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

Feb 22, 2016

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

#### Explanation:

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

$\frac{2 {x}^{3} + 5 {x}^{2} - 8 {x}^{2} - 20 x + 8 x + 20}{{x}^{3} - 2 {x}^{2} - 6 {x}^{2} + 12 x + 9 x - 18}$ or

$\frac{2 {x}^{3} - 3 {x}^{2} - 12 x + 20}{{x}^{3} - 8 {x}^{2} + 21 x - 18}$

As the ratio between highest degrees is $2 {x}^{3} / {x}^{3}$ 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]}