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

1 Answer
Dec 16, 2016

Answer:

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+6##∣##x+2#

#color(white)(aaaa)##2x^2+4x##color(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]}