Answer
asymptotes: $y=\pm\frac{2}{3}(x+1)-1$
foci: $(-1\pm\frac{\sqrt {13}}{6},-1)$
Work Step by Step
Step 1. Rewriting the equation as
$4(x^2+2x+1)-9(y^2+2y+1)=6+4-9=1$
or
$\frac{(x+1)^2}{1/4}-\frac{(y+1)^2}{1/9}=1$
we have $a=\frac{1}{2},b=\frac{1}{3}, c=\sqrt {a^2+b^2}=\frac{\sqrt {13}}{6}$ centered at $(-1,-1)$ with a horizontal transverse axis.
Step 2. We can find the vertices as $(-\frac{3}{2},-1),(-\frac{1}{2},-1)$ and asymptotes as $y+1=\pm\frac{b}{a}(x+1)$ or $y=\pm\frac{2}{3}(x+1)-1$
Step 3. We can graph the equation as shown in the figure with foci at $(-1\pm\frac{\sqrt {13}}{6},-1)$
