How do you find the asymptotes for #y = (8 x^2 + x - 2)/(x^2 + x - 72)#?

1 Answer
Jul 13, 2018

Answer:

vertical asymptotes at #x = 8, x = -9#

horizontal asymptote at #y = 8#

no oblique/slant asymptotes

Explanation:

Given: #(8x^2 + x - 2)/(x^2 + x - 72)#

This type of equation is called a rational function:

#(N(x))/(D(x)) = (a_n x^n + ...)/(b_m x^m + ...)#

First both the numerator and denominator must be factored to see if there are any holes (when factors can be canceled from both the numerator and the denominator).

The numerator can't be factored easily unless you use the quadratic formula. This means there will not be any hole.

#y = (8x^2 + x - 2)/((x-8)(x+9))#

Find vertical asymptotes (D(x) = 0 :

#(x-8)(x+9) = 0 => " "x = 8, x = -9#

Find horizontal asymptotes by comparing the degrees #n " & " m#:

If #n < m => " horizontal asymptote": " "y = 0#

If #n = m => " horizontal asymptote": " "y = (a_n)/(b_m) #

If #n > m " "# no horizontal asymptote

In the given equation #n = m = 2 => " "y = 8/1 " or " y = 8#

Find oblique/slant asymptotes when #n = m+1#

none in the given equation