Answer
asymptotes: $y=\pm\frac{4}{5}(x+3)$
foci: $(-3\pm\sqrt {41},0)$
Work Step by Step
Step 1. From the given equation $\frac{(x+3)^2}{25}-\frac{(y)^2}{16}=1$, we have $a=5, b=4, c=\sqrt {a^2+b^2}=\sqrt {41}$ centered at $(-3,0)$ with a horizontal transverse axis.
Step 2. We can find the vertices as $(-8,0),(2,0),(-3,-4),(-3,4)$ and asymptotes as $y=\frac{b}{a}(x+3)$ or $y=\pm\frac{4}{5}(x+3)$
Step 3. We can graph the equation as shown in the figure with foci at $(-3\pm\sqrt {41},0)$
