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

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Answer 1 · 2016-12-01T05:55:28

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

an oblique asymptote $y = x + 1$ $y=x+1$

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

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

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

Further $f (x) = \frac{x^{2} + 2 x}{x + 1} = \frac{x^{2} + x + x + 1 - 1}{x + 1}$ $f(x)=(x^2+2x)/(x+1)=(x^2+x+x+1-1)/(x+1)$

= $x + 1 - \frac{1}{x + 1}$ $x+1-1/(x+1)$

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

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

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)x2+2xx+1(x^2 + 2x)/ (x +1)?