### Lesson 12: End Behavior of Rational Functions

```Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
Lesson 12: End Behavior of Rational Functions
Student Outcomes
ο§
Students describe the end behavior of rational functions.
Lesson Notes
This lesson offers students opportunities to use tables to analyze the end behavior of rational functions and the behavior
of rational functions as they approach restricted input values. This prepares students for subsequent lessons in which
they graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing
end behavior (F-IF.C.7d).
Classwork
Opening Exercise (3 minutes)
The work in Algebra II showed students how to analyze the end behavior of polynomials. This lesson begins with a set of
exercises that provides an opportunity to recall those skills, and then the end behavior of rational functions is analyzed.
Opening Exercise
Analyze the end behavior of each function below. Then, choose one of the functions, and explain
how you determined the end behavior.
a.
π(π) = ππ
As π β β, π(π) β β. As π β ββ, π(π) β β.
We have ππ = ππ, and ππ = ππ. In general, as we use larger and larger inputs, this
function produces larger and larger outputs, exceeding all bounds.
π is an even function, so the same remarks apply to negative inputs.
b.
π(π) = βππ
As π β β, π(π) β ββ. As π β ββ, π(π) β ββ.
We have βππ = βππ, and βππ = βππ. In general, as we use larger and larger
inputs, this function produces larger and larger negative outputs, exceeding all
bounds.
π is an even function, so the same remarks apply to negative inputs.
c.
Scaffolding:
ο§ As necessary, remind
students that analyzing
the end behavior of a
function entails finding
what value π(π₯)
approaches as π₯ β β or
as π₯ β ββ.
ο§ Also consider showing
students the graphs of
π(π₯) = π₯ 2 and π(π₯) = π₯ 3
describe the end behavior
verbally.
π(π) = ππ
As π β β, π(π) β β. As π β ββ, π(π) β ββ.
We have ππ = π, and ππ = ππ. In general, as we use larger and larger inputs, this function produces larger
and larger outputs, exceeding all bounds.
We also have (βπ)π = βπ, and (βπ)π = βππ. In general, as we use lesser and lesser inputs, this function
produces larger and larger negative outputs, exceeding all bounds.
Lesson 12:
End Behavior of Rational Functions
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M3
d.
π(π) = βππ
As π β β, π(π) β ββ. As π β ββ, π(π) β β.
We have βππ = βπ, and βππ = βππ. In general, as we use larger and larger inputs, this function produces
larger and larger negative outputs, exceeding all bounds.
We also have β(βπ)π = π, and β(βπ)π = ππ. In general, as we use lesser and lesser inputs, this function
produces larger and larger outputs, exceeding all bounds.
Discussion (3 minutes): Power Functions
The following Discussion builds on the material developed in the Opening Exercises. Students analyze simple power
functions π(π₯) = π β π₯ π . This analysis is essential to understanding the end behavior of polynomials as well as that of
rational functions.
ο§
Letβs take a close look at one of these functions to make sure the reasoning is clear. For β(π₯) = π₯ 3 , do you
think the outputs grow without bound? For example, do the outputs ever exceed one trillion? Think about
this question for a moment, and then share your response with a partner. Try to be as specific as possible.
οΊ
ο§
One trillion is 1012 . If we take π₯ = 104 as an input, we get π(104 ) = (104 )3 = 1012 . So, any input
larger than 104 produces an output that is larger than one trillion.
Good. Now letβs focus on the sign of the output. Take π(π₯) = π₯ 13 and π(π₯) = π₯ 14 . When π₯ = β1000, is the
output positive or negative? Explain your thinking to a partner, being as specific as possible about how you
οΊ
Letβs consider π(β1000) = (β1000)13 . The expression (β1000)13 represents the product of 13
negative numbers. Each pair of negative numbers has a positive product, but since there is an odd
number of factors, the final product must be negative.
οΊ
Now letβs consider π(β1000) = (β1000)14 . This time, there are 7 pairs of negative numbers, each of
which has a positive product. Therefore, the final product must be positive.
Discussion (8 minutes): Reciprocals of Power Functions
ο§
We see that analyzing the end behavior of a power function is straightforward. Now letβs turn our attention to
some very simple rational functions.
ο§
What do you know about the function π(π₯) = ? Give several examples of input-output pairs associated with
1
π₯
this function, and then share them with a partner.
οΊ
1
2
This function takes a number and returns its reciprocal. For example, π(2) = , π(10) =
1
3
1
5
1
,
10
π(β3) = β , π(1) = 1, and π ( ) = 5. At π₯ = 0, π is undefined.
ο§
1
π₯
Describe the end behavior for π(π₯) = .
οΊ
ο§
As π₯ β β, π(π₯) β 0, and as π₯ β ββ, π(π₯) β 0.
Can you confirm this using a table? Organize several input-output pairs in a table to confirm the behavior of
the graph of π(π₯) =
1
as π₯ approaches 0 from the positive side and as π₯ β β.
π₯
Lesson 12:
End Behavior of Rational Functions
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π
0.001
0.01
0.1
1
10
100
1,000
π(π)
1
= 1000
0.001
1
= 100
0.01
1
= 10
0.1
1
=1
1
1
= 0.1
10
1
= 0.01
100
1
= 0.001
1000
1
π₯
What does the table indicate about π(π₯) = as π₯ β β and as π₯ β 0 from the positive side?
ο§
οΊ
ο§
It approaches 0 as π₯ β β, and it approaches β as π₯ β 0 from the positive side.
Now letβs examine the behavior of the function when the input is a negative number.
π
β0.001
π(π)
β0.01
β0.1
β1
β10
β100
β1,000
1
1
1
1
1
1
1
= β1000
= β100
= β10
= β0.1
= β0.01
= β0.001
= β1
β0.001
β0.01
β0.1
β10
β100
β1000
β1
1
π₯
What does the table indicate about π(π₯) = as π₯ β ββ and as π₯ β 0 from the negative side?
ο§
οΊ
ο§
It approaches 0 as π₯ β ββ, and it approaches ββ as π₯ β 0 from the negative side.
What pattern do you notice between the tables?
οΊ
ο§
Scaffolding:
It appears that π(βπ₯) = βπ(π₯).
ο§ Recall that even functions
What type of function displays the type of pattern shown in the tables?
οΊ
such as π(π₯) =
Odd functions
symmetry with respect to
the π¦-axis.
How can you prove that π is odd?
ο§
οΊ
π is odd if π(βπ₯) = βπ(π₯). We have π(βπ₯) =
1
1
and βπ(π₯) = β .
βπ₯
π₯
ο§ Odd functions such as
These expressions are indeed equivalent, so π is odd.
Good. Now letβs try analyzing π(π₯) =
ο§
1
exhibit
π₯2
β(π₯) =
1
. What can we say about the end
π₯2
1
exhibit
π₯3
symmetry with respect to
the origin.
behavior of this function? How does the function behave as π₯ approaches 0
from the right and left sides?
π
β100
β10
β1
β1
10
β1
100
1
100
1
10
1
10
100
π(π)
1
10,000
1
100
1
100
10,000
10,000
100
1
1
100
1
10,000
ο§
What does the table indicate about π(π₯) =
οΊ
ο§
It approaches 0 as π₯ β β, and it approaches β as π₯ β 0 from the positive side.
What does the table indicate about π(π₯) =
οΊ
1
as π₯ β β and as π₯ β 0 from the positive side?
π₯2
1
as π₯ β ββ and as π₯ β 0 from the negative side?
π₯2
It approaches 0 as π₯ β ββ, and it approaches β as π₯ β 0 from the negative side.
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
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M3
ο§
What pattern do you notice between the tables?
οΊ
ο§
What type of function displays the type of symmetry shown in the tables?
οΊ
ο§
ο§
ο§
MP.7
&
MP.8
Even functions
How can you prove that π(π₯) =
1
is even?
π₯2
Scaffolding:
1
οΊ A function is even if π(βπ₯) = π(π₯). We have π(βπ₯) = ( )2, and
βπ₯
1
π(π₯) = 2. These expressions are equivalent, so π is even.
π₯
1
1
Now letβs analyze the end behavior of π(π₯) = 3 and π(π₯) = 4.
π₯
π₯
οΊ
As π₯ β β, π(π₯) β 0, and as π₯ β ββ, π(π₯) β 0.
οΊ
As π₯ β β, π(π₯) β 0, and as π₯ β ββ, π(π₯) β 0.
ο§ Recall that even functions
such as π(π₯) =
1
exhibit
π₯2
symmetry with respect to
the π¦-axis.
ο§ Odd functions such as
β(π₯) =
What about the behavior of π(π₯) and π(π₯) as the functions approach 0 from the
right and left sides?
οΊ
οΊ
οΊ
οΊ
ο§
It appears that there is symmetry.
1
exhibit
π₯3
symmetry with respect to
the origin.
π(π₯) β β as π₯ β 0 from the right side.
π(π₯) β ββ as π₯ β 0 from the left side.
π(π₯) β β as π₯ β 0 from the right side.
π(π₯) β β as π₯ β 0 from the left side.
Can you see how to generalize your results?
1
οΊ
If π is any whole number, then as π₯ β β,
οΊ
If π is any odd whole number, then as π₯ β 0 from the positive side,
negative side,
οΊ
1
π₯π
π₯π
β 0, and as π₯ β ββ,
π₯π
1
π₯π
β 0.
β β, and as π₯ β 0 from the
β ββ.
If π is any even whole number, then as π₯ β 0 from the positive side,
negative side,
1
π₯π
1
1
π₯π
β β, and as π₯ β 0 from the
β β.
Discussion: General Polynomials (5 minutes)
ο§
1
. What can we say about the end
π₯π
5π₯3 β 2π₯2 + 4π₯ β 16
behavior of a more complex rational function such as π(π₯) =
2π₯3 + 10π₯2 β π₯ + 4
So far, weβve examined only the simplest rational functions π(π₯) =
we need to review some principles regarding the end behavior of polynomials.
ο§
MP.3
Let π(π₯) = 5π₯ 3 β 2π₯ 2 + 4π₯ β 16 and β(π₯) = 2π₯ 3 + 10π₯ 2 β π₯ + 4. Explain why the functions have similar
end behavior.
οΊ
The functions are polynomials of the same degree, and both have positive leading coefficients.
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
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M3
ο§
We saw in the Opening Exercises that functions of the form π¦ = π₯ π are simple to analyze, but recall that things
get a little more complicated when there are several terms involved. For instance, consider
π(π₯) = π₯ 3 β 5π₯ 2 β 3π₯ β 1. See what you can learn by computing π(4) and π(6) term by term. You may use
a calculator for this purpose.
οΊ
π(4) = 64 β 80 β 12 β 1 = β29
οΊ
π(6) = 216 β 180 β 18 β 1 = 17
ο§
In the case where π₯ = 4, the second term is larger than the first, and so you ended up subtracting more than
you started with, making the answer negative. But in the case where π₯ = 6, the first term was large enough
that the result was still positive even after performing three subtractions.
ο§
More than anything else, itβs the size of the exponent that matters in this analysis. You may recall that the end
behavior of a polynomial is determined by the term with the largest exponent. For instance, in the example
above, π has the same end behavior as the simple power function π(π₯) = π₯ 3 . Do you recall how to
demonstrate this formally? Try factoring out π₯ 3 from each term of π to see why its end behavior is similar to
π.
οΊ
ο§
ο§
3
1
β )
π₯2 π₯3
Use this new form for π to compare π(104 ) and π(104 ).
οΊ
π(104 ) = (104 )3 = 1012
οΊ
π(104 ) = 1012 β (1 β
5
3
1
4β
8β
12)
10
10
10
Notice that for this large input, the output of π is about 99.9% of the value of π. If you use an even larger
input, the relative value of π grows even closer to that of π. Without actually doing the calculations, can you
see intuitively why this is the case? Explain your thinking to a neighbor.
οΊ
ο§
5
π₯
π(π₯) = π₯ 3 β 5π₯ 2 β 3π₯ β 1 = π₯ 3 β (1 β β
Except for the first term, the terms in parentheses are extremely small, so the multiplier has a value
that is very close to 1. This explains why the relative values of the two functions are very close when
the input is a large number.
5
π₯
Letβs examine each component in the expression π₯ 3 β (1 β β
οΊ
As π₯ β β, π₯ 3 β β, 1 β 1,
οΊ
Thus, 1 β β
5
π₯
5
π₯
β 0,
3
π₯2
β 0, and
1
π₯3
3
1
β ). What happens when π₯ β β?
π₯2 π₯3
β 0.
3
1
β β 1 β 0 β 0 β 0 = 1.
π₯2 π₯ 3
ο§
This confirms our observation that π and π have similar end behavior. We can say that the function π(π₯) = π₯ 3
is an end-behavior model for π(π₯) = π₯ 3 β 5π₯ 2 β 3π₯ β 1.
ο§
As it happens, knowing how to analyze the end behavior of a polynomial is really all that is needed to
understand the end behavior of a rational function. Letβs look at some examples.
Example 1 (6 minutes)
In this example, students use what they know about polynomials to analyze a rational function.
ο§
MP.3
Based on our reasoning with π and π above, what would be an end-behavior model for
π(π₯) = π₯ 4 β 2π₯ 2 + 6π₯ β 3?
οΊ
π(π₯) = π₯ 4
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
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M3
ο§
Consider the function π(π₯) =
5π₯3 β 2π₯2 + 4π₯ β 16
. What can we say about the end behavior of this function?
2π₯3 + 10π₯2 β π₯ + 4
Well, letβs break this problem up into two smaller problems: What if we consider the numerator and the
denominator separately? Find an end-behavior model for the numerator and a model for the denominator,
and then take turns sharing your results with a partner.
ο§
οΊ
The function 5π₯ 3 is an end-behavior model for the numerator.
οΊ
The function 2π₯ 3 is an end-behavior model for the denominator.
Now letβs combine these observations: When we form the rational expression π(π₯), does it seem reasonable
that its end behavior is well approximated by the quotient
5
5π₯ 3
5π₯ 3
2π₯
2π₯ 3
? And since
3
5
= , it looks as though the
2
outputs of π must get close to when π₯ is very large. Use a calculator to verify this claim. Try, for example,
2
using 1010 as an input to π.
οΊ
π(1010 ) β 2.499β999β999
5
ο§
You have to admit that is pretty darn close to .
ο§
The general principle here is to focus on the term with the highest power, ignoring all of the other terms.
Looking at the quotient of the two terms with the largest power gives you an idea of what the end behavior of
a rational function is like.
ο§
Letβs take one last look at this example using factoring. Factor out the highest power of π₯ from the numerator
2
and denominator of π(π₯) =
5π₯3 β 2π₯2 + 4π₯ β 16
. This should help you to see how all of the pieces in this
2π₯3 + 10π₯2 β π₯ + 4
discussion fit together.
οΊ
π₯3(5 β 2π₯ + 42 β 163)
5π₯3 β 2π₯2 + 4π₯ β 16
π₯
π₯ .
We have π(π₯) =
= 3
1
4
2π₯3 + 10π₯2 β π₯ + 4
π₯ (2 + 10
π₯ β 2 + 3)
π₯
οΊ
The quotient of
π₯3
π₯3
2
expression
as π₯ β β.
π₯
is 1. The fractions in the expression are all approaching 0 as π₯ β β, so the
4
16
5β + 2β 3
π₯ π₯
π₯
10
1
4
2+ β 2+ 3
π₯
π₯
π₯
5
5
2
2
is approaching . This confirms our view that the function is approaching
Example 2 (5 minutes)
ο§
Now that weβve seen that we can use our knowledge of polynomials to develop end-behavior models for
rational functions, letβs get some additional practice with this technique.
ο§
Consider the function π(π₯) =
2π₯3 + 3π₯ β 1
. Find an end-behavior model for this function.
π₯2 + π₯ + 1
οΊ
The function 2π₯ 3 is an end-behavior model for the numerator.
οΊ
The function π₯ 2 is an end-behavior model for the denominator.
οΊ
Thus, the expression
Lesson 12:
2π₯ 3
π₯2
= 2π₯ is an end-behavior model for π.
End Behavior of Rational Functions
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This file derived from PreCal-M3-TE-1.3.0-08.2015
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Lesson 12
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M3
ο§
ο§
What conclusions can you draw about the end behavior of π based on your analysis?
οΊ
We know that as π₯ β β, 2π₯ β β, so we conclude that π(π₯) β β as well.
οΊ
Similarly, we know that as π₯ β ββ, 2π₯ β ββ, so we conclude that π(π₯) β ββ as well.
Use factoring to confirm that your analysis of the end behavior of π is correct.
οΊ
π(π₯) =
3
1
3
2 + 32 β 13
2π₯3 + 3π₯ β 1 π₯ β(2 + π₯2 β π₯3)
π₯
π₯
=
=
π₯
β
π₯2 + π₯ + 1
π₯2β(1 +1π₯ + 12)
1 + 1π₯ + 12
π₯
οΊ
π₯
2
Each fraction is approaching 0 as π₯ β β, so the expression on the right is approaching . This confirms
1
that π¦ = 2π₯ is an end-behavior model for π¦ = π(π₯).
ο§
So far, so good. Now letβs use technology to further our understanding of this function. When you enter the
οΊ
π
π = π(π)
100
198
500
398
1,000
1,998
5,000
9,998
π
π = π(π)
β100
β202
β500
β1,002
β1,000
β2,002
β5,000
β10,002
The numerical data confirms our views about the end behavior of π.
Example 3 (4 minutes)
ο§
Letβs look at one final example together. What happens if we switch the numerator and the denominator in
the previous example? We get the expression π(π₯) =
π₯2 + π₯ + 1
. What do you suppose the end behavior of
2π₯3 + 3π₯ β 1
this function is like? Make a conjecture, and share it with a neighbor.
οΊ
ο§
ο§
ο§
In the original function, the outputs got very large. Since this function is the reciprocal of the previous
example, perhaps the outputs are very small.
Find an end-behavior model for this function.
οΊ
The function π₯ 2 is an end-behavior model for the numerator.
οΊ
The function 2π₯ 3 is an end-behavior model for the denominator.
οΊ
Thus, the expression
π₯2
2π₯ 3
=
1
π₯
is an end-behavior model for π.
What conclusions can you draw about the end behavior of π based on your analysis?
1
οΊ
We know that as π₯ β β,
οΊ
Similarly, we know that as π₯ β ββ,
π₯
β 0, so we conclude that π(π₯) β 0 as well.
1
π₯
β 0, so we conclude that π(π₯) β 0 as well.
π
π = π(π)
οΊ
β200
β0.0025
β100
β0.0049
100
0.0050
200
0.0025
The numerical data confirms our views about the end behavior of π.
Lesson 12:
End Behavior of Rational Functions
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Exercises (5 minutes)
Give students time to work on the following exercises. Encourage students to compare their answers with a partner.
Select three students to present their work to the class and to explain their thinking.
Exercises
Determine the end behavior of each rational function below.
π(π) =
1.
πππ β ππ + π
πππ + π
This function has the same end behavior as
πππ π π
= π .
πππ π
Using this model as a guide, we conclude that π(π) β β as π β β and that π(π) β β as π β ββ.
π(π) =
2.
π
ππ β ππ + π
πππ + π
This function has the same end behavior as
πππ π
= .
πππ π
Using this model as a guide, we conclude that π(π) β
π(π) =
3.
π
π
as π β β and that π(π) β as π β ββ.
π
π
π
ππ + π
π
ππ β ππ + π
This function has the same end behavior as
πππ
π
=
.
πππ πππ
Using this model as a guide, we conclude that π(π) β π as π β β and that π(π) β π as π β ββ.
Closing (1 minutes)
ο§
Take a minute to write a summary in your notebook of what you learned today.
οΊ
We can use what we know about the end behavior of polynomials to analyze the end behavior of
rational functions. The key point to understand is that the term with the highest exponent determines
the end behavior of a polynomial.
οΊ
We also explored how to describe the behavior of rational functions as they approach restricted input
values.
Exit Ticket (5 minutes)
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
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M3
Name
Date
Lesson 12: End Behavior of Rational Functions
Exit Ticket
Given π(π₯) =
π₯+2
, find the following, and justify your findings.
π₯2β1
a.
The end-behavior model for the numerator
b.
The end-behavior model for the denominator
c.
The end-behavior model for π
d.
What is the value of π(π₯) as π₯ β β?
e.
What is the value of π(π₯) as π₯ β ββ?
f.
What is the value of π(π₯) as π₯ β 1 from the positive and negative sides?
g.
What is the value of π(π₯) as π₯ β β1 from the positive and negative sides?
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
M3
Exit Ticket Sample Solutions
Given π(π) =
a.
π+π
, find the following, and justify your findings.
ππ β π
The end-behavior model for the numerator
π
The end-behavior model is π(π) = ππ where π is the greatest power of the expression. Here, the numerator is
π + π, so π = π, and the model is π(π) = π.
b.
The end-behavior model for the denominator
ππ
The end-behavior model is π(π) = ππ where π is the greatest power of the expression. Here, the
denominator is ππ β π, so π = π, and the model is π(π) = ππ .
c.
The end-behavior model for π
π
π
=
ππ π
I replaced the numerator and denominator with the end-behavior models of each and simplified.
d.
What is the value of π(π) as π β β?
π(π) β π
As
π β β, π(π) β
e.
π
= π.
β
What is the value of π(π) as π β ββ?
π(π) β π
As
π β ββ, π(π) β
f.
π
= π.
ββ
What is the value of π(π) as π β π from the positive and negative sides?
π(π) β β as π β π from the positive side. As π nears π from the positive side, the numerator of the function
approaches π, and the denominator becomes a tiny positive number. The ratio of these is a very large
positive number that exceeds any bounds.
π(π) β ββ as π β π from the negative side. As π nears π from the negative side, the numerator of the
function approaches π, and the denominator becomes a tiny negative number. The ratio of these is a very
large negative number that exceeds any bounds.
g.
What is the value of π(π) as π β βπ from the positive and negative sides?
π(π) β ββ as π β βπ from the positive side. As π nears βπ from the positive side, the numerator of the
function approaches π, and the denominator becomes a tiny negative number. The ratio of these is a very
large negative number that exceeds any bounds.
π(π) β β as π β βπ from the negative side. As π nears βπ from the negative side, the numerator of the
function approaches π, and the denominator becomes a tiny positive number. The ratio of these is a very
large positive number that exceeds any bounds.
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
M3
Problem Set Sample Solutions
1.
Analyze the end behavior of both functions.
a.
π(π) = π, π(π) =
π
π
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β β as π β π, and π(π) β ββ as π β π.
b.
π(π) = ππ , π(π) =
π
ππ
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β β as π β π, and π(π) β ββ as π β π.
c.
π(π) = ππ , π(π) =
π
ππ
For π(π): π(π) β β as π β β, and π(π) β β as π β ββ.
For π(π): π(π) β β as π β π, and π(π) β β as π β π.
d.
π(π) = ππ , π(π) =
π
ππ
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β β as π β π, and π(π) β ββ as π β π.
e.
π(π) = π β π, π(π) =
π
πβπ
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β β as π β π, and π(π) β ββ as π β π.
f.
π(π) = π + π, π(π) =
π
π+π
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β π as π β β, and π(π) β π as π β ββ.
g.
π(π) = ππ β π, π(π) =
π
ππ β π
For π(π): π(π) β β as π β β, and π(π) β ββ as π β ββ.
For π(π): π(π) β β a π β π, and π(π) β ββ as π β π.
Lesson 12:
End Behavior of Rational Functions
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M3
2.
For the following functions, determine the end behavior. Confirm your answer with a table of values.
a.
π(π) =
ππ β π
π+π
π has the same end behavior as
ππ
= π. Using this model as a guide, we conclude that
π
π(π) β π as π β β, and π(π) β π as π β ββ.
b.
π
π = π(π)
πππ
π. πππ
πππ
π. πππ
π, πππ
π. πππ
π, πππ
π. πππ
π
π = π(π)
βπππ
βπ. πππ
βπππ
π. πππ
βπ, πππ
π. πππ
βπ, πππ
π. πππ
π(π) =
ππ + π
ππ β π β π
π has the same end behavior as
ππ
ππ
π
= . Using this model as a guide, we conclude that
π
π(π) β π as π β β, and π(π) β π as π β ββ.
c.
π
π = π(π)
πππ
π. πππ
πππ
π. πππ
π, πππ
π. πππ
π, πππ
π. πππ
π
π = π(π)
βπππ
βπ. πππ
βπππ
βπ. πππ
βπ, πππ
βπ. πππ
βπ, πππ
βπ. πππ
π(π) =
ππ β π
ππ β π
π has the same end behavior as
ππ
ππ
= π. Using this model as a guide, we conclude that
π(π) β β as π β β, and π(π) β ββ as π β ββ.
d.
π
π = π(π)
πππ
πππ
πππ
πππ
π, πππ
π, πππ
π, πππ
π, πππ
π
π = π(π)
βπππ
βπππ
βπππ
βπππ
βπ, πππ
βπ, πππ
βπ, πππ
βπ, πππ
π(π) =
ππ β π
ππ β π
π has the same end behavior as
ππ
ππ
π
= . Using this model as a guide, we conclude that
π
π(π) β π as π β β, and π(π) β π as π β ββ.
π
π = π(π)
πππ
π. πππ
πππ
π. πππ
π, πππ
π. πππ
π, πππ
π. πβπππ
π
π = π(π)
βπππ
βπ. πππ
βπππ
βπ. πππ
βπ, πππ
βπ. πππ
βπ, πππ
βπ. πβπππ
Lesson 12:
End Behavior of Rational Functions
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M3
e.
π(π) =
π
(ππ + π)
π
(ππ β π)
π has the same end behavior as
πππ
ππ
π
= . Using this model as a guide, we conclude that
π
π(π) β π as π β β, and π(π) β π as π β ββ.
3.
π
π = π(π)
πππ
π. πππ
πππ
π. πππ
π, πππ
π. πππ
π, πππ
π. πβπππ
π
π = π(π)
βπππ
βπ. πππ
βπππ
βπ. πππ
βπ, πππ
βπ. πππ
βπ, πππ
βπ. πβπππ
For the following functions, determine the end behavior.
a.
π(π) =
πππ β πππ + π β π
πππ β πππ + π β π
π has the same end behavior as
πππ
πππ
= ππ . Using this model as a guide, we conclude that
π(π) β β as π β β, and π(π) β β as π β ββ.
b.
π(π) =
πππ β πππ + π β π
πππ β πππ + π β π
π has the same end behavior as
πππ
πππ
=
π
ππ
. Using this model as a guide, we conclude that
π(π) β π as π β β, and π(π) β π as π β ββ.
c.
π(π) =
πππ β πππ + π β π
πππ β πππ + π β π
π has the same end behavior as
πππ
πππ
= π. Using this model as a guide, we conclude that
π(π) β π as π β β, and π(π) β π as π β ββ.
d.
π(π) =
βπππ
+π+π
ππ + π
π has the same end behavior as
βπππ
βπ
=
π. Using this model as a guide, we conclude that
ππ
π
π(π) β β as π β β, and π(π) β ββ as π β ββ.
e.
π(π) =
π
ππ β ππ β π
πππ + π β π
π has the same end behavior as
πππ
πππ
π
= .
π
π(π) β π as π β β, and π(π) β π as π β ββ.
Lesson 12:
End Behavior of Rational Functions
This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from PreCal-M3-TE-1.3.0-08.2015
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
4.
Determine the end behavior of each function.
a.
π(π) =
π¬π’π§(π)
π
βπ β€ π¬π’π§ π β€ π,
π(π) β π as π β β, and π(π) β π as π β ββ.
b.
π(π) =
ππ¨π¬(π)
π
βπ β€ ππ¨π¬ π β€ π,
π(π) β π as π β β, and π(π) β π as π β ββ.
c.
π(π) =
π
π
π
π(π) β β as π β β, and π(π) β π as π β ββ.
d.
π(π) =
π
π
π
π(π) β π as π β β, and π(π) β ββ as π β ββ.
e.
π(π) =
π
βπ
π+π
π(π) β π as π β β, and π(π) β π as π β ββ.
f.
π(π) =
ππ
βπ
π+π
π(π) β ππ as π β β, and π(π) β π as π β ββ.
5.
Consider the functions π(π) = π! and π(π) = ππ for natural numbers π.
a.
What are the values of π(π) and π(π) for π = π, ππ, ππ, ππ, ππ?
π(π) = {πππ, πβπππβπππ, πβπππβπππβπππβπππ, π. π × ππππ , π. π × ππππ }
π(π) = {ππππ, πππβπππ, πππβπππ, πβπππβπππ, πβπππβπππ}
b.
What is the end behavior of π(π) as π β β?
π(π) β β
c.
What is the end behavior of π(π) as π β β?
π(π) β β
d.
Make an argument for the end behavior of
π(π)
π(π)
as π β β.
Since π = π(π) increases so much faster than π = π(π), π(π) overpowers the division by π(π). By π = ππ,
π(π)
π(π)
β π. π × ππππ . Thus,
Lesson 12:
π(π)
π(π)
β β.
End Behavior of Rational Functions
This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from PreCal-M3-TE-1.3.0-08.2015
216
Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
e.
Make an argument for the end behavior of
π(π)
as π β β.
π(π)
For the same reasons as in part (d), division by π(π) overpowers π(π) in the numerator. By π = ππ,
π(π)
π(π)
6.
β π. πππβπππβπππβπππβπππ. Thus,
π(π)
π(π)
β π.
Determine the end behavior of the functions.
a.
π(π) =
π
π
, π(π) =
π
ππ
π(π) β π as π β β, and π(π) β π as π β ββ.
π(π) β π as π β β, and π(π) β π as π β ββ.
b.
π(π) =
π+π
π
, π(π) =
πβπ
ππ β π
π(π) β π as π β β, and π(π) β π as π β ββ.
π(π) β π as π β β, and π(π) β π as π β ββ.
c.
π(π) =
πβπ
π
, π(π) =
π+π
ππ β π β π
π(π) β π as π β β, and π(π) β π as π β ββ.
π(π) β π as π β β, and π(π) β π as π β ββ.
d.
π(π) =
ππ β π
, π(π) = ππ + π + π
πβπ
π(π) β β as π β β, and π(π) β β as π β ββ.
π(π) β β as π β β, and π(π) β β as π β ββ.
7.
Use a graphing utility to graph the following functions π and π. Explain why they have the same graphs. Determine
the end behavior of the functions and whether the graphs have any horizontal asymptotes.
a.
π(π) =
π+π
π
, π(π) = π +
πβπ
πβπ
The graphs are the same because the expressions are equivalent: π +
π β  π.
π
πβπ
π
π+π
=
+
=
for all
πβπ πβπ πβπ πβπ
π(π) β π as π β β, and π(π) β π as π β ββ.
π(π) β π as π β β, and π(π) β π as π β ββ.
b.
π(π) =
βππ + π
π
, π(π) =
βπ
π+π
π+π
The graphs are the same because the expressions are equivalent:
π
π+π
βπ=
π
π+π
β
π(π + π)
π+π
=
π β ππ β π
π+π
=
βππ +π
π+π
for all π β  βπ
π(π) β βπ as π β β, and π(π) β βπ as π β ββ.
π(π) β βπ as π β β, and π(π) β βπ as π β ββ.
Lesson 12:
End Behavior of Rational Functions
This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from PreCal-M3-TE-1.3.0-08.2015
217