Answer
See below.
Work Step by Step
a) The y-intercepts are when $x=0$, thus $\frac{y^2}{9}=1\\y^2=9\\y=\pm3$
Thus the y-intercepts are $y=\pm3$
b) The x-intercepts are when $y=0$, thus $\frac{-x^2}{16}=1\\x^2=-16$
This has no real solutions because $x^2\geq0$, so there are no x-intercepts.
