Answer
see graph; domain $(-\infty,\infty)$, range $(-\infty,-4]\cup[4,\infty)$
Work Step by Step
Step 1. From the given equation $\frac{(y)^2}{16}-\frac{(x)^2}{9}=1$, we have $a=4, b=3, c=\sqrt {a^2+b^2}=5$ centered at $(0,0)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(0,\pm4)$, foci as $(0,\pm5)$, and asymptotes as $y=\frac{a}{b}(x)$ or $y=\pm\frac{4}{3}(x)$
Step 3. We can graph the equation as shown in the figure with domain $(-\infty,\infty)$ and range $(-\infty,-4]\cup[4,\infty)$
