Answer
see below
Work Step by Step
$ y = \sec (\frac{1}{4}x)$
We begin with graph of $y = \cos (\frac{1}{4}x)$
Amplitude = 1,
For one cycle
$0 \leq \frac{1}{4}x \leq 2\pi$
$0 \leq x \leq 8\pi$
So period is $8\pi$
Now
To sketch the graph of the secant function, we note that the zeros of the cosine
graph correspond to the vertical asymptotes of the secant graph, and the peaks and
valleys of the cosine graph correspond to the valleys and peaks of the secant graph,
respectively.
the range of the function is $y \geq 1$ or $y \leq -1$
Using graph of $y = \cos (\frac{1}{4}x)$ as aid we can draw graph of $y = \sec (\frac{1}{4}x)$
