# How do you find the slant asymptote of (x^2)/(x-3)?

Dec 27, 2015

Divide the numerator ${x}^{2}$ by the denominator $\left(x - 3\right)$ to obtain a quotient $\left(x + 3\right)$ and remainder $9$. The quotient polynomial is the slant asymptote (also called an oblique asymptote).

#### Explanation:

You can divide the polynomials in several different ways.

Here's a long division of the coefficients: Note the $0$'s in the dividend for the missing $x$ and constant terms.

Equivalently, you can add and subtract terms to separate out multiples of the divisor like this:

${x}^{2} / \left(x - 3\right)$

$= \frac{{x}^{2} - 3 x + 3 x}{x - 3}$

$= \frac{x \left(x - 3\right) + 3 x}{x - 3}$

$= x + \frac{3 x}{x - 3}$

$= x + \frac{3 x - 9 + 9}{x - 3}$

$= x + \frac{3 \left(x - 3\right) + 9}{x - 3}$

$= x + 3 + \frac{9}{x - 3}$

In either case, we find that the quotient is $x + 3$ and the remainder is $9$.

As $x \to \pm \infty$ the term $\frac{9}{x - 3} \to 0$, so the asymptote is:

$y = x + 3$