## Rational Functions |

**Contents**: This page corresponds to **§ 3.5 (p. 289)** and **§
3.6 (p. 299) **of the text.

Suggested Problems from Text:

p. 295 #1-12, 15-17, 21, 24, 37

p 304 #35, 37, 49, 52

## Introduction

## Asymptotes

## Application

A **rational function** is one that can be written as a polynomial divided by a polynomial. Since polynomials
are defined everywhere, the domain of a rational function is the set of all numbers *except* the zeros of
the denominator.

__Example 1__.

f(x) = x / (x - 3). The denominator has only one zero, x = 3. So the domain of f is the set of all numbers other than 3.

Domain of f: (-inf, 3) union (3, inf).

The graph of f is pictured below.

Look again at the graph of f(x) = x / (x - 3) shown above. Since 3 is not in the domain of f, there is no point on the graph with first coordinate 3, so there has to be a break in the graph. In fact, we can see that the function values become unbounded (go to infinity or negative infinity) as x approaches 3. The following tables of function values illustrate this behavior.

x |
4 |
3.5 |
3.1 |
3.01 |
3.001 |

f(x) = x / (x - 3) |
4 |
7 |
31 |
301 |
3001 |

We say that f (x) approaches infinity as x approaches 3 *from the right*, or

f(x) -> inf as x -> 3^{+}.

The phrase* from the right* is important. It means that we are using values for x that are *larger than*
3 and getting close to 3. The next table shows the behavior of f as x approaches 3 from the left.

x |
2 |
2.5 |
2.9 |
2.99 |
2.999 |

f(x) = x / (x - 3) |
-2 |
-5 |
-29 |
-299 |
-2999 |

We say that f(x) approaches negative infinity as x approaches 3 from the left. Here we will use a superscript - to indicate approaching from the left.

f(x) -> - inf as x -> 3^{-}.

In the last section we pointed out that polynomial graphs either rise to the right or fall to the right. The graph of the rational function f does neither of these. It appears from the picture that the points on the graph of f approach the horizontal line y = 1 as x goes right and as x goes left. The tables below provide further evidence that this is the case.

x |
5 |
10 |
100 |
1000 |
10000 |

f(x) = x / (x - 3) |
2.5 |
1.4285715 |
1.0309278 |
1.0030091 |
1.0003 |

x |
-5 |
-10 |
-100 |
-1000 |
-10000 |

f(x) = x / (x - 3) |
0.625 |
0.7692308 |
0.9708738 |
0.997009 |
0.99970007 |

We write

f(x) -> 1 as x -> inf and f(x) -> 1 as x -> - inf,

or

f(x) -> 1 as x -> ± inf.

The vertical line x = 3 and the horizontal line y = 1 are examples of **asymptotes**. An asymptote is a line
that a graph *gets close to* as x goes to plus or minus infinity or a particular number.

**Definition of Horizontal and Vertical Asymptotes**

The line x = a is a

vertical asymptoteof the graph of f if f(x) -> ± infinity as x -> a from the left or the right.The line y = b is a horizontal asymptote of the graph of f if f(x) -> b as x -> ± infinity.

In general, the graph of a rational function will have a vertical asymptote at a zero of the denominator. The exception to this rule is the case where the numerator and denominator share a zero.

__Example 2__.

g(x) = (x

^{2}- 4) / (x - 2). g is a rational function and g is not defined at 2 because 2 is a zero of the denominator. However, 2 is also a zero of the numerator, and we can simplify the quotient.(x

^{2}- 4) / (x - 2) = (x - 2)(x + 2)/(x - 2) = x + 2.However, the function g is

not equal tothe function h(x) = x + 2, because h is defined at 2 while g is not! (This point is usually mentioned in a College Algebra class and then promptly forgotten.)The graph of g does not have a vertical asymptote through 2. The graph of g is the line y = x + 2 with a hole where the point (2, 4) would be.

You can usually see from a graphing utility whether or not the graph of a rational function has a horizontal asymptote. However it is generally not clear from the picture exactly what number the asymptote goes through on the y-axis. Generating a table as we did above gives you a much better idea of where the horizontal asymptote is.

__Example 3__.

f(x) = x

^{2}/ (x - 3). The graph of f is pictured below.We see from the graph that f(x) -> -inf as x -> 3

^{-}, and f(x) -> inf as x -> 3^{+}. It is also clear from the picture that the graphhas no horizontal asymptote. The graph rises to the right and falls to the left.The dotted line is a

slant asymptoteof the graph of f. Slant asymptotes are discussed on page 302 of the text.

It is possible to tell whether or not a rational function has a horizontal asymptote, and if so, exactly where it is, by analyzing the leading terms of the numerator and denominator. This procedure is described in detail on page 291 of the text.

It is certainly possible to be fooled by a graph. Consider the graph pictured below.

This appears to be a graph with the y-axis as a vertical asymptote. In fact this is the graph of f(x) = 5x /
(x^{2} + 0.01), as rendered by the Java Grapher. The denominator has no (real) zeros, so the graph has
no vertical asymptotes.

Use a graphing utility to graph f(x) = 3x

^{2}/ (x^{2}- 16). Find all asymptotes. Answer

__Example 4__.

We are going to enclose a corral adjacent to a river as in the diagram below. No fence is needed on the river side. The enclosed area needs to be 800 square yards. Find the dimensions x and y that require the least amount of fence.

Since we are given that the area must be 800 sq yds, and the area of a rectangle is the product of the two dimensions, this gives us an equation that x and y must satisfy.

x y = 800.

Solving for y yields

y = 800 / x.

Let F stand for the length of fence used, in yards. Since there is one side of length x and two sides of length y, we have

F = x + 2y.

Substituting for y yields

F = x + 2(800 / x).

F = x + 1600 / x.

F is a rational function. Its graph has the y-axis (x = 0) as a vertical asymptote, because F is not defined at 0. The graph of F has no horizontal asymptote, but the line y = x is a slant asymptote.

F = x + 1600 / xNote that since the variable x in this problems stands for a length, we are only interested in values x > 0, so we may focus on quadrant I of the graph of F.

We need to find the first coordinate of the relative minimum point in quadrant I, for that is the length x corresponding to the smallest values of F, and our goal is to use the smallest possible amount of fence.

One of the graded assignments for the course will be to complete this problem by using a graphing utility to approximate the coordinates of the relative minimum point and find the x and y dimensions of the corral that require the least amount of fence.