# How do you find the slant asymptote of y=(x^2+12)/(x-2)?

May 3, 2018

The slant asymptote is $y = x + 2$

#### Explanation:

Given: $y = \frac{{x}^{2} + 12}{x - 2}$

The slant asymptote occurs when the degree in the numerator is one greater than the degree in the denominator.

Since the denominator is a linear factor, you can use either long division or synthetic division to find the slant asymptote.

Using long division:
First set up the divisor and dividend as follows. Make sure the dividend has a term for each degree even if it is zero. Select $x$ as the first monomial since $x \cdot x = {x}^{2}$ Multiply this monomial by each monomial in the divisor:

" "ul(" "x" ")
$x - 2 | {x}^{2} + 0 x + 12$
$\text{ } \underline{{x}^{2} - 2 x}$

Subtract and bring down the next monomial in the dividend:

" "ul(" "x" ")
$x - 2 | {x}^{2} + 0 x + 12$
$\text{ } \underline{{x}^{2} - 2 x}$
$\text{ } 2 x + 12$

What monomial can we add to the quotient that will eliminate the $2 x$ term? $\text{ } 2$

Multiply this monomial in the quotient by each monomial in the divisor, then subtract:

" "ul(" "x + 2" ")
$x - 2 | {x}^{2} + 0 x + 12$
$\text{ } \underline{{x}^{2} - 2 x}$
$\text{ } 2 x + 12$
" "ul(2x -4" ")
$\text{ } 16$

The slant asymptote is $y = x + 2$