Domain and intercepts

The denominator $x^2-1=(x-1)(x+1)$ is zero at $x=1$ and $x=-1$, so the domain is all real numbers except $x=\pm1$.

Setting the numerator $2x^2-8=0$ gives $x^2=4$, so $x=2$ and $x=-2$ are the $x$-intercepts (both valid since the denominator is nonzero there). Evaluating $f(0)=\dfrac{-8}{-1}=8$ gives the $y$-intercept $(0,8)$.

Since the numerator is nonzero at $x=\pm1$ (it equals $-6$ there), the denominator's zeros give vertical asymptotes at $x=1$ and $x=-1$.

As $x\to\pm\infty$, the expression is $\infty/\infty$, so applying L'Hôpital's Rule: $\lim \dfrac{4x}{2x}=\lim 2 = 2$. Thus the horizontal asymptote is $y=2$ in both directions.

By the quotient rule, $f'(x) = \dfrac{(4x)(x^2-1) - (2x^2-8)(2x)}{(x^2-1)^2} = \dfrac{4x^3-4x-4x^3+16x}{(x^2-1)^2} = \dfrac{12x}{(x^2-1)^2}$.

Since $(x^2-1)^2>0$ everywhere in the domain, the sign of $f'(x)$ matches the sign of $12x$. So $f'(x)<0$ for $x<0$ (excluding $x=-1$) and $f'(x)>0$ for $x>0$ (excluding $x=1$), meaning $f$ decreases on $(-\infty,-1)\cup(-1,0)$ and increases on $(0,1)\cup(1,\infty)$.

The only critical point in the domain is $x=0$, where $f$ changes from decreasing to increasing, giving a local minimum at $(0,8)$.