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

Feb 17, 2016

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

Explanation:

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

Horizontal asymptotes at ${\lim}_{x \to \pm \infty} f \left(x\right) = \infty$.

The graph verifies this :

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

Feb 17, 2016

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

Explanation:

Do the division to get

${x}^{2} / \left(x + 1\right) = x - 1 + \frac{2}{x + 1}$.

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

${x}^{2} / \left(x + 1\right) - \left(x - 1\right) = \frac{2}{x + 1}$.

As x increases without bound, the difference, $\frac{2}{x + 1}$ goes to $0$.