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#.