How do you find the asymptotes for #F(x)=((x-2)^2(2x+5) )/( (x-3)^2(x-2))#?

Question

1252 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-22T12:52:43

Three asymptotes are #x=3#, #x=2# and #y=2#.

As denominator is #(x−3)^2(x−2)#, two vertical asymptotes are #x-3=0# and #x-2=0#.

Let us expand numerator and denominator.

#F(x)=((x−2)^2(2x+5))/((x−3)^2(x−2))# or

#((x^2−4x+4)(2x+5))/((x^2−6x+9)(x−2))# or

#(2x^3+5x^2-8x^2-20x+8x+20)/(x^3-2x^2-6x^2+12x+9x-18)# or

#(2x^3-3x^2-12x+20)/(x^3-8x^2+21x-18)#

As the ratio between highest degrees is #2x^3/x^3# or #2#,

we have a horizontal asymptote #y=2#.

So three asymptotes are #x=3#, #x=2# and #y=2#.

graph{(2x^3-3x^2-12x+20)/(x^3-8x^2+21x-18) [-10, 10, -5, 5]}