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

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Discussion Overview

The discussion revolves around finding the horizontal asymptote of the function f(x) = (x^2 - 9)/(x - 3). Participants seek clarification on the steps involved in determining horizontal asymptotes, while also exploring related concepts and examples.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Some participants assert that the function f(x) = (x^2 - 9)/(x - 3) has no horizontal asymptote.
  • Others propose that a standard rule for determining horizontal asymptotes involves comparing the degrees of the numerator and denominator.
  • A participant questions whether the term "degree" refers to power or exponent in this context.
  • Another example, y = (x^2 + 3)/(x^2 + 5), is presented, with a claim that its horizontal asymptote is y = 1 due to the coefficients of x^2 being equal in both the numerator and denominator.
  • A later reply provides a mathematical observation that as x approaches infinity, the expression approaches 1, supporting the claim about the horizontal asymptote for the second example.

Areas of Agreement / Disagreement

Participants express differing views on whether the original function has a horizontal asymptote, with some asserting it does not and others discussing rules that might suggest otherwise. The discussion remains unresolved regarding the original function's asymptotic behavior.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the degrees of polynomials and the conditions under which horizontal asymptotes exist. The mathematical steps leading to conclusions about asymptotes are not fully resolved.

mathdad
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Find the horizontal asymptote of f(x) = (x^2 - 9)/(x - 3). I need the steps not the solution.
 
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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.
 

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