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Transcript
MATH 1314 College Algebra
Unit 3
Polynomial Functions
Section: 5.1
Polynomial Functions
β€’ This section studies the Polynomial Function.
Your Goal - learn to identify the function components
that lead to its characteristic graph behavior.
β€’ Polynomial function: refer to page 327.
Leading coefficient
degree of 𝑓(π‘₯) is the highest power 𝑛.
𝑓 π‘₯ = 𝒂𝒏 π‘₯ 𝒏 + π‘Žπ‘›βˆ’1 π‘₯ π‘›βˆ’1 + β‹― + π‘Ž1 π‘₯ 1 + π’‚πŸŽ
Leading term: determines
the end behavior of graph
& number of complex zeros
Constant term
(y-intercept)
οƒ˜ any combination of variable terms and constant term can exist.
οƒ˜ all coefficients are real numbers
οƒ˜ exponents are non-negative integers
οƒ˜ variable cannot be in a denominator
οƒ˜ graphs are smooth and continuous
Polynomial Functions
β€’ Practice: According to the description of a
polynomial function, which of the following are
considered polynomial functions?
P
O
P
O
O
? 𝑓 π‘₯
? 𝑓 π‘₯
? 𝑓 π‘₯
? 𝑓 π‘₯
? 𝑓 π‘₯
1
= 5 Why? Follows Polynomial form.
π‘₯ β†’ π‘₯2
= π‘₯ 2 + 3 π‘₯ βˆ’ 4 Why not? Exponents must be
=
1 3
βˆ’ π‘₯
5
1
+ π‘₯
4
=
π‘₯ 2 +2π‘₯βˆ’15
π‘₯ 2 βˆ’4
+7
nonnegative integers.
Why? Follows Polynomial form.
Can’t have a variable
Why not?
in denominator.
= π‘₯ βˆ’2 + 4π‘₯ + 1 Why not?
Exponents must be
nonnegative integers.
Polynomial Functions
β€’ Exercise: Graph. Determine end behavior rules for 𝑓 π‘₯ = π‘Žπ‘₯ 𝑛 .
1
5
 π‘Œ1 = π‘₯ 2
π‘Œ1 = 3π‘₯ 4
π‘Œ1 = π‘₯ 6
up
 Conclusion: if degree is even and π‘Ž > 0, both ends go _______.
P
1
5
 π‘Œ1 = βˆ’ π‘₯ 2
π‘Œ1 = βˆ’3π‘₯ 4
π‘Œ1 = βˆ’π‘₯ 6
down
 Conclusion: if degree is even and π‘Ž < 0, both ends go _______.
P
Polynomial Functions
β€’ Exercise: Graph. Determine end behavior rules for 𝑓(π‘₯) = π‘Žπ‘₯ 𝑛 .
 π‘Œ1 = 3π‘₯
1
4
π‘Œ1 = π‘₯ 3
π‘Œ1 = π‘₯ 5
down
 Conclusion: if degree is odd and π‘Ž > 0, left end goes ________
P
up
right end goes ________.
 π‘Œ1 = βˆ’3π‘₯
π‘Œ1 =
1 3
βˆ’ π‘₯
4
π‘Œ1 = βˆ’π‘₯ 5
up
 Conclusion: if degree is odd and π‘Ž < 0, left end goes _______
P
right end goes _______.
down
Polynomial Functions
β€’ Theorems and Definitions
 Real Zeros: A real number π‘₯ = π‘Ÿ that makes 𝑓 π‘₯
P
called a real zero of 𝑓(π‘₯). This means that:
= 0 is
 π‘₯ = π‘Ÿ is an x-intercept of 𝑓(π‘₯) and a solution to 𝑓(π‘₯) = 0.
 π‘₯ βˆ’ π‘Ÿ is a factor of 𝑓(π‘₯) .
 Multiplicity: If π‘₯ βˆ’ π‘Ÿ occurs in 𝑓(π‘₯) an even or odd number of
times, then π‘₯ = π‘Ÿ is a zero having even or odd multiplicity.
 even multiplicity: graph touches x-axis at π‘₯ = π‘Ÿ.
 odd multiplicity: graph crosses x-axis at π‘₯ = π‘Ÿ.
 Turning Points: If a polynomial function has degree 𝑛,
then its graph has at most 𝑛 βˆ’ 1 turning points, and vice-versa.
 End Behavior: ends of the graph of polynomial 𝑓(π‘₯)
resembles the graph of its leading term π‘Žπ‘› π‘₯ 𝑛 .
P
P
P
Polynomial Functions
 For the polynomial
function below, find its
real zeros and multiplicity.
𝑓(π‘₯) = (π‘₯ + 5)2 (π‘₯ βˆ’ 2)
factors (π‘₯ βˆ’ 𝑐) are:
(π‘₯ + 5), (π‘₯ + 5), (π‘₯ βˆ’ 2)
zeros π‘₯ = 𝑐 are:
π‘₯ + 5 = 0, π‘₯ + 5 = 0, π‘₯ βˆ’ 2 = 0
π‘₯ = βˆ’5, π‘₯ = βˆ’5
multiplicity 2
π‘₯=2
multiplicity 1
**Find the polynomial
function having the given
real zeros and multiplicity.
degree: 3
zeros: βˆ’4 multiplicity 1,
3 multiplicity 2
zeros π‘₯ = 𝑐 are:
π‘₯ = βˆ’4,
π‘₯ = 3, π‘₯ = 3
factors (π‘₯ βˆ’ 𝑐) are:
(π‘₯ + 4), (π‘₯ βˆ’ 3), (π‘₯ βˆ’ 3)
𝑓(π‘₯) = (π‘₯ + 4)(π‘₯ βˆ’ 3)(π‘₯ βˆ’ 3)
= (π‘₯ + 4)(π‘₯ βˆ’ 3)2
= π‘₯ 3 βˆ’ 2π‘₯ 2 βˆ’ 15π‘₯ + 36
P
Polynomial Functions
Your turn. Practice.
Find the polynomial function 𝑓(π‘₯) whose
degree is 4
and zeros are: βˆ’1 with multiplicity 2 Occurs twice
7 with multiplicity 2 Occurs twice
zeros π‘₯ = 𝑐
factors (π‘₯ βˆ’ 𝑐)
βˆ’1, βˆ’1, 7, 7
(π‘₯ + 1),(π‘₯ + 1) ,(π‘₯ βˆ’ 7),(π‘₯ βˆ’ 7)
P
P
So 𝑓(π‘₯) = (π‘₯ + 1)(π‘₯ + 1)(π‘₯ βˆ’ 7)(π‘₯ βˆ’ 7)
= (π‘₯ + 1)2 (π‘₯ βˆ’ 7)2
= π‘₯ 4 βˆ’ 12π‘₯ 3 + 22π‘₯ 2 + 84π‘₯ + 49
P
Polynomial Functions
Practice.
If 𝑓 π‘₯ = π‘₯ 4 + 2π‘₯ 3 + π‘₯ 2 , find the following:
0, βˆ’1
a) real zeros: _______________
P
Factor, then solve each factor for π‘₯.
π‘₯ 4 + 2π‘₯ 3 + π‘₯ 2 = 0
π‘₯ 2 (π‘₯
π‘₯ 2 + 2π‘₯ + 1)
1 =0
π‘₯2 π‘₯ + 1 π‘₯ + 1 = 0
π‘₯2 = 0 ,
π‘₯+1=0,π‘₯+1=0
π‘₯ = βˆ’1 , π‘₯ = βˆ’1
π‘₯ =0,π‘₯ =0
Polynomial Functions
Practice ….continued.
If 𝑓 π‘₯ = π‘₯ 4 + 2π‘₯ 3 + π‘₯ 2 , find the following:
b) Graph crosses or touches x-axis at:
𝟎 largest,
largest zero _________
touches
occurs twice
multiplicity
βˆ’πŸ smallest,
occurs twice
multiplicity
P
touches P
smallest zero _________
Degree 𝑛 βˆ’ 1
c) Maximum number of turning points
on the graph of the function:
_____
3
**Remember Turning Points Theorem
P
Polynomial Functions
Practice.
If 𝑓 π‘₯ = 5(π‘₯ 2 βˆ’ 3)(π‘₯ + 4)2 , find the following:
3, βˆ’ 3, βˆ’4
a) real zeros: _______________
P
Solve each factor for π‘₯.
π‘₯2 βˆ’ 3 = 0
π‘₯+4=0 π‘₯+4=0
π‘₯ = βˆ’4
π‘₯ = βˆ’4
π‘₯2 = 3
π‘₯=± 3
3
βˆ’4 βˆ’ 3
∞
= 3, βˆ’ 3 βˆ’βˆž
0
πŸ‘ largest,
1
multiplicity of largest zero is: ____
occurs once
βˆ’ πŸ‘ middle,
1
multiplicity of middle zero is: ____
occurs once
βˆ’πŸ’ smallest,
2
multiplicity of smallest zero is: ____
occurs twice
P
P
P
Polynomial Functions
Practice ….continued.
If 𝑓 π‘₯ = 5(π‘₯ 2 βˆ’ 3)(π‘₯ + 4)2 , find the following:
b) Graph crosses or touches x-axis at:
πŸ‘ largest,
crosses
occurs once
largest zero _________
βˆ’ πŸ‘ middle,
crosses
occurs once
middle zero _________
βˆ’πŸ’ middle,
touches
occurs twice
smallest zero _________
P
P
P
multiplicity
multiplicity
multiplicity
c) Maximum number of turning points Degree 𝑛 βˆ’ 1
3
on the graph of the function:
_____
P
d) Power function that graph resembles
5π‘₯ P
for large values of π‘₯ :
_____
4
Practice and
Complete HW 5.1
MATH 1314 College Algebra
Unit 3
Real Zeros of
Polynomial Functions
Section: 5.2
Real Zeros
 Section 5.1 asked: Find real zeros of polynomial 𝑓(π‘₯).
 Helpful:
𝑓(π‘₯) was already factored
or
𝑓(π‘₯) was easily factored.
𝑓 π‘₯ = π‘₯ + 1 2 (π‘₯ βˆ’ 3)
𝑓 π‘₯ = π‘₯ 3 βˆ’ 6π‘₯ 2 + 9π‘₯
 Section 5.2 explains how to find real zeros in:
 polynomial 𝑓(π‘₯) that are not already factored
or
 polynomial 𝑓(π‘₯) that are not easily factored.
Real Zeros
β€’ Some Theorems that can be used to find real zeros
and factors of polynomial functions
 Remainder Theorem(p347):
 If polynomial 𝑓(π‘₯) is divided by linear factor π‘₯ βˆ’ 𝑐,
then 𝑓(𝑐) = remainder.
 Factor Theorem(p348):
 If 𝑓 𝑐 = 0, then π‘₯ βˆ’ 𝑐 is a linear factor of 𝑓(π‘₯).
 If π‘₯ βˆ’ 𝑐 is a linear factor of 𝑓(π‘₯), then 𝑓 𝑐 = 0.
 Number of Real Zeros Theorem(p349):
 A polynomial cannot have more real zeros than its degree.
 Intermediate Value Theorem(p355):
 For a continuous function 𝑓(π‘₯),
if π‘Ž < 𝑏 and if 𝑓(π‘Ž) and 𝑓(𝑏) have opposite signs,
then 𝑓(π‘₯) has at least one zero between π‘₯ = π‘Ž and π‘₯ = 𝑏.
Real Zeros
β€’ Practice: How to factor polynomial functions
using the previous theorems.
 Factor 𝑓 π‘₯ = 2π‘₯ 3 βˆ’ 11π‘₯ 2 + 10π‘₯ + 8:
 Is π‘₯ + 1 a factor of 𝑓(π‘₯)?
If π‘₯ + 1 is a factor, then π‘₯ = βˆ’1 is a zero of 𝑓(π‘₯).
Using Remainder Theorem, since 𝑓 βˆ’1 = βˆ’15 β‰  0,
you know that π‘₯ + 1 is not a factor of 𝑓 π‘₯ .
 Is π‘₯ βˆ’ 4 a factor of 𝑓(π‘₯)?
If π‘₯ βˆ’ 4 is a factor, then π‘₯ = 4 is a zero of 𝑓(π‘₯).
Using Remainder Theorem, since 𝑓 4 = 0,
you know that π‘₯ βˆ’ 4 is a factor of 𝑓 π‘₯ .
Real Zeros
 Synthetic Division will use a divisor 𝑐 to find zeros,
remainder, and factors of polynomial function 𝑓(π‘₯).
 Practice: Coefficient writing
 Polynomial function 𝑓 π‘₯ = π‘₯ 3 βˆ’ 7π‘₯ 2 + 4π‘₯ + 1
has real coefficients:
1
βˆ’7
4
1
 Polynomial function 𝑓 π‘₯ = π‘₯ 4 + 5π‘₯ 2 βˆ’ 3
𝑓 π‘₯ = π‘₯ 4 + 0π‘₯ 3 + 5π‘₯ 2 + 0π‘₯ βˆ’ 3
has real coefficients:
1
0
5
0 βˆ’3
*Insert a zero coefficient for any missing terms.
Real Zeros
 Practice: Use Synthetic Division to find remainder and
quotient if 𝑓 π‘₯ = π‘₯ 3 + 3π‘₯ 2 βˆ’ 6π‘₯ βˆ’ 8 is divided by π‘₯ + 4.
divisor
π‘₯+4=0
π‘₯ = βˆ’4
𝑐
βˆ’4
multiply
multiply
1coefficients
3 add
βˆ’ 6 add
of 𝑓(π‘₯)
1
βˆ’4
βˆ’1
4
βˆ’2
 coefficients in last row make Quotient and Remainder:
1 βˆ’1 βˆ’2
0
Quotient is π‘₯ 2 βˆ’ π‘₯ βˆ’ 2
Remainder is 0
βˆ’8
8
0
add
Real Zeros
 Since π‘₯ + 4 produced a zero remainder, π‘₯ = βˆ’4 is a zero.
To find remaining zeros in quotient π‘₯ 2 βˆ’ π‘₯ βˆ’ 2:
 Factor or use Quadratic Formula.
βˆ’(𝑏) ± (𝑏)2 βˆ’4 π‘Ž 𝑐
π‘₯=
2(π‘Ž)
βˆ’(βˆ’1) ± (βˆ’1)2 βˆ’4 1 βˆ’2
π‘₯=
2(1)
π‘₯=
1± 9
1±3
1 3
=
=
= ±
2
2
2 2
Zeros of 𝑓(π‘₯):
Factors of 𝑓 π‘₯ =
π‘₯=
1
2
1
2
3
+
2
3
βˆ’
2
=2
= βˆ’1
P
(π‘₯ + 4)(π‘₯ βˆ’ 2)(π‘₯ + 1) P
βˆ’4,
an
d
2,
an
d
βˆ’1
an
d
Real Zeros
 Practice: Does 𝑓 π‘₯ = 2π‘₯ 3 + 5π‘₯ 2 βˆ’ 3π‘₯ + 8
have a real zero between π‘₯ = βˆ’6 and π‘₯ = βˆ’1?
 Use Intermediate Value Theorem: Find 𝑓(βˆ’6) and 𝑓(βˆ’1).
𝑓 βˆ’6 = βˆ’226
𝑓 βˆ’1 = 14
P
P
P
𝑓(βˆ’6) and 𝑓(βˆ’1) have opposite signs. YES.
Practice and
Complete HW 5.2
MATH 1314 College Algebra
Unit 3
Complex Zeros of
Polynomial Functions
Section: 5.3
Real Zeros
 Section 5.1 asked: Find real zeros of polynomial 𝑓(π‘₯).
factored or was ____________.
easily factored
οƒΌ 𝑓(π‘₯) was either _______
 Section 5.2 asked: Find real zeros of polynomial 𝑓(π‘₯).
factored or was not
easily factored
οƒΌ 𝑓(π‘₯) was either not
__________
_______________.
οƒΌ To find real zeros of 𝑓(π‘₯), you learned to use:
β€’ theorems, such as Remainder
_________ Theorem
and Factor
_____ Theorem
Synthetic Division
β€’ and a form of division called _______________.
 Section 5.3: Find Complex
_______ zeros of polynomial 𝑓(π‘₯).
 Complex zeros include non-real or imaginary
________ zeros
 additional theorems will be used.
Complex Zeros
 Complex numbers: Standard Form
π‘Ž + 𝑏𝑖 π‘Ž, 𝑏 are real numbers
real part
imaginary part
 To prepare for this section, review:
οƒ˜ Operations on complex numbers(p120-124)
οƒ˜ Definition of imaginary number: 𝑖 = βˆ’1 2
𝑖 2 = βˆ’1
𝑖 2 = βˆ’1
?
OO
? ?
Is βˆ’4 = 2 = βˆ’2
? 2 ?
2
Only
if βˆ’2
Only
if 2
= βˆ’4= βˆ’4
βˆ’4 =
4 βˆ™ βˆ’1 = 4 βˆ™ βˆ’1
βˆ’4 = 2𝑖
οƒ˜ Complex solutions of Quadratic Equations(p124-126)
Complex Zeros
β€’
Some Theorems you will use in this section:
 Fundamental Theorem of Algebra: (textbook p360)
β€’ Every polynomial function of degree 𝑛 β‰₯ 1,
has at least one complex zero.
 Conjugate Pairs Theorem: (textbook p360)
β€’ Let 𝑓(π‘₯) be a polynomial function whose
coefficients are real. If π‘Ÿ = π‘Ž + 𝑏𝑖 is a zero of f,
then the complex conjugate π‘Ÿ = π‘Ž βˆ’ 𝑏𝑖 is also.
*(The Quadratic Formula is the basis for this.)
 Corollary: (textbook p361)
β€’ A polynomial function 𝑓(π‘₯) of odd degree with
real coefficients has at least one real zero.
*(This guarantees something important that we can use.)
Complex Zeros
 Identifying Conjugate Pairs:
 the complex conjugate of 3 + 4𝑖 is: 3 βˆ’ 4𝑖
 the complex conjugate of βˆ’1 βˆ’ 𝑖 is: βˆ’1 + 𝑖
 the complex conjugate of 2𝑖 is: βˆ’2𝑖
Remember, non-real or imaginary numbers are
the result of an even root of a negative number.
Example:
βˆ’16 = (16)(βˆ’1)
= 16 βˆ™ βˆ’1
= 4𝑖
Complex Zeros
 Practice: If 3𝑖 is a zero of 𝑓 π‘₯ = π‘₯ 3 + 2π‘₯ 2 + 9π‘₯ + 18,
find the remaining zeros of 𝑓 π‘₯ .
 3 complex zeros exist. Why? degree 3
 If 3𝑖 is a zero, then ____
βˆ’3𝑖 is also. Why? Conjugate Pairs Theorem
 Third zero must be real. Why? Corollary(p361): Odd degree
οƒ˜ Use Synthetic/Long Division to find third zero
3𝑖
1
1
βˆ’3𝑖
1
1
2
9
18
2 βˆ’18
3𝑖 6𝑖 βˆ’9
+9(βˆ’1)
+9𝑖
0
2 + 3𝑖 6𝑖
2 + 3𝑖 6𝑖
βˆ’3𝑖 βˆ’6𝑖
2
0
π‘₯ + 2 = 0 Solve for third zero.
π‘₯ = βˆ’2
Remaining zeros: βˆ’3𝑖, βˆ’2
P
Complex Zeros
 Practice: If 3𝑖 is a zero of 𝑓 π‘₯ = π‘₯ 3 + 2π‘₯ 2 + 9π‘₯ + 18,
find the remaining zeros of 𝑓 π‘₯ .
οƒ˜ Or use TABLE/GRAPH on TI-83/84, if possible.
οƒΌ Remaining zeros are : βˆ’3𝑖, βˆ’2
Complex Zeros
 Exercise: Find the complex zeros of 𝑓(π‘₯), then use the
zeros to factor 𝑓(π‘₯). 𝑓 π‘₯ = 2π‘₯ 3 βˆ’ 12π‘₯ 2 + 13π‘₯ βˆ’ 15
 Degree is odd, one real zero exists.
Find with Rational Zeros Theorem(p349)
 or with TI-83 GRAPH/TABLE if possible.
…continued
Complex Zeros
 Exercise: Find the complex zeros of 𝑓(π‘₯), then use the
zeros to factor 𝑓(π‘₯). 𝑓 π‘₯ = 2π‘₯ 3 βˆ’ 12π‘₯ 2 + 13π‘₯ βˆ’ 15
 First real zero: π‘₯ = 5
*Since TI-83 GRAPH/TABLE
doesn’t show 2 more real zeros,
they may be nonreal.
 Use Synthetic Division to divide 𝑓(π‘₯) by π‘₯ = 5 to find
quotient.
2 βˆ’ 12
13
βˆ’ 15
5
10 βˆ’10
15
0
2
3
βˆ’2
 Quotient is 2π‘₯ 2 βˆ’ 2π‘₯ + 3. Two complex zeros remain.
…continued
Complex Zeros
 Use Quadratic Formula to find remaining 2 zeros in
quotient 2π‘₯ 2 βˆ’ 2π‘₯ + 3.
βˆ’(𝑏) ± (𝑏)2 βˆ’4 π‘Ž 𝑐
π‘₯=
2(π‘Ž)
βˆ’(βˆ’2) ± (βˆ’2)2 βˆ’4 2 3
π‘₯=
2(2)
1 𝑖 5
2 2𝑖 5
2 ± 2𝑖 5
2 ± βˆ’4 βˆ™ 5
2 ± βˆ’20
= ±
= ±
=
=
=
2
2
4
4
4
4
4
οƒΌ Zeros of 𝑓 π‘₯ :
5,
1
𝑖 5
+
,
2
2
1
𝑖 5
βˆ’
2
2
1
𝑖 5
1
𝑖 5
)(π‘₯ βˆ’ +
)
οƒΌ Factors of 𝑓 π‘₯ = (π‘₯ βˆ’ 5)(π‘₯ βˆ’ βˆ’
2
2
2
2
Practice and
Complete HW 5.3
MATH 1314 College Algebra
Unit 3
Properties & Graphs
of Rational Functions
Section(s): 5.4. – 5.5
Rational Functions
 A Rational function is a ratio of Polynomial Functions:
𝑅 π‘₯ =
𝑝(π‘₯)
π‘ž(π‘₯)
, where π‘ž π‘₯ β‰  0.
 Practice factoring (p50)
 Domain: all real numbers π‘₯, where π‘ž π‘₯ β‰  0.
 Asymptote: a boundary line in which the graph of 𝑅(π‘₯)
may possibly intersect or never intersect.
 Types of Asymptotes(p367-368)
 Vertical Asymptotes, Horizontal Asymptotes, Oblique Asymptotes
(Which one can never be intersected by the graph of 𝑅(π‘₯)?)
Rational Functions
 Graph of 𝑅 π‘₯ =
 𝑓 π‘₯ =
𝑝(π‘₯)
π‘ž(π‘₯)
π‘₯βˆ’1
(π‘₯βˆ’3)(π‘₯+2)
π‘₯ + 2 β‰  0,
so π‘₯ β‰  βˆ’2
π‘₯ βˆ’ 3 β‰  0,
so π‘₯ β‰  3
π‘₯βˆ’1
π‘₯ 2 βˆ’π‘₯βˆ’6
 Remember, if 𝑓(π‘₯) looked like this : 𝑓 π‘₯ =
then what Algebraic skill is needed to help
determine the restricted domain values of 𝑓(π‘₯)? Factoring
P
Rational Functions
 To find Domain
 𝑓(π‘₯) must be factored completely.
 find restricted values from denominator.
 Practice: Domain
𝑓 π‘₯ =
2
(π‘₯ + 1)(π‘₯ βˆ’ 5)
5π‘₯ + 10
𝑓 π‘₯ = 2
π‘₯ βˆ’π‘₯βˆ’6
5(π‘₯ + 2)
=
(π‘₯ βˆ’ 3)(π‘₯ + 2)
Find restricted values
π‘₯ + 1 β‰  0, π‘₯ βˆ’ 5 β‰  0
𝑓 π‘₯ is factored
factor 𝑓 π‘₯
π‘₯ β‰  βˆ’1, π‘₯ β‰  5
Domain of 𝑓(π‘₯).
P
Find restricted values
π‘₯ βˆ’ 3 β‰  0, π‘₯ + 2 β‰  0
π‘₯ β‰  3, π‘₯ β‰  βˆ’2
Domain of 𝑓(π‘₯).
P
Rational Functions
 To find Vertical Asymptote(VA) after Domain
οƒ˜ 𝑓(π‘₯) must be factored and reduced.
οƒ˜ find zeros of denominator.
 Practice: Vertical Asymptotes.
𝑓 π‘₯ =
2
(π‘₯ + 1)(π‘₯ βˆ’ 5)
𝑓 π‘₯ is factored,
does not reduce.
5π‘₯ + 10
factor 𝑓 π‘₯
𝑓 π‘₯ = 2
and reduce.
π‘₯ βˆ’π‘₯βˆ’6
5(π‘₯ + 2)
5
=
=
(π‘₯ βˆ’ 3)(π‘₯ + 2)
π‘₯βˆ’3
Find zeros of denominator.
π‘₯ + 1 = 0, π‘₯ βˆ’ 5 = 0
P
π‘₯ = βˆ’1, π‘₯ = 5
Vertical Asymptotes
Find zeros of denominator.
π‘₯βˆ’3=0
π‘₯=3
Vertical Asymptote
P
Important: Normally, a VA should have occurred when π‘₯ + 2 = 0 or π‘₯ = βˆ’2,
but when the factor π‘₯ + 2 reduces out of the function, the VA is lost at π‘₯ = βˆ’2
5
and instead, a hole occurs in the graph of 𝑓 π‘₯ = π‘₯βˆ’3 at π‘₯ = βˆ’2.
Rational Functions
 To find Horizontal Asymptote(HA)
οƒ˜ if degree of numerator < degree of denominator,
then 𝑦 = 0 is equation of HA,
οƒ˜ or if degree of numerator = degree of denominator,
then 𝑦 =
π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ 𝑝(π‘₯)
π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘ž(π‘₯)
is equation of HA.
 Practice: Horizontal Asymptotes
𝑓 π‘₯ =
4π‘₯
numerator degree is 1
π‘₯ 2 βˆ’ 9 denominator degree is 2.
3π‘₯ 2
βˆ’π‘₯
𝑓 π‘₯ = 2
1π‘₯ + 4π‘₯ βˆ’ 5
numerator degree is 2
denominator degree is 2.
P
𝑦=0
Horizontal Asymptote
P
3
3
𝑦=
=3
1
1
Horizontal Asymptote
Rational Functions
 To find Oblique Asymptote(OA)
οƒ˜ if degree of numerator > degree of denominator,
then 𝑦 = π‘žπ‘’π‘œπ‘‘π‘–π‘’π‘›π‘‘ of
𝑝(π‘₯)
π‘ž(π‘₯)
is equation of OA.
 Synthetic Division used to find π‘žπ‘’π‘œπ‘‘π‘–π‘’π‘›π‘‘ of
 Practice: Oblique Asymptote
𝑝(π‘₯)
π‘ž(π‘₯)
.
2π‘₯ 2 βˆ’ 5 numerator degree is 2
Oblique Asymptote
𝑓 π‘₯ =
𝑝(π‘₯)
denominator
degree
is
1.
𝑦
=
quotient
of
.
π‘₯βˆ’4
π‘ž(π‘₯)
2π‘₯ 2 + 0π‘₯ βˆ’ 5
=
π‘₯βˆ’4
4
2
0 βˆ’5
8 32
2 8 27
OA at 𝑦 = 2π‘₯ + 8
P
∞
Rational Functions
Practice:
Domain: π‘₯ β‰  4
Range: 𝑦 β‰  βˆ’3
∞
βˆ’βˆž
VA: π‘₯ = 4
HA: 𝑦 = βˆ’3
OA: π‘›π‘œπ‘›π‘’
Intercepts: π‘₯ = 0
𝑦=0
βˆ’βˆž
Rational Functions
 Practice: Find the following for 𝑓(π‘₯).
2π‘₯ 2 + 3π‘₯ βˆ’ 5 (2π‘₯ + 5)(π‘₯ βˆ’ 1) 2π‘₯ + 5
𝑓 π‘₯ = 2
=
=
π‘₯ + 3π‘₯ βˆ’ 4
(π‘₯ + 4)(π‘₯ βˆ’ 1)
π‘₯+4
P
 Domain: π‘₯ β‰  βˆ’4 , π‘₯ β‰  1
Factor 𝑓(π‘₯), find restricted values in denominator.
 Vertical Asymptote:
P
π‘₯ = βˆ’4
Must reduce 𝑓(π‘₯) if possible, then find zeros of denominator.
 Horizontal Asymptote:
Degrees are equal, use 𝑦 =
 Oblique Asymptote:
2
1
P
𝑦= =2
π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
No Oblique Asymptote
.
P
Degrees are equal, so 𝑓(π‘₯) cannot have Oblique Asymptote.
Complete and
Practice HW 5.4-5.5
Rational Functions
 Practice: Find the following for 𝑓(π‘₯).
4π‘₯
4π‘₯
𝑓 π‘₯ = 2
=
π‘₯ βˆ’ 11π‘₯ + 18 (π‘₯ βˆ’ 9)(π‘₯ βˆ’ 2) π‘₯ βˆ’ 9 β‰  0, π‘₯ βˆ’ 2 β‰  0
P
π‘₯ β‰  9, π‘₯ β‰  2
 Domain: π‘₯ β‰  9 , π‘₯ β‰  2
Factor 𝑓(π‘₯), find restricted values in denominator.
 Vertical Asymptote equation(s):
π‘₯ βˆ’ 9 = 0, π‘₯ βˆ’ 2 = 0
π‘₯ = 9, π‘₯ = 2
π‘₯ = 9, π‘₯ = 2
P
Must reduce 𝑓(π‘₯) if possible, then find zeros of denominator.
P
 Horizontal Asymptote equation: 𝑦 = 0
Numerator degree 1 < denominator degree 2.
P
 Oblique Asymptote equation: No Oblique Asymptote
𝑓(π‘₯) cannot have Oblique and Horizontal Asymptote.
Rational Functions
 Practice: Find the following for 𝑓(π‘₯).
2π‘₯ 2 + 3π‘₯ βˆ’ 5 (2π‘₯ + 5)(π‘₯ βˆ’ 1) 2π‘₯ + 5
𝑓 π‘₯ = 2
=
=
π‘₯ + 3π‘₯ βˆ’ 4
(π‘₯ + 4)(π‘₯ βˆ’ 1)
π‘₯+4
P
 Domain: π‘₯ β‰  βˆ’4 , π‘₯ β‰  1
Factor 𝑓(π‘₯), find restricted values in denominator.
 Vertical Asymptote:
P
π‘₯ = βˆ’4
Must reduce 𝑓(π‘₯) if possible, then find zeros of denominator.
 Horizontal Asymptote:
Degrees are equal, use 𝑦 =
 Oblique Asymptote:
2
1
P
𝑦= =2
π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
No Oblique Asymptote
.
P
Degrees are equal, so 𝑓(π‘₯) cannot have Oblique Asymptote.
Rational Functions
 Practice: Find the following for 𝑓(π‘₯).
5π‘₯ 2 + 7π‘₯ βˆ’ 6 (5π‘₯ βˆ’ 3)(π‘₯ + 2)
𝑓 π‘₯ =
= 5π‘₯ βˆ’ 3
=
π‘₯+2
π‘₯+2
 Domain: π‘₯ β‰  βˆ’2
P
Factor 𝑓(π‘₯), find restricted values of denominator.
P
 Vertical Asymptote equations(s): None
Must reduce 𝑓(π‘₯) if possible, then find zeros of denominator.
𝑓(π‘₯) reduces, no zeros from denominator.
P
 Horizontal Asymptote equation: None
Numerator degree 2 > denominator degree 1.
𝑓(π‘₯) has Oblique but no Horizontal Asymptote.
 Oblique Asymptote:
𝑦 = 5π‘₯ βˆ’ 3
P
Use Long Division or reduced 𝑓(π‘₯) to get quotient
Complete and
Practice HW 5.4-5.5
MATH 1314 College Algebra
Unit 3
Polynomial & Rational
Inequalities
Section: 5.6
Polynomial & Rational Inequalities
β€’ This section focuses on learning how to find
Rational
the solutions to Polynomial
_________ & _______
Inequalities.
β€’ Solving inequalities follows the same
solving equations
procedure as _______________,
with one exception:
reverse the inequality when multiplying or
 always __________________
negative constant
dividing by a _______________.
βˆ’2π‘₯ + 1 < 9
Example: Solve for x.
βˆ’2 βˆ’5 + 1 < 9
βˆ’2 βˆ’1 + 1 < 9
11 < 9
3<9
O
(
O
βˆ’βˆž
)(
βˆ’4
P
βˆ’1
βˆ’2π‘₯ < 8
)
∞
0
βˆ’1
βˆ’2
βˆ’2
< βˆ’4
π‘₯>
P
Polynomial & Rational Inequalities
β€’ When you solve a polynomial inequality,
like the one below, you are trying to find
coordinates on the
x-values that lead to __________
_______
𝑦<0
polynomial’s graph that have ______.
π‘₯2 βˆ’ π‘₯ βˆ’ 2 < 0
β€’ There are two ways to do this:
οƒ˜Algebraically
οƒ˜Graphically
P
The graph method is at a disadvantage
if exact x-values are required.
Polynomial & Rational Inequalities
β€’ When solving Polynomial & Rational Inequalities,
always rewrite inequality in Standard Form:
𝑓 π‘₯ < 0 , 𝑓(π‘₯) > 0, 𝑓(π‘₯) ≀ 0, 𝑓(π‘₯) β‰₯ 0
 What each inequality means relative to its graph:
𝑓 π‘₯ < 0 means β€œfor which intervals is graph below π‘₯-axis without touching it?”
𝑓(π‘₯) > 0 means β€œfor which intervals is graph above π‘₯-axis without touching it?”
𝑓(π‘₯) ≀ 0 means β€œfor which intervals is graph below π‘₯-axis and touches it?”
𝑓(π‘₯) β‰₯ 0 means β€œfor which intervals is graph above π‘₯-axis and touches it?”
 To solve Polynomial or Rational Inequalities(Standard Form):
1)
Polynomial: Find real zeros to form intervals on number line.
Rational:
Find real zeros of Numerator & Denominator*
to form intervals on number line.

If < or >, intervals use ( only.
If ≀ or β‰₯, intervals use [ only at numerator zeros, use ( everywhere else.
2)
3)
Test a number in each interval using inequality in Standard Form.
Choose interval that satisfies inequality in Standard Form.
Poly. & Rational Inequalities
Practice: Reading inequalities & graphs.
∞
𝑦>0
𝑦=3
𝑦 β‰… 4.8?
β‰… 4.75?
𝑦=2
𝑦 = 0.5
𝑦<0
Poly. & Rational Inequalities
Practice: Reading graph intervals.
 Find 𝑓 π‘₯ < 0. Use graph of 𝑓(π‘₯) below.
P
Answer: (βˆ’βˆž, βˆ’3)
(
βˆ’βˆž
P)
𝑓 π‘₯ <0
∞
Poly. & Rational Inequalities
Practice: Reading graph intervals.
 Find 𝑓 π‘₯ > 0. Use graph of 𝑓(π‘₯) below.
P
Answer: (βˆ’3,0) βˆͺ (0, ∞)
𝑓 π‘₯ >0
βˆ’βˆž
( P)(
P
)
∞
Poly. & Rational Inequalities
Practice: Use the graph of 𝑓(π‘₯) to solve the inequality.
P
P
Inequality
to use
parentheses
Answer:indicates
(βˆ’4,0)
βˆͺ (6,
∞) only.
Inequality
use βˆͺ
brackets
Answer:indicates
βˆ’βˆž, to
βˆ’4
[0,6)only at x-intercepts.
𝑓 π‘₯ β‰₯0
(
βˆ’βˆž
𝑓 π‘₯ β‰₯0
)( )[ )(
..
𝑓 π‘₯ <0
..
)
∞
𝑓 π‘₯ <0
Poly. & Rational Inequalities
Practice: Solve 𝑓(π‘₯) < 0, where 𝑓 π‘₯ = π‘₯(π‘₯ βˆ’ 5)2
1) Rewrite in Standard Form. Find zeros.

𝑓(π‘₯) < 0
r
r
οƒΌ
Interval I Interval II Interval III
 π‘₯(π‘₯ βˆ’ 5)2 < 0
βˆ’βˆž
 π‘₯ = 0, π‘₯ βˆ’ 5 = 0, π‘₯ βˆ’ 5 = 0
βˆ’1 1
6
0
5
 zeros are π‘₯ = 0, π‘₯ = 5
2) Test intervals using Standard Form. π‘₯(π‘₯ βˆ’ 5)2 < 0
(βˆ’1)(βˆ’1 βˆ’ 5)2 < 0
 Interval I : Test π‘₯ = βˆ’1
βˆ’36 < 0
(
 Interval II : Test π‘₯ = 1
 Interval III : Test π‘₯ = 6
3)
οƒΌ
Solution: (βˆ’βˆž, 0)
)(
)(
οƒΌ
(1)(1 βˆ’ 5)2 < 0
16 < 0 r
(6)(6 βˆ’ 5)2 < 0
6<0r
)
∞
Poly. & Rational Inequalities
Practice: Solve
π‘₯
π‘₯βˆ’3
β‰₯ 1.
1) Rewrite in Standard Form. Find zeros.

π‘₯
(π‘₯ βˆ’ 3)
βˆ’
β‰₯0
π‘₯βˆ’3
π‘₯βˆ’3
π‘₯
π‘₯βˆ’3
β‰₯
π‘₯βˆ’3 π‘₯βˆ’3
 Solve denominator π‘₯ βˆ’ 3 = 0
π‘₯=3
βˆ’βˆž
( or [ at π‘₯ = 3?
(
2)
Test intervals using Standard Form.
 Interval I : Test π‘₯ = 2
 Interval II : Test π‘₯ = 4
3)
Solution:
3, ∞
οƒΌ
3
β‰₯0
π‘₯βˆ’3
r
Interval I
)3(
2
3
π‘₯βˆ’3
οƒΌ
Interval II
)
∞
4
β‰₯0
3
β‰₯0
2 βˆ’3
3
β‰₯0
4 βˆ’3
βˆ’3 β‰₯ 0 r
3β‰₯0
οƒΌ
End - Unit 3
Complete assignments
before Ch5 Exam Due Date