Graphing simple fractional function I am trying to graph 
$$f(x) = \frac{x^2-4}{x^2+4}$$
It seems pretty simple to me but I can't finish it correctly.
I know that there is a horizontal asymptote at $1$ beacuse the degrees on the variable at the same and $\frac{x}{x}$ is 1.
I know that it is a negative function until zero, and then positive because the only critical number of the derivative 
$$\frac{16x}{(x^2+4)^2}$$
is going to be zero since the denominator can't be zero and the only root of the top is $0$.
This then tells me that there is no local max but only a minimum which is at $0$ which gives me $-1$.
Trying to find concavity 
$$\frac{16(x^2+4)^2 - 16x (4x(x^2+4))}{(x^2+4)^4}=
\frac {-64x^5 + 16x^4 + 256x^3 + 128x^2 + 256}{(x^2+4)^4}$$
Which I have no idea how to really work with I can't see to get anything workable out of that.

Correct second derivative:
$$\begin{align*}
f(x) &= \frac{x^2-4}{x^2+4}\\
f'(x) &= \frac{(x^2+4)(x^2-4)' - (x^2-4)(x^2+4)'}{(x^2+4)^2}\\
&= \frac{(x^2+4)(2x) - (x^2-4)(2x)}{(x^2+4)^2}\\
&= \frac{2x(x^2+4-x^2+4)}{(x^2+4)^2}\\
&= \frac{16x}{(x^2+4)^2}.\\
f''(x) &= \frac{(x^2+4)^2(16x)' - 16x\Bigl( (x^2+4)^2\Bigr)'}{\Bigl((x^2+4)^2\Bigr)^2}\\
&= \frac{16(x^2+4)^2 - 16x(2(x^2+4)(x^2+4)')}{(x^2+4)^4}\\
&= \frac{16(x^2+4)^2 - 16x(2)(x^2+4)(2x)}{(x^2+4)^4}\\
&= \frac{16(x^2+4)\Bigl( x^2+4 - 4x^2\Bigr)}{(x^2+4)^4}\\
&= \frac{16(4-3x^2)}{(x^2+4)^3}.
\end{align*}$$
The denominator is never zero; the numerator is zero exactly when $4-3x^2=0$. This is a parabola that opens down, so it will be positive between the roots and negative before and after both roots.
