Answer
See below
Work Step by Step
Given: $\frac{(x+3)^2}{9}-\frac{(y-4)^2}{16}=1$
Compare the given equation to the standard form of an equation of a hyperbola. You can see that the graph is a hyperbola with its center at $(h,k)=(-3,4)$ and $a=3,b=4$
Hence, the center of the hyperbola is at $(-3,4)$
Find asymptotes: $y-k=\pm \frac{b}{a}(x-h)\\y-4=\pm\frac{4}{3}(x+3)$
Thus, $y_1=\frac{4}{3}x+8\\y_2=-\frac{4}{3}x$
Draw a hyperbola.
