the function graphed to the right is of the form y = a sec bx + c or y = a csc bx + c for some a ≠ 0, b > 0…

Explanation:

Step1: Identify the type of function

The graph has vertical asymptotes at (x = 0,\pm\pi,\pm2\pi,\cdots), so the function is of the form (y = a\sec bx + c). The standard - secant function (y=\sec x) has vertical asymptotes at (x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}), and the given function has vertical asymptotes at (x = n\pi,n\in\mathbb{Z}). For (y = a\sec bx + c), the period (T=\frac{2\pi}{b}). Since the period of the given function is (\pi), we have (\frac{2\pi}{b}=\pi), so (b = 2).

Step2: Find the value of (a) and (c)

The mid - line of the secant function (y=a\sec bx + c) is (y = c). The mid - line of the given graph is (y = 1), so (c = 1). The amplitude of the secant function (distance from the mid - line to the minimum or maximum) is (|a|). The minimum value of the function is (y=-1) and the mid - line is (y = 1), so (|a|=2). Since the secant function opens upwards (the minimum values are below the mid - line), (a = 2).

Answer:

(y = 2\sec(2x)+1)