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

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Answer 1 · 2016-02-17T18:19:00

Vertical asymptote at $x=-1$ .
No horizontal asymptotes.

Vertical asymptote at places which make the denominator zero, ie when $x=-1$ .

Horizontal asymptotes at $lim_(x->+-oo)f(x)=oo$ .

The graph verifies this :

graph{x^2/(x+1) [-20.2, 20.39, -10.18, 10.08]}

Answer 2 · 2016-02-17T19:19:18

In you also need slant (oblique) asymptotes, this function has $y=x-1$ as an asymptote.

Do the division to get

$x^2/(x+1) = x-1+2/(x+1)$ .

The difference between the graph of the function and the line is

$x^2/(x+1) - (x-1) = 2/(x+1)$ .

As x increases without bound, the difference, $2/(x+1)$ goes to $0$ .

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