Answer
See below.
Work Step by Step
a) The x-intercepts are when $y=0$, thus $\frac{x^2}{16}=1\\x^2=16\\x=\pm4$
b) The y-intercepts are when $x=0$, thus $\frac{-y^2}{9}=1\\y^2=-9$
This has no real solutions because $y^2\geq0$, so there are no y-intercepts.
