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 = k\pi), (k\in\mathbb{Z}), which is characteristic of (y = a\csc(bx)+c). For (y = a\sec(bx)+c), vertical - asymptotes are at (x=(2k + 1)\frac{\pi}{2},k\in\mathbb{Z}). So the function is of the form (y=a\csc(bx)+c).

Step2: Find the period

The period of (y = a\csc(bx)+c) is (T=\frac{2\pi}{b}). From the graph, the period (T = 2\pi). Since (T=\frac{2\pi}{b}=2\pi), then (b = 1).

Step3: Find the vertical shift (c)

The mid - line of the graph is (y = 1), so (c = 1).

Step4: Find the amplitude (a)

The distance from the mid - line (y = 1) to the maximum or minimum value of the function is (a). The maximum value is (y = 3) and the mid - line is (y = 1), so (a=2).

Answer:

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