# Math Help - Show the graph of a function intersects its horizontal asymptote at one point

1. ## Show the graph of a function intersects its horizontal asymptote at one point

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

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

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

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

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

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

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

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

Is this correct?

2. ## Re: Show the graph of a function intersects its horizontal asymptote at one point

Hey Jonroberts74.

It all looks good.