the function is $\displaystyle \frac{2(x^2+1)}{(x-1)(x+2)}$

So it will have a vertical asymptote $\displaystyle x\rightarrow 1^+, \,\,\, x\rightarrow 1^-, \,\,\, x\rightarrow 2^+, \,\,\, x\rightarrow 2^-$ at $\displaystyle x =1\,\,\, and\,\,\, x = -2 $

And it will have a horizontal asymptote at the line $\displaystyle y=2$

and to determine whether the graph approaches the horizontal asymptote from above or below

$\displaystyle f(x) - 2 = \frac{-2x+6}{(x-1)(x+2)}$

as $\displaystyle x\rightarrow+\infty, \,\, f(x)<2$ and $\displaystyle x\rightarrow -\infty, \,\,\,f(x) >2$

And because $\displaystyle \lim_{x\rightarrow1^+}f(x) = +\infty$ but as $\displaystyle x\rightarrow+\infty,\,\, f(x) < 2$

Thus the graph would intersect the horizontal asymptote at $\displaystyle y=2$ being the point $\displaystyle (3,2)$

Is this correct?