# Properties of Rational Functions in Precalculus

These are my notes on properties of rational functions in precalculus.

Ratios of integers are called rational numbers. Ratios of polynomial functions

are called rational functions. A rational function is of the form

\(r(x)=\frac{p(x)}{q(x)}\). P and Q are polynomial functions and q is not the

zero polynomial. The domain of a rational function is the set of all real

numbers except those for which the denominator is zero.**Example 1**

Find the domain of a rational function

1. The domain of \(r(x)=\frac{2x^2-4}{x+5}\) is the set of all real numbers

except -5. The domain is \({x|x\not=-5}\).

2. The domain of \(r(x)=\frac{1}{x^2-4}\) is the set of all real numbers except

-2 so the domain is \({x|x\not=-2,x\not=2}\).

3. The domain of \(r(x)=\frac{x^3}{x^2+1}\) is the set of all real numbers.

4. The domain of \(r(x)=\frac{3}{x^2+2}\) is the set of all real numbers.

5. The domain of \(r(x)=\frac{x^2-1}{x-1}\) is the set of all real numbers

except 1 so the domain is \({x|x\not=1}\).

If \(r(x)=\frac{p(x)}{q(x)}\) is a rational function and if p and q have no

common factors, then the rational function is said to be in lowest terms. For a

rational function in lowest terms, the real zeros, if any, of the numerator are

the x-intercepts of the graph will play a major role in the graph. The real

zeros of the denominator also play a major role in the graph. **Example 2**

Graphing \(y=\frac{1}{x^2}\)

Solution:

The domain is the set of all real numbers except 0. The graph has no y-intercept

because x can never equal 0. The graph has no x-intercept because the equation

\(y=0\) has no solution. Therefore, the graph will not cross either of the

coordinate axes. Because:

\(h(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = h(x)\)

It is an even function, so its graph is symmetric with respect to the y-axis. **Example 3**

Using transformations to graph a rational function

Graph the rational function \(r(x) = \frac{1}{(x-2)^2 + 1}\)

Solution:

First, we take note of the fact that the domain of r is the set of all real

numbers except x=2. To graph r, we start with the graph of \(y=\frac{1}{x^2}\).

Use a calculator to graph the function.**Asymptotes**

In the previous graph, notice that as the values of x become more negative as x

approaches negative infinity, the values approach 1. In fact, we can conclude

the following:

1. As x approaches negative infinity, the values of r(x) approach 1

2. As x approaches 2, the values of r(x) approach infinity

3. As x approaches infinity, the values of r(x) approach 1

This behavior of the graph is shown by the vertical line x=2 and the horizontal

line y=1. These lines are called asymptotes of the graph.

A horizontal asymptote, when it occurs, describes the end behavior of the graph

as x approaches infinity or as x approaches negative infinity. The graph of a

function may intersect a horizontal asymptote. A vertical asymptote, when it

occurs, describes the behavior of the graph when x is close to some number. The

graph of a function will never intersect a vertical asymptote.

There is a third possibility. If the value of a rational function approaches a

linear expression, then the line y=ax+b is an oblique asymptote. An oblique

asymptote, when it occurs, describes the end behavior of the graph. The graph of

a function may intersect an oblique asymptote.**Find the Vertical Asymptotes of a Rational Function**

The vertical asymptotes of a rational function, in lowest terms, are located at

the real zeros of the denominator. Suppose that r is a real zero of q, so x-r is

a factor of q. As x approaches r, the values of x-r approach 0, causing the

ratio to become unbounded, or head towards infinity. Based on the definition, we

conclude that the line x=r is a vertical asymptote. **Example 4**

Finding vertical asymptotes

Find the vertical asymptotes of the graph of each rational function

1. \(r(x)=\frac{x}{x^2-4}\)

2. \(r(x)=\frac{x+3}{x-1}\)

3. \(r(x)=\frac{x^2}{x^2+1}\)

4. \(r(x)=\frac{x^2-9}{x^2+4x-21}\)

Solution:

1. The function is in lowest terms and the zeros of the denominator are -2 and

The lines x=-2 and x=2 are the vertical asymptotes of the graph.

2. The function is in lowest terms and the only zero of the denominator is 1.

The line x=1 is the vertical asymptote of the graph.

3. The function is in lowest terms and the denominator has no real zeros,

because the equation has no real solutions. The graph has no vertical asymptotes

4. Factor to get it to lowest terms.

\(r(x)=\frac{x^2-9}{x^2+4x-21} = \frac{(x+3)(x-3)}{(x+7)(x-3)} =

\frac{x+3}{x+7}\) The only zero of the denominator in lowest terms is -7. The

line x=-7 is the only vertical asymptote of the graph.

Rational functions can have no vertical asymptotes, one vertical asymptote, or

multiple vertical asymptotes. However, the graph of a rational function will

never intersect any of its vertical asymptotes. **Find the Horizontal or Oblique Asymptotes of a Rational Function**

The procedure for finding horizontal and oblique asymptotes is somewhat more

involved. To find such asymptotes, we need to know how the values of a function

behave as x a[[roaches negative infinity or as x approaches infinity.

If a rational function is proper, that is, if the degree of the numerator is

less than the degree of the denominator, then as x approaches negative infinity

or as x approaches infinity the value of the function approaches 0.

Consequently, the line y=0(x-axis) is a horizontal asymptote of the graph. If a

rational function is proper, the line y=0 is a horizontal asymptote of its graph.**Example 5**

Finding Horizontal Asymptotes

Find the horizontal asymptotes, if any, of the graph of:

\(f(x)=\frac{x-12}{4x^2+x+1}\)

Solution:

Since the degree of the numerator is 1, it is less than the degree of the

denominator which is 2. So this is a proper function. That means that the line

y=0 is a horizontal asymptote of the graph.

To see why y=0 is a horizontal asymptote of the function, we need to investigate

the behavior as x approaches negative infinity and as x approaches infinity.

When |x| is unbounded, the numerator, which is x-12, can be approximated by the

power function y=x, while the denominator, which is \(4x^2+x+1\), can be

approximated by the power function \(y=4x^2\). We find:

\(f(x)=\frac{x-12}{4x^2+x+1} = \frac{x}{4x^2} = \frac{1}{4x}\) as it approaches

0. This shows that the line y=0 is a horizontal asymptote of the graph.

If a rational function is improper, or if the degree of the numerator is greater

than the degree of the denominator, we must use long division to write the

rational function as the sum of a polynomial plus a proper rational function.

1. If f(x)=b, a constant, the line y=b is a horizontal asymptote of the graph

2. If f(x)=ax+b, the line y=ax+b is an oblique asymptote of the graph

3. In all other cases, the graph of r approaches the graph of f, and there are

no horizontal or oblique asymptotes.**Example 6**

Finding Horizontal or Oblique Asymptotes

Find the horizontal or oblique asymptote of the graph of:

\(f(x)=\frac{3x^4-x^2}{x^3-x^2+1}\)

Solution:

Since the degree of the numerator, 4, is larger than the degree of the

denominator, 3, the rational function is improper. To find any horizontal or or

oblique asymptotes, use long division to get 3x+3, which is the oblique

asymptote.**Example 7**

Finding Horizontal or Oblique Asymptotes

Find asymptotes of:

\(f(x)=\frac{8x^2-x+2}{4x^2-1}\)

Solution:

Since the degree of the numerator, 2, equals the degree of the denominator,2,

the rational function is improper. To find any horizontal or oblique asymptotes,

use long division. We get y=2 as a horizontal asymptote of the graph.

We note that the quotient 2 obtained by long division is the quotient of the

leading coefficients of the numerator polynomial and the denominator polynomial.

This means that we can avoid the long division process for rational functions

whose numerator and denominator are of the same degree and conclude that the

quotient of the leading coefficients will give us the horizontal asymptote.**Example 8**

Finding the Horizontal or Oblique Asymptotes

\(f(x)=\frac{2x^5-x+2}{x^3-1}\)

Solution:

Since the degree of the denominator, 5, is larger than the degree of the

denominator, 3, the rational function is improper. To find any horizontal or

oblique asymptotes, use long division. \(2x^2-1\). Since this is not a linear

function, the function has no horizontal or oblique asymptotes.

1. If the degree of the numerator is less than the degree of the denominator, it

is a proper rational function, and the graph will have the horizontal asymptote

y=0.

2. If the degree of the numerator is greater than or equal to the degree of the

denominator, then it is an improper function. Use long division.

A. if the degree of the numerator equals the degree of the denominator, the

quotient obtained will be the number\(\frac{a_{n}}{b_{m}}\) and the line

\(y=\frac{a_{n}}{b_{m}}\) is a horizontal asymptote.

B. If the degree of the numerator is more than the degree of the

denominator, the quotient obtained is of the form ax+b and the line y=ax+b

is an oblique asymptote.

C. If the degree of the numerator is two or more than the degree of the

denominator, the quotient obtained is a polynomial of degree 2 or higher,

and the function has neither a horizontal nor an oblique asymptote. In this

case, for |x| unbounded, the graph will behave like the graph of the

quotient.