How do you find the slant asymptote of (x^2) / (x+1)?

Jan 4, 2016

Re-express as the sum of a linear polynomial and a rational expression that tends to zero as $x \to \pm \infty$ to find slant asymptote:

$y = x - 1$

Explanation:

${x}^{2} / \left(x + 1\right)$

$= \frac{{x}^{2} - 1 + 1}{x + 1}$

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

$= x - 1 + \frac{1}{x + 1}$

Then $\frac{1}{x + 1} \to 0$ as $x \to \pm \infty$

So the slant asymptote (a.k.a. oblique asymptote) is $y = x - 1$