Answer
(a) domain $(-\infty,\infty)$, range $[-\frac{3}{2},\infty)$.
(b) $f(x)=\frac{1}{2}x^2-\frac{3}{2}$, $f(-2)= \frac{1}{2}$
Work Step by Step
(a) Given $x^2-2y=3$, we have $y=\frac{1}{2}x^2-\frac{3}{2}$, as $y^2\geq0$, we can find its domain as $(-\infty,\infty)$, range as $[-\frac{3}{2},\infty)$.
(b) We can rewrite the equation as $y=\frac{1}{2}x^2-\frac{3}{2}$ which represents a function $f(x)=\frac{1}{2}x^2-\frac{3}{2}$, and we can find $f(-2)=\frac{1}{2}(-2)^2-\frac{3}{2}=\frac{1}{2}$
