How to find the asymptotes of #f(x) = (x+6)/(2x+1)# ?

Question

Christian Antunez · Accepted Answer

This function has vertical asymptote #x=-1/2# and horizontal asymptote #y=1/2#

To check if a rational function has a vertical asymtote(s) you have to look for zeroes of the denominator.

In this case there is one zero #x_0=-1/2#. So #x=-1/2# is a vertical asymptote.

To look for the horizontal asymptotes you have to calculate

#lim_{x->-oo}f(x)# and #lim_{x->+oo}f(x)#. If the limits are finite and

equal to #l#, then the line #y=l# is the asymptote.

In this example we have:

#lim_{x->-oo}(x+6)/(2x-1)=lim_{x->-oo}(1+6/x)/(2-1/x)=1/2#

#lim_{x->+oo}(x+6)/(2x-1)=lim_{x->+oo}(1+6/x)/(2-1/x)=1/2#

The limits are finite and equal, so #y=1/2# is a horizontal asymptote.

Theodore Amidon · Answer

undefined To find the asymptotes of $ f(x) = \frac{x+6}{2x+1} $, we examine the behavior of the function as $ x $ approaches positive and negative infinity.

1. Vertical Asymptote:
   Set the denominator equal to zero and solve for $ x $. Any $ x $ value that makes the denominator zero will create a vertical asymptote.
   $$ 2x + 1 = 0 $$
   $$ x = -\frac{1}{2} $$
   Thus, there is a vertical asymptote at $ x = -\frac{1}{2} $.

2. Horizontal Asymptote:
   To find the horizontal asymptote, we look at the degrees of the numerator and denominator.
   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 is equal to the degree of the denominator, divide the leading coefficients.
   If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
   In this case, the degree of the numerator and denominator are both 1. So, we divide the leading coefficients:
   $$ \lim_{x \to \pm \infty} \frac{x + 6}{2x + 1} = \frac{1}{2} $$
   Therefore, the horizontal asymptote is $ y = \frac{1}{2} $.

Hence, the asymptotes of $ f(x) = \frac{x+6}{2x+1} $ are:
- Vertical asymptote: $ x = -\frac{1}{2} $
- Horizontal asymptote: $ y = \frac{1}{2} $