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 the denominator $\cos(x)=0$. We know that $\cos(x) = 0$ when $x=\frac{\pi}{2}+n\pi$, where $n\in\mathbb{Z}$.

Step2: Determine the range of secant function

Since $y = \sec(x)=\frac{1}{\cos(x)}$, and the range of $\cos(x)$ is $[- 1,1]$. When $\cos(x)$ approaches $0$ from the positive side, $\sec(x)\to+\infty$, and when $\cos(x)$ approaches $0$ from the negative side, $\sec(x)\to-\infty$. Also, the maximum value of $\vert\cos(x)\vert$ is $1$, so the minimum value of $\vert\sec(x)\vert$ is $1$. The range of $y = \sec(x)$ is $y\leq - 1$ or $y\geq1$.

Answer:

The vertical asymptotes of the secant function are at $x=\frac{\pi}{2}+n\pi$ ($n\in\mathbb{Z}$). The range of the secant function is $y\leq - 1$ or $y\geq1$.