Learning Objectives
In this section, you will:
- Use arrow notation.
- Identify vertical asymptotes.
- Identify horizontal asymptotes.
Let’s Get Started…
Suppose we know that the cost of making a product is dependent on the number of items, [latex]x,[/latex], 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 [latex]x[/latex] items, we would divide the cost function by the number of items, [latex]x[/latex]. The average cost function, which yields the average cost per item for [latex]x[/latex] items produced, is
[latex]f\left(x\right)=\frac{15,000x-0.1{x}^{2}+1000}{x}[/latex].
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 Section 1.1. Examine these graphs, as shown in Figure 1, and notice some of their features.
Figure 1
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 [latex]x-[/latex]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 [latex]x-[/latex]axis [latex]\left(y=0\right)\; \text{as} \; x\to \infty[/latex].
To summarize, we use arrow notation to show that [latex]x[/latex] or [latex]f\left(x\right)[/latex] is approaching a particular value. See Table 1.
| Symbol | Meaning |
|---|---|
| [latex]x\to {a}^{-}[/latex] | [latex]x[/latex] approaches [latex]a[/latex] from the left ( [latex]x\lt a[/latex] but close to [latex]a[/latex] ) |
| [latex]x\to {a}^{+}[/latex] | [latex]x[/latex] approaches [latex]a[/latex] from the right ( [latex]x \gt a[/latex] but close to [latex]a[/latex] ) |
| [latex]x\to \infty[/latex] | [latex]x[/latex] approaches infinity ( [latex]x[/latex] increases without bound) |
| [latex]x\to -\infty[/latex] | [latex]x[/latex] approaches negative infinity ( [latex]x[/latex] 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 [latex]a[/latex] |
Table 1
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 Table 2.
| [latex]x[/latex] | –0.1 | –0.01 | –0.001 | –0.0001 |
| [latex]f\left(x\right)=\frac{1}{x}[/latex] | –10 | –100 | –1000 | –10,000 |
Table 2
We write in arrow notation [latex]\text{as }x\to {0}^{-}, \; f\left(x\right)\to -\infty[/latex].
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.
| [latex]x[/latex] | 0.1 | 0.01 | 0.001 | 0.0001 |
| [latex]f\left(x\right)=\frac{1}{x}[/latex] | 10 | 100 | 1000 | 10,000 |
Table 3
We write in arrow notation [latex]\text{as }x\to {0}^{+}, \; f\left(x\right)\to \infty[/latex]. 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 [latex]x=0[/latex] as the input becomes close to zero. See Figure 3.
Figure 3
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 [latex]a.[/latex] We write
[latex]\text{As }x\to a, \; f\left(x\right)\to \infty,\; \text{or as } \; x\to a, \; f\left(x\right)\to -\infty .[/latex]
End Behavior of [latex]\,f\left(x\right)=\frac{1}{x}[/latex]
As the values of [latex]x[/latex] approach infinity, the function values approach 0. As the values of [latex]x[/latex] approach negative infinity, the function values approach 0. See Figure 4. 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]
Figure 4
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 [latex]y=0.[/latex] See Figure 5.
Figure 5
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
[latex]\text{As }x\to \infty \; \text{or} \; x\to -\infty, \; \text{ }f\left(x\right)\to b.[/latex]
EXAMPLE 1
Using Arrow Notation
Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 6.
Figure 6
Show/Hide Solution
Solution
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].
[latex]\text{As }x\to {2}^{-}, \; f\left(x\right)\to -\infty, \; \; \text{and as} \; \; x\to {2}^{+}, \; \text{ }f\left(x\right)\to \infty .[/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.
[latex]\text{As }x\to \infty, \; \text{ }f\left(x\right)\to 4 \; \; \text{and as} \; \; x\to -\infty, \; \text{ }f\left(x\right)\to 4.[/latex]
Try It #1
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.
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Solution
Shifting the graph left 2 and up 3 would result in the function
[latex]f\left(x\right)=\frac{1}{x+2}+3[/latex]
or equivalently, by giving the terms a common denominator,
[latex]f\left(x\right)=\frac{3x+7}{x+2}[/latex]
The graph of the shifted function is displayed in Figure 7.
Figure 7
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]
[latex]\text{As }x\to -{2}^{-}, \; f\left(x\right)\to -\infty, \; \text{and as} \; x\to -{2}^{+}, \; f\left(x\right)\to \infty.[/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]
[latex]\text{As }x\to ±\infty , \; f\left(x\right)\to 3.[/latex]
Analysis
Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function.
Try It #2
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.
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.
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, or “holes.”
EXAMPLE 3
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].
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Solution
First, factor the numerator and denominator.
[latex]\begin{array}{ccc}\hfill k\left(x\right)& =& \frac{5+2{x}^{2}}{2-x-{x}^{2}}\hfill \\ & =& \frac{5+2{x}^{2}}{\left(2+x\right)\left(1-x\right)}\hfill \end{array}[/latex]
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero
[latex]\begin{array}{ccc}\hfill \left(2+x\right)\left(1-x\right)& =& 0\hfill \\ \hfill x& =& -2,1\hfill \end{array}[/latex]
Neither [latex]x=–2[/latex] nor [latex]x=1[/latex] are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure 8 confirms the location of the two vertical asymptotes.
Figure 8
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 [latex]\,y=0.[/latex]
[latex]\text{Example: }f\left(x\right)=\frac{4x+2}{{x}^{2}+4x-5}[/latex]
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 [latex]y=0.[/latex] See Figure 9. Note that this graph crosses the horizontal asymptote.
Figure 9
Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote.
[latex]\text{Example: }f\left(x\right)=\frac{3{x}^{2}-2x+1}{x-1}[/latex]
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 [latex]f[/latex] will behave similarly to [latex]g[/latex], it will approach a line close to [latex]y=3x[/latex]. This line is a slant asymptote.
To find the equation of the slant asymptote, divide [latex]\frac{3{x}^{2}-2x+1}{x-1}[/latex]. The quotient is [latex]3x+1[/latex], and the remainder is 2. The slant asymptote is the graph of the line [latex]g\left(x\right)=3x+1[/latex]. See Figure 10.
Figure 10
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].
[latex]\text{Example: }f\left(x\right)=\frac{3{x}^{2}+2}{{x}^{2}+4x-5}[/latex]
In this case, the end behavior is [latex]f\left(x\right)\approx \frac{3{x}^{2}}{{x}^{2}}=3[/latex]. This tells us that as the inputs grow large, this function will behave like the function [latex]g\left(x\right)=3[/latex], which is a horizontal line. As [latex]x\to ±\infty, \; f\left(x\right)\to 3[/latex], resulting in a horizontal asymptote at [latex]y=3[/latex]. See Figure 11. Note that this graph crosses the horizontal asymptote.
Figure 11
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
[latex]f\left(x\right)=\frac{3{x}^{5}-{x}^{2}}{x+3}[/latex]
with end behavior
[latex]f\left(x\right)\approx \frac{3{x}^{5}}{x}=3{x}^{4},[/latex]
the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient.
[latex]x\to ±\infty, \; f\left(x\right)\to \infty[/latex]
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 [latex]y=0.[/latex]
- 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 4
Identifying Horizontal and Slant Asymptotes
For the functions listed, 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]
Show/Hide Solution
Solution
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].
- [latex]h\left(x\right)=\frac{{x}^{2}-4x+1}{x+2}:[/latex] The degree of [latex]p=2[/latex] and degree of [latex]q=1[/latex]. Since [latex]p \gt q[/latex] by 1, there is a slant asymptote found at [latex]\frac{{x}^{2}-4x+1}{x+2}[/latex].
The quotient is [latex]x–6[/latex] and the remainder is 13. There is a slant asymptote at [latex]y=x–6[/latex].
- [latex]k\left(x\right)=\frac{{x}^{2}+4x}{{x}^{3}-8}:[/latex] The degree of [latex]p=2\text{ } \lt[/latex] degree of [latex]q=3[/latex], so there is a horizontal asymptote [latex]y=0[/latex].
EXAMPLE 5
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.
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Solution
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 [latex]t[/latex], with coefficient 1. In the denominator, the leading term is [latex]10t[/latex], with coefficient 10. The horizontal asymptote will be at the ratio of these values:
[latex]t\to \infty, \; C\left(t\right)\to \frac{1}{10}.[/latex]
This function will have a horizontal asymptote at [latex]y=\frac{1}{10}[/latex].
This tells us that as the values of [latex]t[/latex] increase, the values of [latex]C[/latex] 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 6
Identifying Horizontal and Vertical Asymptotes
Find the horizontal and vertical asymptotes of the function
[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]
Show/Hide Solution
Solution
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 ±\infty, \; f\left(x\right)\to 0[/latex]. This function will have a horizontal asymptote at [latex]y=0[/latex]. See Figure 12.
Figure 12
Try It #3
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]
Section 3.4 Exercises
[Answers to odd problem numbers are provided at the end of the problem set. Just scroll down!]
Verbal
1. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function?
2. Can a graph of a rational function have no vertical asymptote? If so, how?
Algebraic
For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions.
3. [latex]f\left(x\right)=\frac{2}{5x+2}[/latex]
4. [latex]f\left(x\right)=\frac{x}{{x}^{2}-9}[/latex]
5. [latex]f\left(x\right)=\frac{x}{{x}^{2}+5x-36}[/latex]
6. [latex]f\left(x\right)=\frac{3+x}{{x}^{3}-27}[/latex]
7. [latex]f\left(x\right)=\frac{3x-4}{{x}^{3}-16x}[/latex]
8. [latex]f\left(x\right)=\frac{{x}^{2}-1}{{x}^{3}+9{x}^{2}+14x}[/latex]
9. [latex]f\left(x\right)=\frac{x+5}{{x}^{2}-25}[/latex]
10. [latex]f\left(x\right)=\frac{x-4}{x-6}[/latex]
11. [latex]f\left(x\right)=\frac{4-2x}{3x-1}[/latex]
12. [latex]f\left(x\right)=\frac{4}{x-1}[/latex]
Answers to Section 3.4 Odd Problems
1. The rational function will be represented by a quotient of polynomial functions.
3. V.A. at [latex]\,x=–\frac{2}{5};\,[/latex] H.A. at [latex]\,y=0;\,[/latex] Domain is all reals [latex]\,x\ne –\frac{2}{5}[/latex]
5. V.A. at [latex]\,x=4, –9;\,[/latex] H.A. at [latex]\,y=0;\,[/latex] Domain is all reals [latex]\,x\ne 4, –9[/latex]
7. V.A. at [latex]\,x=0, 4, -4;\,[/latex] H.A. at [latex]\,y=0;[/latex] Domain is all reals [latex]\,x\ne 0,4, –4[/latex]
9. V.A. at [latex]\,x=-5;\,[/latex] H.A. at [latex]\,y=0;\,[/latex] Domain is all reals [latex]\,x\ne 5,-5[/latex]
11. V.A. at [latex]\,x=\frac{1}{3};\,[/latex] H.A. at [latex]\,y=-\frac{2}{3};\,[/latex] Domain is all reals [latex]\,x\ne \frac{1}{3}.[/latex]
