Question

What is the equation of the oblique asymptote `f(x) = (x^2-5x+6)/(x-4)`?

Answer 1

The equation of the asymptote is `y=x-1`

Explanation:

Perform a long divoision

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

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

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

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

`color(white)(aaaaaaaaa)` `0+2`

Therefore,

`f(x)=(x^2-5x+6)/(x-4)=(x-1)+2/(x-4)`

To determine the slant asymptotes, determine the limits

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

The curve is above the asymptote

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

The curve is below the asymptote

The equation of the asymptote is `y=x-1`

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)=0 [-23.34, 22.27, -8.58, 14.23]}