Answer
Vertices: $(0,\pm 4)$
Co-vertices: $(\pm 3,0)$
Foci: $(0,\pm \sqrt{7}$)
Work Step by Step
Rewrite in the form of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$:
$16x^2+9y^2=144$ (Divide by 144)
$\frac{16x^2}{144}+\frac{9y^2}{144}=\frac{144}{144}$ (Simplify)
$\frac{x^2}{9}+\frac{y^2}{16}=1$
We obtain $a=3$ and $b=4$.
Since $b>a$, the ellipse has the vertical major axis.
Now, find $c$:
$c=\sqrt{b^2-a^2}=\sqrt{4^2-9^2}=\sqrt{16-9}=\sqrt{7}$
Thus, the vertices are $(0,\pm 4)$, the co-vertices are $(\pm 3,0)$, and the foci are $(0,\pm \sqrt{7}$).
