Study the derivatives of the function
$y={(x^3-4x)^{1\over2}}$
and sketch its graph on the real line.
My approach: 1. range of $x$ is $(-2,0)\bigcup (2,\infty)$
2 . Slope of $f(x)\to\infty$ as $x\to \infty$
3 . No idea about the increasingness/decreasingness of $f$
You correctly defined the range of $a$ $(-2,0)\bigcup (2,\infty)$.
Now, what you need to do is to analyze the derivative of the function $$y={(x^3-4x)^{1\over2}}$$ which, for simplicity, we could write as $y={u(x)^{1\over2}}$; then, applying the classical rules, $y'={1\over2}u'(x){u(x)^{-1\over2}}$. So, since $u(x)=x^3-4x$,$u'(x)=3x^2-4$ and then $$y'=\frac{3 x^2-4}{2 \sqrt{x^3-4 x}}$$ The numerator cancels for $3x^2-4=0$, that is to say $x=\pm \frac{2}{\sqrt{3}}$ but the positive root must be discarded because it is between $0$ and $2$ where the function is not defined. So, the derivative only cancels for $x=- \frac{2}{\sqrt{3}}$; for this value of $x$, $y=\frac{4}{3^{3/4}}$. If required, the second derivative test would show that this point corresponds to a maximum of the function.
So, now, we have a quite good idea of $y$ when $-2<x<0$: the function starts at $(-2,0)$ with a vertical tangent (because of the denominator of $y'$), grows up to the point $(- \frac{2}{\sqrt{3}},\frac{4}{3^{3/4}})$ and decreases to $(0,0)$ where it has again a vertical tangent (for the same reason as before).
For the second branch, the curve starts at $(2,0)$, again with a vertical tangent, and increases permanently since the derivative remains positive.
Concerning the behaviour of $y$ for large values of $x$, you could write $$y=\sqrt{x^3(1-\frac {4}{x^2}})=x^{\frac{3}{2}} \sqrt {1-\frac {4}{x^2}}$$ and if you remember that, when $z$ is small compared to $1$, $\sqrt {1+z} \simeq 1 +\frac {z}{2}$, then for large value of $x$, we can approximate $y$ according to $$y \simeq x^{\frac{3}{2}} (1-\frac {2}{x^2})=x^{\frac{3}{2}}-\frac{2}{\sqrt x}$$ So, the asymptote of the curve is $y=x^{\frac{3}{2}}$ and because of the second term, we can conclude that, for any value of $x$, the curve is below its asymptote.
I hope and wish that this clarifies your problem.