Here is a picture for clarification of the question:
So far, I have gotten this: the root is all non-negative numbers; x is greater than or equal to 0.
In order to find the asymptotes I set the denominator to zero and so the asymptotic is zero?? I don't understand
You can see that the domain is $\{x\in \mathbb{R} : x\geq 0 \}$. To find the max/min (which eventually gets you the range), we find out the first derivative.
$f'(x)=\frac{4-x}{2\sqrt(x)(x+4)^2}$. Here the critical point is $x=4$.
Also $f''(x)=-\frac{1}{4x^{3/2}}-\frac{4-x}{x^{1/2}(x+4)^2}$, then $f''(4)=-\frac{1}{32}$. Thus the function has maximum at $x=4$, and the maximum value is $f(4)=\frac{1}{4}$. Obviously the minimum value of $f$ is $0$.
If you want to draw its graph then try to come up with the graph of $\sqrt(x)$, consider the range $0\leq y \leq \frac{1}{4}$. Since $x+4$ is in the denominator the curve decreases as $x$ gets bigger, so the curve moves closer to the $x-$axis as $x$ gets bigger. I hope this will help.