How to find the asymptotes of $f(x) = (x+6)/(2x+1)$ ?

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Answer 1 · 2016-01-21T16:54:07

This function has vertical asymptote $x=-1/2$ and horizontal asymptote $y=1/2$

To check if a rational function has a vertical asymtote(s) you have to look for zeroes of the denominator.

In this case there is one zero $x_0=-1/2$ . So $x=-1/2$ is a vertical asymptote.

To look for the horizontal asymptotes you have to calculate

$lim_{x->-oo}f(x)$ and $lim_{x->+oo}f(x)$ . If the limits are finite and

equal to $l$ , then the line $y=l$ is the asymptote.

In this example we have:

$lim_{x->-oo}(x+6)/(2x-1)=lim_{x->-oo}(1+6/x)/(2-1/x)=1/2$

$lim_{x->+oo}(x+6)/(2x-1)=lim_{x->+oo}(1+6/x)/(2-1/x)=1/2$

The limits are finite and equal, so $y=1/2$ is a horizontal asymptote.

How to find the asymptotes of f(x) = (x+6)/(2x+1)f(x) = (x+6)/(2x+1) ?