Question

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

Answer 1

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]}