Answer
a) $\frac{7}{4}$; b) $\frac{3}{4}$; c) undefined; d) $0$;
Work Step by Step
We are given the rational function:
$$\begin{align}f(x)=\dfrac{x^2+2x-3}{x^2-4}=\dfrac{(x-1)(x+3)}{(x-2)(x+2)}.\end{align}\tag1$$
The domain of the function is the set of all real numbers from which we exclude the zeros of the denominator: $-2$ and $2$:
$$\text{Domain}=(-\infty,-2)\cup(-2,2)\cup(2,\infty).$$
$\textbf{a)}$ We calculate $f(4)$ by substituting $x=4$ in Eq. $(1)$:
$$\begin{align*}
f(4)=\dfrac{4^2+2(4)-3}{4^2-4}=\dfrac{21}{12}=\dfrac{7}{4}.
\end{align*}$$
$\textbf{b)}$ We calculate $f(0)$ by substituting $x=0$ in Eq. $(1)$:
$$\begin{align*}
f(0)=\dfrac{0^2+2(0)-3}{0^2-4}=\dfrac{-3}{-4}=\dfrac{3}{4}.
\end{align*}$$
$\textbf{c)}$ We cannot calculate $f(-2)$ because $-2$ does not belong to the function's domain.
$\textbf{d)}$ We calculate $f(-3)$ by substituting $x=-3$ in Eq. $(1)$:
$$\begin{align*}
f(-3)=\dfrac{(-3)^2+2(-3)-3}{(-3)^2-4}=\dfrac{0}{5}=0.
\end{align*}$$
