Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2-5x+6)/(x-4)`?

Answer 1

The vertical asymptote is `x=4`
The oblique asymptote is `y=x-1`
No horizontal asymptote

Explanation:

As you cannot divide by `0`, `=>`, `x!=4`

The vertical asymptote is `x=4`

The degree of the numerator is `>` than the degree of the denominator, there is an oblique asymptote.

Let `f(x)=(x^2-5x+6)/(x-4)`

Let's do a long division

`color(white)(aaaa)` `x^2-5x+6` `color(white)(aaaa)` `|` `x-4`

`color(white)(aaaa)` `x^2-4x` `color(white)(aaaaaaaa)` `|` `x-1`

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

`color(white)(aaaaaaa)` `-x+4`

`color(white)(aaaaaaa)` `-0+2`

Therefore,

`f(x)=(x-1)+2/(x-4)`

`lim_(x->-oo)(f(x)-(x-1))=lim_(x->-oo)2/(x-4)=0^-`

`lim_(x->+oo)(f(x)-(x-1))=lim_(x->+oo)2/(x-4)=0^+`

The oblique asymptote is `y=x-1`

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)(y-100(x-4))=0 [-18.3, 17.74, -6.74, 11.28]}