Question

How do you find oblique asymptote of `(x^2 + x + 3) /(x-1)`?

Answer 1

Partially divide the numerator by the denominator to get:

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

Hence the oblique asymptote is `y = x+2`

Explanation:

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

with exclusion `x != 1`

`5/(x-1) -> 0` as `x->+-oo`

So the oblique asymptote is `y = x + 2`

graph{(y - (x^2+x+3) / (x-1))(y - x - 2) = 0 [-40, 40, -20, 20]}