Answer
(a) domain $(-\infty,\infty)$ and range $[-\frac{3}{2},\infty)$.
(b) $f(x)=\frac{1}{2}x^2-\frac{3}{2}$ and $f(-2)=\frac{1}{2}$
Work Step by Step
(a) Given $x^2-2y=3$, we have $y=\frac{1}{2}x^2-\frac{3}{2}$. We can find the domain as $(-\infty,\infty)$ and range as $[-\frac{3}{2},\infty)$.
(b) We have $f(x)=\frac{1}{2}x^2-\frac{3}{2}$ and $f(-2)=\frac{1}{2}(-2)^2-\frac{3}{2}=\frac{1}{2}$
