Answer
hyperbola, center $(-1,0)$,
vertices $(-3,0),(1,0)$, foci $(-1-\sqrt {13},0),(-1+\sqrt {13},0)$, and asymptotes $y=\pm\frac{3}{2}(x+1)$
Work Step by Step
1. Given $\frac{(x+1)^2}{4}-\frac{y^2}{9}=1$, we can identify the equation as a hyperbola with a horizontal transverse axis, center $(-1,0)$,
2. We have $a=2, b=3$ and $c=\sqrt {a^2+b^2}=\sqrt {13}$, thus vertices $(-3,0),(1,0)$, foci $(-1-\sqrt {13},0),(-1+\sqrt {13},0)$, and asymptotes $y=\pm\frac{3}{2}(x+1)$
