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

1 Answer
Dec 27, 2015

Answer:

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:enter image source here

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#