What Are the Steps to Find the Horizontal Asymptote of f(x) = (x^2 - 9)/(x - 3)?

  • Context: MHB 
  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    Asymptote Horizontal
Click For Summary
SUMMARY

The discussion centers on determining the horizontal asymptote of the function f(x) = (x^2 - 9)/(x - 3). It is established that this function does not have a horizontal asymptote due to the differing degrees of the numerator and denominator. The standard rule states that a horizontal asymptote exists when the degrees of both the numerator and denominator are equal. An example is provided with the function y = (x^2 + 3)/(x^2 + 5), which confirms that the horizontal asymptote is y = 1, as both the leading coefficients of x^2 are equal.

PREREQUISITES
  • Understanding of polynomial degrees in rational functions
  • Familiarity with horizontal asymptotes in calculus
  • Knowledge of limits and their application in asymptotic analysis
  • Basic algebraic manipulation of rational expressions
NEXT STEPS
  • Study the concept of polynomial long division for rational functions
  • Learn about vertical asymptotes and their relationship to horizontal asymptotes
  • Explore the application of limits in determining asymptotic behavior
  • Investigate other types of asymptotes, such as oblique asymptotes
USEFUL FOR

Students and educators in calculus, mathematicians analyzing rational functions, and anyone seeking to deepen their understanding of asymptotic behavior in functions.

mathdad
Messages
1,280
Reaction score
0
Find the horizontal asymptote of f(x) = (x^2 - 9)/(x - 3). I need the steps not the solution.
 
Physics news on Phys.org
The given function has no horizontal asymptote
 
RTCNTC said:
Find the horizontal asymptote of f(x) = (x^2 - 9)/(x - 3). I need the steps not the solution.

Standard Rule: Numerator and Denominator have the same "Degree". THAT will get you an Horizontal Asymptote.
 
skeeter said:
The given function has no horizontal asymptote

Good to know but why?

- - - Updated - - -

tkhunny said:
Standard Rule: Numerator and Denominator have the same "Degree". THAT will get you an Horizontal Asymptote.

By "Degree" you mean POWER or EXPONENT, right?

What about y = (x^2 + 3)/(x^2 + 5)?

Can we say the horizontal asymptote is y = 1?

The coefficient of x^2 is 1 for the numerator and denominator.

y = (x^2)/(x^2)

y = 1

True?
 
Observe that:

$$\frac{x^2+3}{x^2+5}=\frac{x^2+5-2}{x^2+5}=1-\frac{2}{x^2+5}$$

As $x\to\pm\infty$, the rational term vanishes (gets smaller and smaller), and the entire expression therefore approaches 1. :)

In general, you are correct...it can be shown that:

$$\lim_{x\to\pm\infty}\left(\frac{\sum\limits_{k=0}^n\left(a_kx^k\right)}{\sum\limits_{k=0}^n\left(b_kx^k\right)}\right)=\frac{a_n}{b_n}$$
 
Good to know.
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K