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

Question

1459 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]}

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