Lesson 12: End Behavior of Rational Functions

Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
Lesson 12: End Behavior of Rational Functions
Student Outcomes

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.
𝒇(𝒙) = π’™πŸ’
As 𝒙 β†’ ∞, 𝒇(𝒙) β†’ ∞. As 𝒙 β†’ βˆ’βˆž, 𝒇(𝒙) β†’ ∞.
We have πŸπŸ’ = πŸπŸ”, and πŸ‘πŸ’ = πŸ–πŸ. In general, as we use larger and larger inputs, this
function produces larger and larger outputs, exceeding all bounds.
𝒇 is an even function, so the same remarks apply to negative inputs.
b.
π’ˆ(𝒙) = βˆ’π’™πŸ’
As 𝒙 β†’ ∞, π’ˆ(𝒙) β†’ βˆ’βˆž. As 𝒙 β†’ βˆ’βˆž, π’ˆ(𝒙) β†’ βˆ’βˆž.
We have βˆ’πŸπŸ’ = βˆ’πŸπŸ”, and βˆ’πŸ‘πŸ’ = βˆ’πŸ–πŸ. In general, as we use larger and larger
inputs, this function produces larger and larger negative outputs, exceeding all
bounds.
π’ˆ is an even function, so the same remarks apply to negative inputs.
c.
Scaffolding:
 As necessary, remind
students that analyzing
the end behavior of a
function entails finding
what value 𝑓(π‘₯)
approaches as π‘₯ β†’ ∞ or
as π‘₯ β†’ βˆ’βˆž.
 Also consider showing
students the graphs of
𝑓(π‘₯) = π‘₯ 2 and 𝑔(π‘₯) = π‘₯ 3
and then asking them to
describe the end behavior
verbally.
𝒉(𝒙) = π’™πŸ‘
As 𝒙 β†’ ∞, 𝒉(𝒙) β†’ ∞. As 𝒙 β†’ βˆ’βˆž, 𝒉(𝒙) β†’ βˆ’βˆž.
We have πŸπŸ‘ = πŸ–, and πŸ‘πŸ‘ = πŸπŸ•. In general, as we use larger and larger inputs, this function produces larger
and larger outputs, exceeding all bounds.
We also have (βˆ’πŸ)πŸ‘ = βˆ’πŸ–, and (βˆ’πŸ‘)πŸ‘ = βˆ’πŸπŸ•. 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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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
d.
π’Œ(𝒙) = βˆ’π’™πŸ‘
As 𝒙 β†’ ∞, π’Œ(𝒙) β†’ βˆ’βˆž. As 𝒙 β†’ βˆ’βˆž, π’Œ(𝒙) β†’ ∞.
We have βˆ’πŸπŸ‘ = βˆ’πŸ–, and βˆ’πŸ‘πŸ‘ = βˆ’πŸπŸ•. In general, as we use larger and larger inputs, this function produces
larger and larger negative outputs, exceeding all bounds.
We also have βˆ’(βˆ’πŸ)πŸ‘ = πŸ–, and βˆ’(βˆ’πŸ‘)πŸ‘ = πŸπŸ•. 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 𝑔(π‘₯) = π‘Ž βˆ™ π‘₯ 𝑛 . This analysis is essential to understanding the end behavior of polynomials as well as that of
rational functions.

Let’s take a close look at one of these functions to make sure the reasoning is clear. For β„Ž(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.
οƒΊ

One trillion is 1012 . If we take π‘₯ = 104 as an input, we get 𝑓(104 ) = (104 )3 = 1012 . So, any input
larger than 104 produces an output that is larger than one trillion.
Good. Now let’s focus on the sign of the output. Take 𝑓(π‘₯) = π‘₯ 13 and 𝑔(π‘₯) = π‘₯ 14 . When π‘₯ = βˆ’1000, is the
output positive or negative? Explain your thinking to a partner, being as specific as possible about how you
reached your conclusion.
οƒΊ
Let’s consider 𝑓(βˆ’1000) = (βˆ’1000)13 . The expression (βˆ’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.
οƒΊ
Now let’s consider 𝑔(βˆ’1000) = (βˆ’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

We see that analyzing the end behavior of a power function is straightforward. Now let’s turn our attention to
some very simple rational functions.

What do you know about the function 𝑓(π‘₯) = ? Give several examples of input-output pairs associated with
1
π‘₯
this function, and then share them with a partner.
οƒΊ
1
2
This function takes a number and returns its reciprocal. For example, 𝑓(2) = , 𝑓(10) =
1
3
1
5
1
,
10
𝑓(βˆ’3) = βˆ’ , 𝑓(1) = 1, and 𝑓 ( ) = 5. At π‘₯ = 0, 𝑓 is undefined.

1
π‘₯
Describe the end behavior for 𝑓(π‘₯) = .
οƒΊ

As π‘₯ β†’ ∞, 𝑓(π‘₯) β†’ 0, and as π‘₯ β†’ βˆ’βˆž, 𝑓(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 𝑓(π‘₯) =
1
as π‘₯ approaches 0 from the positive side and as π‘₯ β†’ ∞.
π‘₯
Lesson 12:
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
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
𝒙
0.001
0.01
0.1
1
10
100
1,000
𝒇(𝒙)
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 𝑓(π‘₯) = as π‘₯ β†’ ∞ and as π‘₯ β†’ 0 from the positive side?

οƒΊ

It approaches 0 as π‘₯ β†’ ∞, and it approaches ∞ as π‘₯ β†’ 0 from the positive side.
Now let’s examine the behavior of the function when the input is a negative number.
𝒙
βˆ’0.001
𝒇(𝒙)
βˆ’0.01
βˆ’0.1
βˆ’1
βˆ’10
βˆ’100
βˆ’1,000
1
1
1
1
1
1
1
= βˆ’1000
= βˆ’100
= βˆ’10
= βˆ’0.1
= βˆ’0.01
= βˆ’0.001
= βˆ’1
βˆ’0.001
βˆ’0.01
βˆ’0.1
βˆ’10
βˆ’100
βˆ’1000
βˆ’1
1
π‘₯
What does the table indicate about 𝑓(π‘₯) = as π‘₯ β†’ βˆ’βˆž and as π‘₯ β†’ 0 from the negative side?

οƒΊ

It approaches 0 as π‘₯ β†’ βˆ’βˆž, and it approaches βˆ’βˆž as π‘₯ β†’ 0 from the negative side.
What pattern do you notice between the tables?
οƒΊ

Scaffolding:
It appears that 𝑓(βˆ’π‘₯) = βˆ’π‘“(π‘₯).
 Recall that even functions
What type of function displays the type of pattern shown in the tables?
οƒΊ
such as 𝑔(π‘₯) =
Odd functions
symmetry with respect to
the 𝑦-axis.
How can you prove that 𝑓 is odd?

οƒΊ
𝑓 is odd if 𝑓(βˆ’π‘₯) = βˆ’π‘“(π‘₯). We have 𝑓(βˆ’π‘₯) =
1
1
and βˆ’π‘“(π‘₯) = βˆ’ .
βˆ’π‘₯
π‘₯
 Odd functions such as
These expressions are indeed equivalent, so 𝑓 is odd.
Good. Now let’s try analyzing 𝑔(π‘₯) =

1
exhibit
π‘₯2
β„Ž(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?
𝒙
βˆ’100
βˆ’10
βˆ’1
βˆ’1
10
βˆ’1
100
1
100
1
10
1
10
100
π’ˆ(𝒙)
1
10,000
1
100
1
100
10,000
10,000
100
1
1
100
1
10,000

What does the table indicate about 𝑔(π‘₯) =
οƒΊ

It approaches 0 as π‘₯ β†’ ∞, and it approaches ∞ as π‘₯ β†’ 0 from the positive side.
What does the table indicate about 𝑔(π‘₯) =
οƒΊ
1
as π‘₯ β†’ ∞ and as π‘₯ β†’ 0 from the positive side?
π‘₯2
1
as π‘₯ β†’ βˆ’βˆž and as π‘₯ β†’ 0 from the negative side?
π‘₯2
It approaches 0 as π‘₯ β†’ βˆ’βˆž, and it approaches ∞ as π‘₯ β†’ 0 from the negative side.
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
205
This work is licensed under a
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS

What pattern do you notice between the tables?
οƒΊ

What type of function displays the type of symmetry shown in the tables?
οƒΊ



MP.7
&
MP.8
Even functions
How can you prove that 𝑔(π‘₯) =
1
is even?
π‘₯2
Scaffolding:
1
οƒΊ A function is even if 𝑔(βˆ’π‘₯) = 𝑔(π‘₯). We have 𝑔(βˆ’π‘₯) = ( )2, and
βˆ’π‘₯
1
𝑔(π‘₯) = 2. These expressions are equivalent, so 𝑔 is even.
π‘₯
1
1
Now let’s analyze the end behavior of 𝑓(π‘₯) = 3 and 𝑔(π‘₯) = 4.
π‘₯
π‘₯
οƒΊ
As π‘₯ β†’ ∞, 𝑓(π‘₯) β†’ 0, and as π‘₯ β†’ βˆ’βˆž, 𝑓(π‘₯) β†’ 0.
οƒΊ
As π‘₯ β†’ ∞, 𝑔(π‘₯) β†’ 0, and as π‘₯ β†’ βˆ’βˆž, 𝑔(π‘₯) β†’ 0.
 Recall that even functions
such as 𝑔(π‘₯) =
1
exhibit
π‘₯2
symmetry with respect to
the 𝑦-axis.
 Odd functions such as
β„Ž(π‘₯) =
What about the behavior of 𝑓(π‘₯) and 𝑔(π‘₯) as the functions approach 0 from the
right and left sides?
οƒΊ
οƒΊ
οƒΊ
οƒΊ

It appears that there is symmetry.
1
exhibit
π‘₯3
symmetry with respect to
the origin.
𝑓(π‘₯) β†’ ∞ as π‘₯ β†’ 0 from the right side.
𝑓(π‘₯) β†’ βˆ’βˆž as π‘₯ β†’ 0 from the left side.
𝑔(π‘₯) β†’ ∞ as π‘₯ β†’ 0 from the right side.
𝑔(π‘₯) β†’ ∞ as π‘₯ β†’ 0 from the left side.
Can you see how to generalize your results?
1
οƒΊ
If 𝑛 is any whole number, then as π‘₯ β†’ ∞,
οƒΊ
If 𝑛 is any odd whole number, then as π‘₯ β†’ 0 from the positive side,
negative side,
οƒΊ
1
π‘₯𝑛
π‘₯𝑛
β†’ 0, and as π‘₯ β†’ βˆ’βˆž,
π‘₯𝑛
1
π‘₯𝑛
β†’ 0.
β†’ ∞, and as π‘₯ β†’ 0 from the
β†’ βˆ’βˆž.
If 𝑛 is any even whole number, then as π‘₯ β†’ 0 from the positive side,
negative side,
1
π‘₯𝑛
1
1
π‘₯𝑛
β†’ ∞, and as π‘₯ β†’ 0 from the
β†’ ∞.
Discussion: General Polynomials (5 minutes)

1
. What can we say about the end
π‘₯𝑛
5π‘₯3 βˆ’ 2π‘₯2 + 4π‘₯ βˆ’ 16
behavior of a more complex rational function such as 𝑓(π‘₯) =
? To answer this question,
2π‘₯3 + 10π‘₯2 βˆ’ π‘₯ + 4
So far, we’ve examined only the simplest rational functions 𝑓(π‘₯) =
we need to review some principles regarding the end behavior of polynomials.

MP.3
Let 𝑔(π‘₯) = 5π‘₯ 3 βˆ’ 2π‘₯ 2 + 4π‘₯ βˆ’ 16 and β„Ž(π‘₯) = 2π‘₯ 3 + 10π‘₯ 2 βˆ’ π‘₯ + 4. Explain why the functions have similar
end behavior.
οƒΊ
The functions are polynomials of the same degree, and both have positive leading coefficients.
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
206
This work is licensed under a
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS

We saw in the Opening Exercises that functions of the form 𝑦 = π‘₯ 𝑛 are simple to analyze, but recall that things
get a little more complicated when there are several terms involved. For instance, consider
𝑓(π‘₯) = π‘₯ 3 βˆ’ 5π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 1. See what you can learn by computing 𝑓(4) and 𝑓(6) term by term. You may use
a calculator for this purpose.
οƒΊ
𝑓(4) = 64 βˆ’ 80 βˆ’ 12 βˆ’ 1 = βˆ’29
οƒΊ
𝑓(6) = 216 βˆ’ 180 βˆ’ 18 βˆ’ 1 = 17

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.

More than anything else, it’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, 𝑓 has the same end behavior as the simple power function 𝑔(π‘₯) = π‘₯ 3 . Do you recall how to
demonstrate this formally? Try factoring out π‘₯ 3 from each term of 𝑓 to see why its end behavior is similar to
𝑔.
οƒΊ


3
1
βˆ’ )
π‘₯2 π‘₯3
Use this new form for 𝑓 to compare 𝑓(104 ) and 𝑔(104 ).
οƒΊ
𝑔(104 ) = (104 )3 = 1012
οƒΊ
𝑓(104 ) = 1012 βˆ™ (1 βˆ’
5
3
1
4βˆ’
8βˆ’
12)
10
10
10
Notice that for this large input, the output of 𝑓 is about 99.9% of the value of 𝑔. If you use an even larger
input, the relative value of 𝑓 grows even closer to that of 𝑔. Without actually doing the calculations, can you
see intuitively why this is the case? Explain your thinking to a neighbor.
οƒΊ

5
π‘₯
𝑓(π‘₯) = π‘₯ 3 βˆ’ 5π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 1 = π‘₯ 3 βˆ™ (1 βˆ’ βˆ’
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‘₯
Let’s examine each component in the expression π‘₯ 3 βˆ™ (1 βˆ’ βˆ’
οƒΊ
As π‘₯ β†’ ∞, π‘₯ 3 β†’ ∞, 1 β†’ 1,
οƒΊ
Thus, 1 βˆ’ βˆ’
5
π‘₯
5
π‘₯
β†’ 0,
3
π‘₯2
β†’ 0, and
1
π‘₯3
3
1
βˆ’ ). What happens when π‘₯ β†’ ∞?
π‘₯2 π‘₯3
β†’ 0.
3
1
βˆ’ β†’ 1 βˆ’ 0 βˆ’ 0 βˆ’ 0 = 1.
π‘₯2 π‘₯ 3

This confirms our observation that 𝑓 and 𝑔 have similar end behavior. We can say that the function 𝑔(π‘₯) = π‘₯ 3
is an end-behavior model for 𝑓(π‘₯) = π‘₯ 3 βˆ’ 5π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 1.

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. Let’s look at some examples.
Example 1 (6 minutes)
In this example, students use what they know about polynomials to analyze a rational function.

MP.3
Based on our reasoning with 𝑓 and 𝑔 above, what would be an end-behavior model for
𝑓(π‘₯) = π‘₯ 4 βˆ’ 2π‘₯ 2 + 6π‘₯ βˆ’ 3?
οƒΊ
𝑔(π‘₯) = π‘₯ 4
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
207
This work is licensed under a
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS

Consider the function 𝑓(π‘₯) =
5π‘₯3 βˆ’ 2π‘₯2 + 4π‘₯ βˆ’ 16
. What can we say about the end behavior of this function?
2π‘₯3 + 10π‘₯2 βˆ’ π‘₯ + 4
Well, let’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.

οƒΊ
The function 5π‘₯ 3 is an end-behavior model for the numerator.
οƒΊ
The function 2π‘₯ 3 is an end-behavior model for the denominator.
Now let’s combine these observations: When we form the rational expression 𝑓(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 𝑓 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 𝑓.
οƒΊ
𝑓(1010 ) β‰ˆ 2.499 999 999
5

You have to admit that is pretty darn close to .

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.

Let’s take one last look at this example using factoring. Factor out the highest power of π‘₯ from the numerator
2
and denominator of 𝑓(π‘₯) =
5π‘₯3 βˆ’ 2π‘₯2 + 4π‘₯ βˆ’ 16
. This should help you to see how all of the pieces in this
2π‘₯3 + 10π‘₯2 βˆ’ π‘₯ + 4
discussion fit together.
οƒΊ
π‘₯3(5 βˆ’ 2π‘₯ + 42 βˆ’ 163)
5π‘₯3 βˆ’ 2π‘₯2 + 4π‘₯ βˆ’ 16
π‘₯
π‘₯ .
We have 𝑓(π‘₯) =
= 3
1
4
2π‘₯3 + 10π‘₯2 βˆ’ π‘₯ + 4
π‘₯ (2 + 10
π‘₯ βˆ’ 2 + 3)
π‘₯
οƒΊ
The quotient of
π‘₯3
π‘₯3
2
expression
as π‘₯ β†’ ∞.
π‘₯
is 1. The fractions in the expression are all approaching 0 as π‘₯ β†’ ∞, so the
4
16
5βˆ’ + 2βˆ’ 3
π‘₯ π‘₯
π‘₯
10
1
4
2+ βˆ’ 2+ 3
π‘₯
π‘₯
π‘₯
5
5
2
2
is approaching . This confirms our view that the function is approaching
Example 2 (5 minutes)

Now that we’ve seen that we can use our knowledge of polynomials to develop end-behavior models for
rational functions, let’s get some additional practice with this technique.

Consider the function 𝑓(π‘₯) =
2π‘₯3 + 3π‘₯ βˆ’ 1
. Find an end-behavior model for this function.
π‘₯2 + π‘₯ + 1
οƒΊ
The function 2π‘₯ 3 is an end-behavior model for the numerator.
οƒΊ
The function π‘₯ 2 is an end-behavior model for the denominator.
οƒΊ
Thus, the expression
Lesson 12:
2π‘₯ 3
π‘₯2
= 2π‘₯ is an end-behavior model for 𝑓.
End Behavior of Rational Functions
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS


What conclusions can you draw about the end behavior of 𝑓 based on your analysis?
οƒΊ
We know that as π‘₯ β†’ ∞, 2π‘₯ β†’ ∞, so we conclude that 𝑓(π‘₯) β†’ ∞ as well.
οƒΊ
Similarly, we know that as π‘₯ β†’ βˆ’βˆž, 2π‘₯ β†’ βˆ’βˆž, so we conclude that 𝑓(π‘₯) β†’ βˆ’βˆž as well.
Use factoring to confirm that your analysis of the end behavior of 𝑓 is correct.
οƒΊ
𝑓(π‘₯) =
3
1
3
2 + 32 βˆ’ 13
2π‘₯3 + 3π‘₯ βˆ’ 1 π‘₯ βˆ™(2 + π‘₯2 βˆ’ π‘₯3)
π‘₯
π‘₯
=
=
π‘₯
βˆ™
π‘₯2 + π‘₯ + 1
π‘₯2βˆ™(1 +1π‘₯ + 12)
1 + 1π‘₯ + 12
π‘₯
οƒΊ
π‘₯
2
Each fraction is approaching 0 as π‘₯ β†’ ∞, so the expression on the right is approaching . This confirms
1
that 𝑦 = 2π‘₯ is an end-behavior model for 𝑦 = 𝑓(π‘₯).

So far, so good. Now let’s use technology to further our understanding of this function. When you enter the
function into your calculator, do the numerical results confirm your analysis?
οƒΊ
𝒙
π’š = 𝒇(𝒙)
100
198
500
398
1,000
1,998
5,000
9,998
𝒙
π’š = 𝒇(𝒙)
βˆ’100
βˆ’202
βˆ’500
βˆ’1,002
βˆ’1,000
βˆ’2,002
βˆ’5,000
βˆ’10,002
The numerical data confirms our views about the end behavior of 𝑓.
Example 3 (4 minutes)

Let’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 𝑓(π‘₯) =
π‘₯2 + π‘₯ + 1
. What do you suppose the end behavior of
2π‘₯3 + 3π‘₯ βˆ’ 1
this function is like? Make a conjecture, and share it with a neighbor.
οƒΊ



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.
οƒΊ
The function π‘₯ 2 is an end-behavior model for the numerator.
οƒΊ
The function 2π‘₯ 3 is an end-behavior model for the denominator.
οƒΊ
Thus, the expression
π‘₯2
2π‘₯ 3
=
1
π‘₯
is an end-behavior model for 𝑓.
What conclusions can you draw about the end behavior of 𝑓 based on your analysis?
1
οƒΊ
We know that as π‘₯ β†’ ∞,
οƒΊ
Similarly, we know that as π‘₯ β†’ βˆ’βˆž,
π‘₯
β†’ 0, so we conclude that 𝑓(π‘₯) β†’ 0 as well.
1
π‘₯
β†’ 0, so we conclude that 𝑓(π‘₯) β†’ 0 as well.
Use your calculator to confirm your results numerically.
𝒙
π’š = 𝒇(𝒙)
οƒΊ
βˆ’200
βˆ’0.0025
βˆ’100
βˆ’0.0049
100
0.0050
200
0.0025
The numerical data confirms our views about the end behavior of 𝑓.
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
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.
𝒇(𝒙) =
1.
πŸ•π’™πŸ“ βˆ’ πŸ‘π’™ + 𝟏
πŸ’π’™πŸ‘ + 𝟐
This function has the same end behavior as
πŸ•π’™πŸ“ πŸ• 𝟐
= 𝒙 .
πŸ’π’™πŸ‘ πŸ’
Using this model as a guide, we conclude that 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞ and that 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’βˆž.
𝒇(𝒙) =
2.
πŸ‘
πŸ•π’™ βˆ’ πŸ‘π’™ + 𝟏
πŸ’π’™πŸ‘ + 𝟐
This function has the same end behavior as
πŸ•π’™πŸ‘ πŸ•
= .
πŸ’π’™πŸ‘ πŸ’
Using this model as a guide, we conclude that 𝒇(𝒙) β†’
𝒇(𝒙) =
3.
πŸ•
πŸ•
as 𝒙 β†’ ∞ and that 𝒇(𝒙) β†’ as 𝒙 β†’ βˆ’βˆž.
πŸ’
πŸ’
πŸ‘
πŸ•π’™ + 𝟐
πŸ“
πŸ’π’™ βˆ’ πŸ‘π’™ + 𝟏
This function has the same end behavior as
πŸ•π’™πŸ‘
πŸ•
=
.
πŸ’π’™πŸ“ πŸ’π’™πŸ
Using this model as a guide, we conclude that 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞ and that 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
Closing (1 minutes)

Take a minute to write a summary in your notebook of what you learned today.
οƒΊ
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.
οƒΊ
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
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
Name
Date
Lesson 12: End Behavior of Rational Functions
Exit Ticket
Given 𝑓(π‘₯) =
π‘₯+2
, find the following, and justify your findings.
π‘₯2βˆ’1
a.
The end-behavior model for the numerator
b.
The end-behavior model for the denominator
c.
The end-behavior model for 𝑓
d.
What is the value of 𝑓(π‘₯) as π‘₯ β†’ ∞?
e.
What is the value of 𝑓(π‘₯) as π‘₯ β†’ βˆ’βˆž?
f.
What is the value of 𝑓(π‘₯) as π‘₯ β†’ 1 from the positive and negative sides?
g.
What is the value of 𝑓(π‘₯) as π‘₯ β†’ βˆ’1 from the positive and negative sides?
Lesson 12:
End Behavior of Rational Functions
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 12
M3
PRECALCULUS AND ADVANCED TOPICS
Exit Ticket Sample Solutions
Given 𝒇(𝒙) =
a.
𝒙+𝟐
, find the following, and justify your findings.
π’™πŸ βˆ’ 𝟏
The end-behavior model for the numerator
𝒙
The end-behavior model is 𝒇(𝒙) = 𝒙𝒏 where 𝒏 is the greatest power of the expression. Here, the numerator is
𝒙 + 𝟐, so 𝒏 = 𝟏, and the model is 𝒇(𝒙) = 𝒙.
b.
The end-behavior model for the denominator
π’™πŸ
The end-behavior model is 𝒇(𝒙) = 𝒙𝒏 where 𝒏 is the greatest power of the expression. Here, the
denominator is π’™πŸ βˆ’ 𝟏, so 𝒏 = 𝟐, and the model is 𝒇(𝒙) = π’™πŸ .
c.
The end-behavior model for 𝒇
𝒙
𝟏
=
π’™πŸ 𝒙
I replaced the numerator and denominator with the end-behavior models of each and simplified.
d.
What is the value of 𝒇(𝒙) as 𝒙 β†’ ∞?
𝒇(𝒙) β†’ 𝟎
As
𝒙 β†’ ∞, 𝒇(𝒙) β†’
e.
𝟏
= 𝟎.
∞
What is the value of 𝒇(𝒙) as 𝒙 β†’ βˆ’βˆž?
𝒇(𝒙) β†’ 𝟎
As
𝒙 β†’ βˆ’βˆž, 𝒇(𝒙) β†’
f.
𝟏
= 𝟎.
βˆ’βˆž
What is the value of 𝒇(𝒙) as 𝒙 β†’ 𝟏 from the positive and negative sides?
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ 𝟏 from the positive side. As 𝒙 nears 𝟏 from the positive side, the numerator of the function
approaches πŸ‘, and the denominator becomes a tiny positive number. The ratio of these is a very large
positive number that exceeds any bounds.
𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟏 from the negative side. As 𝒙 nears 𝟏 from the negative side, the numerator of the
function approaches πŸ‘, 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 𝒇(𝒙) as 𝒙 β†’ βˆ’πŸ from the positive and negative sides?
𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’πŸ from the positive side. As 𝒙 nears βˆ’πŸ from the positive side, the numerator of the
function approaches 𝟏, and the denominator becomes a tiny negative number. The ratio of these is a very
large negative number that exceeds any bounds.
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’πŸ from the negative side. As 𝒙 nears βˆ’πŸ from the negative side, the numerator of the
function approaches 𝟏, 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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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 12
M3
PRECALCULUS AND ADVANCED TOPICS
Problem Set Sample Solutions
1.
Analyze the end behavior of both functions.
a.
𝒇(𝒙) = 𝒙, π’ˆ(𝒙) =
𝟏
𝒙
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟎.
b.
𝒇(𝒙) = π’™πŸ‘ , π’ˆ(𝒙) =
𝟏
π’™πŸ‘
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟎.
c.
𝒇(𝒙) = π’™πŸ , π’ˆ(𝒙) =
𝟏
π’™πŸ
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎.
d.
𝒇(𝒙) = π’™πŸ’ , π’ˆ(𝒙) =
𝟏
π’™πŸ’
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟎.
e.
𝒇(𝒙) = 𝒙 βˆ’ 𝟏, π’ˆ(𝒙) =
𝟏
π’™βˆ’πŸ
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟎.
f.
𝒇(𝒙) = 𝒙 + 𝟐, π’ˆ(𝒙) =
𝟏
𝒙+𝟐
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
g.
𝒇(𝒙) = π’™πŸ βˆ’ πŸ’, π’ˆ(𝒙) =
𝟏
π’™πŸ βˆ’ πŸ’
For 𝒇(𝒙): 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
For π’ˆ(𝒙): π’ˆ(𝒙) β†’ ∞ a 𝒙 β†’ 𝟎, and π’ˆ(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ 𝟎.
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
2.
For the following functions, determine the end behavior. Confirm your answer with a table of values.
a.
𝒇(𝒙) =
πŸ‘π’™ βˆ’ πŸ”
𝒙+𝟐
𝒇 has the same end behavior as
πŸ‘π’™
= πŸ‘. Using this model as a guide, we conclude that
𝒙
𝒇(𝒙) β†’ πŸ‘ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ πŸ‘ as 𝒙 β†’ βˆ’βˆž.
b.
𝒙
π’š = 𝒇(𝒙)
𝟏𝟎𝟎
𝟐. πŸ–πŸ–πŸ
πŸ“πŸŽπŸŽ
𝟐. πŸ—πŸ•πŸ”
𝟏, 𝟎𝟎𝟎
𝟐. πŸ—πŸ–πŸ–
πŸ“, 𝟎𝟎𝟎
𝟐. πŸ—πŸ—πŸ–
𝒙
π’š = 𝒇(𝒙)
βˆ’πŸπŸŽπŸŽ
βˆ’πŸ‘. 𝟏𝟐𝟐
βˆ’πŸ“πŸŽπŸŽ
πŸ‘. πŸŽπŸπŸ’
βˆ’πŸ, 𝟎𝟎𝟎
πŸ‘. 𝟎𝟏𝟐
βˆ’πŸ“, 𝟎𝟎𝟎
πŸ‘. 𝟎𝟎𝟐
𝒇(𝒙) =
πŸ“π’™ + 𝟏
π’™πŸ βˆ’ 𝒙 βˆ’ πŸ”
𝒇 has the same end behavior as
πŸ“π’™
π’™πŸ
πŸ“
= . Using this model as a guide, we conclude that
𝒙
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
c.
𝒙
π’š = 𝒇(𝒙)
𝟏𝟎𝟎
𝟎. πŸŽπŸ“πŸ
πŸ“πŸŽπŸŽ
𝟎. 𝟎𝟏𝟎
𝟏, 𝟎𝟎𝟎
𝟎. πŸŽπŸŽπŸ“
πŸ“, 𝟎𝟎𝟎
𝟎. 𝟎𝟎𝟏
𝒙
π’š = 𝒇(𝒙)
βˆ’πŸπŸŽπŸŽ
βˆ’πŸŽ. πŸŽπŸ’πŸ—
βˆ’πŸ“πŸŽπŸŽ
βˆ’πŸŽ. 𝟎𝟏𝟎
βˆ’πŸ, 𝟎𝟎𝟎
βˆ’πŸŽ. πŸŽπŸŽπŸ“
βˆ’πŸ“, 𝟎𝟎𝟎
βˆ’πŸŽ. 𝟎𝟎𝟏
𝒇(𝒙) =
π’™πŸ‘ βˆ’ πŸ–
π’™πŸ βˆ’ πŸ’
𝒇 has the same end behavior as
π’™πŸ‘
π’™πŸ
= 𝒙. Using this model as a guide, we conclude that
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
d.
𝒙
π’š = 𝒇(𝒙)
𝟏𝟎𝟎
𝟏𝟎𝟎
πŸ“πŸŽπŸŽ
πŸ“πŸŽπŸŽ
𝟏, 𝟎𝟎𝟎
𝟏, 𝟎𝟎𝟎
πŸ“, 𝟎𝟎𝟎
πŸ“, 𝟎𝟎𝟎
𝒙
π’š = 𝒇(𝒙)
βˆ’πŸπŸŽπŸŽ
βˆ’πŸπŸŽπŸŽ
βˆ’πŸ“πŸŽπŸŽ
βˆ’πŸ“πŸŽπŸŽ
βˆ’πŸ, 𝟎𝟎𝟎
βˆ’πŸ, 𝟎𝟎𝟎
βˆ’πŸ“, 𝟎𝟎𝟎
βˆ’πŸ“, 𝟎𝟎𝟎
𝒇(𝒙) =
π’™πŸ‘ βˆ’ 𝟏
π’™πŸ’ βˆ’ 𝟏
𝒇 has the same end behavior as
π’™πŸ‘
π’™πŸ’
𝟏
= . Using this model as a guide, we conclude that
𝒙
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
𝒙
π’š = 𝒇(𝒙)
𝟏𝟎𝟎
𝟎. 𝟎𝟏𝟎
πŸ“πŸŽπŸŽ
𝟎. 𝟎𝟎𝟐
𝟏, 𝟎𝟎𝟎
𝟎. 𝟎𝟎𝟏
πŸ“, 𝟎𝟎𝟎
𝟎. πŸŽβ€†πŸŽπŸŽπŸ
𝒙
π’š = 𝒇(𝒙)
βˆ’πŸπŸŽπŸŽ
βˆ’πŸŽ. 𝟎𝟏𝟎
βˆ’πŸ“πŸŽπŸŽ
βˆ’πŸŽ. 𝟎𝟎𝟐
βˆ’πŸ, 𝟎𝟎𝟎
βˆ’πŸŽ. 𝟎𝟎𝟏
βˆ’πŸ“, 𝟎𝟎𝟎
βˆ’πŸŽ. πŸŽβ€†πŸŽπŸŽπŸ
Lesson 12:
End Behavior of Rational Functions
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Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
e.
𝒇(𝒙) =
πŸ‘
(πŸπ’™ + 𝟏)
𝟐
(π’™πŸ βˆ’ 𝒙)
𝒇 has the same end behavior as
πŸ–π’™πŸ‘
π’™πŸ’
πŸ–
= . Using this model as a guide, we conclude that
𝒙
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
3.
𝒙
π’š = 𝒇(𝒙)
𝟏𝟎𝟎
𝟎. πŸŽπŸ–πŸ‘
πŸ“πŸŽπŸŽ
𝟎. πŸŽπŸπŸ”
𝟏, 𝟎𝟎𝟎
𝟎. πŸŽπŸŽπŸ–
πŸ“, 𝟎𝟎𝟎
𝟎. πŸŽβ€†πŸŽπŸπŸ”
𝒙
π’š = 𝒇(𝒙)
βˆ’πŸπŸŽπŸŽ
βˆ’πŸŽ. πŸŽπŸ•πŸ•
βˆ’πŸ“πŸŽπŸŽ
βˆ’πŸŽ. πŸŽπŸπŸ”
βˆ’πŸ, 𝟎𝟎𝟎
βˆ’πŸŽ. πŸŽπŸŽπŸ–
βˆ’πŸ“, 𝟎𝟎𝟎
βˆ’πŸŽ. πŸŽβ€†πŸŽπŸπŸ”
For the following functions, determine the end behavior.
a.
𝒇(𝒙) =
πŸ“π’™πŸ” βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
πŸ“π’™πŸ’ βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
𝒇 has the same end behavior as
πŸ“π’™πŸ”
πŸ“π’™πŸ’
= π’™πŸ . Using this model as a guide, we conclude that
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’βˆž.
b.
𝒇(𝒙) =
πŸ“π’™πŸ’ βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
πŸ“π’™πŸ” βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
𝒇 has the same end behavior as
πŸ“π’™πŸ’
πŸ“π’™πŸ”
=
𝟏
π’™πŸ
. Using this model as a guide, we conclude that
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
c.
𝒇(𝒙) =
πŸ“π’™πŸ’ βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
πŸ“π’™πŸ’ βˆ’ πŸ‘π’™πŸ‘ + 𝒙 βˆ’ 𝟐
𝒇 has the same end behavior as
πŸ“π’™πŸ’
πŸ“π’™πŸ’
= 𝟏. Using this model as a guide, we conclude that
𝒇(𝒙) β†’ 𝟏 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟏 as 𝒙 β†’ βˆ’βˆž.
d.
𝒇(𝒙) =
βˆšπŸπ’™πŸ
+𝒙+𝟏
πŸ‘π’™ + 𝟏
𝒇 has the same end behavior as
βˆšπŸπ’™πŸ
√𝟐
=
𝒙. Using this model as a guide, we conclude that
πŸ‘π’™
πŸ‘
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
e.
𝒇(𝒙) =
𝟐
πŸ’π’™ βˆ’ πŸ‘π’™ βˆ’ πŸ•
πŸπ’™πŸ‘ + 𝒙 βˆ’ 𝟐
𝒇 has the same end behavior as
πŸ’π’™πŸ
πŸπ’™πŸ‘
𝟐
= .
𝒙
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
Lesson 12:
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
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
4.
Determine the end behavior of each function.
a.
𝒇(𝒙) =
𝐬𝐒𝐧(𝒙)
𝒙
βˆ’πŸ ≀ 𝐬𝐒𝐧 𝒙 ≀ 𝟏,
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
b.
𝒇(𝒙) =
𝐜𝐨𝐬(𝒙)
𝒙
βˆ’πŸ ≀ 𝐜𝐨𝐬 𝒙 ≀ 𝟏,
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
c.
𝒇(𝒙) =
𝒙
𝟐
𝒙
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
d.
𝒇(𝒙) =
𝒙
𝒙
𝟐
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’βˆž as 𝒙 β†’ βˆ’βˆž.
e.
𝒇(𝒙) =
πŸ’
βˆ’π’™
𝟏+𝒆
𝒇(𝒙) β†’ πŸ’ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
f.
𝒇(𝒙) =
𝟏𝟎
βˆ’π’™
𝟏+𝒆
𝒇(𝒙) β†’ 𝟏𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
5.
Consider the functions 𝒇(𝒙) = 𝒙! and π’ˆ(𝒙) = π’™πŸ“ for natural numbers 𝒙.
a.
What are the values of 𝒇(𝒙) and π’ˆ(𝒙) for 𝒙 = πŸ“, 𝟏𝟎, πŸπŸ“, 𝟐𝟎, πŸπŸ“?
𝒇(𝒙) = {𝟏𝟐𝟎, πŸ‘β€†πŸ”πŸπŸ–β€†πŸ–πŸŽπŸŽ, πŸβ€†πŸ‘πŸŽπŸ•β€†πŸ”πŸ•πŸ’β€†πŸ‘πŸ”πŸ–β€†πŸŽπŸŽπŸŽ, 𝟐. πŸ’ × πŸπŸŽπŸπŸ– , 𝟏. πŸ” × πŸπŸŽπŸπŸ“ }
π’ˆ(𝒙) = {πŸ‘πŸπŸπŸ“, πŸπŸŽπŸŽβ€†πŸŽπŸŽπŸŽ, πŸ•πŸ“πŸ—β€†πŸ‘πŸ•πŸ“, πŸ‘β€†πŸπŸŽπŸŽβ€†πŸŽπŸŽπŸŽ, πŸ—β€†πŸ•πŸ”πŸ“β€†πŸ”πŸπŸ“}
b.
What is the end behavior of 𝒇(𝒙) as 𝒙 β†’ ∞?
𝒇(𝒙) β†’ ∞
c.
What is the end behavior of π’ˆ(𝒙) as 𝒙 β†’ ∞?
π’ˆ(𝒙) β†’ ∞
d.
Make an argument for the end behavior of
𝒇(𝒙)
π’ˆ(𝒙)
as 𝒙 β†’ ∞.
Since π’š = 𝒇(𝒙) increases so much faster than π’š = π’ˆ(𝒙), 𝒇(𝒙) overpowers the division by π’ˆ(𝒙). By 𝒙 = πŸπŸ“,
𝒇(𝒙)
π’ˆ(𝒙)
β‰ˆ 𝟏. πŸ” × πŸπŸŽπŸπŸ– . Thus,
Lesson 12:
𝒇(𝒙)
π’ˆ(𝒙)
β†’ ∞.
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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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 12
NYS COMMON CORE MATHEMATICS CURRICULUM
M3
PRECALCULUS AND ADVANCED TOPICS
e.
Make an argument for the end behavior of
π’ˆ(𝒙)
as 𝒙 β†’ ∞.
𝒇(𝒙)
For the same reasons as in part (d), division by 𝒇(𝒙) overpowers π’ˆ(𝒙) in the numerator. By 𝒙 = πŸπŸ“,
π’ˆ(𝒙)
𝒇(𝒙)
6.
β‰ˆ 𝟎. πŸŽπŸŽπŸŽβ€†πŸŽπŸŽπŸŽβ€†πŸŽπŸŽπŸŽβ€†πŸŽπŸŽπŸŽβ€†πŸŽπŸŽπŸŽ. Thus,
π’ˆ(𝒙)
𝒇(𝒙)
β†’ 𝟎.
Determine the end behavior of the functions.
a.
𝒇(𝒙) =
𝒙
𝟏
, π’ˆ(𝒙) =
𝒙
π’™πŸ
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
b.
𝒇(𝒙) =
𝒙+𝟏
𝟏
, π’ˆ(𝒙) =
π’™βˆ’πŸ
π’™πŸ βˆ’ 𝟏
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
c.
𝒇(𝒙) =
π’™βˆ’πŸ
𝟏
, π’ˆ(𝒙) =
𝒙+𝟏
π’™πŸ βˆ’ 𝒙 βˆ’ 𝟐
𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ 𝟎 as 𝒙 β†’ βˆ’βˆž.
d.
𝒇(𝒙) =
π’™πŸ‘ βˆ’ 𝟏
, π’ˆ(𝒙) = π’™πŸ + 𝒙 + 𝟏
π’™βˆ’πŸ
𝒇(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ ∞ as 𝒙 β†’ βˆ’βˆž.
7.
Use a graphing utility to graph the following functions 𝒇 and π’ˆ. Explain why they have the same graphs. Determine
the end behavior of the functions and whether the graphs have any horizontal asymptotes.
a.
𝒇(𝒙) =
𝒙+𝟏
𝟐
, π’ˆ(𝒙) = 𝟏 +
π’™βˆ’πŸ
π’™βˆ’πŸ
The graphs are the same because the expressions are equivalent: 𝟏 +
𝒙 β‰  𝟏.
𝟐
π’™βˆ’πŸ
𝟐
𝒙+𝟏
=
+
=
for all
π’™βˆ’πŸ π’™βˆ’πŸ π’™βˆ’πŸ π’™βˆ’πŸ
𝒇(𝒙) β†’ 𝟏 as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ 𝟏 as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ 𝟏 as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ 𝟏 as 𝒙 β†’ βˆ’βˆž.
b.
𝒇(𝒙) =
βˆ’πŸπ’™ + 𝟏
πŸ‘
, π’ˆ(𝒙) =
βˆ’πŸ
𝒙+𝟏
𝒙+𝟏
The graphs are the same because the expressions are equivalent:
πŸ‘
𝒙+𝟏
βˆ’πŸ=
πŸ‘
𝒙+𝟏
βˆ’
𝟐(𝒙 + 𝟏)
𝒙+𝟏
=
πŸ‘ βˆ’ πŸπ’™ βˆ’ 𝟐
𝒙+𝟏
=
βˆ’πŸπ’™ +𝟏
𝒙+𝟏
for all 𝒙 β‰  βˆ’πŸ
𝒇(𝒙) β†’ βˆ’πŸ as 𝒙 β†’ ∞, and 𝒇(𝒙) β†’ βˆ’πŸ as 𝒙 β†’ βˆ’βˆž.
π’ˆ(𝒙) β†’ βˆ’πŸ as 𝒙 β†’ ∞, and π’ˆ(𝒙) β†’ βˆ’πŸ as 𝒙 β†’ βˆ’βˆž.
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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Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.