Answer
Please see the graph.
Work Step by Step
$\frac{y^{2}}{16}-\frac{x^{2}}{9}>1$
This graph is for a hyperbola, and we have three regions to test (for different values of y): $(−∞,−4]$, $[−4,4]$, and $[4,∞)$. We can use the same $x$ value for the three chosen values of $y$ to test.
Points to use to test: $(0,-5)$, $(0,0)$, $(0,5)$
$(0,-5)$
$\frac{y^{2}}{16}-\frac{x^{2}}{9}>1$
$\frac{-5^{2}}{16}-\frac{0^{2}}{9}>1$
$\frac{25}{16}-\frac{0}{9}>1$
$25/16 -0 > 1$
$25/16 >1$ (true, so we shade this region)
$(0,0)$
$\frac{y^{2}}{16}-\frac{x^{2}}{9}>1$
$\frac{0^{2}}{16}-\frac{0^{2}}{9}>1$
$\frac{0}{16}-\frac{0}{9}>1$
$0 - 0 > 1$
$0 > 1$ (false)
$(0,5)$
$\frac{y^{2}}{16}-\frac{x^{2}}{9}>1$
$\frac{5^{2}}{16}-\frac{0^{2}}{9}>1$
$\frac{25}{16}-\frac{0}{9}>1$
$25/16 -0 > 1$
$25/16 >1$ (true, so we shade this region)
