Question

the function graphed to the right is of the form $y = a\sec bx + c$ or $y = a\csc bx + c$ for some $a\
eq0$, $b > 0$. determine the equation of the function. an equation of the function shown is $y=$.

Explanation:

Step1: Determine the type of function

The general form of \(y = a\sec(bx)+c\) or \(y=a\csc(bx)+c\). The vertical asymptotes of \(y = \sec(x)\) are \(x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}\), and for \(y=\csc(x)\) are \(x = n\pi,n\in\mathbb{Z}\). The vertical asymptotes of the given graph are \(x=\pm\frac{\pi}{2}\), which is in the form of the vertical asymptotes of \(y = \sec(x)\) (when \(n = 0,\pm1\)). So the function is of the form \(y=a\sec(bx)+c\).

Step2: Find the value of \(c\)

The mid - line of the function \(y = a\sec(bx)+c\) is \(y = c\). The mid - line of the given graph is \(y = 6\) (average of the minimum and the horizontal asymptote behavior). So \(c = 6\).

Step3: Find the value of \(a\)

The distance from the mid - line \(y = c\) to the minimum value of \(y=a\sec(bx)+c\) is \(|a|\). The minimum value of the function is \(y = 4\). Since \(c - |a|=4\) and \(c = 6\), then \(|a|=2\). The graph of \(y=\sec(x)\) has a "U - shape" opening upwards when \(a>0\). So \(a = 2\).

Step4: Find the value of \(b\)

The period of \(y=\sec(bx)\) is \(T=\frac{2\pi}{b}\). The distance between two consecutive vertical asymptotes of \(y = \sec(bx)\) is \(\frac{\pi}{b}\). The distance between \(x =-\frac{\pi}{2}\) and \(x=\frac{\pi}{2}\) is \(\pi\). So \(\frac{\pi}{b}=\pi\), which gives \(b = 1\).

Answer:

\(y = 2\sec(x)+6\)