Question

How do I find the asymptotes of `f(x) = (3x^2 + 2x - 1 )/ (x + 1)`? Are there even any asymptotes?

Answer 1

no asymptotes

Explanation:

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

This is a rational function of the form `(N(x))/(D(x))`.
Factor the numerator.

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

There is a removable discontinuity (a hole) at `x = -1` since both the numerator and the denominator have the same factor `x + 1`

There are no vertical, horizontal or slant asymptotes because of the hole.

The graph of the function is a line `y = 3x - 1` with a hole at `x = -1`.

Typically you will not see the hole unless you perform a Trace on a graphing calculator.

You won't have a `y` value : `X = -1, Y =`

graph{(3x^2 + 2x -1)/(x+1) [-5, 5, -5, 5]}