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

May 28, 2018

no asymptotes

#### Explanation:

Given: $f \left(x\right) = \frac{3 {x}^{2} + 2 x - 1}{x + 1}$

This is a rational function of the form $\frac{N \left(x\right)}{D \left(x\right)}$.
Factor the numerator.

$f \left(x\right) = \frac{3 {x}^{2} + 2 x - 1}{x + 1} = \frac{\left(3 x - 1\right) \left(x + 1\right)}{x + 1}$

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 $y = 3 x - 1$ 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]}