Answer
See the explanation below.
Work Step by Step
a) To find the $x$ interpret , we put $y=0$
Thus, $\dfrac{x^2}{16}-\dfrac{0^2}{9}=1$
or, $x^2 =16$
or, $x = \pm 4$
Therefore, $x$ intercepts are: {$-4,4$}
b)To find the $y$ interpret , we put $x=0$
Thus, $\dfrac{0^2}{16}-\dfrac{y^2}{9}=1$
or, $y^2 = -9$
or, $y = \pm \sqrt{-9}$
Therefore, there are no $y$ intercepts because of of no real numbers or a non-real solution.