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

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To find the horizontal asymptote of the function f(x) = (x^2 - 9)/(x - 3), it's important to note that the degrees of the numerator and denominator must be compared. The function has no horizontal asymptote because the numerator's degree (2) is greater than the denominator's degree (1). For functions where the degrees are the same, such as y = (x^2 + 3)/(x^2 + 5), the horizontal asymptote can be determined by the leading coefficients, resulting in y = 1. As x approaches infinity, the lower-order terms become negligible, confirming the horizontal asymptote. Understanding these principles is crucial for analyzing rational functions.
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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?

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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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