Answer
$\frac{(x+1)^2}{9}-\frac{(y-2)^2}{7}=1$
See graph.
Work Step by Step
1. Based on the given conditions, the hyperbola has a horizontal transverse axis with center $(-1,2)$, thus we have a general form $\frac{(x+1)^2}{a^2}-\frac{(y-2)^2}{b^2}=1$
2. With $c=4, a=3$, we have $b^2=c^2-a^2=7$, thus the equation is $\frac{(x+1)^2}{9}-\frac{(y-2)^2}{7}=1$ with asymptotes $y-2=\pm\frac{b}{a}(x+1)=\pm\frac{\sqrt 7}{3}(x+1)$
3. See graph.
