## Precalculus (6th Edition) Blitzer

Step 1. Given $f(x)=2sec(x)$ and $h(x)=2x-\frac{\pi}{2}$, we have $y=(f\circ h)(x)=2sec(2x-\frac{\pi}{2})$ Step 2. Starting from $y=cos(2x)$ (purple), we can obtain $y=sec(2x)$ (green) using reciprocal properties. Step 3. Start from $y=sec(2x)$, shift the curve horizontally $\frac{\pi}{4}$ to the right, and then stretch vertically by a factor of 2 to get the red curve for $y=(f\circ h)(x)=2sec(2x-\frac{\pi}{2})$ See graph and explanations.