Answer
Vertices: $(-3,0),(3,0)$
Foci: $(-\sqrt{5},0),(\sqrt{5},0)$
See graph
Work Step by Step
We are given the ellipse:
$\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$
The equation is in the form:
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
Determine $h,k,a,b,c$:
$h=0$
$k=0$
$a^2=9\Rightarrow a=\sqrt{9}=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2-b^2=9-4=5$
$c=\sqrt{5}$
The center is:
$(h,k)=(0,0)$
Find the vertices:
$(h-a,k)=(0-3,0)=(-3,0)$
$(h+a,k)=(0+3,0)=(3,0)$
Find the foci:
$(h-c,k)=(0-\sqrt{5},0)=(-\sqrt{5},0)$
$(h+c,k)=(0+\sqrt{5},0)=(\sqrt{5},0)$
Use the value of $b=2$ to find the two points above and below the center:
$(h,k-b)=(0,0-2)=(0,-2)$
$(h,k+b)=(0,0+2)=(0,2)$
Plot the center, the vertices, the two points above and below the center, and the foci. Then, graph the ellipse:
