How do you find the asymptotes of $f(x)= (x^2+1)/(x+1)$ ?

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Answer 1 · 2015-10-06T16:58:41

$y=x-1$ is an oblique asymptote.

$(x^2+1)/(x+1) = x-1+2/(x+1)$ $" "$ (by division)

So, as $x$ increases or decreases without bound, the difference between $f(x)$ and the $y$ value of $y=x-1$ approaches $0$ .

So the graph of $f(x)$ approaches the line $y=x-1$ .

How do you find the asymptotes of f(x)= (x^2+1)/(x+1)f(x)= (x^2+1)/(x+1)?