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

Question

Caleb Alexander · Accepted Answer

The slant asymptote is #y = x+2#

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#

Elijah Allen · Answer

undefined To find the slant asymptote of the function $ y = \frac{x^2 + 12}{x - 2} $, you can perform polynomial long division to divide $ x^2 + 12 $ by $ x - 2 $. The slant asymptote occurs when the degree of the numerator is one greater than the degree of the denominator after division.

Perform the division as follows:

1. Divide $ x^2 $ by $ x $ to get $ x $.
2. Multiply $ x $ by $ x - 2 $ to get $ x^2 - 2x $.
3. Subtract $ x^2 - 2x $ from $ x^2 + 12 $ to get $ 2x + 12 $.
4. Divide $ 2x $ by $ x $ to get $ 2 $.
5. Multiply $ 2 $ by $ x - 2 $ to get $ 2x - 4 $.
6. Subtract $ 2x - 4 $ from $ 2x + 12 $ to get $ 16 $.

After performing the division, you get $ x + 2 + \frac{16}{x - 2} $. The slant asymptote is the linear term $ x + 2 $ in this expression. Therefore, the slant asymptote of the function $ y = \frac{x^2 + 12}{x - 2} $ is $ y = x + 2 $.