Question

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

Answer 1

Simplify:

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

to find the oblique asymptote `y = x + 2` and vertical asymptote `x = 1`

Explanation:

Let `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`, `5/(x-1)->0`

So `y = x+2` is an oblique asymptote.

`f(x)` also has a vertical asymptote `x=1`, where the denominator of `5/(x-1)` becomes `0` while the numerator is non-zero.

graph{(y - (x^2+x+3)/(x-1))(y - x - 2)(0.999x - 1) = 0 [-21.19, 24.42, -9.2, 13.63]}