Honors ALGEBRA II - Suffolk City Public Schoolsstar.spsk12.net/math/Alg...
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STANDARDS OF LEARNING
CONTENT REVIEW NOTES
Honors ALGEBRA II
3rd Nine Weeks, 2018-2019
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OVERVIEW
Algebra II Content Review Notes are designed by the High School Mathematics Steering Committee as a
resource for students and parents. Each nine weeksβ Standards of Learning (SOLs) have been identified and a
detailed explanation of the specific SOL is provided. Specific notes have also been included in this document
to assist students in understanding the concepts. Sample problems allow the students to see step-by-step models
for solving various types of problems. A β β section has also been developed to provide students with the
opportunity to solve similar problems and check their answers. Supplemental online information can be
accessed by scanning QR codes throughout the document. These will take students to video tutorials and online
resources. In addition, a self-assessment is available at the end of the document to allow students to check their
readiness for the nine-weeks test.
The document is a compilation of information found in the Virginia Department of Education (VDOE)
Curriculum Framework, Enhanced Scope and Sequence, and Released Test items. In addition to VDOE
information, Prentice Hall Textbook Series and resources have been used. Finally, information from various
websites is included. The websites are listed with the information as it appears in the document.
Supplemental online information can be accessed by scanning QR codes throughout the document. These will
take students to video tutorials and online resources. In addition, a self-assessment is available at the end of the
document to allow students to check their readiness for the nine-weeks test.
To access the database of online resources scan this QR code.
The Algebra II Blueprint Summary Table is listed below as a snapshot of the reporting categories, the number
of questions per reporting category, and the corresponding SOLs.
It is the Mathematics Instructorsβ desire that students and parents will use this document as a tool toward the
studentsβ success on the end-of-year assessment.
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Radical Functions
AII/T.1 The student, given rational, radical, or polynomial expressions, will
b) add, subtract, multiply, divide, and simplify radical expressions containing rational numbers and variables, and expressions containing rational exponents;
c) write radical expressions as expressions containing rational exponents and vice versa
Simplifying Radicals
To simplify a radical, you will pull out any perfect square factors (i.e. 4, 9, 16, 25, etc.)
β18 = β9 β 2
The square root of 9 is equal to 3, so you can pull the square root of 9 from underneath
the radical sign to find the simplified answer 3 β2 , which means 3 times the square
root of 2. You can check this simplification in your calculator by verifying that
β18 = 3β2 .
Another way to simplify radicals, if you donβt know the factors of a number, is to create
a factor tree and break the number down to its prime factors. When you have broken
the number down to all of its prime factors you can pull out pairs of factors for square
roots, which will multiply together to make perfect squares.
Example 1: Simplify β128 β2 β 2 β 2 β 2 β 2 β 2 β 2
2 β 2 β 2 β2 = 8β2
Example 2: Simplify β32π₯3π¦
2 64
8 8
4 2 2 4
2 2 2 2
Scan this QR code to go to a video tutorial on
simplifying radicals.
16 2 x x x y
4 4
2 2 2 2
β2 β 2 β 2 β 2 β 2 β π₯ β π₯ β π₯ β π¦
2 β 2 β π₯β2π₯π¦ = 4π₯β2π₯π¦
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To simplify a root of a higher index, you pull out factors that occurs the same number
of times as the index of the radical. As an example, if you are simplifying β64π75 you
would only pull out factors that occurred 5 times, since 5 is the index of the root.
Example 3: Simplify β64π74
β2 β 2 β 2 β 2 β 2 β 2 β π β π β π β π β π β π β π4
2π β4π34
Radical Operations
In order to multiply or divide radical expressions, they must share the same index. This
means that the power of the root must be the same.
β312π₯π¦54 β β125π₯85
could not be simplified because one is a 4th root and one is a 5th
5π¦ β27π₯43Γ· 3 β80π¦43
could be simplified because they are both cube roots.
If βππ
and βππ
are real numbers, then βππ
β βππ
= βπππ
If βππ
and βππ
are real numbers, and π β 0, then βπ
π
βππ = β
π
π
π
Example 1: Simplify 2π₯ β16π₯43β β6π¦3
Because these radicals have the same index, they can be multiplied and simplified.
2π₯ β96π₯4π¦3
2π₯ β2 β 2 β 2 β 2 β 2 β 3 β π₯ β π₯ β π₯ β π₯ β π¦3
2π₯ β 2 β π₯ β2 β 2 β 3 β π₯ β π¦3 = 4π₯2 β12π₯π¦3
Example 2: Simplify 5π¦ β27π¦3 Γ· 3 β90π₯π¦5
5π¦
3β
27π¦3
90π₯π¦5 = 5π¦
3β
3
10π₯π¦2 =5π¦
3β
3
2β5βπ₯βπ¦βπ¦ =
5π¦
3π¦ β
3
10π₯ =
5
3β
3
10π₯
Because this is a 4th root, pull out things that occur 4 times.
Because this is a cube root, pull out things that occur 3 times.
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Radical Functions
1. Simplify β135ππ4π63
2. Simplify 3π₯π¦ β100π₯2π¦5
3. Simplify 5π¦ β8π¦3 β 3 β18π₯π¦4
4. Simplify 2π₯ β16π₯43 Γ· β8π₯33
Rational Exponents
Properties of Exponents
Property Example
π₯π β π₯π = π₯π+π 5
12 β 5 2 = 5
12
+2 = 5 52
(π₯π)π = π₯π β π (8 3)5 = 8 3 β 5 = 8 15
(π₯π¦)π = π₯π β π¦π (10π₯)2 = 100π₯2
π₯βπ = 1
π₯π 3β3 =
1
33=
1
27
π₯π
π₯π= π₯ πβπ
75
77= 7 5β7 = 7β2 =
1
72=
1
49
( π₯
π¦ )π =
π₯π
π¦π (
2
3 )3 =
23
33=
8
27
βππ
= π 1π β50
3= 50
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βπππ= π
ππ βπ35
= π 35
Example 1: Simplify 162 1
4
162 1
4 = β1624
= β2 β 3 β 3 β 3 β 34
3 β24
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Example 2: Write πβ1.5 in radical form.
πβ3
2 = 1
π 32
= 1
βπ3=
1
βπβπβπ=
1
π βπ
Example 3: Write in simplest form βπ54
βπ3
π
54
π 3 2
= π 5
4 β
3
2 = π 5
4 β
6
4 = πβ 1
4 = 1
π 14
= 1
βπ4
Radical Functions
5. Write π 7
4 in radical form.
6. Write βπ₯5 in exponential form.
7. Simplify βπ₯6
βπ₯23
Write each radical in exponential form, then use
the properties of exponents to simplify.
Scan this QR code to go to a video tutorial on rational
exponents.
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Radical Equations and Graphs
AII/T.4 The student will solve, algebraically and graphically,
d) equations containing radical expressions.
AII/T.6 The student will recognize the general shape of function (absolute value, square root,
cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions. A transformational approach to graphing will be employed. Graphing calculators will be used as a tool to investigate the shapes and behaviors of these functions.
Radical Equations may require you to square both sides of the equation. This has the potential to introduce extraneous solutions. For this reason, it is very important to always check your solutions to radical equations.
Example 1: Solve for y β5 β 4π₯3
+ 6 = 5
β5 β 4π₯3
+ 6 = 5 β6 β 6
β5 β 4π₯3
= β1
(β5 β 4π₯3
)3 = (β1)3 5 β 4π₯ = β1 β5 β 5 β4π₯ = β6 Γ· (β4) Γ· (β4)
π₯ = 3
2
β5 β 4(3
2)
3+ 6 = β5 β 6
3+ 6 = ββ1
3+ 6 = β1 + 6 = 5
To solve equations with a fractional exponent, raise each side of the equation to the reciprocal power. This is because when you raise a power to a power you multiply the exponents, and multiplying by the reciprocal will make the exponent 1.
Cube both sides of the equation to cancel out the cube root.
Donβt forget to check!
Scan this QR code to go to a video tutorial on
radical equations.
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Example 2: Solve for x: (2π₯ β 6)3
2 = 8
(2π₯ β 6)3
2 = 8
[(2π₯ β 6)32 ]
23 = 8
23
2π₯ β 6 = 4 +6 + 6 2π₯ = 10 Γ· 2 Γ· 2 π₯ = 5
(2(5) β 6) 32 = (10 β 6)
32 = 4
32 = 8
Sometimes it might be helpful to rewrite a radical using a rational exponent to solve.
Example 3: What is the solution of 2 + β(π₯ β 1)53= 34
2 + β(π₯ β 1)53= 34
β2 β 2
β(π₯ β 1)53= 32
(π₯ β 1) 5
3 = 32
[(π₯ β 1)53 ]
35 = 32
35
π₯ β 1 = 8 π₯ = 9
2 + β(9 β 1)53= 2 + β853
= 2 + β327683
= 2 + 32 = 34 Graphs of Radical Functions
π¦ = βπ₯ π¦ = βπ₯3
Only in the 1st quadrant, because square roots are only defined for
positive values of x.
In the 1st and 3rd quadrant because when x is positive y will be positive, and when x is negative y will also be
negative.
Donβt forget to check!
Raise both sides to the reciprocal
power. 3
2 β
2
3= 1
Isolate the radical expression.
Rewrite in exponential form.
Raise both sides to the reciprocal
power. 5
3 β
3
5= 1
Donβt forget to check!
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Example 4: The maximum walking speed of a large animal can be modeled by the
equation: π = βππΏ, where S is the speed (ft/sec), g is a constant 32 ππ‘/π ππ2,
and L is the length (ft) of the animalβs leg. Use this model to predict how long an animalβs leg would need to be in order to travel 15 ft/sec.
This equation could be solved algebraically or graphically. We will show both examples here.
Algebraically: Plug in the given values for S and g
15 = β32πΏ
(15)2 = (β32πΏ)2 225 = 32πΏ Γ· 32 Γ· 32 7.03 = πΏ
Graphically: Graph the equation π = βππΏ in Y1 as π1 = β32π₯.
The y value that we want to consider is 15 ft/sec, so graph π¦ = 15 in Y2.
Press 2nd then TRACE to take you to the CALC menu. Scroll down to 5: intersect.
This again shows your answer as X = 7.03 feet.
Radical Equations
1. 4 β2π₯ + 2 β 6 = 10
2. (4π₯ + 1)1
4 + 5 = 2
3. 2 β(2π₯ β 4)23 = 32
4. β3 β β2π₯5
= β1 5. You can model the population, π, of St. Louis, Missouri between 1900 and 1940 by
the function π(π₯) = 52,000 βπ₯ β 18903
, where π₯ is the year. Using this model, in what year was the population of St. Louis 175,000?
Square both sides of the equation to cancel out the square root.
The animalβs leg would need to be just over 7 feet long in order
to travel 15 ft/sec.
You will need to adjust your zoom in order to see the intersection. You can
do this by pressing ZOOM then 0 for ZoomFit.
Your graph should look like this --->
The calculator will ask for first curve?, second curve?
and guess?. Just scroll near the intersection and press ENTER until your intersection is displayed.
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Rational Expressions AII/T.1 The student, given rational, radical, or polynomial expressions, will
a) add, subtract, multiply, divide, and simplify rational algebraic expressions;
A rational expression is the quotient of two polynomials (or a polynomial fraction). To simplify rational expressions you need to factor both the numerator and denominator, and then divide out common factors.
Example 1: Simplify 9π₯3β12π₯2
3π₯2
3π₯2 (3π₯β4)
3π₯2 (1)=
3π₯β4
1= 3π₯ β 4 , π₯ β 0
Example 2: Simplify π₯2β7π₯+12
(π₯+3)(π₯β4)
Factor the numerator! π β π = 1 β 12 = 12 So, we are looking for factors of 12 that add up to β7 !
To multiply rational expressions, you multiply the numerators and multiply the denominators. You can use factoring to simplify.
Example 3: Multiply and Simplify 9π₯2β1
3π₯2 β 3π₯2
3π₯2β2π₯β1
(3π₯+1)(3π₯β1)
3π₯2 β 3π₯2
(3π₯+1)(π₯β1)
3π₯ β 1
π₯ β 1 , π₯ β 1, 0, β
1
3
π₯2 β3π₯
β4π₯ 12
The GCF is 3π₯2. If you remove this from both the numerator and the denominator, you can divide it out. This is because
3π₯2
3π₯2 = 1 where π₯ β 0. If x were 0, then
the original problem would be undefined (the denominator would equal 0).
Find the greatest common factor in each row and each column. These will give you your two binomials! Put the now factored trinomial back into the original problem and divide out factors that are in both the numerator and denominator.
π₯
β4
π₯ β3
(π₯β4) (π₯β3)
(π₯+3)(π₯β4)=
π₯β3
π₯+3 , π₯ β 4, πππ π₯ β β3
Divide out common factors that occur in both the numerator and denominator. Your answer is the product of what remains after common factors are divided out.
Scan this QR code to go to a video tutorial on simplifying rational
expressions.
Donβt forget: anytime the π term is negative and the π term is positive, your answer will have two minus signs!
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To divide rational expressions, you can simply multiply the first expression by the reciprocal of the second.
Example 4: Divide and Simplify 12(π₯β5)
π₯2β1 Γ·
4π₯β12
π₯2β2π₯β3
12(π₯β5)
π₯2β1 β
π₯2β2π₯β3
4π₯β12
12(π₯β5)
(π₯+1)(π₯β1) β
(π₯+1)(π₯β3)
4(π₯β3)
12(π₯ β 5)
4(π₯ β 1) =
3(π₯ β 5)
π₯ β 1 =
3π₯ β 15
π₯ β 1 , π₯ β 1, β1, 3
If you want to add or subtract rational expressions, you must first find a common denominator. Once both of your fractions have a common denominator, you can add or subtract the numerators and simplify. You can find the least common denominator (LCD) by finding the least common multiple (LCM) of each denominator. Do this by writing the prime factorization of each expression, and then write the product of the prime factors raised to the highest power in either expression.
Example 5: Find the LCM of 4π₯ (π₯2 β 36) πππ 2π₯3(2π₯2 + 15π₯ + 18) Start by writing the prime factorization of each expression
4π₯(π₯2 β 36) = 22π₯ ( π₯ + 6 )( π₯ β 6)
2π₯3(2π₯2 + 15π₯ + 18) = 2π₯3(2π₯ + 3)(π₯ + 6) Then, look at each factor individually. If the factor occurs in both expressions, choose the one that has the highest power.
22π₯3 (π₯ + 6) (π₯ β 6) (2π₯ + 3)
The LCM would be 4π₯3 (π₯ + 6) (π₯ β 6) (2π₯ + 3)
Multiply by the reciprocal. Factor all polynomials and divide out common factors. Simplify
Scan this QR code to go to a video tutorial on
multiplying and dividing
rational expressions.
Scan this QR code to go to a video tutorial on adding and subtracting rational
expressions.
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Example 6: Subtract and simplify π₯+8
2π₯+8β
4π₯β11
π₯2+11π₯+28
Start by factoring the denominators and finding the least common denominator (LCD)
2π₯ + 8 = 2 (π₯ + 4) π₯2 + 11π₯ + 28 = (π₯ + 4)(π₯ + 7) In this case, the LCD is 2(x + 4)(x + 7). This means that you will need to multiply the
first fraction by (x+7)
(x+7), and the second fraction by
2
2. This will give each fraction the same
denominator.
( π₯ + 8
2(π₯ + 4) ) β (
π₯ + 7
π₯ + 7 ) β (
4π₯ β 11
(π₯ + 4)(π₯ + 7) ) β (
2
2 )
π₯2 + 15π₯ + 56
2(π₯ + 4)(π₯ + 7)β
8π₯ β 22
2(π₯ + 4)(π₯ + 7)
π₯2 + 15π₯ + 56 β 8π₯ + 22
2(π₯ + 4)(π₯ + 7)
π₯2 + 7π₯ + 78
2(π₯ + 4)(π₯ + 7) , π₯ β β4, β7
A complex fraction is a rational expression that has a fraction in either the numerator, denominator or both (basically this is a fraction in a fraction). There are two methods to simplify a complex fraction. Only one will be shown in these notes. The second method will be explained in a video tutorial.
Example 7: Simplify 5+
1
π₯π¦π₯
π₯2+2
π¦
The LCD of all of the rational expressions is π₯2π¦. If we multiply both the numerator and denominator by this LCD, it will help us to simplify.
(5+1
π₯π¦ )(π₯2π¦)
( π₯
π₯2+2
π¦ )(π₯2π¦)
=5π₯2π¦+π₯
π₯π¦+2π₯2 π₯(5π₯π¦+1)
π₯(π¦+2π₯)=
5π₯π¦+1
π¦+2π₯
Multiply both fractions to create LCD. Rewrite as one fraction over LCD Simplify numerator and check for common factors between numerator and denominator.
We can now check both the numerator and denominator for common factors. In this
case, we can factor an x out of both. π₯ β 0, π¦ β 0
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Rational Expressions
1. π₯2β3π₯β10
π₯2β10π₯+25
2. π₯2βπ₯β12
2 β
π₯+4
π₯2β16
3. 3π₯2β15π₯
π₯2+2π₯+1 Γ·
6π₯2β30π₯
π₯+1
4. 2π₯
π₯2βπ₯β6+
4
π₯2+4π₯+4
5. π₯
3π₯+9β
3
π₯2+3π₯
6. 2+
1
π¦
π₯2
π¦2+2
π¦
Graphing Rational Functions AII/T.6 The student will recognize the general shape of function (absolute value, square
root, cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions. A transformational approach to graphing will be employed. Graphing calculators will be used as a tool to investigate the shapes and behaviors of these functions.
AII/T.7 The student will investigate and analyze functions algebraically and graphically.
a) domain and range, including limited and discontinuous domains and ranges; b) zeros; e) asymptotes; f) end behavior;
AII/T.10 The student will identify, create, and solve real-world problems involving inverse
variation, joint variation, and a combination of direct and inverse variations. To graph a rational function, first check for horizontal and vertical asymptotes. Find x and y-intercepts, look for holes (where the graph is undefined), and then use a few points to determine the general shape of the graph.
Scan this QR code to go to a video tutorial on complex
fractions.
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Vertical Asymptote
A rational function π(π₯) = π(π₯)
π(π₯), has a
vertical asymptote at each real zero of π(π₯) if π(π₯) and π(π₯) have no
common zeros. If they do have a common zero [i.e. (π₯ β π)], and the
degree of the term in the denominator is greater, then π(π₯) also has a vertical
asymptote at π₯ = π.
π(π₯) = π₯
π₯ β 1
This graph has a
vertical asymptote at x = 1
Horizontal Asymptote
The horizontal asymptote of a rational function is found by comparing the degree of the numerator (m) and
denominator (n). If m < n, the horizontal asymptote is
π¦ = 0. If m > n, there is no horizontal
asymptote. If m = n, the horizontal asymptote
is π¦ = π
π, where a is the coefficient of
the highest degree term in the numerator and b is the coefficient of
the highest degree term in the denominator.
π(π₯) = π₯ β 1
2π₯
This graph has a
horizontal asymptote at π¦ = 1/2
The numerator and denominator have
the same degree (1), so the asymptote is
found by dividing the coefficients Β½.
The x-intercepts are found by determining where the numerator is equal to zero. The y-intercepts are found by setting x = 0.
Example 1: Graph π(π₯) = π₯β2
π₯2β2π₯β3
Start by factoring the numerator and denominator: π(π₯) = π₯β2
(π₯+1)(π₯β3)
Vertical Asymptote: Because the numerator and the denominator do not share any common zeros, the function will have vertical asymptotes at any value that makes the denominator zero π = βπ πππ π = π Horizontal Asymptote: The degree of the denominator (2) is higher than the degree of the numerator (1) therefore the horizontal asymptote is π = π
X-Intercepts: The x-intercepts are where the numerator is equal to zero. (π, π)
Y-Intercepts: Set x = 0. Therefore, 0β2
0β2(0)β3=
β2
β3=
2
3 (π,
π
π )
Now you can sketch your asymptotes, and plot your intercepts.
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You can solve a rational equation algebraically or graphically. To solve a rational equation graphically, graph both sides of the equation in your calculator and check for intersection points. This procedure will be shown in the video tutorial. Example 2 will show you how to solve algebraically.
Example 2: Solve. 7
π₯+2+
5
π₯β2=
10π₯β2
π₯2β4 Be sure to check for extraneous solutions!
7
π₯ + 2+
5
π₯ β 2=
10π₯ β 2
(π₯ + 2)(π₯ β 2)
( 7
π₯+2+
5
π₯β2 ) (π₯ + 2)(π₯ β 2 ) =
10π₯β2
(π₯+2)(π₯β2)( π₯ + 2)(π₯ β 2)
7(π₯ β 2) + 5 (π₯ + 2) = 10π₯ β 2
7π₯ β 14 + 5π₯ + 10 = 10π₯ β 2
12π₯ β 4 = 10π₯ β 2
2π₯ = 2
π₯ = 1
7
1 + 2+
5
1 β 2β
10(1) β 2
12 β 4
7
3+
5
β1β
8
β3
β8
3= β
8
3
If you chose to plot a few more points, such as
(4,2
5), πππ (β2, β
4
5)
your graph would start to take
shape.
Scan this QR code to go to a video tutorial on graphing
rational functions.
Rewrite with factored denominators so that we can find the LCD. (x+2)(x-2)
Multiply both sides of the equation by the
LCD to clear the denominators.
Solve the equation!
Check for extraneous solutions!
This is not an extraneous solution. Therefore, the
solution to the equation is:
x = 1
Scan this QR code to go to a video tutorial on solving
rational equations.
17
Direct and Inverse Variation
Direct Variation Inverse Variation
π = ππ (π€βπππ π¦
π₯= π) π =
π
π (π€βπππ π₯π¦ = π)
Direct variations are always linear and
pass through the origin.
Inverse variations are rational functions and will never pass through the origin.
If the ratio between two variables is a constant, then a direct variation exists. A direct variation can be written in the form π¦ = ππ₯, where π is the constant of variation. If the product of two variables is a constant, then an inverse variation exists. An
inverse variation can be written in the form = π
π₯ .
Example 3: Determine if each relation is a direct variation, inverse variation, or neither.
x y
1 3
2 4
3 5
x y
-1 2
0 0
1 -2
x y
3 4
1 12
-2 -6
First check the ratios:
Does the ratio 1
3=
2
4 ? NO! Therefore this is NOT a direct variation!
Next check the products: Does 1 β 3 = 2 β 4 ? NO! Therefore this is NOT an inverse variation!
First check the ratios:
Does the ratio 3
4=
1
12 ? NO! Therefore this is NOT a direct variation!
Next check the products: Does 3 β 4 = 1 β 12? Does this also equal β2 β β6 ? YES! Therefore, this IS an inverse variation!
First check the ratios:
Does the ratio β1
2=
1
β2 ? YES! Therefore this IS a direct variation!
Notice that we did not use the ordered pair (0,0) to check the ratios. It is impossible to divide by zero. Therefore, we used the other point.
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To write an equation of a direct variation, use the given point plugged into π¦ = ππ₯, to
solve for π. For an inverse variation, plug into π¦ =π
π₯ .
Example 4: Suppose π¦ varies inversely with π₯, and π¦ = 10 when π₯ = 25. What inverse variation equation relates π₯ and π¦?
Start with π¦ = π
π₯.
We are given a value for π₯ and π¦, so plug those in and solve for π.
10 =π
25
β 25 β 25 250 = π
π¦ =250
π₯ ππ π₯π¦ = 250
Once you have a direct or inverse variation equation, you can use that equation to determine other values. Example 5: The distance that you jog, π, varies directly with the amount of time you jog, π‘. If you can jog 4 miles in 0.9 hours, how long will it take you to jog 6.5 miles? Jogging varies directly with time β π = ππ‘
Now we need to solve for π in order to write a direct variation equation.
4 = π(0.9) Γ· 0.9 Γ· 0.9
40
9= π
π =40
9π‘
Now, we can use this equation to solve for the time it takes to jog 6.5 miles. We are given that you jog 6.5 miles. This will be plugged in for π. Then, solve for π‘.
6.5 = 40
9π‘
Γ·40
9 Γ·
40
9
1.4625 = π‘ Therefore, it would take you almost 1.5 hours to jog 6.5 miles.
This is the constant of variation!
This is the inverse variation equation!
This is the constant of variation!
This is the direct variation equation!
Scan this QR code to go to a video tutorial on direct,
inverse, and joint
variations.
19
Sometimes, a quantity varies with respect to more than one quantity. This creates a combined variation. If the quantities vary directly or jointly, those quantities will be in the numerator (with k). If the quantities vary inversely, those quantities will be in the denominator. Example 6: The amount of cake, π, that it takes to feed a birthday party varies directly with the number of children, π, and inversely with the age of the children, π. If a party of sixteen 8 year-old children eat 112 inΒ² of birthday cake, how much cake would a party of ten 6 year-olds require? Start by setting up the formula. π varies directly with π and inversely with π.
π = ππ
π
Substitute for π, π, and π. π = 112 inΒ² of cake, π = 16 children and π = 8 years old.
112 = 16π
8
π = 56
Write your new combined variation equation with π substituted in.
π = 56π
π
Substitute in your new values for n and a to solve for c.
π = 56(10)
6
π = 93.33 ππ2 ππ ππππ
Graphing Rational Functions
1. State all asymptotes, x and y-intercepts, and restrictions of π(π₯) = π₯2β36
2π₯2+11π₯β6
2. Sketch the graph of π(π₯) = (π₯+2)(π₯+5)
(π₯2β4)
3. Solve for x. 2
π₯+2β
3
π₯=
6
π₯(π₯+2)
4. y varies inversely with x. If y is 240 when x is 3, what is the constant of variation?
5. The amount of hats knitted by a group is directly proportional to the number of
people in the group. If 5 people can knit 12 hats, how many hats can 12 people
knit?
6. π₯ varies jointly with π¦ and π§ and inversely with both π€ and π£. Write a formula relating
these variables.
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Exponential and Logarithmic Functions AII/T.6 The student will recognize the general shape of function (absolute value, square
root, cube root, rational, polynomial, exponential, and logarithmic) families and will convert between graphic and symbolic forms of functions.
AII/T.7 The student will investigate and analyze functions algebraically and graphically.
a) domain and range, including limited and discontinuous domains and ranges; e) asymptotes; f) end behavior;
Rules of Exponents
Multiplying When multiplying two
exponential terms with the same base, you add the exponents.
π₯4 β π₯3 = π₯4+3 = π₯7
Dividing When dividing two exponential terms with the same base, you
subtract the exponents.
π8
π2= π8β2 = π6
Raising a power to a
power
When raising an exponential term to a power, you multiply
the exponents. (π2)5 = π2β5 = π10
Negative Exponents
To simplify negative exponents, write the term as its reciprocal
with a positive exponent. πβ4 =
1
π4
5
π₯β2=
5π₯2
1
Example 1: Simplify (β2π₯Β³π¦π§Β²)Β³(2π₯Β³π¦π§Β²)β΄
(β8π₯9π¦3π§6) β (16π₯12π¦4π§8)
β128π₯21π¦7π§14
Example 2: (π2ππ5
π2π2 )3
(π2ππ5
π2π2 )
3
= (π2β2ππ5β2)3 = (π0ππ3)3 = 13π3π9 = π3π9
Example 3: (2π₯2π¦β4
3π₯π¦5 )
β3
(2π₯2π¦β4
3π₯π¦5)
β3
= (3π₯π¦5
2π₯2π¦β4)
3
= (3π₯π¦5π¦4
2π₯2 )
3
= (3π¦9
2π₯ )
3
= 27π¦27
8π₯3
Remember that anything to the zero power equals 1!
Scan this QR code to go to a video tutorial on
rules of exponents.
21
Exponential Growth
π¦ = πππ₯ π€βπππ π > 0 πππ π > 1
The graph crosses the y-axis
at (0, π). The asymptote is π¦ = 0
Exponential Decay
π¦ = πππ₯ π€βπππ π > 0 πππ
π ππ πππ‘π€πππ 0 πππ 1
The graph crosses the y-axis at (0, π).
The asymptote is π¦ = 0
Logarithmic Functions
π¦ = logπ π₯ π€βπππ π > 0 πππ π β 1
A logarithmic function is the inverse of an exponential
function.
You can model exponential growth and decay with the function
π΄(π‘) = π(1 + π)π‘ where π΄(π‘) is the final amount, π is the initial amount, π is the rate of growth or decay,
and π‘ is the number of time periods. Example 4: If you invested $5,000 in a savings account today that pays 3.5% interest annually, how much will you have in 10 years?
π΄(π‘) = π(1 + π)π‘ π = ππππ‘πππ ππππ’ππ‘ = $5000 π = πππ‘ππππ π‘ πππ‘π = 0.035 π‘ = π‘πππ πππππππ = 10 π¦ππππ
π΄(π‘) = 5000(1 + .035)10
π΄(π‘) = 5000(1.035)10 π΄(π‘) = 5000(1.41) π΄(π‘) = $7050
Plug in your given values and solve for π΄(π‘).
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Continuously compounded interest can be calculated using the formula π΄(π‘) = π β πππ‘ where A(t) is the final amount, P is the initial amount, r is the annual interest rate, and t is the number of years. Example 5: If you invested $2,000 in a savings account today that pays 2.75% interest compounded continuously, in how many years would you have $5,000?
π΄(π‘) = ππππ‘ π΄(π‘) = πππππ ππππ’ππ‘ = $5000 π = ππππ‘πππ ππππ’ππ‘ = $2000 π = πππ‘ππππ π‘ πππ‘π = 0.0275
Exponential and Logarithmic Functions
1. (π₯4π¦2
2π₯π¦ )
3
2. π4π2πβ5
π6πβ2π0
3. A population of 800,000 decreases at a rate of 2.2% annually. After how many
years will the population go below 600,000? (remember that your growth is negative now!)
4. If $1000 is invested in a savings account with 5.1% interest compounded
continuously, how much money will you have after 3 years?
To find the number of years, letβs graph the function and look at the table
of values in our calculator. We will graph π¦ = 2000π .0275π₯
Once youβve graphed the
function, press 2nd and GRAPH to show the table
of values.
We want to know where the final
value (Y) equals 5000, so scroll
through the table until you find it.
It would take 34 years for your investment of
$2000 to grow to $5000.
34 years
Scan this QR code to go to a video tutorial on exponential
growth and decay.
23
Properties of Logarithms
logπ π₯ = π¦ ππ πππ ππππ¦ ππ ππ¦ = π₯
πΉππ πππ¦ πππ ππ‘ππ£π ππ’πππππ π, π, πππ π π€βπππ π β 1, π‘βπ ππππππ€πππ πππππππ‘πππ πππππ¦
Product Property
logπ ππ = logπ π + logπ π log6 36π₯ = log6 36 + log6 π₯
Quotient Property
logπ
π
π= logπ π β logπ π log2
π¦
32= log2 π¦ β log2 32
Power Property
logπ ππ = π logπ π log4 π₯2 = 2 log4 π₯
πΉππ πππ¦ πππ ππ‘ππ£π ππ’πππππ π, π, πππ π π€βπππ π β 1 πππ π β 1, π‘βπ ππππππ€πππ πππππππ‘π¦ πππππππ
Change of Base Formula logπ π = logπ π
logπ π log4 32 =
log2 32
log2 4=
5
2
Example 6: Write log6 12 + log6 18 as a single logarithm. Using the properties of logarithms, this can be condensed to log6(12 β 18)
log6 216 = log6 63 = 3
Example 7: Write log381
π₯5 in expanded form.
log381
π₯5= log3 81 β log3 π₯5
log3 34 β 5 log3 π₯
4 β 5 log3 π₯ Example 8: Evaluate log6 49. (Hint: Use Change of Base)
log6 49 = log 49
log 6 β 2.17
Exponential and Logarithmic Functions 5. Write as a single logarithm: log5 600 β log5 24
6. Write as a single logarithm: log8 12 + log8 9
7. Expand: log4 16π¦2
8. Expand: log2π₯
64
9. Evaluate: log25 125
10. Evaluate: log9 100
Use a calculator to approximate the answer.
Scan this QR code to go to a video tutorial on properties of
logarithms.
24
Exponential and Logarithmic Equations
AII/T.4 The student will solve, algebraically and graphically,
c) equations containing rational algebraic expressions;
You can use the properties of exponents and logarithms to help you solve exponential and logarithmic equations. One way to solve an exponential equation is to rewrite terms on both sides of the equation with a common base.
Example 1: Solve 16 = 2 2π₯β1
24 = 22π₯β1
4 = 2π₯ β 1
π₯ = 5
2
If you cannot find like bases, you can solve by taking the logarithm of both sides. This is only true as long as both bases are positive.
Example 2: Solve 73π₯ = 45
log 73π₯ = log 45
3π₯ log 7 = log 45
π₯ = log 45
3 log 7
π₯ β 0.652
You can also solve an exponential or logarithmic equation by graphing both sides of the equation in your calculator and finding where the two graphs intersect.
Exponential and Logarithmic Equations
1. 93π₯ = 27π₯β5
2. 124π₯ = 25 3. log (2π₯ + 2) = 3 4. log π₯ + log(π₯ + 9) = 1 (Remember that logarithms are only defined for positive numbers)
Rewrite each term with a common base
Set exponents equal to each other (Power Property of Exponents)
Solve
Take the logarithm of both sides
Power Property of Logarithms
Use your calculator to estimate the answer
Scan this QR code to go to a video tutorial on logarithmic and exponential equations.
25
Honors Algebra II
Natural Log and Log Applications
AII/T.4 The student will solve, algebraically and graphically,
c) equations containing rational algebraic expressions;
AII/T.9 The student will collect and analyze data, determine the equation of the curve of best
fit, make predictions, and solve real-world problems, using mathematical models. Mathematical models will include polynomial, exponential, and logarithmic functions.
The same properties that apply to logarithms also apply to natural logarithms. The natural logarithm is a logarithm with base π.
π₯ = ln π¦ ππ πππ ππππ¦ ππ π¦ = ππ₯
Example 1: ln(3π₯ + 4) = 3
3π₯ + 4 = π3
π₯ = π3 β 4
3
π₯ β 5.362
Example 2: 4 β π3π₯ = 2
βπ3π₯ = β2
π3π₯ = 2 3π₯ = ln 2
π₯ = ln 2
3
π₯ β 0.231
Natural Log and Log Applications
1. Simplify 3 ln 5 + ln 2
2. ln(π₯ + 1)2 = 2
3. 2ππ₯
3 + 5 = 37
4. A bacteria population π at time π‘ (in months) is given by π = π0π2π‘ where π0 is the initial population. If the initial population was 100, how long does it take for the population to reach one million?
Rewrite in exponential form. Remember that the natural log uses base e.
Solve for x.
Use your calculator to estimate the answer
Isolate the ππ₯ term.
Rewrite in logarithmic form.
Solve for x.
Use your calculator to estimate the answer
Scan this QR code to go to a video tutorial on natural log
and log applications.
26
Honors Algebra II- Sequences and Series AII/T.2 The student will investigate and apply the properties of arithmetic and geometric
sequences and series to solve real-world problems, including writing the first n terms, finding the nth term, and evaluating summation formulas. Notation will include Ζ© and an.
Arithmetic Sequence Geometric Sequence
A sequence where the difference between consecutive terms is a constant.
(You add or subtract a constant value)
A sequence where the difference between consecutive terms is a common ratio.
(You multiply or divide a constant value)
Examples: 3, 5, 7, 9, 11β¦ (constant is +2) 25, 20, 15, 10β¦ (constant is β5)
Examples: 3, 6, 12, 24, 48β¦ (ratio is 2
1 )
6, 9, 13.5, 20.25, 30.375β¦ (ratio is 3
2 )
Formula
ππ = π + (π β 1)π π is the starting value, π is the common difference and π is the number of terms.
Formula
ππ = π β ππβ1 π is the starting value, π is the common
ratio and π is the number of terms.
Example 1: What is the 35th term of the arithmetic sequence that begins 7, 4β¦ ππ = π + (π β 1)π
ππ = 7 + (35 β 1)(β3)
ππ = β95
Example 2: What is the 20th term of the geometric sequence that begins 1, 2, 4β¦
ππ = π β ππβ1
ππ = 1 β 220β1
ππ = 524,288
Example 3: What is the missing term in this geometric sequence 9, β , 1 β¦
ππ = π β ππβ1
1 = 9 β π3β1
1
9= π2
π = 1
3
The missing term is 9 β 1
3= 3
Substitute your values (π = 7, π = 35, π = β3)
Simplify
Substitute your values (π = 1, π = 2, π = 20)
Simplify
Substitute your values (ππ = 1, π = 9, π = 3)
Simplify
Solve for the common ratio, π.
27
A series is the sum of a geometric or arithmetic sequence.
Sum of a Finite Arithmetic Series
Sum of a Finite Geometric Series
Sum of an Infinite Geometric Sequence
(Only applicable for |π| < 1)
ππ =π
2 (π1 + ππ)
Where π1 is the first term, ππ is
the ππ‘β term, and n is the number of terms.
ππ =π1(1 β ππ)
1 β π
Where π1 is the first term, π is the common ratio, and n is the
number of terms.
ππ =π1
1 β π
. Where π1 is the first term, and π is
the common ratio.
You may see series written in Summation Notation You can write the series 7+9+11+ β¦+89 as
β (2π + 5
42
π = 1
)
Example 4: Evaluate Because the explicit formula is linear, this will be an arithmetic series. In order to evaluate an arithmetic series we need to know the first and last term and number of terms.
ππ =π
2 (π1 + ππ)
ππ =42
2 (7 + 89)
ππ = 2016
Sequences and Series 1. What is the 9th term of the geometric sequence that begins 2, 1, β¦ 2. What is the missing term in this arithmetic sequence 12, β , 25 β¦ 3. Write the series in summation notation 120 + 115 + 110 + 105 + 100 + 95 4. Find the sum of the geometric series
β 4π
10
π=1
n = 1 is the lower limit 42 is the upper limit 2n +5 is the explicit formula for each term in the series.
β(2π + 5)
42
π=1
Substitute your values
(π = 42, π1 = 2(1) + 5 = 7 , π42 = 2(42) + 5 = 89)
Simplify
Scan this QR code to go to a video tutorial on sequences
and series.
28
Statistics AII/T.12 The student will compute and distinguish between permutations and combinations
and use technology for applications.
You can use multiplication to quickly determine the number of ways a certain event can
happen. You can use Permutations and Combinations to determine the number of
ways. You use Combinations when order is not important and Permutations when the
order is important. These formulas are on your formula sheet, but your formula sheet
does not tell you when to use each formula.
Permutation
π 28
The number of permutations of 8
items of a set arranged 2 items at a time.
π 28 =
8!
(8 β 2)!=
8!
6!=
40320
720= 56
Combination
πΆ 28
The number of
combinations of 8
items taken 2 at a time.
πΆ 28 =
8!
2! (8 β 2)!=
8!
2! (6!)=
40320
1440= 28
Example 1: In a diving meet, the top 8 divers advance to the finals and scores are
cleared. In how many ways can a field of 14 divers qualify to the finals?
Because all of the top 8 advance to the finals, the order that they finish isnβt important
as long as they are in the top 8. This means we will use a combination!
πΆ 814 =
14!
8! (14 β 8)!=
14!
8! (6!)=
8.71782912 Γ 1010
29030400= 3,003
There are 3,003 possible combinations of divers that can advance to the finals.
29
Example 2: In the finals of the diving meet referenced in Example 1, the top 3
finishers score points for their team. First place receives 10 points, 2nd place receives 8
points, and 3rd place receives 6 points. In how many ways can the 8 finalists finish in
the top 3?
Now, the order is important because 1st place gets more points than 2nd place. We will
use a permutation!
π 38 =
8!
(8 β 3)!=
8!
5!=
40320
120= 336
There are 336 possible ways that the top 8 divers can finish in the top 3.
Statistics 1. A teacher is making a multiple choice quiz. She wants to give each student the
same questions, but have each student's questions appear in a different order. If there are twenty-seven students in the class, what is the least number of questions the quiz must contain?
2. A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters?
30
Answers to the problems: Radical Functions
1. 3ππ2 β5ππ3
2. 30π₯2π¦3 βπ¦
3. 180π¦4 βπ₯π¦
4. 2π₯ β2π₯3
5. βπ74
6. π₯ 5
2
7. 1
π₯ 12
ππ 1
βπ₯
Radical Equations 1. π₯ = 7 2. ππ ππππ’π‘πππ ππ { } 3. π₯ = 34 4. π₯ = β16 5. π₯ = 1928 Rational Expressions
1. π₯+2
π₯β5 , π₯ β 5
2. π₯+3
2 , π₯ β 4, β4
3. 1
2(π₯+1) , π₯ β β1, 0, 5
4. 2(π₯2+4π₯β6)
(π₯β3)(π₯+2)2 , π₯ β 3, β2
5. π₯β3
3π₯ , π₯ β β3, 0
6. π¦(2π¦+1)
π₯2+2π¦
Graphing Rational Functions
1. vertical asymptote: π₯ = 1
2
horizontal asymptote: π¦ = 1
2
x-intercept: (6, 0) y-intercept: (0, 6)
π₯ β β6,1
2
2.
Graphing Rational Functions 3. π₯ = β12 4. π = 720 5. β = 28.8 ππ 28 βππ‘π
6. π₯ = ππ¦π§
π€π£
Exponential and Logarithmic Functions
1. π₯9π¦3
8
2. 1
π3
3. 13 π¦ππππ 4. $1165.32 5. 2
6. log8 108 7. 2 + 2 log4 π¦ 8. log2 π₯ β 6
9. 1.5 10. β 2.1 Exponential and Logarithmic Equations 1. π₯ = β5 2. π₯ β 0.324
3. π₯ = 499 4. π₯ = 1 Natural Log and Log Applications
1. ln 250 β 5.521
2. π₯ β β0.757 ππ 3.11 3. π₯ = 3 ln 16 β 8.318 4. 4.605 ππππ‘βπ Algebra II Honors
Sequences and Series
1. 1
128
2. 18.5
3. β (β5π + 125)6π=1
4. 1,398,100
Statistics
1. 5 ππ’ππ π‘ππππ π€ππ’ππ πππ£π 120 πππππ πππ π πππππ‘πππ 2. 792