where are the vertical asymptotes on the graph of the secant function? x = π/2 + nπ x = nπ x = 2nπ complete…

Explanation:

Step1: Recall secant - cosine relationship

The secant function is defined as (y = \sec(x)=\frac{1}{\cos(x)}). Vertical asymptotes occur where (\cos(x) = 0). We know that (\cos(x)=0) when (x=\frac{\pi}{2}+n\pi), where (n\in\mathbb{Z}) (set of integers).

Step2: Recall period of secant function

The period of the cosine function (y = \cos(x)) is (2\pi). Since (\sec(x)=\frac{1}{\cos(x)}), the period of the secant function is also (2\pi).

Step3: Recall range of secant function

Since (- 1\leqslant\cos(x)\leqslant1), then (\sec(x)=\frac{1}{\cos(x)}). When (-1\leqslant\cos(x)\lt0), (\sec(x)\leqslant - 1); when (0\lt\cos(x)\leqslant1), (\sec(x)\geqslant1). So the range of (y = \sec(x)) is (y\leqslant - 1) or (y\geqslant1).

Answer:

1. Vertical asymptotes: (x=\frac{\pi}{2}+n\pi)
2. Period: (2\pi)
3. Range: (y\leqslant - 1) or (y\geqslant1)