rational function problem solving examples

Rational Function Problems

These lessons are compiled to help PreCalculus students learn about rational function problems and applications.

Related Pages Simplifying Rational Expressions Graphing Rational Functions PreCalculus Lessons

The following figure shows how to solve rational equations . Scroll down the page for examples and solutions on how to solve rational function problems and applications.

Rational Function Problems - Work And Tank

The video explains application problems that use rational equations. Part 1 of 2.

Martin can pour a concrete walkway in 6 hours working alone. Victor has more experience and can pour the same walkway in 4 hours working alone. How long will it take both people to pour the concrete walkway working together?
An inlet pipe can fill a water tank in 12 hours. An outlet pipe can drain the tank in 20 hours. If both pipes are mistakenly left open, how long will it take to fill the tank?

Rational Function Applications - Work And Rate

The video explains application problems that use rational equations. Part 2 of 2.

One person can complete a task 8 hours sooner than another person. Working together, both people can perform the task in 3 hours. How many hours does it take each person to complete the task working alone?
The speed of a passenger train is 12 mph faster than the speed of the freight train. The passenger train travels 330 miles in the same time it takes the freight train to travel 270 miles. Find the speed of each train.

Rational Functions Word Problems - Work, Tank And Pipe

Here are a few examples of work problems that are solved with rational equations.

Sam can paint a house in 5 hours. Gary can do it in 4 hours. How long will it take the two working together?
Joy can file 100 claims in 5 hours. Stephen can file 100 claims in 8 hours. If they work together, how long will it take to file 100 claims?
A water tank is emptied through two drains in 50 minutes. If only the larger drain is used, the tank will empty in 85 minutes. How long would it take to empty if only the smaller drain is used?
One computer can run a sorting algorithm in 24 minutes. If a second computer is used together with the first, it takes 13 minutes. How long would it take the second computer alone?
Two pipes are filling a tank. One pipe fills three times as fast as the other. With both pipes working, the tank fills in 84 minutes. How long would each pipe take working alone?

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5.6 Rational Functions

Learning objectives.

In this section, you will:

Use arrow notation.
Solve applied problems involving rational functions.
Find the domains of rational functions.
Identify vertical asymptotes.
Identify horizontal asymptotes.
Graph rational functions.

Suppose we know that the cost of making a product is dependent on the number of items, x , x , produced. This is given by the equation C ( x ) = 15,000 x − 0.1 x 2 + 1000. C ( x ) = 15,000 x − 0.1 x 2 + 1000. If we want to know the average cost for producing x x items, we would divide the cost function by the number of items, x . x .

The average cost function, which yields the average cost per item for x x items produced, is

Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.

In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.

Using Arrow Notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure 1 , and notice some of their features.

Several things are apparent if we examine the graph of f ( x ) = 1 x . f ( x ) = 1 x .

On the left branch of the graph, the curve approaches the x -axis ( y = 0 ) as x → – ∞ . ( y = 0 ) as x → – ∞ .
As the graph approaches x = 0 x = 0 from the left, the curve drops, but as we approach zero from the right, the curve rises.
Finally, on the right branch of the graph, the curves approaches the x- axis ( y = 0 ) as x → ∞ . ( y = 0 ) as x → ∞ .

To summarize, we use arrow notation to show that x x or f ( x ) f ( x ) is approaching a particular value. See Table 1 .

Symbol	Meaning
	approaches from the left ( but close to )
	approaches from the right ( but close to )
	approaches infinity ( increases without bound)
	approaches negative infinity ( decreases without bound)
	the output approaches infinity (the output increases without bound)
	the output approaches negative infinity (the output decreases without bound)
	the output approaches

Local Behavior of f ( x ) = 1 x f ( x ) = 1 x

Let’s begin by looking at the reciprocal function, f ( x ) = 1 x . f ( x ) = 1 x . We cannot divide by zero, which means the function is undefined at x = 0 ; x = 0 ; so zero is not in the domain . As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table 2 .

	–0.1	–0.01	–0.001	–0.0001
	–10	–100	–1000	–10,000

We write in arrow notation

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table 3 .

	0.1	0.01	0.001	0.0001
	10	100	1000	10,000

See Figure 2 .

This behavior creates a vertical asymptote , which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 x = 0 as the input becomes close to zero. See Figure 3 .

Vertical Asymptote

A vertical asymptote of a graph is a vertical line x = a x = a where the graph tends toward positive or negative infinity as the input approaches a a from either the left or the right. We write

End Behavior of f ( x ) = 1 x f ( x ) = 1 x

As the values of x x approach infinity, the function values approach 0. As the values of x x approach negative infinity, the function values approach 0. See Figure 4 . Symbolically, using arrow notation

As x → ∞ , f ( x ) → 0 , and as x → − ∞ , f ( x ) → 0. As x → ∞ , f ( x ) → 0 , and as x → − ∞ , f ( x ) → 0.

Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote , a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line y = 0. y = 0. See Figure 5 .

Horizontal Asymptote

A horizontal asymptote of a graph is a horizontal line y = b y = b where the graph approaches the line as the inputs increase or decrease without bound. We write

Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 6 .

Notice that the graph is showing a vertical asymptote at x = 2 , x = 2 , which tells us that the function is undefined at x = 2. x = 2.

And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at y = 4. y = 4. As the inputs increase without bound, the graph levels off at 4.

Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.

Using Transformations to Graph a Rational Function

Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

Shifting the graph left 2 and up 3 would result in the function

or equivalently, by giving the terms a common denominator,

The graph of the shifted function is displayed in Figure 7 .

Notice that this function is undefined at x = −2 , x = −2 , and the graph also is showing a vertical asymptote at x = −2. x = −2.

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y = 3. y = 3.

Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.

Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

Solving Applied Problems Involving Rational Functions

In Example 2 , we shifted a toolkit function in a way that resulted in the function f ( x ) = 3 x + 7 x + 2 . f ( x ) = 3 x + 7 x + 2 . This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions P ( x ) and Q ( x ) . P ( x ) and Q ( x ) .

Solving an Applied Problem Involving a Rational Function

After running out of pre-packaged supplies, a nurse in a refugee camp is preparing an intravenous sugar solution for patients in the camp hospital. A large mixing tank currently contains 100 gallons of distilled water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of distilled water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the ratio of sugar to water, in pounds per gallon in the tank after 12 minutes. Is that a greater ratio of sugar to water, in pounds per gallon than at the beginning?

Let t t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

The ratio of sugar to water, in pounds per gallon, C C , will be the ratio of pounds of sugar to gallons of water

The ratio of sugar to water, in pounds per gallon after 12 minutes is given by evaluating C ( t ) C ( t ) at t = 12. t = 12.

This means the ratio of sugar to water, in pounds per gallon is 17 pounds of sugar to 220 gallons of water.

At the beginning, the ratio of sugar to water, in pounds per gallon is

Since 17 220 ≈ 0.08 > 1 20 = 0.05 , 17 220 ≈ 0.08 > 1 20 = 0.05 , the ratio of sugar to water, in pounds per gallon is greater after 12 minutes than at the beginning.

There are 1,200 first-year and 1,500 second-year students at a rally at noon. After 12 p.m., 20 first-year students arrive at the rally every five minutes while 15 second-year students leave the rally. Find the ratio of first-year to second-year students at 1 p.m.

Finding the Domains of Rational Functions

A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero.

Domain of a Rational Function

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

Given a rational function, find the domain.

Set the denominator equal to zero.
Solve to find the x -values that cause the denominator to equal zero.
The domain is all real numbers except those found in Step 2.

Finding the Domain of a Rational Function

Find the domain of f ( x ) = x + 3 x 2 − 9 . f ( x ) = x + 3 x 2 − 9 .

Begin by setting the denominator equal to zero and solving.

The denominator is equal to zero when x = ± 3. x = ± 3. The domain of the function is all real numbers except x = ± 3. x = ± 3.

A graph of this function, as shown in Figure 8 , confirms that the function is not defined when x = ± 3. x = ± 3.

There is a vertical asymptote at x = 3 x = 3 and a hole in the graph at x = −3. x = −3. We will discuss these types of holes in greater detail later in this section.

Find the domain of f ( x ) = 4 x 5 ( x − 1 ) ( x − 5 ) . f ( x ) = 4 x 5 ( x − 1 ) ( x − 5 ) .

Identifying Vertical Asymptotes of Rational Functions

By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location.

Vertical Asymptotes

The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors.

Given a rational function, identify any vertical asymptotes of its graph.

Factor the numerator and denominator.
Note any restrictions in the domain of the function.
Reduce the expression by canceling common factors in the numerator and the denominator.
Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities, or “holes.”

Identifying Vertical Asymptotes

Find the vertical asymptotes of the graph of k ( x ) = 5 + 2 x 2 2 − x − x 2 . k ( x ) = 5 + 2 x 2 2 − x − x 2 .

First, factor the numerator and denominator.

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

Neither x = –2 x = –2 nor x = 1 x = 1 are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure 9 confirms the location of the two vertical asymptotes.

Removable Discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity .

For example, the function f ( x ) = x 2 − 1 x 2 − 2 x − 3 f ( x ) = x 2 − 1 x 2 − 2 x − 3 may be re-written by factoring the numerator and the denominator.

Notice that x + 1 x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = −1 , x = −1 , is the location of the removable discontinuity. Notice also that x – 3 x – 3 is not a factor in both the numerator and denominator. The zero of this factor, x = 3 , x = 3 , is the vertical asymptote. See Figure 10 . [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.]

Removable Discontinuities of Rational Functions

A removable discontinuity occurs in the graph of a rational function at x = a x = a if a a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Identifying Vertical Asymptotes and Removable Discontinuities for a Graph

Find the vertical asymptotes and removable discontinuities of the graph of k ( x ) = x − 2 x 2 − 4 . k ( x ) = x − 2 x 2 − 4 .

Factor the numerator and the denominator.

Notice that there is a common factor in the numerator and the denominator, x – 2. x – 2. The zero for this factor is x = 2. x = 2. This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, x + 2. x + 2. The zero for this factor is x = −2. x = −2. The vertical asymptote is x = −2. x = −2. See Figure 11 .

The graph of this function will have the vertical asymptote at x = −2 , x = −2 , but at x = 2 x = 2 the graph will have a hole.

Find the vertical asymptotes and removable discontinuities of the graph of f ( x ) = x 2 − 25 x 3 − 6 x 2 + 5 x . f ( x ) = x 2 − 25 x 3 − 6 x 2 + 5 x .

Identifying Horizontal Asymptotes of Rational Functions

While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the function that is the ratio of the leading terms.

There are three distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y = 0. y = 0.

In this case, the end behavior is f ( x ) ≈ 4 x x 2 = 4 x . f ( x ) ≈ 4 x x 2 = 4 x . This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function g ( x ) = 4 x , g ( x ) = 4 x , and the outputs will approach zero, resulting in a horizontal asymptote at y = 0. y = 0. See Figure 12 . Note that this graph crosses the horizontal asymptote.

Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.

In this case, the end behavior is f ( x ) ≈ 3 x 2 x = 3 x . f ( x ) ≈ 3 x 2 x = 3 x . This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function g ( x ) = 3 x . g ( x ) = 3 x . As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of g ( x ) = 3 x g ( x ) = 3 x looks like a diagonal line, and since f f will behave similarly to g , g , it will approach a line close to y = 3 x . y = 3 x . This line is a slant asymptote.

To find the equation of the slant asymptote, divide 3 x 2 − 2 x + 1 x − 1 . 3 x 2 − 2 x + 1 x − 1 . The quotient is 3 x + 1 , 3 x + 1 , and the remainder is 2. The slant asymptote is the graph of the line g ( x ) = 3 x + 1. g ( x ) = 3 x + 1. See Figure 13 .

Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y = a n b n , y = a n b n , where a n a n and b n b n are the leading coefficients of p ( x ) p ( x ) and q ( x ) q ( x ) for f ( x ) = p ( x ) q ( x ) , q ( x ) ≠ 0. f ( x ) = p ( x ) q ( x ) , q ( x ) ≠ 0.

In this case, the end behavior is f ( x ) ≈ 3 x 2 x 2 = 3. f ( x ) ≈ 3 x 2 x 2 = 3. This tells us that as the inputs grow large, this function will behave like the function g ( x ) = 3 , g ( x ) = 3 , which is a horizontal line. As x → ± ∞ , f ( x ) → 3 , x → ± ∞ , f ( x ) → 3 , resulting in a horizontal asymptote at y = 3. y = 3. See Figure 14 . Note that this graph crosses the horizontal asymptote.

Notice that, while the graph of a rational function will never cross a vertical asymptote , the graph may or may not cross a horizontal or slant asymptote. Also, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote.

It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function

with end behavior

the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.

Horizontal Asymptotes of Rational Functions

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.

Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. y = 0.
Degree of numerator is greater than degree of denominator by one : no horizontal asymptote; slant asymptote.
Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

Identifying Horizontal and Slant Asymptotes

For the functions listed, identify the horizontal or slant asymptote.

ⓐ g ( x ) = 6 x 3 − 10 x 2 x 3 + 5 x 2 g ( x ) = 6 x 3 − 10 x 2 x 3 + 5 x 2
ⓑ h ( x ) = x 2 − 4 x + 1 x + 2 h ( x ) = x 2 − 4 x + 1 x + 2
ⓒ k ( x ) = x 2 + 4 x x 3 − 8 k ( x ) = x 2 + 4 x x 3 − 8

For these solutions, we will use f ( x ) = p ( x ) q ( x ) , q ( x ) ≠ 0. f ( x ) = p ( x ) q ( x ) , q ( x ) ≠ 0.

ⓐ g ( x ) = 6 x 3 − 10 x 2 x 3 + 5 x 2 : g ( x ) = 6 x 3 − 10 x 2 x 3 + 5 x 2 : The degree of p = degree of q = 3 , p = degree of q = 3 , so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at y = 6 2 y = 6 2 or y = 3. y = 3.

The quotient is x – 6 x – 6 and the remainder is 13. There is a slant asymptote at y = x – 6. y = x – 6.

ⓒ k ( x ) = x 2 + 4 x x 3 − 8 : k ( x ) = x 2 + 4 x x 3 − 8 : The degree of p = 2 < p = 2 < degree of q = 3 , q = 3 , so there is a horizontal asymptote y = 0. y = 0.

Identifying Horizontal Asymptotes

In the sugar concentration problem earlier, we created the equation C ( t ) = 5 + t 100 + 10 t . C ( t ) = 5 + t 100 + 10 t .

Find the horizontal asymptote and interpret it in context of the problem.

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t , t , with coefficient 1. In the denominator, the leading term is 10 t , 10 t , with coefficient 10. The horizontal asymptote will be at the ratio of these values:

This function will have a horizontal asymptote at y = 1 10 . y = 1 10 .

This tells us that as the values of t increase, the values of C C will approach 1 10 . 1 10 . In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or 1 10 1 10 pounds per gallon.

Identifying Horizontal and Vertical Asymptotes

Find the horizontal and vertical asymptotes of the function

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x = 1 , – 2 , and 5 , x = 1 , – 2 , and 5 , indicating vertical asymptotes at these values.

The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as x → ± ∞ , f ( x ) → 0. x → ± ∞ , f ( x ) → 0. This function will have a horizontal asymptote at y = 0. y = 0. See Figure 15 .

Find the vertical and horizontal asymptotes of the function:

f ( x ) = ( 2 x − 1 ) ( 2 x + 1 ) ( x − 2 ) ( x + 3 ) f ( x ) = ( 2 x − 1 ) ( 2 x + 1 ) ( x − 2 ) ( x + 3 )

Intercepts of Rational Functions

A rational function will have a y -intercept at f ( 0 ) f ( 0 ) , if the function is defined at zero. A rational function will not have a y -intercept if the function is not defined at zero.

Likewise, a rational function will have x -intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x -intercepts can only occur when the numerator of the rational function is equal to zero.

Finding the Intercepts of a Rational Function

Find the intercepts of f ( x ) = ( x − 2 ) ( x + 3 ) ( x − 1 ) ( x + 2 ) ( x − 5 ) . f ( x ) = ( x − 2 ) ( x + 3 ) ( x − 1 ) ( x + 2 ) ( x − 5 ) .

We can find the y -intercept by evaluating the function at zero

The x -intercepts will occur when the function is equal to zero:

The y -intercept is ( 0 , –0.6 ) , ( 0 , –0.6 ) , the x -intercepts are ( 2 , 0 ) ( 2 , 0 ) and ( –3 , 0 ) . ( –3 , 0 ) . See Figure 16 .

Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x - and y -intercepts and the horizontal and vertical asymptotes.

Graphing Rational Functions

In Example 9 , we see that the numerator of a rational function reveals the x -intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 17 .

When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 18 .

For example, the graph of f ( x ) = ( x + 1 ) 2 ( x − 3 ) ( x + 3 ) 2 ( x − 2 ) f ( x ) = ( x + 1 ) 2 ( x − 3 ) ( x + 3 ) 2 ( x − 2 ) is shown in Figure 19 .

At the x -intercept x = −1 x = −1 corresponding to the ( x + 1 ) 2 ( x + 1 ) 2 factor of the numerator, the graph "bounces", consistent with the quadratic nature of the factor.
At the x -intercept x = 3 x = 3 corresponding to the ( x − 3 ) ( x − 3 ) factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
At the vertical asymptote x = −3 x = −3 corresponding to the ( x + 3 ) 2 ( x + 3 ) 2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function f ( x ) = 1 x 2 . f ( x ) = 1 x 2 .
At the vertical asymptote x = 2 , x = 2 , corresponding to the ( x − 2 ) ( x − 2 ) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side.

Given a rational function, sketch a graph.

Evaluate the function at 0 to find the y -intercept.
For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x -intercepts.
Find the multiplicities of the x -intercepts to determine the behavior of the graph at those points.
For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
Sketch the graph.

Graphing a Rational Function

Sketch a graph of f ( x ) = ( x + 2 ) ( x − 3 ) ( x + 1 ) 2 ( x − 2 ) . f ( x ) = ( x + 2 ) ( x − 3 ) ( x + 1 ) 2 ( x − 2 ) .

We can start by noting that the function is already factored, saving us a step.

Next, we will find the intercepts. Evaluating the function at zero gives the y -intercept:

To find the x -intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x -intercepts at x = –2 x = –2 and x = 3. x = 3. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a y -intercept at ( 0 , 3 ) ( 0 , 3 ) and x -intercepts at ( –2 , 0 ) ( –2 , 0 ) and ( 3 , 0 ) . ( 3 , 0 ) .

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when x + 1 = 0 x + 1 = 0 and when x – 2 = 0 , x – 2 = 0 , giving us vertical asymptotes at x = –1 x = –1 and x = 2. x = 2.

There are no common factors in the numerator and denominator. This means there are no removable discontinuities.

Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. y = 0.

To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x -intercepts between the vertical asymptotes, and the y -intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 20 .

The factor associated with the vertical asymptote at x = −1 x = −1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

For the vertical asymptote at x = 2 , x = 2 , the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure 21 . After passing through the x -intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.

Given the function f ( x ) = ( x + 2 ) 2 ( x − 2 ) 2 ( x − 1 ) 2 ( x − 3 ) , f ( x ) = ( x + 2 ) 2 ( x − 2 ) 2 ( x − 1 ) 2 ( x − 3 ) , use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Writing Rational Functions

Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x -intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x -intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors.

Writing Rational Functions from Intercepts and Asymptotes

If a rational function has x -intercepts at x = x 1 , x 2 , ... , x n , x = x 1 , x 2 , ... , x n , vertical asymptotes at x = v 1 , v 2 , … , v m , x = v 1 , v 2 , … , v m , and no x i = any v j , x i = any v j , then the function can be written in the form:

where the powers p i p i or q i q i on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a a can be determined given a value of the function other than the x -intercept or by the horizontal asymptote if it is nonzero.

Given a graph of a rational function, write the function.

Determine the factors of the numerator. Examine the behavior of the graph at the x -intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
Use any clear point on the graph to find the stretch factor.

Writing a Rational Function from Intercepts and Asymptotes

Write an equation for the rational function shown in Figure 22 .

The graph appears to have x -intercepts at x = –2 x = –2 and x = 3. x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = –1 x = –1 seems to exhibit the basic behavior similar to 1 x , 1 x , with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at x = 2 x = 2 is exhibiting a behavior similar to 1 x 2 , 1 x 2 , with the graph heading toward negative infinity on both sides of the asymptote. See Figure 23 .

We can use this information to write a function of the form

To find the stretch factor, we can use another clear point on the graph, such as the y -intercept ( 0 , –2 ) . ( 0 , –2 ) .

This gives us a final function of f ( x ) = 4 ( x + 2 ) ( x − 3 ) 3 ( x + 1 ) ( x − 2 ) 2 . f ( x ) = 4 ( x + 2 ) ( x − 3 ) 3 ( x + 1 ) ( x − 2 ) 2 .

Access these online resources for additional instruction and practice with rational functions.

Find the Equation of a Rational Function
Determining Vertical and Horizontal Asymptotes
Find the Intercepts, Asymptotes, and Hole of a Rational Function

5.6 Section Exercises

What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?

What is the fundamental difference in the graphs of polynomial functions and rational functions?

If the graph of a rational function has a removable discontinuity, what must be true of the functional rule?

Can a graph of a rational function have no vertical asymptote? If so, how?

Can a graph of a rational function have no x -intercepts? If so, how?

For the following exercises, find the domain of the rational functions.

f ( x ) = x − 1 x + 2 f ( x ) = x − 1 x + 2

f ( x ) = x + 1 x 2 − 1 f ( x ) = x + 1 x 2 − 1

f ( x ) = x 2 + 4 x 2 − 2 x − 8 f ( x ) = x 2 + 4 x 2 − 2 x − 8

f ( x ) = x 2 + 4 x − 3 x 4 − 5 x 2 + 4 f ( x ) = x 2 + 4 x − 3 x 4 − 5 x 2 + 4

For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.

f ( x ) = 4 x − 1 f ( x ) = 4 x − 1

f ( x ) = 2 5 x + 2 f ( x ) = 2 5 x + 2

f ( x ) = x x 2 − 9 f ( x ) = x x 2 − 9

f ( x ) = x x 2 + 5 x − 36 f ( x ) = x x 2 + 5 x − 36

f ( x ) = 3 + x x 3 − 27 f ( x ) = 3 + x x 3 − 27

f ( x ) = 3 x − 4 x 3 − 16 x f ( x ) = 3 x − 4 x 3 − 16 x

f ( x ) = x 2 − 1 x 3 + 9 x 2 + 14 x f ( x ) = x 2 − 1 x 3 + 9 x 2 + 14 x

f ( x ) = x + 5 x 2 − 25 f ( x ) = x + 5 x 2 − 25

f ( x ) = x − 4 x − 6 f ( x ) = x − 4 x − 6

f ( x ) = 4 − 2 x 3 x − 1 f ( x ) = 4 − 2 x 3 x − 1

For the following exercises, find the x - and y -intercepts for the functions.

f ( x ) = x + 5 x 2 + 4 f ( x ) = x + 5 x 2 + 4

f ( x ) = x x 2 − x f ( x ) = x x 2 − x

f ( x ) = x 2 + 8 x + 7 x 2 + 11 x + 30 f ( x ) = x 2 + 8 x + 7 x 2 + 11 x + 30

f ( x ) = x 2 + x + 6 x 2 − 10 x + 24 f ( x ) = x 2 + x + 6 x 2 − 10 x + 24

f ( x ) = 94 − 2 x 2 3 x 2 − 12 f ( x ) = 94 − 2 x 2 3 x 2 − 12

For the following exercises, describe the local and end behavior of the functions.

f ( x ) = x 2 x + 1 f ( x ) = x 2 x + 1

f ( x ) = 2 x x − 6 f ( x ) = 2 x x − 6

f ( x ) = − 2 x x − 6 f ( x ) = − 2 x x − 6

f ( x ) = x 2 − 4 x + 3 x 2 − 4 x − 5 f ( x ) = x 2 − 4 x + 3 x 2 − 4 x − 5

f ( x ) = 2 x 2 − 32 6 x 2 + 13 x − 5 f ( x ) = 2 x 2 − 32 6 x 2 + 13 x − 5

For the following exercises, find the slant asymptote of the functions.

f ( x ) = 24 x 2 + 6 x 2 x + 1 f ( x ) = 24 x 2 + 6 x 2 x + 1

f ( x ) = 4 x 2 − 10 2 x − 4 f ( x ) = 4 x 2 − 10 2 x − 4

f ( x ) = 81 x 2 − 18 3 x − 2 f ( x ) = 81 x 2 − 18 3 x − 2

f ( x ) = 6 x 3 − 5 x 3 x 2 + 4 f ( x ) = 6 x 3 − 5 x 3 x 2 + 4

f ( x ) = x 2 + 5 x + 4 x − 1 f ( x ) = x 2 + 5 x + 4 x − 1

For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.

The reciprocal function shifted up two units.

The reciprocal function shifted down one unit and left three units.

The reciprocal squared function shifted to the right 2 units.

The reciprocal squared function shifted down 2 units and right 1 unit.

For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.

p ( x ) = 2 x − 3 x + 4 p ( x ) = 2 x − 3 x + 4

q ( x ) = x − 5 3 x − 1 q ( x ) = x − 5 3 x − 1

s ( x ) = 4 ( x − 2 ) 2 s ( x ) = 4 ( x − 2 ) 2

r ( x ) = 5 ( x + 1 ) 2 r ( x ) = 5 ( x + 1 ) 2

f ( x ) = 3 x 2 − 14 x − 5 3 x 2 + 8 x − 16 f ( x ) = 3 x 2 − 14 x − 5 3 x 2 + 8 x − 16

g ( x ) = 2 x 2 + 7 x − 15 3 x 2 − 14 x + 15 g ( x ) = 2 x 2 + 7 x − 15 3 x 2 − 14 x + 15

a ( x ) = x 2 + 2 x − 3 x 2 − 1 a ( x ) = x 2 + 2 x − 3 x 2 − 1

b ( x ) = x 2 − x − 6 x 2 − 4 b ( x ) = x 2 − x − 6 x 2 − 4

h ( x ) = 2 x 2 + x − 1 x − 4 h ( x ) = 2 x 2 + x − 1 x − 4

k ( x ) = 2 x 2 − 3 x − 20 x − 5 k ( x ) = 2 x 2 − 3 x − 20 x − 5

w ( x ) = ( x − 1 ) ( x + 3 ) ( x − 5 ) ( x + 2 ) 2 ( x − 4 ) w ( x ) = ( x − 1 ) ( x + 3 ) ( x − 5 ) ( x + 2 ) 2 ( x − 4 )

z ( x ) = ( x + 2 ) 2 ( x − 5 ) ( x − 3 ) ( x + 1 ) ( x + 4 ) z ( x ) = ( x + 2 ) 2 ( x − 5 ) ( x − 3 ) ( x + 1 ) ( x + 4 )

For the following exercises, write an equation for a rational function with the given characteristics.

Vertical asymptotes at x = 5 x = 5 and x = −5 , x = −5 , x -intercepts at ( 2 , 0 ) ( 2 , 0 ) and ( −1 , 0 ) , ( −1 , 0 ) , y -intercept at ( 0 , 4 ) ( 0 , 4 )

Vertical asymptotes at x = −4 x = −4 and x = −1 , x = −1 , x- intercepts at ( 1 , 0 ) ( 1 , 0 ) and ( 5 , 0 ) , ( 5 , 0 ) , y- intercept at ( 0 , 7 ) ( 0 , 7 )

Vertical asymptotes at x = −4 x = −4 and x = −5 , x = −5 , x -intercepts at ( 4 , 0 ) ( 4 , 0 ) and ( −6 , 0 ) , ( −6 , 0 ) , Horizontal asymptote at y = 7 y = 7

Vertical asymptotes at x = −3 x = −3 and x = 6 , x = 6 , x -intercepts at ( −2 , 0 ) ( −2 , 0 ) and ( 1 , 0 ) , ( 1 , 0 ) , Horizontal asymptote at y = −2 y = −2

Vertical asymptote at x = −1 , x = −1 , Double zero at x = 2 , x = 2 , y -intercept at ( 0 , 2 ) ( 0 , 2 )

Vertical asymptote at x = 3 , x = 3 , Double zero at x = 1 , x = 1 , y -intercept at ( 0 , 4 ) ( 0 , 4 )

For the following exercises, use the graphs to write an equation for the function.

For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote

f ( x ) = 1 x − 2 f ( x ) = 1 x − 2

f ( x ) = x x − 3 f ( x ) = x x − 3

f ( x ) = 2 x x + 4 f ( x ) = 2 x x + 4

f ( x ) = 2 x ( x − 3 ) 2 f ( x ) = 2 x ( x − 3 ) 2

f ( x ) = x 2 x 2 + 2 x + 1 f ( x ) = x 2 x 2 + 2 x + 1

For the following exercises, use a calculator to graph f ( x ) . f ( x ) . Use the graph to solve f ( x ) > 0. f ( x ) > 0.

f ( x ) = 2 x + 1 f ( x ) = 2 x + 1

f ( x ) = 4 2 x − 3 f ( x ) = 4 2 x − 3

f ( x ) = 2 ( x − 1 ) ( x + 2 ) f ( x ) = 2 ( x − 1 ) ( x + 2 )

f ( x ) = x + 2 ( x − 1 ) ( x − 4 ) f ( x ) = x + 2 ( x − 1 ) ( x − 4 )

f ( x ) = ( x + 3 ) 2 ( x − 1 ) 2 ( x + 1 ) f ( x ) = ( x + 3 ) 2 ( x − 1 ) 2 ( x + 1 )

For the following exercises, identify the removable discontinuity.

f ( x ) = x 2 − 4 x − 2 f ( x ) = x 2 − 4 x − 2

f ( x ) = x 3 + 1 x + 1 f ( x ) = x 3 + 1 x + 1

f ( x ) = x 2 + x − 6 x − 2 f ( x ) = x 2 + x − 6 x − 2

f ( x ) = 2 x 2 + 5 x − 3 x + 3 f ( x ) = 2 x 2 + 5 x − 3 x + 3

f ( x ) = x 3 + x 2 x + 1 f ( x ) = x 3 + x 2 x + 1

Real-World Applications

For the following exercises, express a rational function that describes the situation.

In the refugee camp hospital, a large mixing tank currently contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t t minutes.

In the refugee camp hospital, a large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t t minutes.

For the following exercises, use the given rational function to answer the question.

The concentration C C of a drug in a patient’s bloodstream t t hours after injection is given by C ( t ) = 2 t 3 + t 2 . C ( t ) = 2 t 3 + t 2 . What happens to the concentration of the drug as t t increases?

The concentration C C of a drug in a patient’s bloodstream t t hours after injection is given by C ( t ) = 100 t 2 t 2 + 75 . C ( t ) = 100 t 2 t 2 + 75 . Use a calculator to approximate the time when the concentration is highest.

For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question.

An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let x x = length of the side of the base.

A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/ square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x x = length of the side of the base.

A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let x x = radius.

A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let x x = radius.

A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let x x = radius.

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Study Guides > Precalculus I

Rational functions, learning objectives.

Use arrow notation.
Solve applied problems involving rational functions.
Find the domains of rational functions.
Identify vertical asymptotes.
Identify horizontal asymptotes.
Graph rational functions.

Suppose we know that the cost of making a product is dependent on the number of items, x , produced. This is given by the equation [latex]C\left(x\right)=15,000x - 0.1{x}^{2}+1000[/latex]. If we want to know the average cost for producing x items, we would divide the cost function by the number of items, x .

The average cost function, which yields the average cost per item for x items produced, is

Use arrow notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs and notice some of their features.

Several things are apparent if we examine the graph of [latex]f\left(x\right)=\frac{1}{x}[/latex].

On the left branch of the graph, the curve approaches the x -axis [latex]\left(y=0\right) \text{ as } x\to -\infty [/latex].
As the graph approaches [latex]x=0[/latex] from the left, the curve drops, but as we approach zero from the right, the curve rises.
Finally, on the right branch of the graph, the curves approaches the x- axis [latex]\left(y=0\right) \text{ as } x\to \infty [/latex].

To summarize, we use arrow notation to show that x or [latex]f\left(x\right)[/latex] is approaching a particular value.

Arrow Notation
Symbol	Meaning
[latex]x\to {a}^{-}[/latex]	approaches from the left ( but close to )
[latex]x\to {a}^{+}[/latex]	approaches from the right ( > but close to )
[latex]x\to \infty\\ [/latex]	approaches infinity ( increases without bound)
[latex]x\to -\infty [/latex]	approaches negative infinity ( decreases without bound)
[latex]f\left(x\right)\to \infty [/latex]	the output approaches infinity (the output increases without bound)
[latex]f\left(x\right)\to -\infty [/latex]	the output approaches negative infinity (the output decreases without bound)
[latex]f\left(x\right)\to a[/latex]	the output approaches

Local Behavior of [latex]f\left(x\right)=\frac{1}{x}[/latex]

Let’s begin by looking at the reciprocal function, [latex]f\left(x\right)=\frac{1}{x}[/latex]. We cannot divide by zero, which means the function is undefined at [latex]x=0[/latex]; so zero is not in the domain . As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in the table below.

	–0.1	–0.01	–0.001	–0.0001
	–10	–100	–1000	–10,000

We write in arrow notation

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in the table below.

	0.1	0.01	0.001	0.0001
	10	100	1000	10,000

Graph of f(x)=1/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.

This behavior creates a vertical asymptote , which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 as the input becomes close to zero.

Graph of f(x)=1/x with its vertical asymptote at x=0.

A General Note: Vertical Asymptote

A vertical asymptote of a graph is a vertical line [latex]x=a[/latex] where the graph tends toward positive or negative infinity as the inputs approach a . We write

End Behavior of [latex]f\left(x\right)=\frac{1}{x}[/latex]

As the values of x approach infinity, the function values approach 0. As the values of x approach negative infinity, the function values approach 0. Symbolically, using arrow notation

[latex]\text{As }x\to \infty ,f\left(x\right)\to 0,\text{and as }x\to -\infty ,f\left(x\right)\to 0[/latex].

Graph of f(x)=1/x which highlights the segments of the turning points to denote their end behavior.

Graph of f(x)=1/x with its vertical asymptote at x=0 and its horizontal asymptote at y=0.

A General Note: Horizontal Asymptote

A horizontal asymptote of a graph is a horizontal line [latex]y=b[/latex] where the graph approaches the line as the inputs increase or decrease without bound. We write

Example 1: Using Arrow Notation

Graph of f(x)=1/(x-2)+4 with its vertical asymptote at x=2 and its horizontal asymptote at y=4.

Notice that the graph is showing a vertical asymptote at [latex]x=2[/latex], which tells us that the function is undefined at [latex]x=2[/latex].

And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at [latex]y=4[/latex]. As the inputs increase without bound, the graph levels off at 4.

Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function.

Example 2: Using Transformations to Graph a Rational Function

Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any.

Shifting the graph left 2 and up 3 would result in the function

or equivalently, by giving the terms a common denominator,

The graph of the shifted function is displayed in Figure 7.

Graph of f(x)=1/(x+2)+3 with its vertical asymptote at x=-2 and its horizontal asymptote at y=3.

Notice that this function is undefined at [latex]x=-2[/latex], and the graph also is showing a vertical asymptote at [latex]x=-2[/latex].

As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at [latex]y=3[/latex].

Analysis of the Solution

Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.

Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units.

Solve applied problems involving rational functions

In Example 2, we shifted a toolkit function in a way that resulted in the function [latex]f\left(x\right)=\frac{3x+7}{x+2}[/latex]. This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

A General Note: Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions [latex]P\left(x\right) \text{and} Q\left(x\right)[/latex].

Example 3: Solving an Applied Problem Involving a Rational Function

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C , will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05[/latex], the concentration is greater after 12 minutes than at the beginning.

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

Notice the horizontal asymptote is [latex]y=\text{ }0.1[/latex]. This means the concentration, C , the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.

Find the domains of rational functions

A General Note: Domain of a Rational Function

The domain of a rational function includes all real numbers except those that cause the denominator to equal zero.

How To: Given a rational function, find the domain.

Set the denominator equal to zero.
Solve to find the x -values that cause the denominator to equal zero.
The domain is all real numbers except those found in Step 2.

Example 4: Finding the Domain of a Rational Function

Find the domain of [latex]f\left(x\right)=\frac{x+3}{{x}^{2}-9}[/latex].

Begin by setting the denominator equal to zero and solving.

The denominator is equal to zero when [latex]x=\pm 3[/latex]. The domain of the function is all real numbers except [latex]x=\pm 3[/latex].

A graph of this function confirms that the function is not defined when [latex]x=\pm 3[/latex].

Graph of f(x)=1/(x-3) with its vertical asymptote at x=3 and its horizontal asymptote at y=0.

There is a vertical asymptote at [latex]x=3[/latex] and a hole in the graph at [latex]x=-3[/latex]. We will discuss these types of holes in greater detail later in this section.

Find the domain of [latex]f\left(x\right)=\frac{4x}{5\left(x - 1\right)\left(x - 5\right)}[/latex].

Identify vertical asymptotes

Vertical Asymptotes

How To: Given a rational function, identify any vertical asymptotes of its graph.

Factor the numerator and denominator.
Note any restrictions in the domain of the function.
Reduce the expression by canceling common factors in the numerator and the denominator.
Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.
Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities.

Example 5: Identifying Vertical Asymptotes

Find the vertical asymptotes of the graph of [latex]k\left(x\right)=\frac{5+2{x}^{2}}{2-x-{x}^{2}}[/latex].

First, factor the numerator and denominator.

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

Neither [latex]x=-2[/latex] nor [latex]x=1[/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9 confirms the location of the two vertical asymptotes.

Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.

Removable Discontinuities

Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity .

For example, the function [latex]f\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}-2x - 3}[/latex] may be re-written by factoring the numerator and the denominator.

Notice that [latex]x+1[/latex] is a common factor to the numerator and the denominator. The zero of this factor, [latex]x=-1[/latex], is the location of the removable discontinuity. Notice also that [latex]x - 3[/latex] is not a factor in both the numerator and denominator. The zero of this factor, [latex]x=3[/latex], is the vertical asymptote.

Graph of f(x)=(x^2-1)/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.

A General Note: Removable Discontinuities of Rational Functions

A removable discontinuity occurs in the graph of a rational function at [latex]x=a[/latex] if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value.

Example 6: Identifying Vertical Asymptotes and Removable Discontinuities for a Graph

Find the vertical asymptotes and removable discontinuities of the graph of [latex]k\left(x\right)=\frac{x - 2}{{x}^{2}-4}[/latex].

Factor the numerator and the denominator.

Notice that there is a common factor in the numerator and the denominator, [latex]x - 2[/latex]. The zero for this factor is [latex]x=2[/latex]. This is the location of the removable discontinuity.

Notice that there is a factor in the denominator that is not in the numerator, [latex]x+2[/latex]. The zero for this factor is [latex]x=-2[/latex]. The vertical asymptote is [latex]x=-2[/latex].

Graph of k(x)=(x-2)/(x-2)(x+2) with its vertical asymptote at x=-2 and a removable discontinuity at x=2.

The graph of this function will have the vertical asymptote at [latex]x=-2[/latex], but at [latex]x=2[/latex] the graph will have a hole.

Find the vertical asymptotes and removable discontinuities of the graph of [latex]f\left(x\right)=\frac{{x}^{2}-25}{{x}^{3}-6{x}^{2}+5x}[/latex].

Identify horizontal asymptotes

There are three distinct outcomes when checking for horizontal asymptotes:

Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y = 0.

In this case, the end behavior is [latex]f\left(x\right)\approx \frac{4x}{{x}^{2}}=\frac{4}{x}[/latex]. This tells us that, as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\left(x\right)=\frac{4}{x}[/latex], and the outputs will approach zero, resulting in a horizontal asymptote at y = 0. Note that this graph crosses the horizontal asymptote.

Graph of f(x)=(4x+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=0.

Figure 12. Horizontal Asymptote y = 0 when [latex]f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne{0}\text{ where degree of }p<\text{degree of q}[/latex].

Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.

In this case, the end behavior is [latex]f\left(x\right)\approx \frac{3{x}^{2}}{x}=3x[/latex]. This tells us that as the inputs increase or decrease without bound, this function will behave similarly to the function [latex]g\left(x\right)=3x[/latex]. As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal asymptote. However, the graph of [latex]g\left(x\right)=3x[/latex] looks like a diagonal line, and since f will behave similarly to g , it will approach a line close to [latex]y=3x[/latex]. This line is a slant asymptote.

Graph of f(x)=(3x^2-2x+1)/(x-1) with its vertical asymptote at x=1 and a slant asymptote aty=3x+1.

Figure 13. Slant Asymptote when [latex]f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0[/latex] where degree of [latex]p>\text{ degree of }q\text{ by }1[/latex].

Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at [latex]y=\frac{{a}_{n}}{{b}_{n}}[/latex], where [latex]{a}_{n}[/latex] and [latex]{b}_{n}[/latex] are the leading coefficients of [latex]p\left(x\right)[/latex] and [latex]q\left(x\right)[/latex] for [latex]f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0[/latex].

Graph of f(x)=(3x^2+2)/(x^2+4x-5) with its vertical asymptotes at x=-5 and x=1 and its horizontal asymptote at y=3.

Figure 14. Horizontal Asymptote when [latex]f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)},q\left(x\right)\ne 0\text{where degree of }p=\text{degree of }q[/latex].

with end behavior

the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.

A General Note: Horizontal Asymptotes of Rational Functions

The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator.

Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0.
Degree of numerator is greater than degree of denominator by one : no horizontal asymptote; slant asymptote.
Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients.

Example 7: Identifying Horizontal and Slant Asymptotes

For the functions below, identify the horizontal or slant asymptote.

[latex]g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[/latex]
[latex]h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}[/latex]
[latex]k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}[/latex]

For these solutions, we will use [latex]f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}, q\left(x\right)\ne 0[/latex].

[latex]g\left(x\right)=\frac{6{x}^{3}-10x}{2{x}^{3}+5{x}^{2}}[/latex]: The degree of [latex]p=\text{degree of} q=3[/latex], so we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at [latex]y=\frac{6}{2}[/latex] or [latex]y=3[/latex].

The quotient is [latex]x - 2[/latex] and the remainder is 13. There is a slant asymptote at [latex]y=-x - 2[/latex].

[latex]k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}[/latex]: The degree of [latex]p=2\text{ }<[/latex] degree of [latex]q=3[/latex], so there is a horizontal asymptote y = 0.

Example 8: Identifying Horizontal Asymptotes

In the sugar concentration problem earlier, we created the equation [latex]C\left(t\right)=\frac{5+t}{100+10t}[/latex].

Find the horizontal asymptote and interpret it in context of the problem.

Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t , with coefficient 1. In the denominator, the leading term is 10 t , with coefficient 10. The horizontal asymptote will be at the ratio of these values:

This function will have a horizontal asymptote at [latex]y=\frac{1}{10}[/latex].

This tells us that as the values of t increase, the values of C will approach [latex]\frac{1}{10}[/latex]. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or [latex]\frac{1}{10}[/latex] pounds per gallon.

Example 9: Identifying Horizontal and Vertical Asymptotes

Find the horizontal and vertical asymptotes of the function

First, note that this function has no common factors, so there are no potential removable discontinuities.

The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at [latex]x=1,-2,\text{and }5[/latex], indicating vertical asymptotes at these values.

The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to tend towards zero as the inputs get large, and so as [latex]x\to \pm \infty , f\left(x\right)\to 0[/latex]. This function will have a horizontal asymptote at [latex]y=0[/latex].

Graph of f(x)=(x-2)(x+3)/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5 and its horizontal asymptote at y=0.

Find the vertical and horizontal asymptotes of the function:

[latex]f\left(x\right)=\frac{\left(2x - 1\right)\left(2x+1\right)}{\left(x - 2\right)\left(x+3\right)}[/latex]

A General Note: Intercepts of Rational Functions

A rational function will have a y -intercept when the input is zero, if the function is defined at zero. A rational function will not have a y -intercept if the function is not defined at zero.

Example 10: Finding the Intercepts of a Rational Function

Find the intercepts of [latex]f\left(x\right)=\frac{\left(x - 2\right)\left(x+3\right)}{\left(x - 1\right)\left(x+2\right)\left(x - 5\right)}[/latex].

We can find the y -intercept by evaluating the function at zero

The x -intercepts will occur when the function is equal to zero:

The y -intercept is [latex]\left(0,-0.6\right)[/latex], the x -intercepts are [latex]\left(2,0\right)[/latex] and [latex]\left(-3,0\right)[/latex].

Graph of f(x)=(x-2)(x+3)/(x-1)(x+2)(x-5) with its vertical asymptotes at x=-2, x=1, and x=5, its horizontal asymptote at y=0, and its intercepts at (-3, 0), (0, -0.6), and (2, 0).

Graph rational functions

In Example 9, we see that the numerator of a rational function reveals the x -intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials.

Graph of y=1/x with its vertical asymptote at x=0.

Graph of y=1/x^2 with its vertical asymptote at x=0.

For example, the graph of [latex]f\left(x\right)=\frac{{\left(x+1\right)}^{2}\left(x - 3\right)}{{\left(x+3\right)}^{2}\left(x - 2\right)}[/latex] is shown in Figure 19.

Graph of f(x)=(x+1)^2(x-3)/(x+3)^2(x-2) with its vertical asymptotes at x=-3 and x=2, its horizontal asymptote at y=1, and its intercepts at (-1, 0), (0, 1/6), and (3, 0).

At the x -intercept [latex]x=-1[/latex] corresponding to the [latex]{\left(x+1\right)}^{2}[/latex] factor of the numerator, the graph bounces, consistent with the quadratic nature of the factor.
At the x -intercept [latex]x=3[/latex] corresponding to the [latex]\left(x - 3\right)[/latex] factor of the numerator, the graph passes through the axis as we would expect from a linear factor.
At the vertical asymptote [latex]x=-3[/latex] corresponding to the [latex]{\left(x+3\right)}^{2}[/latex] factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex].
At the vertical asymptote [latex]x=2[/latex], corresponding to the [latex]\left(x - 2\right)[/latex] factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the behavior of the function [latex]f\left(x\right)=\frac{1}{x}[/latex].

How To: Given a rational function, sketch a graph.

Evaluate the function at 0 to find the y -intercept.
For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x -intercepts.
Find the multiplicities of the x -intercepts to determine the behavior of the graph at those points.
For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve.
For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve.
Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes.
Sketch the graph.

Example 11: Graphing a Rational Function

Sketch a graph of [latex]f\left(x\right)=\frac{\left(x+2\right)\left(x - 3\right)}{{\left(x+1\right)}^{2}\left(x - 2\right)}[/latex].

We can start by noting that the function is already factored, saving us a step.

Next, we will find the intercepts. Evaluating the function at zero gives the y -intercept:

To find the x -intercepts, we determine when the numerator of the function is zero. Setting each factor equal to zero, we find x -intercepts at [latex]x=-2[/latex] and [latex]x=3[/latex]. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept.

We have a y -intercept at [latex]\left(0,3\right)[/latex] and x -intercepts at [latex]\left(-2,0\right)[/latex] and [latex]\left(3,0\right)[/latex].

To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when [latex]x+1=0[/latex] and when [latex]x - 2=0[/latex], giving us vertical asymptotes at [latex]x=-1[/latex] and [latex]x=2[/latex].

There are no common factors in the numerator and denominator. This means there are no removable discontinuities.

Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at [latex]y=0[/latex].

Graph of only the middle portion of f(x)=(x+2)(x-3)/(x+1)^2(x-2) with its intercepts at (-2, 0), (0, 3), and (3, 0).

The factor associated with the vertical asymptote at [latex]x=-1[/latex] was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well.

For the vertical asymptote at [latex]x=2[/latex], the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. After passing through the x -intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote.

Graph of f(x)=(x+2)(x-3)/(x+1)^2(x-2) with its vertical asymptotes at x=-1 and x=2, its horizontal asymptote at y=0, and its intercepts at (-2, 0), (0, 3), and (3, 0).

Given the function [latex]f\left(x\right)=\frac{{\left(x+2\right)}^{2}\left(x - 2\right)}{2{\left(x - 1\right)}^{2}\left(x - 3\right)}[/latex], use the characteristics of polynomials and rational functions to describe its behavior and sketch the function.

Writing Rational Functions

A General Note: Writing Rational Functions from Intercepts and Asymptotes

If a rational function has x -intercepts at [latex]x={x}_{1}, {x}_{2}, ..., {x}_{n}[/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\dots ,{v}_{m}[/latex], and no [latex]{x}_{i}=\text{any }{v}_{j}[/latex], then the function can be written in the form:

where the powers [latex]{p}_{i}[/latex] or [latex]{q}_{i}[/latex] on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the x -intercept or by the horizontal asymptote if it is nonzero.

How To: Given a graph of a rational function, write the function.

Determine the factors of the numerator. Examine the behavior of the graph at the x -intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the "simplest" function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.)
Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their powers.
Use any clear point on the graph to find the stretch factor.

Example 12: Writing a Rational Function from Intercepts and Asymptotes

Write an equation for the rational function shown in Figure 22.

The graph appears to have x -intercepts at [latex]x=-2[/latex] and [latex]x=3[/latex]. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at [latex]x=-1[/latex] seems to exhibit the basic behavior similar to [latex]\frac{1}{x}[/latex], with the graph heading toward positive infinity on one side and heading toward negative infinity on the other. The asymptote at [latex]x=2[/latex] is exhibiting a behavior similar to [latex]\frac{1}{{x}^{2}}[/latex], with the graph heading toward negative infinity on both sides of the asymptote.

Graph of a rational function denoting its vertical asymptotes and x-intercepts.

We can use this information to write a function of the form

To find the stretch factor, we can use another clear point on the graph, such as the y -intercept [latex]\left(0,-2\right)[/latex].

This gives us a final function of [latex]f\left(x\right)=\frac{4\left(x+2\right)\left(x - 3\right)}{3\left(x+1\right){\left(x - 2\right)}^{2}}[/latex].

Key Equations

Rational Function

[latex]f\left(x\right)=\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}}, Q\left(x\right)\ne 0[/latex]

Key Concepts

We can use arrow notation to describe local behavior and end behavior of the toolkit functions [latex]f\left(x\right)=\frac{1}{x}[/latex] and [latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex].
A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote.
Application problems involving rates and concentrations often involve rational functions.
The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero.
A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero.
A rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions.
Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior.
If a rational function has x -intercepts at [latex]x={x}_{1},{x}_{2},\dots ,{x}_{n}[/latex], vertical asymptotes at [latex]x={v}_{1},{v}_{2},\dots ,{v}_{m}[/latex], and no [latex]{x}_{i}=\text{any }{v}_{j}[/latex], then the function can be written in the form [latex]f\left(x\right)=a\frac{{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}}{{\left(x-{v}_{1}\right)}^{{q}_{1}}{\left(x-{v}_{2}\right)}^{{q}_{2}}\cdots {\left(x-{v}_{m}\right)}^{{q}_{n}}}[/latex]

Section Exercises

Graph of a rational function with vertical asymptotes at x=-3 and x=4.

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Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/ [email protected] :1/Preface. License: CC BY: Attribution .

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Rational functions

Rational function

Here you will learn about rational functions, including how to simplify rational functions and operations with rational functions.

Students first learn about rational functions in Algebra 1 and expand their knowledge as they progress through high school math.

What are rational functions?

Rational functions or algebraic fractions, are functions that are fractions because they have a numerator and denominator. Rational functions are expressed as the ratio of two polynomials such that the denominator is not equal to 0.

R(x)=\cfrac{p(x)}{q(x)} where q(x) ≠ 0.

Rational functions contain at least one variable. They also have denominators and numerators that can be linear functions, quadratic functions, or any polynomial function.

Here are the three types of rational functions and the graph of the functions

[FREE] Algebra Worksheet (Grade 6 to 8)

Use this quiz to check your grade 6 to 8 students’ understanding of algebra. 10+ questions with answers covering a range of 6th and 8th grade algebra topics to identify areas of strength and support!

Simplifying rational functions is a necessary skill when applying operations to rational functions.

Let’s look at a few examples of how to simplify rational functions.

Remember when you learned how to simplify fractions in elementary school?

For example, if you have the fraction, \cfrac{7}{21}, you look to see if it can be simplified by determining if there are any common factors of the numerator and the denominator.

In this case, there is a common factor of 7 for both the numerator and denominator.

In order to simplify algebraic fractions or rational functions you will use a similar strategy. Look for a common factor to the numerator and the denominator.

Let’s simplify:

First, find the common factor between the numerator and the denominator.

Looking at the factors for both expressions, 3, 5, and x are the common factors.

15x is the common factor.

If you want to learn more on how to simplify rational functions, go to:

Step-by-step guide: Simplifying rational expressions

Adding and subtracting rational expressions

Rational expressions can be added and subtracted. Similar to adding and subtracting fractions, rational expressions need a common denominator before they can be added or subtracted.

Let’s add the rational functions:

Before the expressions can be added together, first you will need to find the common denominator.

The two denominators are:

Looking at the coefficients, the least common multiple of 24 and 6 is 24.

To find the least common multiple of the variables, xy and y^2, you need to look at each variable and take the variable to the higher exponent.

x has to be included in the least common denominator because it is the variable to the highest exponent (x^1 has a higher exponent than x^0)

y^2 has to be included in the least common denominator because y^2 has a higher exponent than y^1.

So the least common denominator of \cfrac{4}{24 x y}+\cfrac{10}{6y^2} is 24 x y^2.

Now, you will have to adjust each rational expression so that it has a denominator of 24 x y^2. Be sure to multiply the numerator and denominator by the missing factors.

After adjusting both rational expressions to have a common denominator, you can add them.

\cfrac{4 y}{24 x y^2}+\cfrac{40 x}{24 x y^2}=\cfrac{4 y+40 x}{24 x y^2} \rightarrow The rational expression can be written as one expression.

The numerators cannot be simplified any further because they are not like terms.

If you want to learn more about adding and subtracting rational functions, go to:

Step-by-step guide: Adding rational expressions

Step-by-step guide: Subtracting rational expressions

Multiplying and dividing rational expressions

Rational expressions can be multiplied and divided. Similar to multiplying and dividing fractions, rational expressions do not need a common denominator to be multiplied or divided. The product or quotient should be fully simplified.

Let’s multiply the rational expressions.

Before multiplying, factor any expression that can be factored.

In this case, the denominators of both expressions can be factored.

When multiplying or dividing, look to cancel any matching expressions as long as they are placed in opposite positions, (one in the numerator and one in the denominator)

The rational expressions can now be multiplied, meaning they can be written as one rational expression.

If you want to learn more about how to multiply and divide rational expressions, go to:

Step-by-step guide: Multiplying rational expressions

Step-by-step guide: Dividing rational expressions

Solving rational equations

The strategy to solve rational equations is similar to the strategy used to solve equations with fractions in them.

Let’s solve the rational equation,

First, find the common denominator, in this case, the common denominator is 2x.

Multiply the equation by the common denominator, 2x.

The new equation after multiplying by 2x is:

Solving for x\text{:}

Check the solution to make sure that 0 is not in the denominator.

Common Core State Standards

How does this relate to high school math?

High School Algebra: Arithmetic with Polynomials and Rational Expressions (HSA-APR.D.6) Rewrite simple rational expressions in different forms; write \cfrac{a(x)}{b(x)} in the form q(x)+ \cfrac{r(x)}{b(x)}, where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
High School Algebra: Arithmetic with Polynomials and Rational Expressions (HSA-APR.D.7) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.

How to simplify and apply operations to rational functions

If you want more detailed steps and practice for simplifying rational expressions and applying the operations to rational expressions, check out the links highlighted above or take a look at the examples below.

Simplify and apply operations of rational functions examples

Example 1: simplify rational expressions without factoring.

Simplify the expression: \cfrac{28 x^2 y}{7 x y^2}

Identify the greatest common factor of the numerator and the denominator.

The numerator, 28 x^2 y and the denominator, 7 x y^2 have a greatest common factor of, 7 x y.

Looking at the coefficients of 28 and 7, the greatest common factor between those numbers is 7.

Looking at the variable x, the greatest common factor between x^2 and x is x.

Lastly, looking at the variable y, the greatest common factor between y and y^2 is y.

So, the greatest common factor between 28 x^2 y and 7 x y^2 is 7 x y.

2 Simplify the expression by the greatest common factor.

3 Write the simplified expression.

Example 2: subtracting rational expressions by factoring

Subtract the rational expressions, \cfrac{8}{(x-2)}-\cfrac{9}{\left(x^2-4\right)}

If possible, factor the denominators before finding the least common denominator.

(x-2) \rightarrow Cannot be factored

\left(x^2-4\right) \rightarrow Can be factored because it the difference of two perfect squares.

Applying the strategy of factoring the difference between two perfect squares, \left(x^2-4\right) factors to be (x+2)(x-2).

Find the least common denominator.

After factoring,

\cfrac{8}{(x-2)}-\cfrac{9}{\left(x^2-4\right)}=\cfrac{8}{(x-2)}-\cfrac{9}{(x+2)(x-2)}

Looking at the factored expression, the denominators are:

So, the least common denominator is (x+2)(x-2), take one of each factor to get the common denominator.

Adjust the rational expressions to have the common denominator.

Subtract and simplify the rational expression.

\cfrac{8 x+7}{(x+2)(x-2)} \rightarrow The difference cannot be simplified further, but the denominator can be multiplied back out.

\cfrac{8 x+7}{(x+2)(x-2)}=\cfrac{8 x+7}{\left(x^2-4\right)}

Example 3: multiplying without factoring

Multiply the rational expressions: \cfrac{8 a}{14 b^2} \times \cfrac{7 b^3}{20 a^2}

Factor the expression if possible.

The expression cannot be factored.

Simplify the expression before multiplying.

In this case, 7 and 14 can simplify by a common factor of 7; b^2 and b^3 simplify by a common factor of b^2.

Also, 8 and 20 can simplify by a factor of 4; a and a^2 can simplify by a factor of a.

Multiply the expressions and simplify the product if necessary.

\cfrac{2}{2} \times \cfrac{b}{5 a}=\cfrac{2 b}{10 a} \rightarrow The product can be simplified further because 2 and 10 have a common factor of 2.

So, \cfrac{2 b}{10 a}=\cfrac{b}{5 a}

Example 4: divide rational expressions with factoring

Divide the rational functions, \cfrac{5}{\left(x^2+5 x+6\right)} \div \cfrac{4}{\left(x^2+6 x+9\right)}

The denominators of both rational expressions can be factored.

With division, take the reciprocal of the second expression and change the operation to multiplication.

The reciprocal of \cfrac{4}{(x+3)(x+3)} is \cfrac{(x+3)(x+3)}{4}

The new expression is:

\cfrac{5}{(x+2)(x+3)} \times \cfrac{(x+3)(x+3)}{4}

Simplify before multiplying.

Look for common factors to cancel.

The only common factor was x+3 .

Multiply the expressions and simplify the answer if possible.

You can leave the answer as \cfrac{5(x+3)}{4(x+2)} or apply the distributive property, \cfrac{5 x+3}{4 x+8}

How to solve rational equations

In order to solve rational equations:

Find the common denominator of all the rational expressions in the equation.

Multiply the entire equation by the common denominator.

Solve the equation.

Check the solution

Solve rational equation examples

Example 5: equation with x in the denominator.

Solve the equation:

The denominators are 2x , x and 1 (whole numbers have a denominator of 1 ).

The common denominator which is also the least common multiple of all expressions, is 2x.

2 x\left(\cfrac{5}{2x}-\cfrac{14}{x}=3\right)

After multiplying by 2x, the new equation is:

Check the solution .

Substitute \cfrac{- \, 23}{6} for x into the original equation.

Example 6: distributing binomials

The denominators are: (x+2), (x-4), and 1

The common denominator is (x+2)(x-4) .

(x+2)(x-4)\left[\cfrac{x+1}{x+2}+\cfrac{3}{x-4}=1\right]

(x+2)(x-4)(1)=(x+2)(x-4)=x^2-2 x-8

After multiplying by the common denominator, (x+2)(x-4) the equation is:

Substitute 5 for x into the original equation.

Teaching tips for rational functions

Make connections between operations with fractions to operations with rational functions so that students can develop understanding.
Use active learning activities such as scavenger hunts or game playing to have students review skills instead of giving them a worksheet.
Encourage students who are struggling to use digital platforms such as Khan Academy so they can view tutorial videos.

Easy mistakes to make

Multiplying the numerator and the denominator by the common denominator When solving rational equations, you have to multiply the entire equation by the common denominator. For example, \cfrac{3}{2x}-\cfrac{1}{5x}=11 The common denominator is 10x, so multiply the entire equation by 10x. 10 x\left(\cfrac{3}{2 x}-\cfrac{1}{5 x}=11\right), when performing this multiplication, 10 x \times \cfrac{3}{2 x} ≠ \cfrac{30 x}{20 x} it should be 10 x \times \cfrac{3}{2 x}=15 So then, 10 x \times \cfrac{1}{5 x}=2 and 10 x \times 11=110 x
Not multiplying all terms by the denominator When multiplying the equation by the common denominator, be sure to multiply every term. \begin{aligned}& 3-\cfrac{5}{x-1}=1 \\\\ & (x-1)\left[3-\cfrac{5}{x-1}=1\right] \end{aligned} 3(x-1)-5=1 \rightarrow The last term was not multiplied by (x-1) So, it should be, 3(x-1)-5=x-1
Adding rational expressions incorrectly For example, thinking that you should add the numerators and the denominators, \cfrac{8}{4 x}+\cfrac{7}{x} ≠ \cfrac{15}{5 x} In order to add them, you must find a common denominator just like when you add fractions. In this case, the common denominator is 4x. \begin{aligned}& \cfrac{8}{4 x}+\cfrac{4}{4} \times \cfrac{7}{x}= \\\\ & \cfrac{8}{4 x}+\cfrac{28}{4 x}=\cfrac{36}{4 x} \end{aligned} The answer can be simplified further to be, \cfrac{36}{4 x}=\cfrac{9}{x}

Practice rational functions questions

1. Simplify the rational expression: \cfrac{(x+7)}{\left(x^2+9 x+14\right)}

Factor the numerator and denominator before simplifying.

The numerator cannot be factored in this case, but the denominator can be factored.

After factoring, cancel the terms in the numerator that are the same as the terms in the denominator.

The simplified expression is: \cfrac{1}{x+2} Remember that there is a 1 in the numerator because \cfrac{x+7}{x+7}=\cfrac{1}{1}

2. Add the expression:

In order to add rational functions, you have to find the common denominator.

The common denominator of \cfrac{6}{(x+2)}+\cfrac{9}{(x-1)} is (x+2)(x-1)

Multiply the numerator and denominator by the common denominator.

3. Multiply the expression:

You can either simplify first and then multiply or multiply first and then simplify. In this case, let’s simplify first:

4. Divide the rational expression:

When dividing rational expressions, remember to take the reciprocal of the second expression and multiply it to the first.

Before multiplying, factor and simplify the expressions by canceling any terms that match in the numerator with terms in the denominator.

Now, multiply: \cfrac{1}{1} \times \cfrac{(x+8)}{4}=\cfrac{x+8}{4}

5. Solve the equation:

In this case, the common denominator is 4x.

Checking the solution, it works in the original equation.

6. Solve the equation:

x=2.5 or x=5

Find the common denominator of the equation and multiply the entire equation by the common denominator. In this case, the common denominator is 5x.

After multiplying, the new equation is quadratic.

Solve the quadratic by first setting the quadratic equation equal to 0 and then using factoring or the quadratic formula to solve.

If you need extra help with solving quadratic equations, check out these links.

How to factor quadratic equations

Quadratic formula

Solving quadratic equations

Checking both values, they work in the original equation.

Rational functions FAQs

The asymptotes of rational functions are horizontal lines, vertical lines, or slanted lines that the graph of a rational function approaches but does not cross. Typically, when sketching the graph of the rational function by hand, the horizontal and vertical asymptotes, as well as the slant asymptotes are noted with a dotted line.

Thinking about the function values that exist in the domain of a rational function and those that do not, helps you find the horizontal and vertical asymptotes. The values for x that make the denominator equal to 0 are the possible values for the vertical asymptote. The horizontal asymptote is defined by the ratio of the leading coefficients of the numerator and denominator.

Yes, you can find the inverse functions of rational functions. This concept will be explored in a precalculus course.

You find the x -intercepts and the y -intercepts of rational functions applying the same strategy as you would with any function. For the x -intercept set the function equal to 0 and find the x values. For the y -intercept substitute 0 in for the x values and find y. The x and y intercepts are real numbers.

The reciprocal function, f(x)=\cfrac{1}{x} is defined to be the reciprocal function, which is also a rational function.

No, the equation of a circle is not a rational function or even a function at all, it’s considered to be a conic section.

Matrices are a rectangular array of numbers and can be used to solve systems of linear equations.

The next lessons are

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Rational Functions

Solve applied problems involving rational functions.

A General Note: Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions [latex]P\left(x\right) \text{and} Q\left(x\right)[/latex].

Example 3: Solving an Applied Problem Involving a Rational Function

The concentration, C , will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05[/latex], the concentration is greater after 12 minutes than at the beginning.

Analysis of the Solution

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

Notice the horizontal asymptote is [latex]y=\text{ }0.1[/latex]. This means the concentration, C , the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].

Rational Function

A rational function is a ratio of polynomials where the polynomial in the denominator shouldn't be equal to zero. Isn't it resembling the definition of a rational number (which is of the form p/q, where q ≠ 0)? Did you know Rational functions find application in different fields in our day-to-day life? Not only do they describe the relationship between speed, distance, and time, but also are widely used in the medical and engineering industry.

Let us learn more about rational functions along with how to graph it, its domain, range, asymptotes, etc along with solved examples.

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What is a Rational Function?

A rational function is a function that is the ratio of polynomials. Any function of one variable, x, is called a rational function if, it can be represented as f(x) = p(x)/q(x), where p(x) and q(x) are polynomials such that q(x) ≠ 0. For example, f(x) = (x 2 + x - 2) / (2x 2 - 2x - 3) is a rational function and here, 2x 2 - 2x - 3 ≠ 0.

We know that every constant is a polynomial and hence the numerators of a rational function can be constants also. For example, f(x) = 1/(3x+1) can be a rational function. But note that the denominators of rational functions cannot be constants. For example, f(x) = (2x + 3) / 4 is NOT a rational function, rather, it is a linear function .

Rational function definition and examples

How to Identify a Rational Function?

By the definition of the rational function (from the previous section), if either the numerator or denominator is not a polynomial, then the fraction formed does NOT represent a rational function. For example, f(x) = (4 + √x)/(2-x), g(x) = (3 + (1/x)) / (2 - x), etc are NOT rational functions as numerators in these examples are NOT polynomials .

Domain and Range of Rational Function

Any fraction is not defined when its denominator is equal to 0. This is the key point that is used in finding the domain and range of a rational function.

Domain of Rational Function

The domain of a rational function is the set of all x-values that the function can take. To find the domain of a rational function y = f(x):

Set the denominator ≠ 0 and solve it for x.
Set of all real numbers other than the values of x mentioned in the last step is the domain.

Example: Find the domain of f(x) = (2x + 1) / (3x - 2).

We set the denominator not equal to zero.

3x - 2 ≠ 0 x ≠ 2/3

Thus, the domain = {x ∈ R | x ≠ 2/3}

Range of Rational Function

The range of a rational function is the set of all outputs (y-values) that it produces. To find the range of a rational function y= f(x):

If we have f(x) in the equation, replace it with y.
Solve the equation for x.
Set the denominator of the resultant equation ≠ 0 and solve it for y.
Set of all real numbers other than the values of y mentioned in the last step is the range.

Example: Find the range of f(x) = (2x + 1) / (3x - 2).

Let us replace f(x) with y. Then y = (2x + 1) / (3x - 2). Now, we will solve this for x.

(3x - 2) y = (2x + 1) 3xy - 2y = 2x + 1 3xy - 2x = 2y + 1 x (3y - 2) = (2y + 1) x = (2y + 1) / (3y - 2)

Now (3y - 2) ≠ 0 y ≠ 2/3

So the range = {y ∈ R | y ≠ 2/3}

Asymptotes of Rational Function

A rational function can have three types of asymptotes : horizontal, vertical, and slant asymptotes. Apart from these, it can have holes as well. Let us see how to find each of them.

Holes of a Rational Function

The holes of a rational function are points that seem that they are present on the graph of the rational function but they are actually not present. They can be obtained by setting the linear factors that are common factors of both numerator and denominator of the function equal to zero and solving for x. We can find the corresponding y-coordinates of the points by substituting the x-values in the simplified function. Every rational function does NOT need to have holes. Holes exist only when numerator and denominator have linear common factors.

Example: Find the holes of the function f(x) = (x 2 + 5x + 6) / (x 2 + x - 2).

Let us factorize the numerator and denominator and see whether there are any common factors.

f(x) = [ (x + 2)(x + 3) ] / [ (x + 2) (x - 1) ] = [ ̶(̶x̶ ̶+̶ ̶2̶)̶(x + 3) ] / [ ̶(̶x̶ ̶+̶ ̶2̶)̶ (x - 1) ] = (x + 3) / (x - 1)

Since (x + 2) was striked off, there is a hole at x = -2. Its y-coordinate is f(-2) = (-2 + 3) / (-2 - 1) = -1/3.

Thus, there is a hole at (-2, -1/3).

Vertical Asymptote of a Rational Function

A vertical asymptote (VA) of a function is an imaginary vertical line to which its graph appears to be very close but never touch. It is of the form x = some number. Here, "some number" is closely connected to the excluded values from the domain. But note that there cannot be a vertical asymptote at x = some number if there is a hole at the same number. A rational function may have one or more vertical asymptotes. So to find the vertical asymptotes of a rational function:

Simplify the function first to cancel all common factors (if any).
Set the denominator = 0 and solve for (x) (or equivalently just get the excluded values from the domain by avoiding the holes).

Example: Find the vertical asymptotes of the function f(x) = (x 2 + 5x + 6) / (x 2 + x - 2).

We have already seen that this function simplifies to f(x) = (x + 3) / (x - 1).

Setting the denominator to 0, we get

x - 1 = 0 x = 1

Thus, there is a VA of the given rational function is, x = 1.

Horizontal Asymptote of a Rational Function

A horizontal asymptote (HA) of a function is an imaginary horizontal line to which its graph appears to be very close but never touch. It is of the form y = some number. Here, "some number" is closely connected to the excluded values from the range. A rational function can have at most one horizontal asymptote. Easy way to find the horizontal asymptote of a rational function is using the degrees of the numerator (N) and denominators (D).

If N < D, then there is a HA at y = 0.
If N > D, then there is no HA.
If N = D, then the HA is y = ratio of the leading coefficients.

Example: Find the horizontal asymptote (if any) of the function f(x) = (x 2 + 5x + 6) / (x 2 + x - 2).

Here the degree of the numerator is, N = 2, and the degree of the denominator is, D = 2.

Since N = D, the HA is y = (leading coefficient of numerator) / (leading coefficient of denominator) = 1/1 = 1.

Thus, the HA is y = 1.

Slant (Oblique) Asymptotes of a Rational Function

A slant asymptote is also an imaginary oblique line to which a part of the graph appears to touch. A rational function has a slant asymptote only when the degree of the numerator (N) is exactly one greater than the degree of the denominator (D). Its equation is y = quotient that is obtained by dividing the numerator by denominator using the long division .

Example: Find the slant asymptote of the function f(x) = x 2 /(x+1).

Here the degree of numerator is 2 and that of denominator = 1. So it has a slant asymptote.

Let us divide x 2 by (x + 1) by long division (or we can use synthetic division as well).

slant asymptote of a rational function using long division

Thus, the slant asymptote is y = x - 1.

Graphing Rational Functions

Here are the steps for graphing a rational function:

Identify and draw the vertical asymptote using a dotted line.
Identify and draw the horizontal asymptote using a dotted line.
Plot the holes (if any)
Find x-intercept (by using y = 0) and y-intercept (by x = 0) and plot them.
Draw a table of two columns x and y and place the x-intercepts and vertical asymptotes in the table. Then take some random numbers in the x-column on either side of each of the x-intercepts and vertical asymptotes.
Compute the corresponding y-values by substituting each of them in the function.
Plot all points from the table and join them curves without touching the asymptotes.

Example: Graph the rational function f(x) = (x 2 + 5x + 6) / (x 2 + x - 2).

We have already identified that its VA is x = 1, its HA is y = 1, and the hole is at (-2, -1/3). We use dotted lines for asymptotes so that we can take care that the graph doesn't touch those lines. Note that, the simplified form of the given function is, f(x) = (x + 3) / (x - 1). Now, we will find the intercepts.

For x-intercept, put y = 0. Then we get 0 = (x + 3) / (x - 1) ⇒ x + 3 = 0 ⇒ x = -3. So the x-intercept is at (-3, 0).
For y-intercept, put x = 0. Then we get y = (0 + 3) / (0 - 1) ⇒ y = -3. So the y-intercept is at (0, -3).

We have the VA at x = 1 and x-intercept is at x = -3. Let us construct a table now with these two values in the column of x and some random numbers on either side of each of these numbers -3 and 1.

x	y
-5	y = (-5 + 3) / (-5 - 1) = 0.33
-4	y = (-4 + 3) / (-4 - 1) = 0.2

-2	y = (-2 + 3) / (-2 - 1) = -0.33
0	-3 (y-int)

2	y = (2 + 3) / (2 - 1) = 5
3	y = (3 + 3) / (3 - 1) = 3

Let us plot all these points on the graph along with all asymptotes, hole, and intercepts.

Inverse of a Rational Function

To find the inverse of a rational function y = f(x):

Replace f(x) with y.
Interchange x and y.
Solve the resultant equation for y.
The result would give the inverse f -1 (x).

Example: Find the inverse of the rational function f(x) = (2x - 1) / (x + 3).

The given function can be written as:

y = (2x - 1) / (x + 3)

Interchanging x and y:

x = (2y - 1) / (y + 3)

Now, we will solve for y.

x(y + 3) = 2y - 1

xy + 3x = 2y - 1

3x + 1 = 2y - xy

3x + 1 = y (2 - x)

y = (3x + 1) / (2 - x) = f -1 (x)

Important Notes on Rational Function:

A rational function equation is of the form f(x) = P(x) / Q(x), where Q(x) ≠ 0.
Every rational function has at least one vertical asymptote.
Every rational function has at most one horizontal asymptote.
Every rational function has at most one slant asymptote.
The excluded values of the domain of a rational function help to identify the VAs.
The excluded values of the range of a rational function help to identify the HAs.
The linear factors that get canceled when a rational function is simplified would give us the holes.

☛ Related Topics:

Graphing Functions
Simplifying Rational Expressions Calculator
Asymptote Calculator
Reciprocal Function

Rational Function Examples

Example 1: Find the horizontal and vertical asymptotes of the rational function: f(x) = (3x 3 - 6x) / (x 2 - 5).

The function is in the simplest form.

For finding VA, set x 2 - 5 = 0. Solving this, we get x = ± √5.

Since the degree of the numerator (3) > degree of the denominator (2), it has no HA.

Answer: VAs are at x = √5 and x = -√5 and there is no HA.

Example 2: Find the x-intercepts of the rational function f(x) = (x 2 + x - 2) / (x 2 - 2x - 3).

f(x) = [ (x + 2)(x - 1) ] / [(x - 3) (x + 1)]

No cancellation is possible.

To find the x-intercepts, substitute f(x) = 0.

[ (x + 2)(x - 1) ] / [(x - 3) (x + 1)] = 0

(x + 2)(x - 1) = 0

x = -2; x = 1

Answer: The x-intercepts are (-2, 0) and (1, 0).

Example 3: Is f(x) = 2 + [1 / (x +3)] a rational function? Justify.

We will add the fractions in the given function by making the common denominators.

f(x) = 2 (x + 3) / (x + 3) + [1 / (x +3)]

= (2x + 6 + 1) / (x + 3)

= (2x + 7) / (x + 3)

= p(x) / q(x), where both p(x) and q(x) are polynomials.

Answer: Hence, f(x) is a rational function.

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rational function problem solving examples

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Practice Questions on Rational Functions

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FAQs on Rational Functions

What is the definition of a rational function.

A rational function is a function that looks like a fraction where both the numerator and denominator are polynomials. It looks like f(x) = p(x) / q(x), where both p(x) and q(x) are polynomials.

What is the End Behaviour of Rational Function?

The end behaviour of the parent rational function f(x) = 1/x is:

f(x) → 0 as x → ∞ or -∞ and this corresponds to the horizontal asymptote.
f(x) → ∞ as x → 0 + and f(x) → -∞ as x → 0 - and these correspond to the vertical asymptote.

How Do You Know If a Function is Rational?

Whenever a function has polynomials in its numerator and denominator then it is a rational function. But remember:

The numerator of a rational function can be a constant. For example: 1 / x 2 is a rational function.
The denominator of a rational function cannot be a constant. For example: x 2 / 1 is NOT a rational function.

How to Graph a Rational Function?

To graph a rational function, first plot all the asymptotes by dotted lines. Plot the x and y-intercepts. Make a table with two columns labeled x and y. Put all x-intercepts and vertical asymptotes in the column of x. Take some random numbers on either side of each of these numbers and compute the corresponding y-values using the function. Plot all these points on the graph and join them by curves without touching the asymptotes.

How to Find the Domain and Range of a Rational Function?

To find the domain and range of a rational function:

First, simplify the function.
For domain, set denominator not equal to zero and solve for x.
For range, solve the simplified equation for x, set the denominator not equal to zero, and solve for y.

How to Find Holes in Rational Functions?

To find holes, first, factorize both numerator and denominator. If any linear factors are getting canceled, just set each of them to 0 and simplify. They will give the x-coordinates of the holes. We can use the function to find the corresponding y-coordinates of holes.

How to Find Asymptotes of Rational Functions?

To find the asymptotes of a rational function:

Simplify the function to its lowest form.
Set the denominator = 0 and solve to find the vertical asymptotes.
Solve the equation for x, set the denominator = 0, and solve to find horizontal asymptotes.

How to Find the Inverse of a Rational Function?

To find the inverse of a rational function y = f(x), just switch x and y first, then solve the resultant equation for y. It will give the inverse of f(x) which is represented as f -1 (x).

What are the Applications of Rational Function?

Rational functions are used to model many real-life scenarios. In particular, they are used in the fields of business, science, and medicine.

Rational Expressions

An expression that is the ratio of two polynomials :

It is just like a fraction, but with polynomials.

Other Examples:

+ 2x − 1

− x

	The top polynomial is "1" which is fine.

2x + 3	Yes it is! As it could also be written: + 3

		the top is not a polynomial (a square root of a variable is not allowed)

		1/x is not allowed in a polynomial

A rational function is the ratio of two polynomials P(x) and Q(x) like this

f(x) = P(x) Q(x)

Except that Q(x) cannot be zero (and anywhere that Q(x)=0 is undefined)

Finding Roots of Rational Expressions

A "root" (or "zero") is where the expression :

To find the roots of a Rational Expression we only need to find the the roots of the top polynomial , so long as the Rational Expression is in "Lowest Terms".

So what does "Lowest Terms" mean?

Lowest Terms

Well, a fraction is in Lowest Terms when the top and bottom have no common factors.

Example: Fractions

2 6 is not in lowest terms, as 2 and 6 have the common factor "2"

1 3 is in lowest terms, as 1 and 3 have no common factors

Likewise a Rational Expression is in Lowest Terms when the top and bottom have no common factors.

Example: Rational Expressions

x 3 +3x 2 2x is not in lowest terms, as x 3 +3x 2 and 2x have the common factor "x"

x 2 +3x 2 is in lowest terms, as x 2 +3x and 2 have no common factors

So, to find the roots of a rational expression:

Reduce the rational expression to Lowest Terms,
Then find the roots of the top polynomial

How do we find roots? Read Solving Polynomials to learn how.

Proper vs Improper

Fractions can be or :

(There is nothing wrong with "Improper", it is just a different type)

And likewise:

A Rational Expression can also be proper or improper !

But what makes a polynomial larger or smaller?

The Degree !

For a polynomial with one variable, the Degree is the largest exponent of that variable.

Examples of Degree:

		The Degree is (a variable without an exponent actually has an exponent of 1)

− x + 3		The Degree is (largest exponent of x)

So this is how to know if a rational expression is proper or improper :

Proper : the degree of the top is less than the degree of the bottom.

Proper:

deg(top) < deg(bottom)

Another Example: x x 3 − 1

Improper : the degree of the top is greater than, or equal to, the degree of the bottom.

Improper:

− 1

deg(top) ≥ deg(bottom)

Another Example: 4x 3 − 3 5x 3 + 1

If the polynomial is improper, we can simplify it with Polynomial Long Division

Rational expressions can have asymptotes (a line that a curve approaches as it heads towards infinity) :

Example: (x 2 -3x)/(2x-2)

The -3x)/(2x-2) has:

A rational expression can have:

any number of vertical asymptotes,
only zero or one horizontal asymptote,
only zero or one oblique (slanted) asymptote

Finding Horizontal or Oblique Asymptotes

It is fairly easy to find them ...

... but it depends on the degree of the top vs bottom polynomial.

The one with the larger degree will grow fastest.

Just like "Proper" and "Improper", but in fact there are four possible cases, shown below.

Let's look at each of those examples in turn:

Degree of Top Less Than Bottom

The bottom polynomial will dominate, and there is a Horizontal Asymptote at zero.

Example: f(x) = (3x+1)/(4x 2 +1)

When x is 1000:

f(1000) = 3001/4000001 = 0.00075...

And as x gets larger, f(x) gets closer to 0

Degree of Top is Equal To Bottom

Neither dominates ... the asymptote is set by the leading terms of each polynomial.

Example: f(x) = (3x+1)/(4x+1)

f(1000) = 3001/4001 = 0.750...

And as x gets larger, f(x) gets closer to 3/4

Why 3/4? Because "3" and "4" are the "leading coefficients" of each polynomial

(Technically the 7 is a constant, but here it is easier to think of them all as coefficients.)

The method is easy:

Divide the leading coefficient of the top polynomial by the leading coefficient of the bottom polynomial.

Here is another example:

Example: f(x) = (8x 3 + 2x 2 − 5x + 1)/(2x 3 + 15x + 2)

The degrees are equal (both have a degree of 3)

Just look at the leading coefficients of each polynomial:

Top is 8 (from 8x 3 )
Bottom is 2 (from 2x 3 )

So there is a Horizontal Asymptote at 8/2 = 4

Degree of Top is 1 Greater Than Bottom

This is a special case: there is an oblique asymptote , and we need to find the equation of the line.

To work it out use polynomial long division : divide the top by the bottom to find the quotient (ignore the remainder).

Example: f(x) = (3x 2 +1)/(4x+1)

The degree of the top is 2, and the degree of the bottom is 1, so there will be an oblique asymptote

We need to divide 3x 2 +1 by 4x+1 using polynomial long division:

Ignoring the remainder we get the solution (from the top of the long division):

Degree of Top is More Than 1 Greater Than Bottom

When the top polynomial is more than 1 degree higher than the bottom polynomial, there is no horizontal or oblique asymptote .

Example: f(x) = (3x 3 +1)/(4x+1)

The degree of the top is 3, and the degree of the bottom is 1.

The top is more than 1 degree higher than the bottom so there is no horizontal or oblique asymptote .

Finding Vertical Asymptotes

There is another type of asymptote, which is caused by the bottom polynomial only .

But First: make sure the rational expression is in lowest terms!

Whenever the bottom polynomial is equal to zero (any of its roots) we get a vertical asymptote.

Read Solving polynomials to learn how to find the roots

From our example above:

The bottom polynomial is 2x-2 , which factors into:

And the factor (x−1) means there is a vertical asymptote at x=1 (because 1−1=0)

A Full Example

Example: sketch (x−1)/(x 2 −9).

First of all, we can factor the bottom polynomial (it is the difference of two squares):

x−1 (x+3)(x−3)

Now we can see:

The roots of the top polynomial are: +1 (this is where it crosses the x-axis )

The roots of the bottom polynomial are: −3 and +3 (these are Vertical Asymptotes )

It crosses the y-axis when x=0, so let us set x to 0:

Crosses y-axis at: 0−1 (0+3)(0−3) = −1 −9 = 1 9

We also know that the degree of the top is less than the degree of the bottom, so there is a Horizontal Asymptote at 0

So we can sketch all of that information:

And now we can sketch in the curve:

(Compare that to the plot of (x-1)/(x 2 -9) )

10 Surprising Real-life Examples Of Rational Functions

It is interesting to find various examples of mathematics around us. Even more interesting is how these concepts influence us daily without our knowledge. If we carefully look around, we will find math influencing us more than any other subject from our school years. That’s the power of mathematics.

Although, there is no hiding that math is one of the trickiest subjects. It required advanced critical thinking and a flexible mindset to adapt and apply the mathematical concepts around us. But there is no going back once we can identify and apply math around us.

This article will take you through some real-life examples of rational functions. Go through them and see if you already know how rational function impacts the mentioned areas.

But for those new to this concept, let’s first understand rational functions.

What are rational functions?

If you think Rational function sounds like a tricky and lengthy concept that will be challenging to grasp, you should reconsider. Rational function, in reality, is one of the simplest concepts that can make challenging and advanced-level problems easy to solve.

A rational function is nothing but a ratio of two polynomial values. The term rational implies that one numerator and one denominator can be divided by each other, just like a ratio.

Now, what are polynomial values? It is an expression that involves constant values, exponents, or variables. These are combined using arithmetic operations, that is, addition, subtraction, multiplication, and division.

Considering the above statements, we can represent a rational function as-

3x² + 2y – 5 ————- 7a² × 4 ÷ 2b

Remember, the denominator can be any value but zero.

Let’s try understanding rational functions using an example.

Suppose you are on a road trip with your friends. You want to find the total time you took to reach your destination, including all the distractions and delays. There was bad traffic where you were stuck for many hours, represented by 5t and the finished fuel, represented by 3t². Let’s say 10t is your total distance.

Using the given equation, we can form a rational function equation as, Time (t) = 10t / 5t + 3t²

Rational function in real life: A few examples

There are many ways to get a hang of mathematical concepts, such as math manipulatives , online games, activities, etc. Real-life examples are one such way that is fun yet brain stimulating. Let’s see some real-life examples of rational functions.

1. Is your speed, distance, and Time calculations…. accurate?

Identifying the examples of rate of change around us is a great way to find rational functions in real-life. Rational function helps to identify the speed, distance, and time of the objects in motion. For instance, when you try to calculate how long it will take you to reach your destination in extreme traffic hours, it is the rational function you are using.

2. Trust your doctor!

Any medical treatment or procedure that involves constant monitoring of the rate of change of any drug in the body or body vitals, such as blood pressure, glucose level, etc., represents rational functions in real-life. For instance, when a doctor injects you with a drug at an interval of a specific time, the doctor constantly tracks your body vitals before and after every dose to maintain a constant and healthy condition. To keep a record, doctors use rational functions.

Similarly, keeping track of the time when some hazardous medical waste will decay, considering different environmental conditions, is an example of a rational function.

3. How sloppy are your Supply and demand curves?

Another real-life example of a rational function is the supply and demand curves and other economically relevant processes. There is a constant cycle of supply of products and their demand in the market. With such high demand, keeping track of all taxes and market dynamics is highly important to prevent any potential economic loss.

Also, fixing a suitable price that is flexible to a particular demographic, but justifying other additional expenses is important. All these processes are an example of rational functions in real-life.

4. Are your costs and revenue bringing you profit?

If you are a businessman, you know how rational functions save the day when calculating the cost of the products and the revenue generated from them. Or you might be a frustrated student trying to calculate the cost and selling prices of products for your math homework.

All you have to do is identify all the factors involved with the overall cost and revenue, and you have your profit, loss, CP, and SP calculated in minutes or maybe hours. And it’s that easy with rational functions.

5. Careful! The Electrical circuits are functional!

Have you ever wondered how current flows in electrical circuits? Or why the voltage impacts the flow of current in the circuits? Or how do physicists assess the rate of flow of current? These are the classic examples of rational functions in real-life. Several factors impact the current flow and electrical circuits’ functioning, which can be calculated using rational functions.

6. Population growth is constantly un-constant!

Our natural resources are rapidly depleting, and the population is constantly increasing. By this observation, how long can we expect the justified distribution of resources among all people? How are external factors impacting the rate of depletion of resources?

Questions like these are answered using a logistic growth model, a great example of rational functions in real-life. It helps to understand and predict how fast, moderate, or slow the resources will deplete.

7. Is your Investment functionally secure?

Have you ever noticed numerous conditions when investing in stocks or fixing a certain amount of money? Based on those conditions, you receive your final amount with earned profit. This is an example of a rational function in real-life.

Investors assess the factors impacting their money and use rational functions to identify and predict the final value. Similarly, the calculation of interests on assets and risk assessment is carried out using rational functions.

8. Is your Chemical concentration relevant?

In real life, performing experiments with chemicals and developing chemical equations are great examples of rational functions. Rational functions help determine how much concentration of a chemical will vary and at what time, keeping external conditions constant or changing. Various factors must be regulated while performing any chemical experiment, and rational functions are a great way to identify and regulate them.

9. How does your guitar produce music?

Have you ever wondered how guitars and other musical instruments produce music? Beating the drum can be a noise, but how do musicians make it a melody? It is because of the vibrations that are produced by those musical instruments. And assessing how the change in the vibrations produces different musical melodies is an example of a rational function in real-life.

10. The temperature control system requires input and output signals.

The temperature control system requires the input and output signals to function properly. Input signal might involve the setpoint temperature and output signal might involve the change in temperature done manually. These signals are an example of rational functions used to assess and analyze different temperature-related aspects such as temperature regulation, automation, and motor control.

Concluding thoughts

For many children, achieving an advanced understanding of rational functions is a mathematical goal. After all, the concept is widely used in numerous domains, making it a must to understand. The above-mentioned real-life examples help us understand how the concept is utilized in our day-to-day activities. Rational functions are an example of adaptive learning and can easily be understood and applied with the right strategies and consistent efforts.

I am Sehjal Goel, a psychology student, and a writer. I am currently pursuing my Masters’s from Banaras Hindu University, Varanasi. Child psychology has always fascinated me and I have a deep interest in learning about disabilities in children and spreading awareness regarding the same. My other areas of interest are neuropsychology and cognitive psychology. Connect me on Linkedin

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