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 =-\frac{\pi}{2}, \frac{\pi}{2}) and looks like a secant - type function. The general form of a secant function is (y = a\sec(bx)+c).

Step2: Determine the value of (c)

The mid - line of the graph is (y = 0), so (c = 0).

Step3: Determine the period

The period (P) of the secant function (y=a\sec(bx)+c) is given by (P=\frac{2\pi}{b}). The distance between two consecutive vertical asymptotes is (\pi). For (y = a\sec(bx)+c), the distance between consecutive vertical asymptotes is (\frac{\pi}{b}). Since the distance between (x =-\frac{\pi}{2}) and (x=\frac{\pi}{2}) is (\pi), then (\frac{\pi}{b}=\pi), so (b = 1).

Step4: Determine the value of (a)

When (x = 0), (y=- 5). Substituting (x = 0), (y=-5), (b = 1) and (c = 0) into (y=a\sec(bx)+c), we get (y=a\sec(0)+0). Since (\sec(0)=1), then (-5=a\times1), so (a=-5).

Answer:

(y=-5\sec(x))