Sketch an exact rational function and two different modulus graphs

$F(x)=-1-3/[(x-3)(x+1)]$. Thus horizontal asymptote y=−1 and vertical asymptotes x=−1,3.

At x=0, y=0. Setting the numerator $x(2-x)=0$ gives x=0,2. Both are allowed, so the axial intercepts are (0,0), (2,0).

$\begin{aligned}&F′(x)\\ & =\frac{3(2x-2)}{(x^2-2x-3)^2}.\end{aligned}$

The denominator is positive on the domain. F′=0 at x=1, where F=−1/4; the sign changes from negative to positive, so this is a minimum (1,−1/4).

$|F(x)|$ keeps the zeros 0,2 and poles −1,3. Its horizontal asymptote is y=1. The negative minimum becomes a positive local maximum (1,1/4), and the zero crossings become sharp minima.

$F(|x|)$ is even. Retain the x≥0 branch and reflect it in the y-axis. Its poles are x=±3, not x=−1; zeros are −2,0,2; minima are (±1,−1/4); horizontal asymptote is y=−1. The origin is a sharp local maximum, not an additional stationary point of the original differentiable F.