Answer
See graph and explanations.
Work Step by Step
Step 1. Rewrite the given equation as $\frac{(x+3)^2}{16}+\frac{(y-2)^2}{4}=1$; we have $a^2=16, b^2=4$ and $c=\sqrt {a^2-b^2}=2\sqrt 3$. The ellipse is centered at $(-3,2)$ with a horizontal major axis.
Step 2. We can graph the equation as shown in the figure with foci at $(-3\pm2\sqrt 3,2)$
