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

1 Answer
May 28, 2018

Answer:

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]}