Question

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

Answer 1

Divide the numerator `x^2` by the denominator `(x-3)` to obtain a quotient `(x+3)` 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/(x-3)`

`=(x^2-3x+3x)/(x-3)`

`=(x(x-3)+3x)/(x-3)`

`=x + (3x)/(x-3)`

`=x + (3x-9+9)/(x-3)`

`=x + (3(x-3)+9)/(x-3)`

`=x + 3 + 9/(x-3)`

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

As `x->+-oo` the term `9/(x-3)->0`, so the asymptote is:

`y = x+3`