
## SectionLong-Run Behavior of Rational Functions

A rational function is the quotient of two polynomials. (As with rational numbers, the word rational refers to a ratio.)

###### Rational Functions

A rational function is one of the form

\begin{equation*} f(x)=\frac{P(x)}{Q(x)} \end{equation*}

where $P(x)$ and $Q(x)$ are polynomials and $Q(x)\neq 0\text{.}$

The graphs of rational functions can be quite different from the graphs of polynomials.

###### Example329

Queueing theory is used to predict your waiting time in a line, or queue. For example, suppose the attendant at a toll booth can process $6$ vehicles per minute. The average total time spent by a motorist negotiating the toll booth depends on the rate, $r\text{,}$ at which vehicles arrive, according to the formula

\begin{equation*} T=g(r)=\frac{12-r}{12(6-r)}. \end{equation*}
1. What is the average time spent at the toll booth if vehicles arrive at a rate of $3$ vehicles per minute?

2. Graph the function on the domain $[0, 6)\text{.}$

Solution
1. If the vehicles arrice at a rate of $3$ vehicles per minute, then $r=3$ so

\begin{equation*} g(3)=\dfrac{12-3}{12(6-3)}=\dfrac{9}{36}=0.25 \end{equation*}

The average total time is $0.25$ minutes, or $15$ seconds.

###### Example330

EarthCare decides to sell T-shirts to raise money. The company makes an initial investment of $100 to pay for the design of the T-shirt and to set up the printing process. After that, the T-shirts cost$5 each for labor and materials.

1. Express the average cost per T-shirt as a function of the number of T-shirts EarthCare produces.

2. Make a table of values for the function.

3. Graph the function and explain what it tells you about the cost of the T-shirts.

Solution
1. If EarthCare produces $x$ T-shirts, the total costs will be $100 + 5x$ dollars. To find the average cost per T-shirt, we divide the total cost by the number of T-shirts produced, to get

\begin{equation*} C = g(x) = \frac{100 + 5x}{x}. \end{equation*}
2. We evaluate the function for several values of $x\text{,}$ as shown in the table

 $x$ $1$ $2$ $4$ $5$ $10$ $20$ $C$ $105$ $55$ $40$ $25$ $15$ $10$

If EarthCare makes only one T-shirt, its cost is $105. But if more than one T-shirt is made, the cost of the original$100 investment is distributed among them. For example, the average cost per T-shirt for $\alert{2}$ T-shirts is

and the average cost for $\alert{5}$ T-shirts is

3. The graph is shown in Figure331. You can use a graphing utility with the window

\begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 470\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 30. \end{align*}

to verify the graph.

The graph shows that as the number of T-shirts increases, the average cost per shirt continues to decrease, but not as rapidly as at first. Eventually the average cost levels off and approaches $5 per T-shirt. For example, if EarthCare produces $\alert{400}$ T-shirts, the average cost per shirt is \begin{equation*} \frac{100+5(\alert{400})}{\alert{400}}=5.25. \end{equation*} The horizontal line $C = 5$ on the graph of $C = \displaystyle{\frac{100 + 5x}{x}}$ is a horizontal asymptote. As $x$ increases, the graph approaches the line $C = 5\text{.}$ The average price per T-shirt will always be slightly more than$5. Horizontal asymptotes are also important in sketching the graphs of rational functions.

###### Example332

Delbert prepares a $20\%$ saturated glucose solution of by mixing $2$ mL of saturated glucose with $8$ mL of water. If he adds $x$ ml of saturated glucose to the solution, its concentration is given by

\begin{equation*} C(x) = \frac{2 + x}{8+x}. \end{equation*}
1. How many milliliters of saturated glucose should Delbert add to increase the concentration to $50\%\text{?}$

2. Graph the function on the domain $[0, 100]\text{.}$

3. What is the horizontal asymptote of the graph? What does it tell you about the solution?

Solution
1. To obtain a concentration of $50/%\text{,}$ we solve $C(x)=0.50\text{:}$

\begin{equation*} \begin{aligned} C(x) \amp= 0.50 \\ \dfrac{2+x}{8+x} \amp= 0.50 \\ 2+x \amp= 0.50(8+x) \\ 2+x \amp= 4 + 0.50x \\ 0.5x \amp= 2 \\ x \amp= 4 \end{aligned} \end{equation*}

He needs to add $4$ ml.

2. $C=1\text{.}$ As Delbert adds more glucose to the mixture, its concentration increases toward $100\%\text{.}$

### SubsectionLong Run Behavior for Rational Functions

Recall in the start of this chapter we discussed the long run behavior of polynomial functions. Recall too that given a polynomial

\begin{equation*} f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0 \end{equation*}

the leading term of $f(x)$ is the term $a_nx^n \text{.}$

###### Long Run Behavior

Suppose that $f(x)=\frac{P(x)}{Q(x)}$ is a rational function. Let $ax^n$ be the leading term of $P(x)$ and $bx^m$ be the leading term of $Q(x)\text{.}$ Then the long run behavior of $f(x)$ is the same as the long run behavior of the power function

\begin{equation*} \frac{ax^n}{bx^m}=\frac{a}{b}x^{n-m}. \end{equation*}

If the resulting power function happens to be a constant function (so $n=m$ above) then the rational function has a horizontal asymptote at $y=\frac{a}{b}\text{.}$

###### Example333

Identify the long run behavior of the rational function

\begin{equation*} f(x)=\frac{-2x^7+4x^2-1}{x^3+8x}. \end{equation*}
Solution

First, we note that the leading term of the numerator is $-2x^7$ and the leading term of the denominator is $x^3\text{.}$ So, the long run behavior of $f$ is given by

\begin{equation*} \frac{-2x^7}{x^3}=-2x^4. \end{equation*}

We know that the long run behavior of $x^4$ is

\begin{equation*} x^4\to \infty\ \text{as}\ x\to\pm\infty. \end{equation*}

Since the graph of $-2x^4$ is the graph of $x^4$ stretched vertically by a factor of $2$ and then reflected about the $x$-axis, get that the long run behavior of $-2x^4$ is

\begin{equation*} -2x^4\to -\infty\ \text{as}\ x\to\pm\infty. \end{equation*}

Thus, the long run behavior of $f(x)$ is

\begin{equation*} f(x)\to -\infty\ \text{as}\ x\to\pm\infty. \end{equation*}