Answer
(a) center $(-1,-1)$, vertices $(2,-1), (-4,-1)$, foci $(-1\pm\sqrt {13},-1)$.
asymptotes: $y=\frac{2}{3}x-\frac{1}{3}$ and $y=-\frac{2}{3}x-\frac{5}{3}$.
(b) See graph.
Work Step by Step
(a) Step 1. From the un-shifted equation $\frac{x^2}{9}-\frac{y^2}{4}=1$, we have $a=3, b=2, c=\sqrt {3^2+2^2}=\sqrt {13}$. We can identify the old center as $(0,0)$, vertices as $(\pm3, 0)$, foci as $(\pm\sqrt {13}, 0)$, and asymptotes as $y=\pm\frac{2}{3}x$
Step 2. With the new equation $\frac{(x+1)^2}{9}-\frac{(y+1)^2}{4}=1$, we can identify the shift as 1 unit to the left and 1 unit down, the new center as $(-1,-1)$, new vertices as $(2,-1), (-4,-1)$, new foci as $(-1\pm\sqrt {13},-1)$.
Step 3. To find the new asymptotes, keep the old slopes and use a new point $(-1,-1)$ to determine the y-intercepts. For the asymptote with positive slope, $y=\frac{2}{3}x+b$, plug-in $(-1,-1)$ to get $\frac{2}{3}(-1)+b=-1$ and $b=-\frac{1}{3}$ which gives the equation $y=\frac{2}{3}x-\frac{1}{3}$. For the other asymptote,
$y=-\frac{2}{3}x+c$, plug-in $(-1,-1)$ to get $-\frac{2}{3}(-1)+c=-1$ and $c=-\frac{5}{3}$ which gives the equation $y=-\frac{2}{3}x-\frac{5}{3}$.
(b) See graph.
