How do you find the asymptotes for f(x) =(2x^2-5x+6)/(x+2)?

1 Answer
Dec 16, 2016

The vertical asymptote is x=-2
The slant asymptote is y=2x-9
No horizontal asymptote

Explanation:

As we cannot divide by 0, x!=-2

The degree of the numerator is > than the degree of the denominator, we expect a slant asymptote.

Let's do a long division

color(white)(aaaa)2x^2-5x+6x+2

color(white)(aaaa)2x^2+4xcolor(white)(aaaa)2x-9

color(white)(aaaaa)0-9x+6

color(white)(aaaaaaa)-9x-18

color(white)(aaaaaaaaaaaa)+24

So,

(2x^2-5x+6)/(x+2)=2x-9+24/(x+2)

The slant asymptote is y=2x-9

For the horizontal asymptotes, we calculate the limita as x_>oo

lim_(x->+-oo)f(x)=lim_(x->+-oo)2x^2/x=lim_(x->+-oo)2x=+-oo

There is no horizontal asymptote.

graph{(y-(2x^2-5x+6)/(x+2))(y-2x+9)=0 [-93.7, 93.9, -46.8, 46.9]}