How do you find the asymptotes for #(3x-2) / (x+1) #?

Question

Liam Brennan · Accepted Answer

There are three kinds of asymptotes:

Horizontal asymptotes are particular oblique ones, and so if you find horizontal asymptotes at both #infty# and #-infty#, there's no need to look for oblique ones.

In your case, the only point in which the function is not defined is the one which annihilates the denominator, and #x+1=0 iff x=-1#. So, #x=-1# is a vertical asymptote.

As for the horizontal ones, we have that

#lim_{x\to\pm\infty} \frac{3x-2}{x+1} =lim_{x\to\pm\infty} \frac{cancel(x)(3-2/x)}{cancel(x)(1+1/x)}=lim_{x\to\pm\infty} 3/1 = 3#

So, #y=3# is a horizontal asymptote in both directions.

Sadie Ackerman · Answer

undefined To find the asymptotes of the function $ \frac{3x-2}{x+1} $, you need to check for vertical and horizontal asymptotes.

1. Vertical Asymptote:
   Set the denominator equal to zero and solve for $ x $. Any $ x $ value that makes the denominator zero will result in a vertical asymptote.

$ x + 1 = 0 $
   $ x = -1 $

Therefore, $ x = -1 $ is the vertical asymptote.

2. Horizontal Asymptote:
   Compare the degrees of the numerator and denominator polynomial functions.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $.
   If the degree of the numerator equals the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.

Here, the degree of the numerator (1) is less than the degree of the denominator (1).
   So, the horizontal asymptote is $ y = 0 $.

Therefore, the vertical asymptote is $ x = -1 $ and the horizontal asymptote is $ y = 0 $.