How do you find vertical, horizontal and oblique asymptotes for $(x^2 + 2x)/ (x +1)$ ?

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Answer 1 · 2016-12-01T05:55:28

We have a vertical asymptote at $x=-1$ and

an oblique asymptote $y=x+1$

As in $f(x)=(x^2+2x)/(x+1)$ we have the denominator $x+1$ ,

when $x->-1$ , $(x+1)->0$ and $f(x)->oo$

As such we have a vertical asymptote at $x=-1$

Further $f(x)=(x^2+2x)/(x+1)=(x^2+x+x+1-1)/(x+1)$

= $x+1-1/(x+1)$

Hence when $x->oo$ , $f(x)->x+1$

Hence, we have an oblique asymptote $y=x+1$

There is no other asymptote.
graph{(x^2+2x)/(x+1) [-10, 10, -5, 5]}

How do you find vertical, horizontal and oblique asymptotes for (x^2 + 2x)/ (x +1)(x^2 + 2x)/ (x +1)?