-1\)