Question

How do you identify the oblique asymptote of `f(x) = (2x^2 + 3x + 8)/(x + 3)`?

Answer 1

`f(x) = (2x^2+3x+8)/(x+3) = 2x-3+17/(x+3)`

`17/(x+3)->0` as `x->+-oo`

So the oblique asymptote is the line `y = 2x-3`

Explanation:

`f(x) = (2x^2+3x+8)/(x+3)`

`=(2x^2+6x-3x-9+17)/(x+3)`

`=(2x(x+3)-3(x+3)+17)/(x+3)`

`=2x-3+17/(x+3)` with exclusion `x != -3`

graph{(y - (2x^2+3x+8)/(x+3))(y - (2x-3)) = 0 [-80, 80, -40, 40]}