## Precalculus (6th Edition) Blitzer

Step 1. Graph $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ (red) and $\frac{x\left|x\right|}{16}-\frac{y\left|y\right|}{9}=1$ (blue) as shown in the figure. Step 2. The two curves are not identical because when $x\lt0, x^2\ne x|x|$, and when $y\lt0, y^2\ne y|y|$ resulting in the two curves departing in some regions.