graph the trigonometric function. then give the xmin and xmax values for the viewing window xmin, xmax by…

Explanation:

Step1: Recall period formula for secant

The general form of a secant - type function is $y = A\sec(Bx - C)+D$. For the function $y=\sec(4x)$, $A = 1$, $B = 4$, $C = 0$, and $D = 0$. The period of the secant function $y=\sec(Bx)$ is given by $T=\frac{2\pi}{|B|}$. Here, $T=\frac{2\pi}{4}=\frac{\pi}{2}$.

Step2: Determine the viewing - window for one period

The secant function $y = \sec x$ has vertical asymptotes at $x=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$. For $y=\sec(4x)$, the vertical asymptotes are given by $4x=(2n + 1)\frac{\pi}{2}$, or $x=(2n + 1)\frac{\pi}{8},n\in\mathbb{Z}$. For one - period of the function $y=\sec(4x)$, we can consider the interval between two consecutive vertical asymptotes. For example, when $n = 0$, $x=\frac{\pi}{8}$ and when $n=- 1$, $x =-\frac{\pi}{8}$. A common viewing window for one period of $y=\sec(4x)$ is $\left[-\frac{\pi}{8},\frac{\pi}{8}\right]$.

Answer:

If we assume we want to show one period of the function $y = \sec(4x)$ in the viewing window $[X_{min},X_{max}]$ by $[-10,10]$, then $X_{min}=-\frac{\pi}{8}$ and $X_{max}=\frac{\pi}{8}$