Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) = (x^2-4x+1) /( 2x-3)`?

Answer 1

The vertical asymptote is `x=3/2`
The oblique asymptote is `y=1/2x-5/4`
No horizontal asymptote

Explanation:

As we cannot divide by `0`, `x!=3/2`

The vertical asymptote is `x=3/2`

We perform a long division

`color(white)(aaaa)` `2x-3` `color(white)(aaaa)` `|` `x^2-4x+1` `color(white)(aaaa)` `|` `1/2x-5/4`

`color(white)(aaaaaaaaaaaaaaa)` `x^2-3/2x`

`color(white)(aaaaaaaaaaaaaaaa)` `0-5/2x+1`

`color(white)(aaaaaaaaaaaaaaaaaa)` `-5/2x-15/4`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `-0-11/4`

Therefore,

`f(x)=(x^2-4x+1)/(2x-3)=(1/2x-5/4)-(11/4)/(2x-3)`

`lim_(x->-oo)f(x)-(1/2x-5/4)=lim_(x->-oo)-(11/4)/(2x-3)=0^+`

`lim_(x->+oo)f(x)-(1/2x-5/4)=lim_(x->+oo)-(11/4)/(2x-3)=0^-`

The oblique asymptote is `y=1/2x-5/4`
graph{(y-(x^2-4x+1)/(2x-3))(y-1/2x+5/4)(y-1000(x-3/2))=0 [-14.24, 14.24, -7.12, 7.12]}