Graphing the secant function, $y=2\sec 2\theta$

Question

I am asked to graph the following equation $y = 2 \sec {2\theta}$. Since the equation just has a $ 2\theta$ after the secant, is it correct to say that there is no phase shift? If I would start graphing it from $(0,0)$, would the $y$-axis be my first vertical asymptote? Also then the vertical stretch would just be $2$?

Answer 1

You are right that there is no phase shift. The "first" vertical asymptote of the secant function is $x=\pi/2$ so it would be affected by the compression factor in $2\theta$

Answer 2

Assuming a Cartesian $(\theta,y)$ plane, this is the curve $y=\sec\theta$ (which has no vertical asymptote at $\theta=0$, but rather at $\pm\frac{\pi}{2}$) vertically stretched by a factor of $2$ away from the $\theta$-axis and horizontally compressed by a factor of $\frac{1}{2}$ toward the $y$-axis. There is no phase shift, or shift of any kind away from $y=\sec\theta$.