# Rational Functions

### Reciprocal Function

The reciprocal function $f(x)=\frac{1}{x}$ is the parent function of a family of rational functions.
A rational function is a function whose rule can be expressed as a fraction of two polynomials:
$f(x)=\frac{p(x)}{q(x)}$
The most basic rational function is the reciprocal function:
$f(x)=\frac{1}{x}$
The reciprocal function has both a vertical and a horizontal asymptote. An asymptote is a line that a graph approaches as one of the variables approaches infinity or negative infinity.
• As $x$ approaches positive or negative infinity, $y$ approaches zero. The graph gets closer to the $x$-axis, but does not touch it. The $x$-axis is a horizontal asymptote of the function.
• As $x$ approaches zero from the left side, $y$ approaches negative infinity. As $x$ approaches zero from the right side, $y$ approaches positive infinity. The graph gets closer to the $y$-axis but does not touch it. The $y$-axis is a vertical asymptote of the function.
• The graph is symmetric with respect to the origin. It passes through the points $(1,1)$ and $(-1,-1)$.

### Indirect Variation

Inverse, joint, and combined variation functions show relationships between two or more variables.

Variation describes a relationship between two or more variables and a nonzero constant $k$ known as the constant of proportionality. There are several types of variation:

• Direct variation relationships can be written as $y = kx$ for variables $x$ and $y$ and constant $k$. Since one variable changes in proportion to the other variable, the value of the variables $x$ and $y$ either increase or decrease together.
• Inverse variation is a relationship for variables $x$ and $y$ and constant $k$ that can be written as $y = \frac{k}{x}$ . Since the variables vary inversely with each other, the value of one variable increases while the value of the other variable decreases.
• An example of a joint variation relationship is $z = kxy$ for variables $x$, $y$, and $z$ and constant $k$. Joint variation is similar to direct variation, but involves more than two variables.
• An example of a combined variation relationship is $z = \frac{kx}{y}$ for variables $x$, $y$, and $z$ and constant $k$. Combined variation is a combination of direct or joint variation and inverse variation.

The graph of an indirect variation function is a stretch or compression of the parent rational function. The $x$-axis is a horizontal asymptote, the $y$-axis is a vertical asymptote, and the graph passes through the point $(1,k)$. Variation equations often represent real-world relationships, in which $x$-values are usually positive.

Step-By-Step Example
Understanding Inverse Variation Relationships
Brittany gets $30 to mow a lawn. Since the amount she earns is fixed, Brittany's hourly rate decreases as the time it takes her to mow the lawn increases. In other words, her hourly rate is inversely proportional to the time it takes her to mow. Graph the inverse variation relationship, interpret the meaning of the asymptotes, and find two points on the graph. Step 1 Write the variation equation. Let $x$ represent the time it takes to mow a lawn and $y$ represent Brittany's hourly rate. They are inversely proportional, which means: $y = \frac{k}{x}$ If she mows a lawn in $x=1$ hour, she gets$30. So, her hourly rate is $y=30$ per hour. This means:
$30=\frac{k}{1}$
So, $k=30$. The inverse variation relationship can be written as:
$y=\frac{30}{x}$
Step 2
Graph the relationship. The graph is a vertical stretch of the reciprocal function by a factor of 30. The time to mow a lawn cannot be negative, so the graph is restricted to the first quadrant.
Step 3

Interpret the meaning of the asymptotes.

The vertical asymptote shows that Brittany's hourly rate increases as her time approaches zero. It is impossible to mow the lawn in zero hours, so the hourly rate is not defined for $x=0$.

The horizontal asymptote shows that her hourly rate decreases as her time approaches infinity. The longer she takes to mow the lawn, the less she makes per hour. Her hourly rate cannot be zero. So, the graph never touches the $x$-axis.

Step 4

Identify two points on the graph.

The point $(1,30)$ represents the hourly rate of $30 if she mows the lawn in one hour. If it takes her two hours to mow the lawn, then her hourly rate is$15 per hour:
\begin{aligned}y&=\frac{30}{2}\\&=15\end{aligned}
So, $(2,15)$ is another point on the graph.
Solution
The shape of the graph shows that her hourly rate decreases as the time it takes her to mow the lawn increases. If she mows the lawn in 1 hour, she makes $30 per hour, but if it takes her two hours, she makes only$15 per hour. As the time it takes increases, her rate approaches \$0 per hour.
Step-By-Step Example
Understanding Combined Variation Relationships

If the volume of a pyramid increases, but the area of its base stays the same, the height of the pyramid increases. If the area of the base of a pyramid increases, but the volume of the pyramid stays the same, the height of the pyramid decreases. In other words, the height of a pyramid varies directly as the volume and inversely as the area of the base.

A pyramid with a volume of 100 cubic centimeters (cm3) and a base area of 50 square centimeters (cm2) has a height of 6 centimeters (cm). What is the height of the pyramid if the volume and base area are both doubled?

Step 1

Write the variation equation. Let $h$ represent the height of the pyramid, $V$ represent the volume, and $B$ represent the area of the base. Let $k$ be the constant of variation.

$V$ is in the numerator because the height varies directly as the volume.

$B$ is in the denominator because the height varies inversely as the area of the base.

So, the variation equation is:
$h=\frac{kV}{B}$
Step 2

Use the given values to calculate $k$.

A pyramid with a volume of 100 cm3 and a base area of 50 cm2 has a height of 6 cm.
\begin{aligned}h&=\frac{kV}{B}\\6&=\frac{k(100)}{50}\\6&=2k\\k&=3\end{aligned}
The combined variation equation is:
$h=\frac{3V}{B}$
Solution
Calculate the height if the base area is doubled. Doubling the volume results in 200 cm3. Doubling the base area results in 100 cm2. So, the height is 6 cm:
\begin{aligned}h&=\frac{3(200)}{100}\\&=3(2)\\&=6\end{aligned}
It is the same as the original height. Therefore, if the volume and base area of the pyramid are both doubled, the height stays the same.

### Transformations of the Reciprocal Function

Some rational functions can be graphed as transformations of the parent reciprocal function.

Transformations can be used to make changes to the graph of a parent function. These include translations (shifts), stretches or compressions, and reflections.

For any function $f(x)$, the function can be translated vertically $k$ units and translated horizontally $h$ units. Note that horizontal translations are opposite in direction from the sign: Subtracting $h$ from the input translates the graph in the positive direction, and adding $h$ to the input translates it in the negative direction.

A function $f(x)$ can also be stretched or compressed by a factor of $a$ and reflected across the $x$-axis.

### Translations of the Reciprocal Function

Vertical Translations Horizontal Translations
For the graph of $f(x)=\frac{1}{x}$ and $k>0$:
• The graph of $f(x)+k=\frac{1}{x}+k$ is translated up by $k$ units.
• The graph of $f(x)-k=\frac{1}{x}-k$ is translated down by $k$ units.
For the graph of $f(x)=\frac{1}{x}$ and $h>0$:
• The graph of $f(x-h)=\frac{1}{x-h}$ is translated right by $h$ units.
• The graph of $f(x+h)=\frac{1}{x+h}$ is translated left by $h$ units.

### Stretches, Compressions, and Reflections of the Reciprocal Function

Stretches and Compressions Reflections
For the graph of $f(x)=\frac{1}{x}$ and $a\gt 0$:
• The graph of $af(x)=\frac{a}{x}$ is a vertical stretch of the graph of $f(x)$ by a factor of $a$ if $a\gt 1$.
• The graph of $af(x)=\frac{a}{x}$ is a vertical compression of the graph of $f(x)$ by a factor of $a$ if $0\lt a\lt 1$.
For the graph of $f(x)$:
• The graph of $-f(x)=-\frac{1}{x}$ is a reflection of the graph of $f(x)$ across the $x$-axis.

Step-By-Step Example
Graphing Transformations of the Reciprocal Function
Graph the function:
$f(x) = \frac{-3}{x+1} + 2$
Step 1
Identify the values of $a$, $h$, and $k$.
\begin{aligned}f(x) &= \frac{a}{x+h}+k\\&=\frac{-3}{x+1} + 2\end{aligned}
The value of $a$ is –3. The value of $h$ is 1. The value of $k$ is 2.
Step 2

Perform the stretch and the reflection.

• Since $|a|=3$ and $|a|>1$, the graph is vertically stretched by a factor of 3 compared to the graph of the parent function.
• Since $a<0$, the graph is reflected across the $x$-axis.
The parent function is:
$f(x)=\frac{1}{x}$
Its graph is vertically stretched by a factor of 3 and reflected across the $x$-axis.
Solution

Perform the translation.

• Since $h=1$ and this value is added, the graph is translated 1 unit to the left.
• Since $k=2$ and this value is added, the graph is translated 2 units up.
The resulting graph is translated 1 unit to the left and 2 units up.
The graph is the result after the stretch, reflection, and translation.

### Holes and Asymptotes

Holes or vertical asymptotes of a rational function occur when its denominator equals zero. Horizontal asymptotes of a rational function are determined by the end behavior.
When the denominator of a rational function is zero, the function is not defined. The graph will have either a vertical asymptote or a hole. A hole, or removable discontinuity, is a missing value in a graph such that the graph approaches the same value on both sides. Whether there is a hole or a vertical asymptote depends on the multiplicity of the zero in the numerator and denominator. If $x-c$ is a factor of a polynomial function, the multiplicity of the zero $x=c$ is the number of times that the factor appears, or the exponent of the factor. For example:
$f(x)=(x-2)^3(x+5)$
The polynomial function has a zero of multiplicity 3 at $x=2$ and a zero of multiplicity 1 at $x=-5$.

The end behavior of the graph determines whether there is a horizontal asymptote. If the function approaches a certain value as $x$ approaches positive or negative infinity, then there is a horizontal asymptote at that value. Horizontal asymptotes are determined by comparing the degree of the numerator and denominator. It is possible for a graph to cross a horizontal asymptote, but never a vertical one.

### Determining Holes and Asymptotes

Holes and Vertical Asymptotes Horizontal Asymptotes
Let $q(x)$ have a zero at $x=a$ of multiplicity $k$ in the rational function:
$f(x)=\frac{p(x)}{q(x)}$
There is a vertical asymptote at $x=a$:
• if $p(a)\neq0$ or
• if $a$ is a zero of $p(x)$ with a multiplicity less than $k$.
There is a hole at $x=a$ if $a$ is a zero of $p(x)$ with a multiplicity greater than or equal to $k$.
Let $n$ be the degree of $p(x)$ and $m$ be the degree of $q(x)$.
• If $n\lt m$, there is a horizontal asymptote at $y=0$.
• If $n\gt m$, there is no horizontal asymptote.
• If $m=n$, there is a horizontal asymptote at $y=a$, where $a$ is the leading coefficient of $p(x)$ divided by the leading coefficient of $q(x)$.

To understand when the graph of a rational function will have holes or vertical asymptotes, it is helpful to look at the algebraic rule of the function when the numerator and denominator are in factored form. For example, consider the function:
$f(x)=\frac{4(x+2)(x+4)}{(x+2)(x-4)}$
Values where the denominator of the function is equal to zero can be either holes or vertical asymptotes. The denominator of $f(x)$ equals zero at $x=-2$ and $x=4$.
To determine whether the values are holes or vertical asymptotes, simplify the fraction. Divide the numerator and denominator by the common factor $(x+2)$ to get:
$f(x)=\frac{4(x+4)}{(x-4)}$
• The value $x=-2$ is in the domain of the simplified function, but not in the domain of the original function. Therefore, there is a hole in the graph at $x=-2$. Since this hole is not in the simplified form of the function, it is called a removable discontinuity.
• The value $x=4$ is not in the domain of either the original function or the simplified rule. As $x$ approaches 4, the function approaches positive or negative infinity. Therefore, $f(x)$ has a vertical asymptote at $x=4$.
To find horizontal asymptotes, write the numerator and denominator in standard form.
$f(x)=\frac{4x^2+24x+32}{x^2-2x-8}$
• The end behavior of each polynomial function is determined by the term with the highest degree. So, look at the rational function $g$ made up of only the first term of the numerator and denominator:
$g(x)=\frac{4x^2}{x^2}$
• This can be simplified to $g(x)=4$, or $y=4$.
• The graph of $y=4$ is a horizontal line. Therefore, $f(x)$ has a horizontal asymptote at $y=4$.

### Graphing Rational Functions

Rational functions can be graphed by identifying their vertical asymptotes, horizontal asymptotes, and holes.

To graph a rational function by hand:

1. Analyze the function to determine the location of any holes, vertical asymptotes, horizontal asymptotes, and intercepts.

2. Sketch the asymptotes, and plot any holes or intercepts.

3. Draw curves to complete the graph.

Step-By-Step Example
Graphing a Rational Function with a Horizontal Asymptote
Graph the function:
$f(x)=\frac{x^2-25}{x^2-x-12}$
Step 1
Factor the numerator and denominator.
$f(x)=\frac{(x+5)(x-5)}{(x+3)(x-4)}$
Step 2

Identify any $x$-intercepts.

The function has an $x$-intercept for each real value of $x$ for which the numerator is equal to zero. The numerator is zero at $x=-5$ and $x=5$. These are the $x$-intercepts.

Step 3
Identify the $y$-intercept by evaluating $f(0)$.
\begin{aligned}f(0)&=\frac{(0+5)(0-5)}{(0+3)(0-4)}\\&=\frac{(5)(-5)}{(3)(-4)}\\&=\frac{-25}{-12}\\&=\frac{25}{12}\end{aligned}
The $y$-intercept is $f(0)=\frac{25}{12}$, which is about 2.1.
Step 4

Identify any vertical asymptotes.

The denominator of the function is zero at $x=-3$ and $x=4$. The numerator is not zero at these values. So, there are two vertical asymptotes, one at $x=-3$ and one at $x=4$.

Step 5

Identify any holes.

The graph of the function has no holes because the numerator and denominator have no zeros in common. The zeros of the numerator are –5 and 5, and the zeros of the denominator are –3 and 4.

Step 6

Identify the horizontal asymptote, if any.

Look at the degrees of the numerator and denominator. They have the same degree. So, divide the leading coefficient of the numerator by the leading coefficient of the denominator. Both have a leading coefficient of 1. So, the horizontal asymptote is at:
$y=\frac{1}{1}=1$
Step 7
Sketch the asymptotes, and plot the intercepts. The arrows show the direction of the graph as it approaches the asymptotes. The point above the horizontal axis is part of a U-like shape that points up because the curve must approach all three asymptotes. The points below the horizontal axis each form a curve that approaches one horizontal and one vertical asymptote. It may also be helpful to plot some additional points to fill in the graph.
Solution
Draw curves to complete the graph.
Step-By-Step Example
Graphing a Rational Function with a Hole
Graph the function:
$f(x)=\frac{x^3-2x^2-15x}{x^2-x-12}$
Step 1
Factor the numerator and denominator.
$f(x)=\frac{x(x+3)(x-5)}{(x+3)(x-4)}$
Step 2

Identify any $x$-intercepts.

The numerator is zero at $x=0$, $x=-3$, and $x=5$. The denominator is also zero at $x=-3$. So that value of $x$ cannot be an $x$-intercept. The $x$-intercepts are at zero and 5.

Step 3
Identify the $y$-intercept by evaluating $f(0)$.
\begin{aligned}f(0)&=\frac{0(0+3)(0-5)}{(0+3)(0-4)}\\&=\frac{0(3)(-5)}{(3)(-4)}\\&=\frac{0}{-12}\\&=0\end{aligned}
The $y$-intercept is $f(0)=0$.
Step 4

Identify any vertical asymptotes.

The denominator is zero at $x=-3$ and $x=4$. The numerator is zero at $x=-3$, but not at $x=4$. So, the only vertical asymptote is at $x=4$.

Step 5

Identify any holes.

At $x=-3$, the numerator and denominator both have a zero of multiplicity 1. So, there is a hole at this value of $x$. To see where to plot the hole, simplify the rule for $f$ by dividing out $(x+3)$ in the numerator and denominator to make a new function:
$g(x)=\frac{x(x-5)}{x-4}$
The functions $f$ and $g$ are equivalent for all values of $x$ except –3. At $x=-3$ $g(x)$ is $\frac{24}{7}$:
\begin{aligned}g(x)&=\frac{-3(-3-5)}{-3-4}\\&=\frac{24}{-7}\end{aligned}
$\frac{24}{7}$ is about –3.4.

The graph has a hole at approximately $(-3,-3.4)$.

Step 6

Identify the horizontal asymptote, if any.

Look at the degrees of the numerator and denominator. The numerator has a greater degree. So, there is no horizontal asymptote.

Step 7
Sketch the asymptote, and plot the hole and intercepts. The arrows show the direction of each part of the graph. To the left of the asymptote, the point and the hole will be connected and that the curve will approach the asymptote as $x$ approaches 4. In this case, finding a few more points on the graph will help to fill in the curves, particularly to the right of the asymptote.
Solution
Draw curves to complete the graph. The graph approaches the hole on either side, but the actual value at that point is missing.
Because of the way that graphing calculators and computer programs work, the graphs of rational functions may not show holes or asymptotes correctly. Sometimes, using a special window, such as the Decimal window, or zooming in on the graph can help resolve these issues. However, it may be necessary to use algebraic methods to determine where the holes and asymptotes are to be sure that the graph is shown correctly.
Step-By-Step Example
Graphing a Rational Function Using Technology
Graph the function:
$f(x)=\frac{(x^2+1)(x^2-4)}{x^2(x-3)^2(x-2)}$
Step 1
Factor the numerator and denominator of the function rule.
$f(x)=\frac{(x^2+1)(x+2)(x-2)}{x^2(x-3)^2(x-2)}$
Step 2

Identify any vertical asymptotes.

The denominator is zero at $x=0$, $x=3$, and $x=2$. The numerator is zero at $x=2$, but not at $x=0$ or $x=3$. So, there are vertical asymptotes at $x=0$ and $x=3$.

Step 3

Identify any holes.

At $x=2$, look at the multiplicity of the zeros (the number of times the related factor appears in the polynomial) of the numerator and denominator. Both the numerator and denominator have a zero of multiplicity 1. The multiplicities are equal, so there is a hole at this value of $x$.

Solution
Graph the function using technology. If the vertical asymptotes and the hole are not shown properly, make corrections to the graph as needed.
The graph shows the vertical asymptotes at $x=0$ and $x=3$ and a hole at $(2,5)$.