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# 2.4: Analyzing Rational Functions

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply properties of rational functions to find roots and asymptotes
• Graph rational functions using roots, asymptotes, and behavior of the function around vertical asymptotes

## Summarize Analysis of Rational Functions

Recall that rational functions are defined as r(x)=p(x)q(x)\begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*} where p(x)\begin{align*}p(x)\end{align*} and q(x)\begin{align*}q(x)\end{align*} are polynomials. Rational functions can be slightly more complicated to analyze than polynomials, mostly due to the fact that anything divided by zero is undefined. In this section we review major points about analyzing rational functions, and show how a graphing calculator or other computer tool can be used to analyze rational functions.

## Finding Vertical Asymptotes and Breaks in the Graph of a Rational Function

To find vertical asymptotes and breaks in the domain of a rational function, set the denominator equal to zero and solve for x\begin{align*}x\end{align*}. Given r(x)=p(x)q(x)\begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*}, set q(x)=0\begin{align*}q(x)=0\end{align*} and solve for x\begin{align*}x\end{align*}.

Note, however, that some rational functions do not have vertical asymptotes and they are well-defined for all real numbers x\begin{align*}x\end{align*}. Other rational functions have a break in the function, but no vertical asymptote. This usually happens when one term in the numerator cancels with one term in the denominator. Below we illustrate examples of each of these cases.

r(x)=p(x)q(x)\begin{align*}r(x)=\frac{p(x)}{q(x)}\end{align*}

Example 1

Consider the following rational function. Find all restrictions on the domain and asymptotes

f(x)=x2+2x35x5\begin{align*}f(x)=\frac{x^{2}+2x-35}{x-5}\end{align*}

Solution

Factoring the numerator

f(x)f(x)=(x5)(x+7)x+7=(x5)(x+7)x+7\begin{align*}f(x)&=\frac{(x-5)(x+7)}{x+7}\\ f(x)& =\frac{(x-5)\cancel{(x+7)}}{\cancel{x+7}}\end{align*}

Canceling

f(x)=x+7,x5\begin{align*}f(x)=x+7, x\neq5\end{align*}

Notice that there is no asymptote in this function, but rather a break in the graph at x=5\begin{align*}x=5\end{align*}.

Example 2

Find the restrictions on the domain of

h(x)=3xx225\begin{align*}h(x)=\frac{3x}{x^{2}-25}\end{align*}

Solution

Setting the denominator equal to 0,

x225x2x=0=25=±5\begin{align*}x^{2}-25 & = 0\\ x^2 & = 25\\ x & = \pm 5\end{align*}

Thus, the domain of h(x)\begin{align*}h(x)\end{align*} is the set all all real numbers x\begin{align*}x\end{align*} with the restriction x±5\begin{align*}x\neq\pm 5\end{align*}. h(x)\begin{align*}h(x)\end{align*} has two vertical asymptotes, one at x=5\begin{align*}x=5\end{align*} and one at x=5\begin{align*}x=-5\end{align*}.

Example 3

Find the vertical asymptotes of

g(x)=x3x2+1\begin{align*}g(x)=\frac{x^{3}}{x^{2}+1}\end{align*}

Solution

Setting the denominator equal to zero,

x2+1x2x=0=1=1\begin{align*}x^{2}+1 & = 0\\ x^2 & = -1\\ x & = \sqrt{-1}\end{align*}

There are no real solutions, so there are no vertical asymptotes and no restrictions on the domain of this function.

Example 4

What are the vertical asymptotes of the function

k(x)=x2x2+5\begin{align*}k(x)=\frac{x^{2}}{x^{2}+5}\end{align*}

Solution

Setting the denominator equal to zero,

x2+5x2=0=5\begin{align*}x^{2}+5 & = 0\\ x^2 & = -5\end{align*}

Again there are no real solutions. The horizontal asymptote is y=1\begin{align*}y=1\end{align*}.

## End Behavior of Rational Function

Previously you have analyzed the end behavior of polynomials. You may recall that, as a general rule, as you “zoom out,” the graph of a polynomial looks like the graph of the power function made of the leading term of the polynomial. For instance, g(x)=3x418x315x\begin{align*}g(x)=3x^{4}-18x^{3}-15x\end{align*} looks like the function y=3x4\begin{align*}y=3x^{4}\end{align*} if you choose a sufficiently large range for x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.

There is a similar rule for rational functions. The End Behavior of a rational function can often be identified by the horizontal asymptote. That is, as the values of x\begin{align*}x\end{align*} get very large or very small, the graph of the rational function will approach (but not reach) the horizontal asymptote. In Example 4, we could write as x,k(x)1\begin{align*}x\rightarrow\infty, k(x)\rightarrow 1\end{align*} and as x,k(x)1\begin{align*}x\rightarrow-\infty, k(x)\rightarrow 1\end{align*}. This can be shortened by writing as x±,k(x)1\begin{align*}x\rightarrow\pm\infty, k(x)\rightarrow 1\end{align*}.

## Finding Oblique Asymptotes

Not all asymptotes of rational functions are vertical or horizontal. If we look at the graph of the rational function in Example 3 from above, g(x)=x3x2+1\begin{align*}g(x)=\frac{x^{3}}{x^{2}+1}\end{align*}, we can see that there is no horizontal asymptote of this function.

There is no horizontal asymptote in this function because the degree of the numerator is greater than the degree of the denominator.

As a reminder, the following guidelines can help identify the asymptotes of a rational function r(x)=f(x)D(x)\begin{align*}r(x)=\frac{f(x)}{D(x)}\end{align*}:

• If the degree of the denominator is greater than the degree of the numerator, then the line y=0\begin{align*}y=0\end{align*} is a horizontal asymptote.
• If the degree of the numerator and the denominator are equal, then the line y=ab\begin{align*}y=\frac{a}{b}\end{align*} is a horizontal asymptote, where a\begin{align*}a\end{align*} is the leading coefficient of f(x)\begin{align*}f(x)\end{align*}, the numerator, and b\begin{align*}b\end{align*} is the leading coefficient of D(x)\begin{align*}D(x)\end{align*}, the denominator.
• If the degree of the numerator is larger than the degree of the denominator, then the quotient function, Q(x)\begin{align*}Q(x)\end{align*}, found by dividing the numerator and denominator of the rational function is an oblique asymptote. Recall that for any rational function f(x)D(X)\begin{align*}\frac{f(x)}{D(X)}\end{align*}, you can use polynomial division to re-write that function in the form \begin{align*}\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\end{align*} where \begin{align*}Q(x)\end{align*} is the quotient and \begin{align*}R(x)\end{align*} is the remainder.

To illustrate this last point, look at \begin{align*}g(x)=\frac{x^{3}}{x^{2}+1}\end{align*}. By polynomial division we have,

\begin{align*}& \qquad \quad \ \ x \qquad \qquad \qquad \qquad \ \leftarrow\text{Quotient}\\ & x^2+1 \ \big ) \overline{x^3 + 0x^2 + 0x + 0} \quad \ \leftarrow \text{Dividend}\\ & \qquad \quad \ \underline{x^3 \qquad \quad + x}\\ & \qquad \qquad \qquad \quad \ -x \qquad \quad \ \leftarrow \text{Remainder}\end{align*}

So \begin{align*}g(x)=x-\frac{x}{x^{2}+1}\end{align*}. This tells us that the line \begin{align*}y=x\end{align*} is an oblique asymptote of \begin{align*}g(x)\end{align*}.

Notice that the oblique asymptotes of a rational function also describe the end behavior of the function. That is, as you “zoom out” from the graph of a rational function it looks like a line or the function defined by \begin{align*}Q(x)\end{align*} in \begin{align*}\frac{f(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}\end{align*}.

Example 5

Find the oblique asymptote of \begin{align*}f(x)=\frac{3x^{3}-5x^{2}+2}{x^{2}-3x}\end{align*}. Sketch a graph of \begin{align*}f(x)\end{align*}.

Solution

Using polynomial long division, \begin{align*}f(x)=3x+4+\frac{4x^{2}}{x^{2}-3x}\end{align*}. Thus, the line \begin{align*}y=3x+4\end{align*} is an oblique asymptote of \begin{align*}f(x)\end{align*}.

To sketch the graph we can find the vertical asymptotes by setting the denominator equal to zero,

\begin{align*}x^{2}-3x & = 0\\ x(x-3) & = 0\end{align*}

So the two vertical asymptotes are \begin{align*}x=0\end{align*} and \begin{align*}x=3\end{align*}.

\begin{align*}f(0)\end{align*} is undefined, so there is no \begin{align*}y-\end{align*}intercept. Also, there is no simple way to solve for the roots (setting the numerator equal to zero), but we can see by inspection that \begin{align*}f(1)=0\end{align*}. To get an idea of the shape of the graph we will make a table of a few test points. We used a calculator to evaluate decimal values of \begin{align*}x\end{align*} in \begin{align*}f(x)\end{align*}.

\begin{align*}& x && -2 && -1 && -0.1 && 1 && \ \ \ 2 && \ 4 && 3.5 && \ 5 && \ 6\\ & f(x) && -4.2 && -1.5 && 66.48 && 0 && -3 && 28.5 && 39.6 && 25.2 && 26.1\end{align*}

Finally we use all of this information to make a sketch of the graph of \begin{align*}f(x)\end{align*}:

## Applications, Technological Tools

The last example illustrates the difficulty of graphing any rational function by hand. A graphing calculator can be a valuable tool to help with graphing and analyzing rational functions, provided you know how to use the calculator effectively. Below we illustrate a few of the benefits and potential pitfalls of graphing rational functions using TI-83/84 Calculators.

Most of the calculator analysis tools that we have reviewed for quadratic and polynomial functions also work with rational functions. You can use the \begin{align*}Y=\end{align*} and GRAPH menus to view the graph of a rational function. Once you have graphed a rational function you can use the CALC menu to find the roots and the min/max values of the function.

Potential Pitfalls of Graphing Rational Functions

There are two common pitfalls when using a calculator to graph a rational function:

1. When graphing a rational function by entering the function in the \begin{align*}Y=\end{align*} screen, remember that you need to use parenthesis to group the numerator and denominator of the rational function. Also, the independent variable of a rational function is always \begin{align*}X\end{align*} in the TI-83/84. Thus, for example, you would enter the function \begin{align*}f(t)=\frac{3t^{2}-2t+1}{7t^{3}+4}\end{align*} by typing “\begin{align*}\frac{(3X^{\land} 2-2X+1)}{(7X^{\land} 3+4)}\end{align*}” in the \begin{align*}Y1=\end{align*} slot.
2. Vertical asymptotes are sometimes graphed as vertical lines.
3. Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully.

The following example illustrates the second and third problems.

Example 6

Graph \begin{align*}f(x)=\frac{1}{x-3}\end{align*} on the window \begin{align*}[-10,10] \times [-10,10]\end{align*}. (This means \begin{align*}\text{XMIN}=-10, \text{XMAX}=10, \text{YMIN}=-10,\end{align*} and \begin{align*}\text{YMAX}=10\end{align*})

Solution

((insert screen shot of the graph of \begin{align*}y=\frac{1}{x-3}\end{align*} showing that vertical line at \begin{align*}x=3\end{align*}))

\begin{align*}f(x)\end{align*} is undefined and has a vertical asymptote at \begin{align*}x=3\end{align*}, but the way the graphing calculator draws the graph, it shows a vertical line at \begin{align*}x=3\end{align*}. One way to “fix” this problem is to press MODE and select the option “Dot” rather than “Connected”. However, dot graphs can be hard to interpret as well.

Using the Table to Examine Behavior of a Rational Function

Despite the potential pitfalls of using a graphing calculator to graph rational functions, the TABLE function of the graphing calculator is one very effective tool for analyzing the behavior of a rational function near its vertical asymptotes. Let's consider the function from Example 5 above, \begin{align*}f(x)=\frac{3x^{3}-5x^{2}+2}{x^{2}-3x}\end{align*}. The vertical asymptotes of \begin{align*}f(x)\end{align*} are \begin{align*}x=3\end{align*} and \begin{align*}x=0\end{align*}. To analyze the behavior of \begin{align*}f(x)\end{align*} near these asymptotes, first enter the equation of \begin{align*}f(x)\end{align*} in the \begin{align*}Y=\end{align*} menu. Next, press 2ND WINDOW to go to the TABLSET menu. There, change the setting of Independent to ASK.

((Calculator Image: Show TABLE SETUP menu))

Press 2ND GRAPH to enter the TABLE screen.

In the table you can enter \begin{align*}x\end{align*} values in the first column, and the second column will display corresponding \begin{align*}y\end{align*} values.

To analyze the behavior of \begin{align*}f(x)\end{align*} near \begin{align*}x=0\end{align*}, enter several values for \begin{align*}x\end{align*} getting progressively closer to 0.

\begin{align*}& x && -1 && -0.5 && -0.25 && -0.1 && -0.01 && -0.001\\ & f(x) && -1.5 && 0.214 && \ 2.019 && 6.281 && 66.428 && 666.443\end{align*}

It looks like as \begin{align*}x\rightarrow 0\end{align*} from the left that \begin{align*}f(x)\end{align*} becomes large. (In calculus you will use more rigorous language to describe this situation). Let's look at \begin{align*}x\end{align*} values larger than zero, that is look at how \begin{align*}f(x)\end{align*} behaves as \begin{align*}x\rightarrow 0\end{align*} from the right.

\begin{align*}& x && 1 && \ \ 0.5 && \ \ \ \ 0.25 && \ \ \ \ 0.1 && \ \ \ \ 0.01 && \ \ \ \ 0.001\\ & f(x) && 0 && -0.9 && -2.523 && -6.734 && -66.873 && -666.887\end{align*}

This shows that as \begin{align*}x\rightarrow 0\end{align*} from the right, \begin{align*}f(x)\rightarrow -\infty\end{align*}.

Likewise, you can use the table to show that as \begin{align*}x\rightarrow 3\end{align*} from the left, \begin{align*}f(x)\rightarrow-\infty\end{align*}, and as \begin{align*}x\rightarrow 3\end{align*} from the right, \begin{align*}f(x)\rightarrow\infty\end{align*}.

While using the table to sketch a graph by hand may be redundant (after all, you have a graphing calculator!), this information helps you interpret the output of the graphing tool, which is an important skill to develop as you learn to use a graphing calculator.

## Exercises

For each function in questions 1-5, describe the behavior of the rational function.

a) What are its asymptotes?

b) What is the end behavior of the function?

c) How does the function behave as \begin{align*}x\end{align*} approaches each vertical asymptote form the left and from the right?

d) Sketch a graph of the function an an appropriate domain

1. \begin{align*}g(x)=\frac{4x^{3}-7x}{3x^{3}-x}\end{align*}
2. \begin{align*}h(x)=\frac{x-2}{x+1}\end{align*}
3. \begin{align*}k(x)=\frac{5}{x^{2}-12}\end{align*}
4. \begin{align*}f(x)=\frac{2x^{3}+2x^{2}-1}{x^{2}-5x+4}\end{align*}
5. \begin{align*}p(x)=\frac{3x^{2}-3x}{2x^{2}+1}\end{align*}

1. Notice that \begin{align*}g(x)=\frac{4x^{3}-7x}{3x^{3}-x}=\frac{x(4x^{2}-7)}{x(3x^{2}-1)}=\frac{4x^{2}-7}{3x^{2}-1},\ x\neq0\end{align*}.
1. Vertical asymptotes at \begin{align*}x=\pm\sqrt{\frac{1}{3}}\end{align*}. Horizontal asymptote \begin{align*}y=\frac{4}{3}\end{align*}. \begin{align*}g(x)\end{align*} is undefined at \begin{align*}x=0\end{align*}, but there is not an asymptote there.
2. As \begin{align*}x\rightarrow\pm\infty, g(x)\rightarrow\frac{4}{3}\end{align*}.
3. As \begin{align*}x\rightarrow-\sqrt{\frac{1}{3}}\end{align*} from the left \begin{align*}g(x)\rightarrow-\infty\end{align*} and as \begin{align*}x\rightarrow-\sqrt{\frac{1}{3}}\end{align*} from the right, \begin{align*}g(x)\rightarrow\infty\end{align*}. As \begin{align*}x\rightarrow\sqrt{\frac{1}{3}}\end{align*} from the left, \begin{align*}g(x)\rightarrow\infty\end{align*} and as \begin{align*}x\rightarrow\sqrt{\frac{1}{3}}\end{align*} from the right, \begin{align*}g(x)\rightarrow-\infty\end{align*}.
1. Vertical asymptote at \begin{align*}x=-1\end{align*}. Horizontal asymptote \begin{align*}y=1\end{align*}.
2. As \begin{align*}x\rightarrow\pm\infty, h(x)\rightarrow 1\end{align*}.
3. As \begin{align*}x\rightarrow -1\end{align*} from the left \begin{align*}h(x)\rightarrow\infty\end{align*} and as \begin{align*}x\rightarrow -1\end{align*} from the right, \begin{align*}h(x)\rightarrow-\infty\end{align*}.
1. Vertical asymptotes at \begin{align*}x=\pm2\sqrt{3}\end{align*}. Horizontal asymptote is \begin{align*}y=0\end{align*}.
2. As \begin{align*}x\rightarrow\pm\infty, k(x)\rightarrow 0\end{align*}.
3. As \begin{align*}x\rightarrow-2\sqrt{3}\end{align*} from the left \begin{align*}k(x)\rightarrow\infty\end{align*} and as \begin{align*}x\rightarrow-2\sqrt{3}\end{align*} from the right, \begin{align*}k(x)\rightarrow-\infty\end{align*}. As \begin{align*}x\rightarrow 2\sqrt{3}\end{align*} from the left \begin{align*}k(x)\rightarrow-\infty\end{align*} and as \begin{align*}x\rightarrow 2\sqrt{3}\end{align*} from the right, \begin{align*}k(x)\rightarrow\infty\end{align*}.
1. Vertical asymptotes at \begin{align*}x=4\end{align*} and \begin{align*}x=1\end{align*}. Oblique asymptote \begin{align*}y=2x+12\end{align*}.
2. As \begin{align*}x\rightarrow-\infty, f(x)\rightarrow-\infty\end{align*}. As \begin{align*}x\rightarrow\infty\end{align*}, \begin{align*}f(x)\rightarrow\infty\end{align*}.
3. As \begin{align*}x\rightarrow 1\end{align*} from the left \begin{align*}f(x)\rightarrow\infty\end{align*} and as \begin{align*}x\rightarrow 1\end{align*} from the right, \begin{align*}f(x)\rightarrow-\infty\end{align*}. As \begin{align*}x\rightarrow 4\end{align*} from the left \begin{align*}f(x)\rightarrow-\infty\end{align*} and as \begin{align*}x\rightarrow 4\end{align*} from the right, \begin{align*}f(x)\rightarrow\infty\end{align*}
1. No vertical asymptotes. horizontal asymptote at \begin{align*}y=\frac{3}{2}\end{align*}.
2. As \begin{align*}x\rightarrow\pm\infty, p(x)\rightarrow\frac{3}{2}\end{align*}.
3. N/A: there are no vertical asymptotes.

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