Vertical asymptotes of a given non-rational radical funtion We have that $f$ is a function $f(x) = x\sqrt{x+4}$.

Hence, $f'(x) = \dfrac{3x+8}{2\sqrt{x+4}}$.

Then, $\lim_{x \to -4^+}f'(x) = -\infty$.

This means that $f$ has a vertical slope at $f(-4)$.

It was my first thought to say, by this, $f$ has a vertical asymptote at $x = -4$; I wonder, though, is it so that an asymptote can necessarily only exist when a function is rational?
Rational functions with a zero in the denominator are common causes of vertical asymptotes, but they are not the only ways this can occur.

For example, $f(x)=\tan(x)$ has vertical asymptotes.  Of course you can argue that the reason this occurs is that there is a zero hiding in the denominator since $\tan(x)=\frac{\sin(x)}{\cos(x)}$.

Another example, $f(x)=\ln(x)$ has a vertical asymptote at $x=0$.  I do not see any way (reasonably easy and meaningful) to write $\ln(x)$ as a fraction with a zero appearing in the denominator when $x=0$.

One can construct more functions like $\ln(x)$ (in terms of vertical asymptotes) by taking the inverse function of any one-to-one function with a horizontal asymptote.  In the case of $\ln(x)$, we found the inverse of $e^{x}$ which had a horizontal asymptote at $y=0$.
If the denominator contains a factor going to zero it has infinite value at an asymptote for any function.