Question

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

Answer 1

Vertical asymptote `x = 1`

Oblique asymptote `y = x+2`

Explanation:

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

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

`= ((x+2)(x-1)+5)/(x-1)`

`= x+2+5/(x-1)`

As `x->+-oo` the term `5/(x-1)->0`

So there is an oblique (slant) asymptote `y = x+2`

When `x=1` the denominator of `f(x)` becomes `0` while the numerator is non-zero. So `f(x)` is undefined at `x=1` and has a vertical asymptote there.

graph{(y - (x^2+x+3)/(x-1))(y - (x+2)) = 0 [-39, 41, -17.04, 22.96]}