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MATH 1314 College Algebra
Unit 3
Polynomial Functions
Section: 5.1
Polynomial Functions
• This section studies the Polynomial Function.
Your Goal - learn to identify the function components
that lead to its characteristic graph behavior.
• Polynomial function: refer to page 327.
Leading coefficient
degree of 𝑓(𝑥) is the highest power 𝑛.
𝑓 𝑥 = 𝒂𝒏 𝑥 𝒏 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 1 + 𝒂𝟎
Leading term: determines
the end behavior of graph
& number of complex zeros
Constant term
(y-intercept)
 any combination of variable terms and constant term can exist.
 all coefficients are real numbers
 exponents are non-negative integers
 variable cannot be in a denominator
 graphs are smooth and continuous
Polynomial Functions
• Practice: According to the description of a
polynomial function, which of the following are
considered polynomial functions?
P
O
P
O
O
? 𝑓 𝑥
? 𝑓 𝑥
? 𝑓 𝑥
? 𝑓 𝑥
? 𝑓 𝑥
1
= 5 Why? Follows Polynomial form.
𝑥 → 𝑥2
= 𝑥 2 + 3 𝑥 − 4 Why not? Exponents must be
=
1 3
− 𝑥
5
1
+ 𝑥
4
=
𝑥 2 +2𝑥−15
𝑥 2 −4
+7
nonnegative integers.
Why? Follows Polynomial form.
Can’t have a variable
Why not?
in denominator.
= 𝑥 −2 + 4𝑥 + 1 Why not?
Exponents must be
nonnegative integers.
Polynomial Functions
• Exercise: Graph. Determine end behavior rules for 𝑓 𝑥 = 𝑎𝑥 𝑛 .
1
5
 𝑌1 = 𝑥 2
𝑌1 = 3𝑥 4
𝑌1 = 𝑥 6
up
 Conclusion: if degree is even and 𝑎 > 0, both ends go _______.
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1
5
 𝑌1 = − 𝑥 2
𝑌1 = −3𝑥 4
𝑌1 = −𝑥 6
down
 Conclusion: if degree is even and 𝑎 < 0, both ends go _______.
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Polynomial Functions
• Exercise: Graph. Determine end behavior rules for 𝑓(𝑥) = 𝑎𝑥 𝑛 .
 𝑌1 = 3𝑥
1
4
𝑌1 = 𝑥 3
𝑌1 = 𝑥 5
down
 Conclusion: if degree is odd and 𝑎 > 0, left end goes ________
P
up
right end goes ________.
 𝑌1 = −3𝑥
𝑌1 =
1 3
− 𝑥
4
𝑌1 = −𝑥 5
up
 Conclusion: if degree is odd and 𝑎 < 0, left end goes _______
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right end goes _______.
down
Polynomial Functions
• Theorems and Definitions
 Real Zeros: A real number 𝑥 = 𝑟 that makes 𝑓 𝑥
P
called a real zero of 𝑓(𝑥). This means that:
= 0 is
 𝑥 = 𝑟 is an x-intercept of 𝑓(𝑥) and a solution to 𝑓(𝑥) = 0.
 𝑥 − 𝑟 is a factor of 𝑓(𝑥) .
 Multiplicity: If 𝑥 − 𝑟 occurs in 𝑓(𝑥) an even or odd number of
times, then 𝑥 = 𝑟 is a zero having even or odd multiplicity.
 even multiplicity: graph touches x-axis at 𝑥 = 𝑟.
 odd multiplicity: graph crosses x-axis at 𝑥 = 𝑟.
 Turning Points: If a polynomial function has degree 𝑛,
then its graph has at most 𝑛 − 1 turning points, and vice-versa.
 End Behavior: ends of the graph of polynomial 𝑓(𝑥)
resembles the graph of its leading term 𝑎𝑛 𝑥 𝑛 .
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P
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Polynomial Functions
 For the polynomial
function below, find its
real zeros and multiplicity.
𝑓(𝑥) = (𝑥 + 5)2 (𝑥 − 2)
factors (𝑥 − 𝑐) are:
(𝑥 + 5), (𝑥 + 5), (𝑥 − 2)
zeros 𝑥 = 𝑐 are:
𝑥 + 5 = 0, 𝑥 + 5 = 0, 𝑥 − 2 = 0
𝑥 = −5, 𝑥 = −5
multiplicity 2
𝑥=2
multiplicity 1
**Find the polynomial
function having the given
real zeros and multiplicity.
degree: 3
zeros: −4 multiplicity 1,
3 multiplicity 2
zeros 𝑥 = 𝑐 are:
𝑥 = −4,
𝑥 = 3, 𝑥 = 3
factors (𝑥 − 𝑐) are:
(𝑥 + 4), (𝑥 − 3), (𝑥 − 3)
𝑓(𝑥) = (𝑥 + 4)(𝑥 − 3)(𝑥 − 3)
= (𝑥 + 4)(𝑥 − 3)2
= 𝑥 3 − 2𝑥 2 − 15𝑥 + 36
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Polynomial Functions
Your turn. Practice.
Find the polynomial function 𝑓(𝑥) whose
degree is 4
and zeros are: −1 with multiplicity 2 Occurs twice
7 with multiplicity 2 Occurs twice
zeros 𝑥 = 𝑐
factors (𝑥 − 𝑐)
−1, −1, 7, 7
(𝑥 + 1),(𝑥 + 1) ,(𝑥 − 7),(𝑥 − 7)
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So 𝑓(𝑥) = (𝑥 + 1)(𝑥 + 1)(𝑥 − 7)(𝑥 − 7)
= (𝑥 + 1)2 (𝑥 − 7)2
= 𝑥 4 − 12𝑥 3 + 22𝑥 2 + 84𝑥 + 49
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Polynomial Functions
Practice.
If 𝑓 𝑥 = 𝑥 4 + 2𝑥 3 + 𝑥 2 , find the following:
0, −1
a) real zeros: _______________
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Factor, then solve each factor for 𝑥.
𝑥 4 + 2𝑥 3 + 𝑥 2 = 0
𝑥 2 (𝑥
𝑥 2 + 2𝑥 + 1)
1 =0
𝑥2 𝑥 + 1 𝑥 + 1 = 0
𝑥2 = 0 ,
𝑥+1=0,𝑥+1=0
𝑥 = −1 , 𝑥 = −1
𝑥 =0,𝑥 =0
Polynomial Functions
Practice ….continued.
If 𝑓 𝑥 = 𝑥 4 + 2𝑥 3 + 𝑥 2 , find the following:
b) Graph crosses or touches x-axis at:
𝟎 largest,
largest zero _________
touches
occurs twice
multiplicity
−𝟏 smallest,
occurs twice
multiplicity
P
touches P
smallest zero _________
Degree 𝑛 − 1
c) Maximum number of turning points
on the graph of the function:
_____
3
**Remember Turning Points Theorem
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Polynomial Functions
Practice.
If 𝑓 𝑥 = 5(𝑥 2 − 3)(𝑥 + 4)2 , find the following:
3, − 3, −4
a) real zeros: _______________
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Solve each factor for 𝑥.
𝑥2 − 3 = 0
𝑥+4=0 𝑥+4=0
𝑥 = −4
𝑥 = −4
𝑥2 = 3
𝑥=± 3
3
−4 − 3
∞
= 3, − 3 −∞
0
𝟑 largest,
1
multiplicity of largest zero is: ____
occurs once
− 𝟑 middle,
1
multiplicity of middle zero is: ____
occurs once
−𝟒 smallest,
2
multiplicity of smallest zero is: ____
occurs twice
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P
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Polynomial Functions
Practice ….continued.
If 𝑓 𝑥 = 5(𝑥 2 − 3)(𝑥 + 4)2 , find the following:
b) Graph crosses or touches x-axis at:
𝟑 largest,
crosses
occurs once
largest zero _________
− 𝟑 middle,
crosses
occurs once
middle zero _________
−𝟒 middle,
touches
occurs twice
smallest zero _________
P
P
P
multiplicity
multiplicity
multiplicity
c) Maximum number of turning points Degree 𝑛 − 1
3
on the graph of the function:
_____
P
d) Power function that graph resembles
5𝑥 P
for large values of 𝑥 :
_____
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(𝑥).
 Helpful:
𝑓(𝑥) was already factored
or
𝑓(𝑥) was easily factored.
𝑓 𝑥 = 𝑥 + 1 2 (𝑥 − 3)
𝑓 𝑥 = 𝑥 3 − 6𝑥 2 + 9𝑥
 Section 5.2 explains how to find real zeros in:
 polynomial 𝑓(𝑥) that are not already factored
or
 polynomial 𝑓(𝑥) that are not easily factored.
Real Zeros
• Some Theorems that can be used to find real zeros
and factors of polynomial functions
 Remainder Theorem(p347):
 If polynomial 𝑓(𝑥) is divided by linear factor 𝑥 − 𝑐,
then 𝑓(𝑐) = remainder.
 Factor Theorem(p348):
 If 𝑓 𝑐 = 0, then 𝑥 − 𝑐 is a linear factor of 𝑓(𝑥).
 If 𝑥 − 𝑐 is a linear factor of 𝑓(𝑥), 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(𝑥),
if 𝑎 < 𝑏 and if 𝑓(𝑎) and 𝑓(𝑏) have opposite signs,
then 𝑓(𝑥) has at least one zero between 𝑥 = 𝑎 and 𝑥 = 𝑏.
Real Zeros
• Practice: How to factor polynomial functions
using the previous theorems.
 Factor 𝑓 𝑥 = 2𝑥 3 − 11𝑥 2 + 10𝑥 + 8:
 Is 𝑥 + 1 a factor of 𝑓(𝑥)?
If 𝑥 + 1 is a factor, then 𝑥 = −1 is a zero of 𝑓(𝑥).
Using Remainder Theorem, since 𝑓 −1 = −15 ≠ 0,
you know that 𝑥 + 1 is not a factor of 𝑓 𝑥 .
 Is 𝑥 − 4 a factor of 𝑓(𝑥)?
If 𝑥 − 4 is a factor, then 𝑥 = 4 is a zero of 𝑓(𝑥).
Using Remainder Theorem, since 𝑓 4 = 0,
you know that 𝑥 − 4 is a factor of 𝑓 𝑥 .
Real Zeros
 Synthetic Division will use a divisor 𝑐 to find zeros,
remainder, and factors of polynomial function 𝑓(𝑥).
 Practice: Coefficient writing
 Polynomial function 𝑓 𝑥 = 𝑥 3 − 7𝑥 2 + 4𝑥 + 1
has real coefficients:
1
−7
4
1
 Polynomial function 𝑓 𝑥 = 𝑥 4 + 5𝑥 2 − 3
𝑓 𝑥 = 𝑥 4 + 0𝑥 3 + 5𝑥 2 + 0𝑥 − 3
has real coefficients:
1
0
5
0 −3
*Insert a zero coefficient for any missing terms.
Real Zeros
 Practice: Use Synthetic Division to find remainder and
quotient if 𝑓 𝑥 = 𝑥 3 + 3𝑥 2 − 6𝑥 − 8 is divided by 𝑥 + 4.
divisor
𝑥+4=0
𝑥 = −4
𝑐
−4
multiply
multiply
1coefficients
3 add
− 6 add
of 𝑓(𝑥)
1
−4
−1
4
−2
 coefficients in last row make Quotient and Remainder:
1 −1 −2
0
Quotient is 𝑥 2 − 𝑥 − 2
Remainder is 0
−8
8
0
add
Real Zeros
 Since 𝑥 + 4 produced a zero remainder, 𝑥 = −4 is a zero.
To find remaining zeros in quotient 𝑥 2 − 𝑥 − 2:
 Factor or use Quadratic Formula.
−(𝑏) ± (𝑏)2 −4 𝑎 𝑐
𝑥=
2(𝑎)
−(−1) ± (−1)2 −4 1 −2
𝑥=
2(1)
𝑥=
1± 9
1±3
1 3
=
=
= ±
2
2
2 2
Zeros of 𝑓(𝑥):
Factors of 𝑓 𝑥 =
𝑥=
1
2
1
2
3
+
2
3
−
2
=2
= −1
P
(𝑥 + 4)(𝑥 − 2)(𝑥 + 1) P
−4,
an
d
2,
an
d
−1
an
d
Real Zeros
 Practice: Does 𝑓 𝑥 = 2𝑥 3 + 5𝑥 2 − 3𝑥 + 8
have a real zero between 𝑥 = −6 and 𝑥 = −1?
 Use Intermediate Value Theorem: Find 𝑓(−6) and 𝑓(−1).
𝑓 −6 = −226
𝑓 −1 = 14
P
P
P
𝑓(−6) and 𝑓(−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(𝑥).
factored or was ____________.
easily factored
 𝑓(𝑥) was either _______
 Section 5.2 asked: Find real zeros of polynomial 𝑓(𝑥).
factored or was not
easily factored
 𝑓(𝑥) was either not
__________
_______________.
 To find real zeros of 𝑓(𝑥), you learned to use:
• theorems, such as Remainder
_________ Theorem
and Factor
_____ Theorem
Synthetic Division
• and a form of division called _______________.
 Section 5.3: Find Complex
_______ zeros of polynomial 𝑓(𝑥).
 Complex zeros include non-real or imaginary
________ zeros
 additional theorems will be used.
Complex Zeros
 Complex numbers: Standard Form
𝑎 + 𝑏𝑖 𝑎, 𝑏 are real numbers
real part
imaginary part
 To prepare for this section, review:
 Operations on complex numbers(p120-124)
 Definition of imaginary number: 𝑖 = −1 2
𝑖 2 = −1
𝑖 2 = −1
?
OO
? ?
Is −4 = 2 = −2
? 2 ?
2
Only
if −2
Only
if 2
= −4= −4
−4 =
4 ∙ −1 = 4 ∙ −1
−4 = 2𝑖
 Complex solutions of Quadratic Equations(p124-126)
Complex Zeros
•
Some Theorems you will use in this section:
 Fundamental Theorem of Algebra: (textbook p360)
• Every polynomial function of degree 𝑛 ≥ 1,
has at least one complex zero.
 Conjugate Pairs Theorem: (textbook p360)
• Let 𝑓(𝑥) be a polynomial function whose
coefficients are real. If 𝑟 = 𝑎 + 𝑏𝑖 is a zero of f,
then the complex conjugate 𝑟 = 𝑎 − 𝑏𝑖 is also.
*(The Quadratic Formula is the basis for this.)
 Corollary: (textbook p361)
• A polynomial function 𝑓(𝑥) 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 − 4𝑖
 the complex conjugate of −1 − 𝑖 is: −1 + 𝑖
 the complex conjugate of 2𝑖 is: −2𝑖
Remember, non-real or imaginary numbers are
the result of an even root of a negative number.
Example:
−16 = (16)(−1)
= 16 ∙ −1
= 4𝑖
Complex Zeros
 Practice: If 3𝑖 is a zero of 𝑓 𝑥 = 𝑥 3 + 2𝑥 2 + 9𝑥 + 18,
find the remaining zeros of 𝑓 𝑥 .
 3 complex zeros exist. Why? degree 3
 If 3𝑖 is a zero, then ____
−3𝑖 is also. Why? Conjugate Pairs Theorem
 Third zero must be real. Why? Corollary(p361): Odd degree
 Use Synthetic/Long Division to find third zero
3𝑖
1
1
−3𝑖
1
1
2
9
18
2 −18
3𝑖 6𝑖 −9
+9(−1)
+9𝑖
0
2 + 3𝑖 6𝑖
2 + 3𝑖 6𝑖
−3𝑖 −6𝑖
2
0
𝑥 + 2 = 0 Solve for third zero.
𝑥 = −2
Remaining zeros: −3𝑖, −2
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Complex Zeros
 Practice: If 3𝑖 is a zero of 𝑓 𝑥 = 𝑥 3 + 2𝑥 2 + 9𝑥 + 18,
find the remaining zeros of 𝑓 𝑥 .
 Or use TABLE/GRAPH on TI-83/84, if possible.
 Remaining zeros are : −3𝑖, −2
Complex Zeros
 Exercise: Find the complex zeros of 𝑓(𝑥), then use the
zeros to factor 𝑓(𝑥). 𝑓 𝑥 = 2𝑥 3 − 12𝑥 2 + 13𝑥 − 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(𝑥), then use the
zeros to factor 𝑓(𝑥). 𝑓 𝑥 = 2𝑥 3 − 12𝑥 2 + 13𝑥 − 15
 First real zero: 𝑥 = 5
*Since TI-83 GRAPH/TABLE
doesn’t show 2 more real zeros,
they may be nonreal.
 Use Synthetic Division to divide 𝑓(𝑥) by 𝑥 = 5 to find
quotient.
2 − 12
13
− 15
5
10 −10
15
0
2
3
−2
 Quotient is 2𝑥 2 − 2𝑥 + 3. Two complex zeros remain.
…continued
Complex Zeros
 Use Quadratic Formula to find remaining 2 zeros in
quotient 2𝑥 2 − 2𝑥 + 3.
−(𝑏) ± (𝑏)2 −4 𝑎 𝑐
𝑥=
2(𝑎)
−(−2) ± (−2)2 −4 2 3
𝑥=
2(2)
1 𝑖 5
2 2𝑖 5
2 ± 2𝑖 5
2 ± −4 ∙ 5
2 ± −20
= ±
= ±
=
=
=
2
2
4
4
4
4
4
 Zeros of 𝑓 𝑥 :
5,
1
𝑖 5
+
,
2
2
1
𝑖 5
−
2
2
1
𝑖 5
1
𝑖 5
)(𝑥 − +
)
 Factors of 𝑓 𝑥 = (𝑥 − 5)(𝑥 − −
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 𝑥 =
𝑝(𝑥)
𝑞(𝑥)
, where 𝑞 𝑥 ≠ 0.
 Practice factoring (p50)
 Domain: all real numbers 𝑥, where 𝑞 𝑥 ≠ 0.
 Asymptote: a boundary line in which the graph of 𝑅(𝑥)
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(𝑥)?)
Rational Functions
 Graph of 𝑅 𝑥 =
 𝑓 𝑥 =
𝑝(𝑥)
𝑞(𝑥)
𝑥−1
(𝑥−3)(𝑥+2)
𝑥 + 2 ≠ 0,
so 𝑥 ≠ −2
𝑥 − 3 ≠ 0,
so 𝑥 ≠ 3
𝑥−1
𝑥 2 −𝑥−6
 Remember, if 𝑓(𝑥) looked like this : 𝑓 𝑥 =
then what Algebraic skill is needed to help
determine the restricted domain values of 𝑓(𝑥)? Factoring
P
Rational Functions
 To find Domain
 𝑓(𝑥) must be factored completely.
 find restricted values from denominator.
 Practice: Domain
𝑓 𝑥 =
2
(𝑥 + 1)(𝑥 − 5)
5𝑥 + 10
𝑓 𝑥 = 2
𝑥 −𝑥−6
5(𝑥 + 2)
=
(𝑥 − 3)(𝑥 + 2)
Find restricted values
𝑥 + 1 ≠ 0, 𝑥 − 5 ≠ 0
𝑓 𝑥 is factored
factor 𝑓 𝑥
𝑥 ≠ −1, 𝑥 ≠ 5
Domain of 𝑓(𝑥).
P
Find restricted values
𝑥 − 3 ≠ 0, 𝑥 + 2 ≠ 0
𝑥 ≠ 3, 𝑥 ≠ −2
Domain of 𝑓(𝑥).
P
Rational Functions
 To find Vertical Asymptote(VA) after Domain
 𝑓(𝑥) must be factored and reduced.
 find zeros of denominator.
 Practice: Vertical Asymptotes.
𝑓 𝑥 =
2
(𝑥 + 1)(𝑥 − 5)
𝑓 𝑥 is factored,
does not reduce.
5𝑥 + 10
factor 𝑓 𝑥
𝑓 𝑥 = 2
and reduce.
𝑥 −𝑥−6
5(𝑥 + 2)
5
=
=
(𝑥 − 3)(𝑥 + 2)
𝑥−3
Find zeros of denominator.
𝑥 + 1 = 0, 𝑥 − 5 = 0
P
𝑥 = −1, 𝑥 = 5
Vertical Asymptotes
Find zeros of denominator.
𝑥−3=0
𝑥=3
Vertical Asymptote
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Important: Normally, a VA should have occurred when 𝑥 + 2 = 0 or 𝑥 = −2,
but when the factor 𝑥 + 2 reduces out of the function, the VA is lost at 𝑥 = −2
5
and instead, a hole occurs in the graph of 𝑓 𝑥 = 𝑥−3 at 𝑥 = −2.
Rational Functions
 To find Horizontal Asymptote(HA)
 if degree of numerator < degree of denominator,
then 𝑦 = 0 is equation of HA,
 or if degree of numerator = degree of denominator,
then 𝑦 =
𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑝(𝑥)
𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑞(𝑥)
is equation of HA.
 Practice: Horizontal Asymptotes
𝑓 𝑥 =
4𝑥
numerator degree is 1
𝑥 2 − 9 denominator degree is 2.
3𝑥 2
−𝑥
𝑓 𝑥 = 2
1𝑥 + 4𝑥 − 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)
 if degree of numerator > degree of denominator,
then 𝑦 = 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 of
𝑝(𝑥)
𝑞(𝑥)
is equation of OA.
 Synthetic Division used to find 𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡 of
 Practice: Oblique Asymptote
𝑝(𝑥)
𝑞(𝑥)
.
2𝑥 2 − 5 numerator degree is 2
Oblique Asymptote
𝑓 𝑥 =
𝑝(𝑥)
denominator
degree
is
1.
𝑦
=
quotient
of
.
𝑥−4
𝑞(𝑥)
2𝑥 2 + 0𝑥 − 5
=
𝑥−4
4
2
0 −5
8 32
2 8 27
OA at 𝑦 = 2𝑥 + 8
P
∞
Rational Functions
Practice:
Domain: 𝑥 ≠ 4
Range: 𝑦 ≠ −3
∞
−∞
VA: 𝑥 = 4
HA: 𝑦 = −3
OA: 𝑛𝑜𝑛𝑒
Intercepts: 𝑥 = 0
𝑦=0
−∞
Rational Functions
 Practice: Find the following for 𝑓(𝑥).
2𝑥 2 + 3𝑥 − 5 (2𝑥 + 5)(𝑥 − 1) 2𝑥 + 5
𝑓 𝑥 = 2
=
=
𝑥 + 3𝑥 − 4
(𝑥 + 4)(𝑥 − 1)
𝑥+4
P
 Domain: 𝑥 ≠ −4 , 𝑥 ≠ 1
Factor 𝑓(𝑥), find restricted values in denominator.
 Vertical Asymptote:
P
𝑥 = −4
Must reduce 𝑓(𝑥) if possible, then find zeros of denominator.
 Horizontal Asymptote:
Degrees are equal, use 𝑦 =
 Oblique Asymptote:
2
1
P
𝑦= =2
𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
No Oblique Asymptote
.
P
Degrees are equal, so 𝑓(𝑥) cannot have Oblique Asymptote.
Complete and
Practice HW 5.4-5.5
Rational Functions
 Practice: Find the following for 𝑓(𝑥).
4𝑥
4𝑥
𝑓 𝑥 = 2
=
𝑥 − 11𝑥 + 18 (𝑥 − 9)(𝑥 − 2) 𝑥 − 9 ≠ 0, 𝑥 − 2 ≠ 0
P
𝑥 ≠ 9, 𝑥 ≠ 2
 Domain: 𝑥 ≠ 9 , 𝑥 ≠ 2
Factor 𝑓(𝑥), find restricted values in denominator.
 Vertical Asymptote equation(s):
𝑥 − 9 = 0, 𝑥 − 2 = 0
𝑥 = 9, 𝑥 = 2
𝑥 = 9, 𝑥 = 2
P
Must reduce 𝑓(𝑥) 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(𝑥) cannot have Oblique and Horizontal Asymptote.
Rational Functions
 Practice: Find the following for 𝑓(𝑥).
2𝑥 2 + 3𝑥 − 5 (2𝑥 + 5)(𝑥 − 1) 2𝑥 + 5
𝑓 𝑥 = 2
=
=
𝑥 + 3𝑥 − 4
(𝑥 + 4)(𝑥 − 1)
𝑥+4
P
 Domain: 𝑥 ≠ −4 , 𝑥 ≠ 1
Factor 𝑓(𝑥), find restricted values in denominator.
 Vertical Asymptote:
P
𝑥 = −4
Must reduce 𝑓(𝑥) if possible, then find zeros of denominator.
 Horizontal Asymptote:
Degrees are equal, use 𝑦 =
 Oblique Asymptote:
2
1
P
𝑦= =2
𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
No Oblique Asymptote
.
P
Degrees are equal, so 𝑓(𝑥) cannot have Oblique Asymptote.
Rational Functions
 Practice: Find the following for 𝑓(𝑥).
5𝑥 2 + 7𝑥 − 6 (5𝑥 − 3)(𝑥 + 2)
𝑓 𝑥 =
= 5𝑥 − 3
=
𝑥+2
𝑥+2
 Domain: 𝑥 ≠ −2
P
Factor 𝑓(𝑥), find restricted values of denominator.
P
 Vertical Asymptote equations(s): None
Must reduce 𝑓(𝑥) if possible, then find zeros of denominator.
𝑓(𝑥) reduces, no zeros from denominator.
P
 Horizontal Asymptote equation: None
Numerator degree 2 > denominator degree 1.
𝑓(𝑥) has Oblique but no Horizontal Asymptote.
 Oblique Asymptote:
𝑦 = 5𝑥 − 3
P
Use Long Division or reduced 𝑓(𝑥) 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
• This section focuses on learning how to find
Rational
the solutions to Polynomial
_________ & _______
Inequalities.
• Solving inequalities follows the same
solving equations
procedure as _______________,
with one exception:
reverse the inequality when multiplying or
 always __________________
negative constant
dividing by a _______________.
−2𝑥 + 1 < 9
Example: Solve for x.
−2 −5 + 1 < 9
−2 −1 + 1 < 9
11 < 9
3<9
O
(
O
−∞
)(
−4
P
−1
−2𝑥 < 8
)
∞
0
−1
−2
−2
< −4
𝑥>
P
Polynomial & Rational Inequalities
• 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 ______.
𝑥2 − 𝑥 − 2 < 0
• There are two ways to do this:
Algebraically
Graphically
P
The graph method is at a disadvantage
if exact x-values are required.
Polynomial & Rational Inequalities
• When solving Polynomial & Rational Inequalities,
always rewrite inequality in Standard Form:
𝑓 𝑥 < 0 , 𝑓(𝑥) > 0, 𝑓(𝑥) ≤ 0, 𝑓(𝑥) ≥ 0
 What each inequality means relative to its graph:
𝑓 𝑥 < 0 means “for which intervals is graph below 𝑥-axis without touching it?”
𝑓(𝑥) > 0 means “for which intervals is graph above 𝑥-axis without touching it?”
𝑓(𝑥) ≤ 0 means “for which intervals is graph below 𝑥-axis and touches it?”
𝑓(𝑥) ≥ 0 means “for which intervals is graph above 𝑥-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 ≥, 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 ≅ 4.8?
≅ 4.75?
𝑦=2
𝑦 = 0.5
𝑦<0
Poly. & Rational Inequalities
Practice: Reading graph intervals.
 Find 𝑓 𝑥 < 0. Use graph of 𝑓(𝑥) below.
P
Answer: (−∞, −3)
(
−∞
P)
𝑓 𝑥 <0
∞
Poly. & Rational Inequalities
Practice: Reading graph intervals.
 Find 𝑓 𝑥 > 0. Use graph of 𝑓(𝑥) below.
P
Answer: (−3,0) ∪ (0, ∞)
𝑓 𝑥 >0
−∞
( P)(
P
)
∞
Poly. & Rational Inequalities
Practice: Use the graph of 𝑓(𝑥) to solve the inequality.
P
P
Inequality
to use
parentheses
Answer:indicates
(−4,0)
∪ (6,
∞) only.
Inequality
use ∪
brackets
Answer:indicates
−∞, to
−4
[0,6)only at x-intercepts.
𝑓 𝑥 ≥0
(
−∞
𝑓 𝑥 ≥0
)( )[ )(
..
𝑓 𝑥 <0
..
)
∞
𝑓 𝑥 <0
Poly. & Rational Inequalities
Practice: Solve 𝑓(𝑥) < 0, where 𝑓 𝑥 = 𝑥(𝑥 − 5)2
1) Rewrite in Standard Form. Find zeros.

𝑓(𝑥) < 0
r
r

Interval I Interval II Interval III
 𝑥(𝑥 − 5)2 < 0
−∞
 𝑥 = 0, 𝑥 − 5 = 0, 𝑥 − 5 = 0
−1 1
6
0
5
 zeros are 𝑥 = 0, 𝑥 = 5
2) Test intervals using Standard Form. 𝑥(𝑥 − 5)2 < 0
(−1)(−1 − 5)2 < 0
 Interval I : Test 𝑥 = −1
−36 < 0
(
 Interval II : Test 𝑥 = 1
 Interval III : Test 𝑥 = 6
3)

Solution: (−∞, 0)
)(
)(

(1)(1 − 5)2 < 0
16 < 0 r
(6)(6 − 5)2 < 0
6<0r
)
∞
Poly. & Rational Inequalities
Practice: Solve
𝑥
𝑥−3
≥ 1.
1) Rewrite in Standard Form. Find zeros.

𝑥
(𝑥 − 3)
−
≥0
𝑥−3
𝑥−3
𝑥
𝑥−3
≥
𝑥−3 𝑥−3
 Solve denominator 𝑥 − 3 = 0
𝑥=3
−∞
( or [ at 𝑥 = 3?
(
2)
Test intervals using Standard Form.
 Interval I : Test 𝑥 = 2
 Interval II : Test 𝑥 = 4
3)
Solution:
3, ∞

3
≥0
𝑥−3
r
Interval I
)3(
2
3
𝑥−3

Interval II
)
∞
4
≥0
3
≥0
2 −3
3
≥0
4 −3
−3 ≥ 0 r
3≥0

End - Unit 3
Complete assignments
before Ch5 Exam Due Date