Answer
We can not define a value for $f(-4)$ such that $f$ will become continuous.
Work Step by Step
$\lim\limits_{x \to -4}(x^2-3x)=(-4)^2-3(-4)=28$
$\lim\limits_{x \to -4}(x+4)=-4+4=0$
$\lim\limits_{x \to -4}\frac{x^2-3x}{x+4}=\frac{28}{0}=\infty$
