Question

How do you find the asymptotes for `f(x)= (x-3 )/( 2x-1)`?

Answer 1

Horizontal asymptote: `f(x)=1/2`
Vertical asymptote: `x=-1/2`
There is no oblique asymptote.

Explanation:

Horizontal asymptote
`f(x)=(x-3)/(2x-1)`

`color(white)("XXX")=x/(2x-1)-3/(2x-1)`

`color(white)("XXX")=1/(2-1/x) -3/(2x-1)`

`lim_(xrarr+-oo) f(x) = 1/(2-0)-0 = 1/2`

Vertical asymptote
As `xrarr(1/2)^+`
`color(white)("XXX")(2x-1)rarr +0`
As `xrarr(1/2)^-`
`color(white)("XXX")(2x-1)rarr -0`

When `(2x-1)rarr 0`
`color(white)("XXX")f(x)=(x-3)/(2x-1) rarr +_oo`

Therefore `x=1/2` is a vertical asymptote

Oblique asymptote
When a function is defined as a polynomial divided by another polynomial
an oblique asymptote exists if and only if
`color(white)("XXX")`the degree of the numerator is greater than the degree of the denominator.

graph{(x-3)/(2x-1) [-12.21, 13.09, -5.02, 7.64]}