Answer
Center: $(-4,-2)$
Foci: $(-4-\sqrt 5,-2),(-4+\sqrt 5,-2)$
Vertices: $(-4,0),(-4,4)$
See graph
Work Step by Step
We are given the ellipse:
$\dfrac{(x+4)^2}{9}+\dfrac{(y+2)^2}{4}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Identify $h,k,a,b$:
$h=-4$
$k=-2$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
Determine $c$:
$a^2=b^2+c^2$
$c^2=a^2-b^2$
$c^2=9-4$
$c^2=5$
$c=\sqrt 5$
Determine the center:
$(h,k)=(-4,-2)$
Determine the foci:
$(h-c,k)=(-4-\sqrt 5,-2)$
$(h+c,k)=(-4+\sqrt 5,-2)$
Determine the vertices:
$(h-a,k)=(-4-3,-2)=(-7,-2)$
$(h+a,k)=(-4+3,-2)=(-1,-2)$
Determine the $y$-intercepts:
$(h,k-b)=(-4,-2-2)=(-4,-4)$
$(h,k+b)=(-4,-2+2)=(-4,0)$
Graph the ellipse:
