How do you find the vertical, horizontal and slant asymptotes of: #f(x)= (4x^2+ 4x-24)/(x^4- 2x^3 - 9x^2+ 18x)#?

1 Answer
Jun 18, 2016

Answer:

#f(x)# has vertical asymptotes #x=0# and #x=3#, a horizontal asymptote #y=0# and holes (removable singularities) at #x=2# and #x=-3#.

Explanation:

The numerator factors like this:

#4x^2+4x-24 = 4(x^2+x-6) = 4(x+3)(x-2)#

The denominator factors like this:

#x^4-2x^3-9x^2+18x#

#= x((x^3-2x^2)-(9x-18))#

#= x(x^2(x-2)-9(x-2))#

#= x(x^2-9)(x-2)#

#= x(x-3)(x+3)(x-2)#

Note the common factors #(x-2)# and #(x+3)# in the numerator and denominator, so #f(x)# can be simplified, noting that #x=2# and #x = -3# are excluded from the domain:

#f(x) = (4x^2+4x-24)/(x^4-2x^3-9x^2+18x)#

#= (4color(red)(cancel(color(black)((x+3))))color(red)(cancel(color(black)((x-2)))))/(x(x-3)color(red)(cancel(color(black)((x+3))))color(red)(cancel(color(black)((x-2)))))#

#= (4)/(x(x-3))#

So when #x=0# or #x=3#, the denominator of #f(x)# is #0# and the numerator is non-zero, resulting in a vertical asymptote.

As #x->+-oo# the denominator #->+oo# resulting in a horizontal asymptote #y=0#.

#f(x)# has removable singularities (holes) at #x=2# and #x=-3#

graph{(4x^2+4x-24)/(x^4-2x^3-9x^2+18x) [-10, 10, -5, 5]}