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

1 Answer
May 3, 2018

Answer:

The slant asymptote is #y = x+2#

Explanation:

Given: #y = (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 * x = x^2# Multiply this monomial by each monomial in the divisor:

#" "ul(" "x" ")#
#x - 2|x^2 + 0x + 12#
#" "ul(x^2 -2x)#

Subtract and bring down the next monomial in the dividend:

#" "ul(" "x" ")#
#x - 2|x^2 + 0x + 12#
#" "ul(x^2 -2x)#
#" "2x + 12#

What monomial can we add to the quotient that will eliminate the #2x# term? #" "2#

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

#" "ul(" "x + 2" ")#
#x - 2|x^2 + 0x + 12#
#" "ul(x^2 -2x)#
#" "2x + 12#
#" "ul(2x -4" ")#
#" "16#

The slant asymptote is #y = x+2#