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College Algebra

Section 11.2 End Behavior of Rational Functions

Section 10.3
Remember that with polynomials, we only needed to look at the leading term to find the end behavior. The reason was that since we are pluggin in huge positive or negative values, only the highest power on \(x\) will make a difference. For rational functions, the same logic applies, but we will have a leading term in both the numerator and the denominator.
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There are three possibilities of what can happen in that third step:
  • All of the \(x\)’s cancel out and we are just left with a number.
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  • There are extra \(x\)’s in the denominator.
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  • There are extra \(x\)’s in the numerator.
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Let’s do an example of each to see what happens.
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Example 11.2.2.

Suppose \(f(x)=\dfrac{4x^3-2x^2-9x+3}{7x^3-5x^2-9x+1}\) and we want to find its end behavior. Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{4x^3}{7x^3} = \frac{4}{7} \end{equation*}
In this case, our end behavior is just a number. We don’t have to bother asking what happens when we plug in huge values for \(x\text{,}\) because there are no more \(x\)’s to plug huge numbers into.
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Definition 11.2.3.

When the end behavior of a function is a number, it is a horizontal asymptote: a horizonal line at that number.
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Example 11.2.4.

\(y=\frac{4}{7}\text{.}\)
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Example 11.2.5.

Suppose \(f(x)=\dfrac{4x^3-2x^2-9x+3}{7x^5-5x-9x+1}\) and we want to find its end behavior. Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{4x^3}{7x^5} = \frac{4}{7x^2} \end{equation*}
Since there are leftover \(x\)’s in the denominator, we need to ask: "What happens when we plug in huge numbers for \(x\text{?}\)" In this case, we would be dividing by a huge number, so the whole thing gets very close to 0. Therefore, this function also has a horizontal asymptote, this time at \(y=0\text{.}\)
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Example 11.2.6.

Suppose \(f(x)=\dfrac{4x^7-2x^2-9x+3}{7x^3-5x^2-9x+1}\) and we want to find its end behavior. Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{4x^7}{7x^3} = \frac{4x^4}{7} \end{equation*}
Since there are leftover \(x\)’s in the numerator, we need to ask: "What happens when we plug in huge numbers for \(x\text{?}\)" In this case, we are left with something we’ve seen before: a polynomial! The leading term is \(\frac{4}{7}\text{,}\) which is positive, and the degree is \(4\text{,}\) which is even. So, the end behavior is
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  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow \infty\text{.}\)
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  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)
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Now that we’ve seen all three examples, try some yourself.
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Checkpoint 11.2.7.

Find the end behavior of \(f(x)=\dfrac{-8x^4+2x-1}{3x^2+7x+12}\text{.}\)
Answer.
  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow -\infty\text{.}\)
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  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow -\infty\text{.}\)
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Solution.
Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{-8x^4}{3x^2} = \frac{-8x^2}{3} \end{equation*}
Since there are leftover \(x\)’s in the numerator, we need to ask: "What happens when we plug in huge numbers for \(x\text{?}\)" In this case, we are left with something we’ve seen before: a polynomial! The leading term is \(\frac{-8}{3}\text{,}\) which is negative, and the degree is \(2\text{,}\) which is even. So, the end behavior is
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  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow -\infty\text{.}\)
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  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow -\infty\text{.}\)
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Checkpoint 11.2.8.

Find the end behavior of \(f(x)=\dfrac{16x^7+5x^6-1}{4x^7+7}\text{.}\)
Answer.
This function has a horizontal asymptote at \(y=\frac{4}{7}\text{.}\)
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Solution.
Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{4x^3}{7x^3} = \frac{4}{7} \end{equation*}
In this case, our end behavior is just a number. We don’t have to bother asking what happens when we plug in huge values for \(x\text{,}\) because there are no more \(x\)’s to plug huge numbers into. This function has a horizontal asymptote at \(y=\frac{4}{7}\text{.}\)
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Checkpoint 11.2.9.

Find the end behavior of \(f(x)=\dfrac{14x^5+4x-1}{3x^7+4x-1}\text{.}\)
Answer.
This function also has a horizontal asymptote, this time at \(y=0\text{.}\)
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Solution.
Our first step is to isolate the leading term of top and bottom of the fraction and then simplify it:
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\begin{equation*} \frac{14x^5}{3x^7} = \frac{14}{3x^2} \end{equation*}
Since there are leftover \(x\)’s in the denominator, we need to ask: "What happens when we plug in huge numbers for \(x\text{?}\)" In this case, we would be dividing by a huge number, so the whole thing gets very close to 0. Therefore, this function also has a horizontal asymptote, this time at \(y=0\text{.}\)
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