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.