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Secondary Mathematics 2 Table of Contents
Unit 1: Extending the Number System
Cluster 1: Extending Properties of Exponents
(N.RN.1 and N.RN.2) ......................................................................................... 3
Cluster 2: Using Properties of Rational and Irrational Numbers
(N.RN.3) ........................................................................................................... 10
Cluster 4: Performing Arithmetic Operations on Polynomials
(A.APR.1) ......................................................................................................... 21
Unit 2: Quadratic Functions and Modeling
Cluster 1 and 2: Interpreting and Analyzing Functions
(F.IF.4, F.IF.5, F.IF.7 and F.IF.9) ..................................................................... 26
Cluster 1 and 5: Quadratic Functions and Modeling
(F.IF.6 and F.LE.3) ........................................................................................... 42
Factoring ....................................................................................................................... 48
Cluster 2: Forms of Quadratic Functions
(F.IF.8a, A.SSE.1a, and A.SSE.3a and b) ......................................................... 57
Cluster 3 (Unit 6): Translating between descriptions and equations for a conic section
(G.GPE.2) ......................................................................................................... 67
Honors H.5.1 (G.GPE.3) .................................................................................. 73
Cluster 3: Building Functions that Model Relationships between Two Quantities
(F.BF.1) ............................................................................................................. 90
Cluster 4: Transformations and Inverses
(F.BF.3 and F.BF.4) ........................................................................................ 104
Unit 3: Expressions and Equations
Cluster 1: Interpreting the Structure of Expressions
(A.SSE.2) ........................................................................................................ 120
Cluster 3 (Unit 1): Performing Arithmetic Operations with Complex Numbers
(N.CN.1 and N.CN.2) ..................................................................................... 124
Honors H.2.1: (N.CN.3) ................................................................................. 129
Cluster 4 and 5: Solving Equations in One Variable with Complex Solutions
(A.REI.4 and N.CN.7) .................................................................................... 134
Honors (N.CN.8 and N.CN.9) ........................................................................ 145
Cluster 3: Writing and Solving Equations and Inequalities in One Variable
(A.CED.1 and A.CED.4) ................................................................................ 148
Honors H.1.2 .................................................................................................. 162
Cluster 3: Writing and Graphing Equations in Two Variables
(A.CED.2) ....................................................................................................... 172
Cluster 6: Solving Systems of Equations
(A.REI.7) ......................................................................................................... 181
Honors (A.REI.8 and A.REI.9) ...................................................................... 186
Cluster 2: Forms and Uses of Exponential Functions
(F.IF.8b, A.SSE.1b, and A.SSE.3c) ................................................................ 196
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Secondary Mathematics 2
Unit 4: Applications of Probability
Cluster 1: Understand independence and conditional probability and use them to
interpret data.
(S.CP.1) ..................................................................................................... 213
Cluster 2: Use the rules of probability to compute probabilities of compound events in a
uniform probability model.
(S.CP.2 through S.CP.7) ........................................................................... 215
Cluster 3: Use probability to evaluate outcomes of decisions.
Honors (S.CP.8, S.CP.9, S.MD.6 and S.MD.7) ......................................... 224
Unit 5: Similarity, Right Triangle Trigonometry, and Proof
Cluster 1: Understand similarity in terms of similarity transformations
(G.SRT.1, G.SRT.2, and G.SRT.3) .......................................................... 239
Cluster 2: Prove geometric theorems.
(G.CO.9) ................................................................................................... 248
(G.CO.10) ................................................................................................. 254
(G.CO.11) ................................................................................................. 263
Cluster 3: Prove theorems involving similarity.
(G.SRT.4 and G.SRT.5) ............................................................................ 270
Cluster 4: Use coordintes to prove simple geometric theorems algebraically.
(G.GPE.6) ................................................................................................. 275
Cluster 5: Define trigonometric ratios and solve problems involving right triangles.
(G.SRT.6, G.SRT.7, and G.SRT.8) .......................................................... 282
Honors (N.CN.3, N.CN.4, N.CN.5, and N.CN.6) .................................... 289
Cluster 6: Prove and apply trigonometric identities.
(F.TF.8) ..................................................................................................... 301
Honors (H.5.6) Unit Circle ....................................................................... 304
Honors (H.5.7) Trigonometry Proofs ....................................................... 317
Honors (F.TF.9 and H.5.9) ....................................................................... 324
Unit 6: Circles With and Without Coordinates
Cluster 1: Understand and apply theorems about circles.
(G.C.1, G.C.2, G.C.3, and G.C.4 (Honors)) ............................................. 334
Cluster 2: Find arc length and areas of sectors of circles.
(G.C.5) ...................................................................................................... 353
Cluster 3: Translate between the geometric description and the equation for a conic
section.
(G.GPE.1) ................................................................................................. 360
Cluster 4: Use coordinates to prove simple geometric theorems algebraically.
(G.GPE.4) ................................................................................................. 365
Cluster 5: Explain volume formulas and use them to solve problems.
(G.GMD.1, G.GMD.2 (Honors), and G.GMD.3) .................................... 368
Selected Answers ............................................................................................. 386
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Secondary Mathematics 2
Unit 1
Extending the Number
System
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Secondary Mathematics 2
Unit 1 Cluster 1 (N.RN.1 & N.RN.2):
Extending Properties of Exponents
Cluster 1: Extending properties of exponents
1.1.1 Define rational exponents and extend the properties of integer exponents to
rational exponents
1.1.2 Rewrite expressions, going between rational and radical form
VOCABULARY
If the exponent on a term is of the form
m
, where n 0 , then the number is said to have a
n
1
rational exponent. 8 3 is an example of a constant with a rational exponent.
Properties of
Exponents
(All bases are non-zero)
x a xb x a b
a
x
x a b
b
x
x0 1
x
a
xy
n
Properties of Rational
Exponents
(All bases are non-zero)
p
q
r
s
x x x
x
x
p
q
x
r
s
p r
q s
xn y n
x
3
23
5
x3
1
x 15
xa
x aa x0
a
x
xa
However: a 1
x
Therefore: x 0 1
x
52
1
2
x5
p
q
p
p
xy q
3
5
1
x
19
x 20
2
x3
x
p
q
31
5
1
x4 x5 x4
p r
q s
Students should have this
property memorized.
1
a
x
Examples
p
xy
xq yq
3
4
3
3
x4 y4
r
x
m n
x
m n
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p r
qp s
q s
x
x
Page 4
3
x4
2
3
32
x4 3 x4 x2
2
1
Secondary Mathematics 2
m
p
p
2
x 3 x3
23
y
y
x q xq
p
y
yq
x
x
m
y
y
m
1
xm
xm y n
n n m
y
x
y
xn
a
b
y
m
1
x2
3
4
y
1
x n yb
a m
y b xn
m
a
1
2
x 2 y4
3 1
y 4 x2
3
1
Practice Exercises A
Simplify each expression using only positive exponents.
121/3
2. 1/3
4
1. 5 5
12
2
1/4
4. 81 2 51/3
7.
4
34
2
2
3.
5. y 2 / 3
6
8. 4 2 / 3
9. z 0
y2/3
11. 1 / 3
y
7
10. 1/3
7
k
6. z 2 / 3 z 1 / 2
1/4
1
1/3
12.
x4
x
3
7
Definition
1
n
A radical can also be written as a term with a rational exponent. For example, x n x where n
m
n
is an integer and n 0 . In general, x n x m where m and n are integers and n 0 .
m
n
x nx
The denominator of the rational
exponent becomes the index of the
radical.
m
m
n
x can also be written as
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n
m
x
Secondary Mathematics 2
Rational Exponent
Form
Radical Form
19
20
x 20
3
x19
x3
x2
7
3
a3
a7
Practice Exercises B
Rewrite each expression in radical form
1. 84 3
2. x3/4
3. a 5/9
4. 363 2
5. k 3/2
6. 2x1/5
7.
64
23
8.
8
9. 2x
53
1/5
Practice Exercises C
Rewrite each expression with rational exponents.
1. 3 11
4.
5
4
2.
7
6
x7
42
2
3.
3
10
8
2
7. 7 w5
5.
6. 3 r
8. 3 k 2
9.
z
5
2
Vocabulary
For an integer n greater than 1, if a n k , then a is the nth root of k.
A radical or the principal nth root of k:
k, the radicand, is a real number.
n, the index, is a positive integer greater than one.
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Secondary Mathematics 2
Properties of Radicals
an a
n
ab n a n b
n
n
a na
b nb
Simplifying Radicals:
Radicals that are simplified have:
no fractions left under the radical.
no perfect power factors in the radicand, k.
no exponents in the radicand, k, greater than
the index, n.
Vocabulary
A prime number is a whole number greater
than 1 that is only divisible by 1 and itself. In
other words, a prime number has exactly two
factors: 1 and itself.
Example:
5 1 5
prime
30 2 3 5
not prime
Division Rules For a Few Prime Numbers
A number is divisible by:
2
3
5
7
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If:
The last digit is even
(0, 2, 4, 6, 8)
Example:
256 is
255 is not
381 (3+8+1=12 and 12÷3=4) Yes
The sum of the digits is
2
383 (3+8+3=14 and 14 3 4 ) No
divisible by 3
3
175 is
The last digit is 0 or 5
809 is not
If you double the last digit 672 (Double 2 is 4, 67 4 63 and
and subtract it from the rest 63 7 9 ) Yes
of the number and the
905 (Double 5 is 10, 90 10 80 and
answer is:
3
0
80 7 11 ) No
Divisible by 7
7
Page 7
Secondary Mathematics 2
Simplifying Radicals
Method 1:
Find Perfect Squares Under the Radical
1. Rewrite the radicand as factors that are
perfect squares and factors that are not
perfect squares.
2. Rewrite the radical as two separate
radicals.
3. Simplify the perfect square.
Example:
Method 2:
Use a Factor Tree
1. Work with only the radicand.
2. Split the radicand into two factors.
3. Split those numbers into two factors
until the number is a prime number.
4. Group the prime numbers into pairs.
5. List the number from each pair only
once outside of the radicand.
6. Leave any unpaired numbers inside the
radical.
Note: If you have more than one pair, multiply
the numbers outside of the radical as well as
the numbers left inside.
Method 3:
Divide by Prime Numbers
1. Work with only the radicand.
2. Using only prime numbers, divide each
radicand until the bottom number is a
prime number.
3. Group the prime numbers into pairs.
4. List the number from each pair only
once outside of the radicand.
5. Leave any unpaired numbers inside the
radical.
Note: If you have more than one pair, multiply
the numbers outside of the radical as well as
the numbers left inside.
Example:
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75
25 3
25 3
5 3
75
75
25
5
3
5
5 3
Example:
Page 8
5
75
5
5
75
15
3
5 3
Secondary Mathematics 2
Method 4:
Use Exponent Rules
1. Rewrite the exponent as a rational
exponent.
2. Rewrite the radicand as factors that are
perfect squares and factors that are not
perfect squares.
3. Rewrite the perfect square factors with
an exponent of 2.
4. Split up the factors, giving each the
rational exponent.
5. Simplify.
75
Example:
751/2
25 3
1/2
5 3
5 3
1/2
2
2
1/2
1/2
5 3
1/2
6. Rewrite as a radical
5 3
Method 4 with Variables:
Example:
3
1. Rewrite the exponent as a rational
exponent.
2. Rewrite the radicand as two factors.
One with the highest exponent that is
divisible by the root and the other
factor with an exponent of what is left
over.
3. Split up the factors, giving each the
rational exponent.
4. Rewrite the exponents using exponent
rules.
5. Simplify.
6. Rewrite as a radical
x7
x 7/3
x x
1/3
6
x
6
1/3
x1/3
x6/3 x1/3
x 2 x1/3
x2 3 x
Practice D
Simplify each radical expression.
1.
4.
3
45 p 2
2.
80 p3
3.
3
24x3 y 3
16u 4v3
5.
75x 2 y
6.
3
64m3n3
7. 4 36y 3
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8. 6 150r
Page 9
9. 7 3 96m3
Secondary Mathematics 2
Unit 1 Cluster 2 (N.RN.3):
Using Properties of Rational and Irrational numbers
Cluster 1: Extending properties of exponents
1.2.1 Properties of rational and irrational numbers (i.e. sum of 2 rational numbers is
rational, sum of a rational and irrational number is irrational)
Number Systems
Complex Numbers: all numbers of the form a + bi where a and b
are real numbers. -4 + 3i, 2 – i
Real Numbers
Rational Numbers consist of all
numbers that can be written as the ratio of two
integers 2, 0.125, 1 , 0.91
3
Integers are the whole numbers and
their opposites (-3, -2, -1, 0, 1, 2, 3, …)
Natural
Numbers
(Counting
Numbers)
Whole
Numbers
Imaginary
Numbers: are
of the form of bi
where i 1
-3i, 2i
include zero
and the natural
numbers
Irrational Numbers consist of all
numbers that cannot be written as the ratio of
two integers. , 3, 4 5
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Secondary Mathematics 2
Properties of Real Numbers
Description
Numbers
Commutative Property
7 11 11 7
You can add or multiply real
7 11 11 7
numbers in any order without
changing the result.
Associative Property
The sum or product of three or
5 3 7 5 3 7
more real numbers is the same
regardless of the way the numbers
are grouped.
Distributive Property
When you multiply a sum by a
number, the result is the same
5 2 8 5 2 5 8
whether you add and then
multiply or whether you multiply 2 8 5 2 5 2 8
each term by the number and then
add the products.
Additive Identity Property
The sum of a number and 0, the
3 0 3
additive identity, is the original
number.
Multiplicative Identity Property
2
2
The product of a number and 1,
1
the multiplicative identity, is the
3
3
original number.
Additive Inverse Property
5 5 0
The sum of a number and its
opposite, or additive inverse, is 0.
Multiplicative Inverse Property
1
The product of a non-zero number
8 1
and its reciprocal, or
8
multiplicative inverse
Closure Property
23 5
The sum or product of any two
2 6 12
real numbers is a real number.
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Algebra
ab ba
ab ba
ab c a bc
a b c ab bc
b c a ba ca
n0 0n n
n 1 1 n n
n n 0
1
n 1, n 0
n
a b
ab
Secondary Mathematics 2
Sorry we are
a CLOSED
set!
Why can’t I
come in?????
Closure
When an operation is executed on the members of a set, the result is guaranteed to be in the set.
Addition: If two integers are added together,
2 5 3
Example:
the sum is an integer. Therefore, integers are
closed under addition.
Multiplication: If two integers are multiplied
6 7 42
Example:
together, the product is an integer. Therefore,
integers are closed under multiplication.
Subtraction: If one integer is subtracted from
2 6 4
another, the difference is an integer. Therefore, Example:
integers are closed under subtraction.
Division: If one integer is divided by another
10 2 5 closed
integer, the quotient may or may not be an
1
integer. Therefore, integers are not closed
2 10 not closed
5
under division.
Example:
You Decide
1.
What number systems are closed under addition? Justify your conclusions using the method
of your choice.
2.
What number systems are closed under multiplication? Justify your conclusions using the
method of your choice.
3.
What number systems are closed under subtraction? Justify your conclusions using the
method of your choice.
4.
What number systems are closed under division? Justify your conclusions using the method
of your choice.
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Secondary Mathematics 2
Vocabulary
n
For an integer n greater than 1, if a k , then a is the nth root of k.
A radical or the principal nth root of k:
k, the radicand, is a real number.
n, the index, is a positive integer greater than one.
Properties of Radicals
n
a a
n
ab a b
n
a na
b nb
n
n
n
Simplifying Radicals:
Radicals that are simplified have:
no fractions left under the radical.
no perfect power factors in the radicand, k.
no exponents in the radicand, k, greater than
the index, n.
Vocabulary
A prime number is a whole number greater
than 1 that is only divisible by 1 and itself. In
other words, a prime number has exactly two
factors: 1 and itself.
Example:
5 1 5
prime
30 2 3 5
not prime
Division Rules For a Few Prime Numbers
A number is divisible by:
2
3
5
7
Jordan School District
If:
The last digit is even
(0, 2, 4, 6, 8)
Example:
256 is divisible by 2
255 is not divisible by 2
381 (3+8+1=12 and 12÷3=4) Yes
The sum of the digits is
2
14 3 4
divisible by 3
3 ) No
383 (3+8+3=14 and
175 is divisible by 5
The last digit is 0 or 5
809 is not divisible by 5
67 4 63 and
If you double the last digit 672 (Double 2 is 4,
and subtract it from the rest 63 7 9 ) Yes
of the number and the
answer is:
905 (Double 5 is 10, 90 10 80 and
0
3
80 7 11
Divisible by 7
7 ) No
Page 13
Secondary Mathematics 2
Simplifying Radicals
Method 1:
Find Perfect Squares Under the Radical
4. Rewrite the radicand as factors that are
perfect squares and factors that are not
perfect squares.
5. Rewrite the radical as two separate
radicals .
6. Simplify the perfect square.
Method 2:
Use a Factor Tree
7. Work with only the radicand.
8. Split the radicand into two factors.
9. Split those numbers into two factors
until the number is a prime number.
10. Group the prime numbers into pairs.
11. List the number from each pair only
once outside of the radicand.
12. Leave any unpaired numbers inside the
radical.
Note: If you have more than one pair, multiply
the numbers outside of the radical as well as
the numbers left inside.
Method 3:
Divide by Prime Numbers
6. Work with only the radicand.
7. Using only prime numbers, divide each
radicand until the bottom number is a
prime number.
8. Group the prime numbers into pairs.
9. List the number from each pair only
once outside of the radicand.
10. Leave any unpaired numbers inside the
radical.
Note: If you have more than one pair, multiply
the numbers outside of the radical as well as
the numbers left inside.
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75
Example:
25 3
25 3
5 3
75
Example:
75
25
5
3
5
5 3
Example:
Page 14
5
75
5
5
75
15
3
5 3
Secondary Mathematics 2
Method 4:
Use Exponent Rules
75
Example:
7. Rewrite the exponent as a rational
exponent.
8. Rewrite the radicand as factors that are
perfect squares and factors that are not
perfect squares.
9. Rewrite the perfect square factors with
an exponent of 2.
10. Split up the factors, giving each the
rational exponent.
11. Simplify.
751/2
25 3
1/2
5 3
5 3
1/2
2
2
1/2
1/2
5 3
1/2
12. Rewrite as a radical
5 3
Method 4 with Variables:
Example:
3
7. Rewrite the exponent as a rational
exponent.
8. Rewrite the radicand as two factors.
One with the highest exponent that is
divisible by the root and the other
factor with an exponent of what is left
over.
9. Split up the factors, giving each the
rational exponent.
10. Rewrite the exponents using exponent
rules.
11. Simplify.
12. Rewrite as a radical
x7
x 7/3
x x
1/3
6
x
6
1/3
x1/3
x6/3 x1/3
x 2 x1/3
x2 3 x
Adding and Subtracting Radicals
To add or subtract radicals, simplify first if possible, and then add or subtract “like” radicals.
1.
2.
1.
2.
Example:
2 3 5 3
They both have the same term under
the radical so they are “like” terms.
Add the coefficients of the radicals.
(2 5) 3
They both have the same term under
the radical so they are “like” terms.
Subtract the coefficients of the radicals.
7 3
Example:
4 3 7 3
(4 7) 3
(3) 3
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Secondary Mathematics 2
1.
2.
3.
4.
5.
6.
1.
2.
3.
4.
5.
6.
7.
1.
2.
3.
4.
5.
6.
7.
They are not “like” terms, but one of
them can be simplified.
Rewrite the number under the radical.
Use the properties of radicals to write
the factors as two radicals.
25 is a perfect square and the square
root of it is 5.
Multiply the coefficients of the second
radical.
Now they are “like” terms, add the
coefficients.
None of them are “like” terms.
Simplify if you can.
Factor each number inside the radical.
Use the properties of radicals to
simplify.
4 and 9 are perfect squares; their square
roots are 2 and 3.
Multiply the numbers outside of the
radical.
Only the terms with 2 are “like”
terms.
Simplify.
They are not like terms, but they can be
simplified.
Rewrite the expressions under the
radical.
Use properties of radicals to rewrite the
expressions.
The cube root of 8 is 2 and the cube
root of 27 is 3.
Multiply the coefficients of the last
radical.
Add or subtract the coefficients of the
like terms.
Simplify.
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Example:
5 3 2 75
5 3 2 25 3
5 3 2 25 3
5 3 25 3
5 3 10 3
(5 10) 3
15 3
Example:
5 8 3 18 3
5 42 3 92 3
5 4 2 3 9 2 3
5 2 2 3 3 2 3
10 2 9 2 3
(10 9) 2 3
2 3
Example:
3
40 3 3 5 2 3 135
Page 16
3
5 8 3 3 5 2 3 5 27
3
5 3 8 3 3 5 2 3 5 3 27
3
5 2 33 5 2 33 5
2 3 5 33 5 6 3 5
2 3 6 3 5
53 5
Secondary Mathematics 2
1.
2.
3.
4.
5.
6.
7.
They are not like terms, but one of
them can be simplified.
Rewrite the expression under the
radical.
Use properties of radicals to rewrite the
expression.
3
8 and x are perfect cubes. The cube
Example:
2 3 24 x3 x 3 3
2 3 8 3 x3 x 3 3
2 3 8 3 3 3 x3 x 3 3
22 x3 3 x3 3
3
root of 8 is 2 and x is x.
Multiply the coefficients of the first
radical.
Now they are like terms, add the
coefficients of each.
Simplify.
4x 3 3 x 3 3
4x x 3 3
3x 3 3
Multiplying Radicals
Multiplying radicals with the same index
1.
2.
Multiply the coefficients and multiply
the numbers under the radicand.
If possible, simplify. This is already
simplified.
Example:
2 6 3 7
(2 3) 6 7
6 42
Example:
1.
Use the distributive method to multiply.
2.
Use properties of radicals to simplify.
3.
Simplify any radicals.
4.
Combined “like” terms if possible.
6 3
3 6 3 3 2
32
6 2(3)
18 3 3 2 6 6
92 3 3 2 6 6
3 2 3 3 2 6 6
Example:
5 1
5 4
1.
Use the distributive method to multiply.
5 5 5(4) (1) 5 (1)(4)
2.
Use properties of radicals to simplify.
3.
Simplify and combine “like” terms.
55 4 5 5 4
25 (4 1) 5 4
4.
The square root of 25 is 5.
5.
Combine like terms.
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5 5 5 4
95 5
Page 17
Secondary Mathematics 2
Example:
5 2 3 5 2 3
1.
Use the distributive method to multiply.
2.
3.
4.
5.
Use properties of radicals to simplify.
Simplify and combine like terms.
The square root of 4 is 2.
Simplify.
5 2 5 2 5 2(3) 3 5 2 3(3)
5 5 2 2 3 5 2 3 5 2 9
25 4 (15 15) 2 9
25 2 9
50 9
41
Multiplying Radicals with Different Indices
Note: In order to multiply radicals with different indices the radicands must be the same.
Example:
75 7
1
1
1.
Rewrite each radical using rational
5
2
7
7
exponents.
1 1
2.
Use properties of exponents to
2 5
7
simplify.
7
3.
Combine the fractions by finding a
10
7
common denominator.
Example:
3
1.
2.
3.
Rewrite each radical using rational
exponents.
Use properties of exponents to
simplify.
Combine the fractions by finding a
common denominator.
x2 4 x
2
1
x3 x4
2 1
4
x3
11
x 12
Example:
3
1.
2.
3.
4.
Rewrite the inner radical using rational
exponents.
Rewrite the outer radical using rational
exponents.
Use properties of exponents to
simplify.
Simplify by multiplying fractions.
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x
1
x3
1
13 2
x
11
x3 2
1
x6
Page 18
Secondary Mathematics 2
Practice Exercises A
Add or Subtract
1. 10 7 12 7
3
3
2. 2 3 2 24
3. 12 3 3
3
3
4. 3 2 54
3
3
5. 3 16 3 2
6. 3 20 5
8. 3 18 3 8 24
9. 2 18 2 12 2 18
7.
3
40 2 6 3 3 5
Practice Exercises B
Multiply and simplify the result.
2. 3 3 12 3 6
1. 4 28x 7 x3
4.
7.
10.
6
3 12
5 3
3
5 3
65 6
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5.
8.
11.
3
4
2
4
3
3.
235
32
3
3 5 10 6
6.
3 5
2y 7 2y
Page 19
20 x 2 4 3 20 x
9. 5 4 5 2 5
12.
3 8
z
Secondary Mathematics 2
You Decide:
1.
4
Add: 2 . Can you write the result as the ratio of two numbers? (Use your graphing
5
calculator to change the sum from a decimal to a fraction by pushing the math button and
select FRAC)
1 2
. Can you write the result as the ratio of two numbers?
2 3
2.
Add:
3.
Add: 2 1.5 . Can you write the result as the ratio of two numbers?
4.
Add: 1.75 1.35 . Can you write the result as the ratio of two numbers?
5.
Add: 2 3 . Can you write the result as the ratio of two numbers?
6.
Add:
7.
Write a rule based on your observations with adding rational and irrational numbers.
5
. Can you write the result as the ratio of two numbers?
3
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Page 20
Secondary Mathematics 2
Unit 1 Cluster 4 (A.APR.1): Polynomials
Cluster 4: Perform arithmetic operations on polynomials
1.4.1 Polynomials are closed under addition, subtraction, and multiplication
1.4.1 Add, subtract, and multiply polynomials (NO DIVISION)
VOCABULARY
A term that does not have a variable is called a constant. For example the number 5 is a constant because
it does not have a variable attached and will always have the value of 5.
A constant or a variable or a product of a constant and a variable is called a term. For example
3x 2 are all terms.
2, x , or
Terms with the same variable to the same power are like terms. 2x 2 and 7x 2 are like terms.
An expression formed by adding a finite number of unlike terms is called a polynomial. The variables
can only be raised to positive integer exponents. 4 x3 6 x 2 1 is a polynomial, while x 2 2 x 1 5 is not
a polynomial. NOTE: There are no square roots of variables, no fractional powers, and no variables in the
denominator of any fractions.
3
. A polynomial with two terms is called a
binomial ( 2 x 1 ). A polynomial with three terms is called a trinomial 5 x x 3 .
A polynomial with only one term is called a monomial 6x
4
2
Polynomials are in standard (general) form when written with exponents in descending order and the
constant term last. For example 2 x4 5x3 7 x2 x 3 is in standard form.
The exponent of a term gives you the degree of the term. The term 3x 2 has degree two. For a
polynomial, the value of the largest exponent is the degree of the whole polynomial. The polynomial
2 x4 5x3 7 x2 x 3 has degree 4.
The number part of a term is called the coefficient when the term contains a variable and a number. 6x
has a coefficient of 6 and x 2 has a coefficient of -1.
The leading coefficient is the coefficient of the first term when the polynomial is written in standard
form. 2 is the leading coefficient of 2 x4 5x3 7 x2 x 3 .
Degree n
General Polynomial: f ( x) an x n an1 x n1
Leading
Coefficient an
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Leading
Term
a2 x 2 a1x a0
Constant
Page 21
Secondary Mathematics 2
CLASSIFICATIONS OF POLYNOMIALS
Name
Form
Degree
None
0
1
Zero
Constant
Linear
f ( x) 0
f ( x) a, a 0
f ( x) ax b
Quadratic
f ( x) ax2 bx c
2
Cubic
f ( x) ax3 bx 2 cx d
3
Example
f ( x) 0
f ( x) 5
f ( x) 2 x 1
1
7
f ( x) 3 x 2 x
2
9
3
2
f ( x) x 3x
Practice Exercises A:
Determine which of the following are polynomial functions. If the function is a polynomial, state
the degree and leading coefficient. If it is not, explain why.
1.
2.
3.
f ( x) 3x 5 17
f ( x) 9 2 x
1
f ( x) 2 x 5 x 9
2
4.
f ( x) 13
5.
f ( x) 3 27 x3 8x6
6.
f ( x) 4 x 5 x 2
Operations of Polynomials
Addition/Subtraction: Combine like terms.
Example 1:
Horizontal Method
Vertical Method
2 x 3x 4 x 1 x 2 x 5x 3
2 x x 3x 2 x 4 x 5 x 1 3
3
2
3
3
3
2
2
2
3x3 x 2 x 2
2 x3 3x 2 4 x 1
x3 2 x 2 5 x 3
3x3 x 2 x 2
Example 2:
Horizontal Method
4x
2
3x 4 2 x3 x 2 x 2
Vertical Method
4 x 2 3x 4
2 x3 x 2 x 2
= 4 x 2 3x 4 2 x 3 x 2 x 2
= 2 x3 3x2 4 x 6
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2 x3 3x 2 4 x 6
Page 22
Secondary Mathematics 2
Multiplication: Multiply by a monomial
Example 3:
3x 2 x 2 6 x 5
3x 2 x 2 3x 6 x 3x 5
6 x3 18 x 2 15 x
Example 4:
5 x 2 3x3 2 x 2 6 x 8
15 x5 10 x 4 30 x3 40 x 2
Multiplication: Multiply two binomials 5x 7 2 x 9
Distributive (FOIL) Method
Box Method
5x 7 2 x 9
5x 2 x 9 7 2 x 9
10 x 2 45 x 14 x 63
*combine like terms
7
14x
63
5x
10x 2
45x
2x
9
Vertical Method
10 x 2 14 x
*combine terms on the
diagonals of the unshaded
boxes(top right to lower left)
10 x 2 31x 63
5x 7
2x 9
45 x 63
10 x 2 31x 63
10 x2 31x 63
Multiplication: Multiply a binomial and a trinomial 2 x 3 6 x2 7 x 5
Distributive Method
2 x 3 6 x 2 7 x 5
10 x 18 x
2x 6x2 7 x 5 3 6 x2 7 x 5
12 x3 14 x 2
2
Box Method
21x 15
12 x 14 x 10 x 18 x 21x 15
*combine like terms
3
2
12 x 4 x 31x 15
3
2
2x
3
6x 2
12x3
18x
2
7x
14x 2
Vertical Method
5
10x
6x2 7 x 5
2x 3
18 x 2 21x 15
21x
15
2
12 x 3 14 x 2 10 x
12 x3 4 x 2 31x 15
*combine terms on the
diagonals of the unshaded
boxes(top right to lower left)
12 x3 4 x2 31x 15
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Page 23
Secondary Mathematics 2
Practice Exercises B:
Perform the required operations. Write your answers in standard form and determine if the result
is a polynomial.
1.
2.
3.
4.
5.
6.
x 3x 7 3 x 5 x 3
3x 5 x 7 x 12
4x x 3x x 12x 3
y 2 y 3 5 y 3 y 4
2 x x x 3
y 2 y 3 y 4
2
2
2
2
3
2
3
2
2
2
2
2
7. 3u 4u 1
8.
2 x 3x 5 x
2
x 7 x 3
10. 3x 5 x 2
9.
11. 2 x 3 4 x 1
12. 3x y 3x y
13. 2 x 7
14. 3 5x
2
15. 5 x3 1
2
2
16. 2 x3 3 y 2 x3 3 y
17. x 2 2 x 3 x 4
18. x 2 3x 2 x 3
19. x 2 x 3 x 2 x 1
20. 2 x 2 3x 1 x 2 x 1
YOU DECIDE
Are polynomials closed under addition, subtraction, multiplication? Justify your conclusion
using the method of your choice.
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Page 24
Secondary Mathematics 2
Unit 2
Quadratic Functions
and Modeling
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Page 25
Secondary Mathematics 2
Unit 2 Cluster 1 (F.1F.4, F.1F.5, F.1F.6)
Unit 2 Cluster 2 (F.1F.7, F.1F.9)
Interpret functions that arise in applications in terms of a context
Analyzing functions using different representations
Cluster 1:
2.1.1
2.1.2
2.1.3
Cluster 2:
2.2.1
2.2.1
2.2.3
Interpret key features; intercepts, intervals where increasing and decreasing,
intervals where positive and negative, relative maximums and minimums,
symmetry, end behavior, domain and range, periodicity
Relate the domain of a function to its graph or a context
Average rate of change over an interval: calculate, interpret, and estimate from a
graph.
Graph functions from equations by hand and with technology showing key
features (square roots, cube roots, piecewise-defined functions including step and
functions, and absolute value).
Graph linear and quadratic functions and show intercepts, maxima, and minima
Compare properties (key features) of functions each represented differently (table,
graph, equation or description)
VOCABULARY
The domain is the set of all first coordinates when given a table or a set of ordered pairs. It is the
set of all -coordinates of the points on the graph and is the set of all numbers for which an
equation is defined. The domain it is written from the least value to the greatest value.
The range is the set of all second coordinates when given a table or a set of ordered pairs. It is
the set of all y-coordinates of the points on the graph. When modeling real world situations, the
range is the set of all numbers that make sense in the problem. The range is written from the
least value to the greatest value.
Example:
Find the domain and range of f x 2 x 2 3 .
Domain
1. Find any values for which the function is
undefined.
2. Write the domain in interval notation.
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The square root function has real number
solutions if the expression under the radicand
is positive or zero. This means that x 2 0
therefore x 2 .
The domain is [2, ) .
Page 26
Secondary Mathematics 2
Range
1. Find all values for which the output exists.
2. Write the range in interval notation.
The square root function uses the principal
square root which is a positive number or zero
( y 0 ). However, the function has been
shifted down three units so the range is also
shifted down three units y 3 .
The range is [3, )
Example:
Find the domain and range of the function graphed to the right.
Domain
1. List all the x-values of the function
graphed.
If you were to flatten the function against the
x-axis you would see something like this:
The function is defined for all the x-values
along the x-axis.
2. Write the domain in interval notation.
The domain is , .
Range
1. List all the y-values of the function
graphed.
If you were to flatten the function against the
y-axis you would see something like this:
The function is defined for all the y-values
greater than or equal to -2.
2. Write the range in interval notation.
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The range is [2, ) .
Page 27
Secondary Mathematics 2
Example:
The path of a ball thrown straight up can be modeled by the equation
h t 16t 2 20t 4 where t is the time in seconds that the ball is in the air and h is the
height of the ball. What is the real world domain and range for the situation?
Domain
1. Find all the values that would make sense
for the situation.
2. Write the domain in interval notation.
The domain is (0,1.425) .
Range
1. Find all the values that would make sense
for the situation.
2. Write the range in interval notation.
The domain represents the amount of time that
the ball is in the air. At t = 0 the ball is thrown
and enters the air shortly afterwards so the
domain must be greater than zero. The ball
will hit the ground at 1.425 seconds. Once it is
on the ground it is no longer in the air so the
domain must be less than 1.425 seconds. The
ball is in the air for 0 t 1.425 seconds.
The ball will not go lower than the ground so
the height must be greater than zero. The ball
will go no higher than its maximum height so
the height must be less than or equal to 10.25
feet. The range will be 0 h 10.25 .
The range is (0,10.25] .
Practice Exercises A:
Find the domain and range.
1. f x 3x 2
2. f ( x) 3 x 1
3.
4.
5.
Your cell phone plan charges a flat fee of $10 for up to1000 texts and $0.10 per text over
1000.
6.
The parking lot for a movie theater in the city has no charge for the first hour, but charges
$1.50 for each additional hour or part of an hour with a maximum charge of $7.50 for the
night.
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Page 28
Secondary Mathematics 2
VOCABULARY
The x-intercept is where a graph crosses or touches the -axis. It is the ordered pair a, 0 .
Where a is a real number.
The y -intercept is where a graph crosses or touches the -axis. It is the ordered pair 0,b .
Where b is a real number.
A relative maximum occurs when the y-value is greater than all of the y-values near it. A
function may have more than one relative maximum value. A relative minimum occurs when
the y-value is less than all of the y-values near it. A function may have more than one relative
minimum value.
Example:
Find the intercepts of the function f x 2 x 1 .
x-intercept
1. Substitute y in for f(x).
2. Substitute 0 in for y.
3. Solve for x.
y 2x 1
4. Write the intercept as an ordered pair.
y-intercept
1. Substitute 0 in for x.
2. Solve for y.
y 2(0) 1
y 0 1
y 1
3. Write the intercept as an ordered pair.
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0 2x 1
1 2x
1
x
2
1
,0
2
Page 29
0, 1
Secondary Mathematics 2
Example:
Find the intercepts of the function f x 3x 2 5x 2 .
x-intercept
1. Use your graphing calculator to graph the
function.
2. Use the Calculate Menu (2nd, Trace, Zero) to
find the x-intercepts. (Zero is another name
for the x-intercept)
y-intercept
1. To find the y-intercept, replace each x with 0.
2. Solve the equation for y.
The x-intercepts are
1
, 0 and 2, 0 .
3
y 3 0 5 0 2
y 002
2
y 2
3. Write the intercept as an ordered pair.
0, 2
Example:
Find the maximum of f x x 2 4 x 4 .
To find the maximum use your graphing calculator to graph the
function. Then use the Calculate Menu (2nd, Trace, Maximum).
Enter a number that is to the left of the maximum, for example 0,
then push enter. Then enter a number that is to the right of the
maximum, for example 4, then push enter. You can guess the
value of the maximum or just push enter again and the maximum
will be calculated. The maximum is (2, 8).
Example:
2
Find the minimum of f x x 2 3 .
To find the maximum use your graphing calculator to graph the
function. Then use the Calculate Menu (2nd, Trace, Minimum).
Enter a number that is to the left of the minimum, for example -3,
then push enter. Then enter a number that is to the right of the
minimum, for example -1, then push enter. You can guess the
value of the minimum or just push enter again and the minimum
will be calculated. The minimum is (-2, -3).
Jordan School District
Page 30
Secondary Mathematics 2
Practice Exercises B
Find the x and y-intercepts for each function.
1. 2 x 5 y 10
2. f x 4 x 7
5. f x x 2 3x 18
1
4. f x x 4
3
Find the relative maximums or minimums of each function.
2
2
7. f x 2 x 1 3
8. f x 3 x 2 7
10. f x x 5 4
2
11. f x 3x 2 18x 23
3. f x x 2 x 30
6. f x 2 x 2 3x 4
9. f x 4 x 2 16 x 18
12. f x x 2 8x 14
VOCABULARY
An interval is a set of numbers between two x -values. An open interval is a set of numbers
between two x -values that doesn’t include the two end values. Open intervals are written in the
form x1 , x2 or x1 x x2 . A closed interval is a set of numbers between two -values that
does include the two end values. Closed intervals are written in the form x1 , x2 or x1 x x2 .
A function f is increasing when it is rising (or going up) from left to right and it is decreasing
when it is falling (or going down) from left to right. A constant function is neither increasing nor
decreasing; it has the same y-value for its entire domain.
A function is positive when f x 0 or the y-coordinates are always positive. A function is
negative when f x 0 or the y-coordinates are always negative.
Example:
Find the intervals where the function f x x 2 2 x 3 is:
a. increasing
b. decreasing
c. constant
d. positive
e. negative
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Page 31
Secondary Mathematics 2
Increasing/Decreasing/Constant
1. Find the maximums or minimums.
The minimum is (-1, -4).
2. Determine if the function is rising, falling,
or constant between the maximums and
minimums.
To the left of the minimum the function is
falling or decreasing. To the right of the
minimum the function is rising or increasing.
3. Write the intervals where the function is
increasing, decreasing, or constant using
interval notation.
The function is increasing on the interval
(1, ) . The function is decreasing on the
interval (, 1) . The function is never
constant.
Positive/Negative
1. Find all the x-intercepts of the function.
The x-intercepts are at (-3, 0) and (1, 0).
2. Determine if the function has positive or
negative y-values on the intervals between
each x-intercept by testing a point on the
interval.
3. Write the intervals where the function is
positive or negative using interval notation.
x 3
x 4
f (4) 5
Positive
3 x 1
x0
f (0) 3
Negative
x 1
x2
f (2) 5
Positive
The function is positive on the intervals
(, 3) and (1, ) . The function is negative
on the interval (3,1) .
Example:
Find the intervals where the function f x 3 x 2 1 is:
a. increasing
b. decreasing
c. constant
d. positive
e. negative
Increasing/Decreasing/Constant
1. Find the maximums or minimums.
There are no maximums or minimums.
2. Determine if the function is rising, falling,
or constant on its entire domain.
The function is rising from left to right so it is
increasing on its entire domain.
3. Write the intervals where the function is
increasing, decreasing, or constant using
interval notation.
The function is increasing on the interval
, . The function is never decreasing nor
is it constant.
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Page 32
Secondary Mathematics 2
Positive/Negative
1. Find all the x-intercepts of the function.
The x-intercept is (-1, 0).
2. Determine if the function has positive or
negative y-values on the intervals between
each x-intercept by testing a point on the
interval.
3. Write the intervals where the function is
positive or negative using interval notation.
x 1
x 2
f (2) 1
Negative
x 1
x0
f (0) 0.26
Positive
The function is positive on the interval (1, ).
The function is negative on the interval
(, 1) .
Example:
| x 2 | 3, x 1
Find the intervals where the function f x
is:
x 1
3,
a. increasing
b. decreasing
c. constant
d. positive
e. negative
Increasing/Decreasing/Constant
1. Find the maximums or minimums and any
breaks in the domain.
There is a maximum at (-2, 0) and a break in
the domain at x = -1.
2. Determine if the function is rising, falling,
or constant between each maximum or
minimum and each break in the graph.
The function is rising (increasing) to the left of
the maximum. It is falling (decreasing) to the
right of the maximum. It is constant to the
right of x = -1.
3. Write the intervals where the function is
increasing, decreasing, or constant using
interval notation.
The function is increasing on the interval
, 2 . It is decreasing on the interval
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2, 1 . It is constant on the interval 1, .
Page 33
Secondary Mathematics 2
Positive/Negative
1. Find all the x-intercepts of the function and
any places where there is a break in the
domain.
The x-intercept is (-5, 0). There is a break in
the domain at x = -1.
2. Determine if the function has positive or
negative y-values on the intervals between
each x-intercept by testing a point on the
interval.
x 5
x 6
f (6) 1
Negative
3. Write the intervals where the function is
positive or negative using interval notation.
5 x 1
x 3
f (3) 2
Positive
x 1
x0
f (0) 3
Positive
The function is positive on the intervals
(5, 1) and (1, ) . The function is negative
on the interval (, 5) .
Practice Exercises C
Find the intervals where the function is:
a. increasing
b. decreasing
c. constant
d. positive
e. negative
1. f x
1
x3
2
2. f x 2 x 2 3x 2
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3. f x 2 x 3
5. f x x 4 1
4. f x x 3
x0
2,
6. f x 2
x 1, x 0
3
Page 34
Secondary Mathematics 2
VOCABULARY
GRAPHICALLY
ALGEBRAICALLY
f ( x) x 5
A function is symmetric
with respect to the y-axis if,
for every point x, y on the
f ( x) x 5
graph, the point x, y is
also on the graph. In other
words,
if you substitute –x in for
every x you end up with the
original function. When
looking at the graph, you
could “fold” the graph along
the y-axis and both sides are
the same.
f ( x) f ( x) x 5
A function is symmetric
with respect to the origin
if, for every point x, y on
f ( x) 8 x 3
f ( x) 8 x
3
f ( x) f ( x) 8 x 3
the graph, the point x, y
is also on the graph. In other
words,
if you substitute –x in for
every x you end up with the
opposite of the original
function. When looking at
the graph, there is a mirror
image in Quadrants 1 & 3 or
Quadrants 2 & 4.
f ( x) x 2 2 x
An equation with no
symmetry. If you substitute
–x in for every x you end up
with something that is
neither the original function
nor its opposite. When
looking at the graph, you
could not “fold” the graph
along the y-axis and have
both sides the same. It also
does not reflect a mirror
image in opposite quadrants.
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f ( x) ( x) 2 2( x)
f ( x) x 2 2 x f ( x) f ( x)
Page 35
Secondary Mathematics 2
Example:
Determine what kind of symmetry, if any, f x 2 x 1 has.
Test for y-axis Symmetry
Replace x with –x and see if
the result is the same as the
original equation.
f x 2( x) 1
Test for Origin Symmetry
Replace x with –x and see if
the result is the opposite of the
original equation.
f x 2( x) 1
f ( x) | 2 x 1|
This is not the same as the
original equation.
f ( x) | 2 x 1|
This is not the opposite of the
original equation.
Graph
The function f x 2 x 1 has no symmetry.
Example:
Determine what kind of symmetry, if any, f x 2 x 2
Test for y-axis Symmetry
Replace x with –x and see if
the result is the same as the
original equation.
f x 2( x) 2
Test for Origin Symmetry
Replace x with –x and see if
the result is the opposite of the
original equation.
f x 2( x) 2
f ( x) 2 x 2
This is equal to the original
equation.
f ( x) 2 x 2
This is not the opposite of the
original equation.
Graph
The function f x 2 x 2 has y-axis symmetry.
Example:
Determine what kind of symmetry, if any, the function graphed at
the right has.
Test for y-axis Symmetry
Pick a point x, y on the graph and see if
Test for origin symmetry
Pick a point x, y on the graph and see if
x, y is also on the graph. The point (-2, 2)
x, y is also on the graph. The point (-2, 2)
is on the graph but the point (2, 2) is not. The
is on the graph and the point (2, -2) is also on
function does not have y-axis symmetry.
the graph. The function has origin symmetry.
The function graphed has origin symmetry.
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Page 36
Secondary Mathematics 2
VOCABULARY
End behavior describes what is happening to the y-values of a graph when x goes to the far right
or x goes the far left .
End behavior is written in the following format:
Right End Behavior:
lim f ( x) c
Left End Behavior:
lim f ( x) c
x
x
Example:
Find the end behavior of f x 4 x 3 .
As x gets larger the function is getting more and more negative.
Therefore, the right end behavior is lim f ( x) . As x gets
x
smaller the function is getting more and more positive. Therefore
the left end behavior is lim f ( x) .
x
Example:
Find the end behavior of f x 3x 2 x 1 .
As x gets larger the function is getting more and more negative.
Therefore, the right end behavior is lim f ( x) . As x gets
x
smaller the function is getting more and more negative. Therefore
the left end behavior is lim f ( x) .
x
Example:
Find the end behavior of f x x 2 1 .
As x gets larger the function is getting more and more positive.
Therefore, the right end behavior is lim f ( x) . The domain is
x
restricted to numbers greater than or equal to 2, therefore this graph
has no left end behavior.
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Page 37
Secondary Mathematics 2
Practice Exercises D
Graph each function below and find the:
a. Domain and Range
b. Intercepts, if any
c. Determine whether the function has any symmetry.
d. List the intervals where the function is increasing, decreasing, or constant.
e. List the intervals where the function is positive or negative.
f. Find all the relative maximums and minimums.
g. Find end behavior
1. f ( x) 2 x 5
4. f ( x) x 3
2. f ( x) x 3 1
5. f ( x) 3 x 1 5
3. f ( x) x 2 4
3,
6. f ( x) 1
x 2,
3
2
x 1
1 x 12
VOCABULARY
Periodicity refers to a function with a
repeating pattern. The period of this function is
6 horizontal units. Meaning the pattern will
repeat itself every 6 horizontal units.
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Page 38
Secondary Mathematics 2
You Decide
Mr. Astro’s physics class created rockets for an end of the year competition. There were three
groups who constructed rockets. On launch day the following information was presented for
review to determine a winner.
Group A estimated that their rocket was easily modeled by the equation: y 16 x2 176 x 3 .
height (feet)
Group B presented the following graph of the height of their rocket, in feet, over time.
time (seconds)
Group C recorded their height in the table below.
Time (seconds)
Height (feet)
0
3
2
256
4
381
6
377
8
246
10
0
Who should be the winner of the competition? Use mathematical reasons to support your
conclusion.
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Secondary Mathematics 2
Unit 2 Cluster 2 (F.IF.7b)
Graphing Square Root, Cube Root, and Piecewise-Defined
Functions, Including Step Functions and Absolute Value Functions
Cluster 2: Analyzing functions using different representations
2.2.1b Graph functions from equations by hand and with technology showing key
features (square roots, cube roots, piecewise-defined functions including step
functions, and absolute value)
VOCABULARY
There are several types of functions (linear, exponential, quadratic, absolute value, etc.). Each of
these could be considered a family with unique characteristics that are shared among the
members. The parent function is the basic function that is used to create more complicated
functions.
Square Root Function
Parent Function
f x x x1/2
Domain: 0,
Key Features
Range: 0,
Intercepts: x-intercept 0, 0 , y-intercept 0, 0
Intervals of Increasing/Decreasing: increasing 0,
Intervals where Positive/Negative: 0,
Relative maximums/minimums: minimum at 0, 0
Symmetries: none
End Behavior: right end behavior lim x ; left end
x
behavior lim x 0
x 0
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Secondary Mathematics 2
Cube Root Function
Parent Function
f x 3 x x1/3
Domain: ,
Key Features
Range: ,
Intercepts: x-intercept 0, 0 , y-intercept 0, 0
Intervals of Increasing/Decreasing: increasing ,
Intervals where Positive/Negative: positive 0, ,
negative , 0
Relative maximums/minimums: none
Symmetries: origin
End Behavior: right end behavior lim 3 x ; left end
x
behavior lim
3
x
x
Absolute Value Function
Parent Function
f x x
Domain: ,
Key Features
Range: 0,
Intercepts: x-intercept 0, 0 , y-intercept 0, 0
Intervals of Increasing/Decreasing: increasing 0, ,
decreasing , 0
Intervals where Positive/Negative: positive
,0 0,
Relative maximums/minimums: minimum at 0, 0
Symmetries: y-axis symmetry
End Behavior: right end behavior lim x ; left end
x
behavior lim x
x
Piecewise-Defined Functions
A piecewise-defined function is a function that consists of pieces of two or more functions. For
x 2
x 2,
example f x 1,
2 x 0 is a piecewise-defined function. It has a piece of the
2 x 5,
x0
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Secondary Mathematics 2
function f x x 2 but only the piece where x 2 . It also contains the function f x 1 ,
but only where 2 x 0 . Finally, it contains the function f x 2 x 5 but only where
x 0.
Piecewise-Defined Function
x 2
x 2,
f x 1,
2 x 0
2 x 5,
x0
Domain: ,
Key Features
Range: ,5
Intercepts: x-intercept 2.5, 0 , y-intercept 0,1
Intervals of Increasing/Decreasing: increasing , 2 ,
decreasing 5,
Intervals where Positive/Negative: positive
2,0 and 0, 2.5 , negative , 2 and 2.5,
Relative maximums/minimums: none
Symmetries: none
End Behavior: right end behavior lim 2 x 5 ; left
end behavior lim x 2
x
x
Step Functions are piecewise-defined functions made up of constant functions. It is called a
step function because the graph resembles a staircase.
Step Function
f x int x
Domain: ,
Key Features
Range: y | y is an integer
Intercepts: x-intercept x 0,1 and y 0, y-intercept
0, 0
Intervals of Increasing/Decreasing: neither increasing nor
decreasing
Intervals where Positive/Negative: positive 1, ,
negative , 0
Relative maximums/minimums: none
Symmetries: none
End Behavior: right end behavior lim int x ; left end
x
behavior lim int x
x
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Secondary Mathematics 2
Unit 2 Cluster 1(F.IF.6) and Cluster 5(F.LE.3)
Quadratic Functions and Modeling
Cluster 1: Interpret Functions that Arise in Applications in Terms of a Context
2.1.3 Average rate of change over an interval: calculate, interpret, and estimate from a
graph.
Cluster 5: Constructing and comparing linear, quadratic, and exponential models; solve problems
2.5.1 Exponential functions will eventually outgrow all other functions
VOCABULARY
The average rate of change of a function over an interval is the ratio of the difference (change)
in y over the difference (change) in x.
average rate of change
y y2 y1
x x2 x1
Example:
Find the average rate of change for f x 2 x 2 3x 1 on the interval [0, 2].
First, find the value of the function at each end point of the interval.
f 2 2(2)2 3(2) 1
f 0 2(0)2 3(0) 1
f 0 0 0 1
f 2 2 4 6 1
f 0 1
f 2 8 6 1
2,3
0,1
f 2 3
Next, find the slope between the two points (0, 1) and (2, 3).
m
3 1 2
1
20 2
The average rate of change of f x 2 x 2 3x 1 on the interval [0, 2] is 1 .
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Secondary Mathematics 2
Example:
The per capita consumption of ready-to-eat and ready-to-cook breakfast cereal is shown
below. Find the average rate of change from 1992 to 1995 and interpret its meaning.
Years since 1990
0
1
2
3
4
5
6
7
8
9
Cereal Consumption
15.4 16.1 16.6 17.3 17.4 17.1 16.6 16.3 15.6 15.5
(pounds)
The year 1992 is two years since 1990 and 1995 is 5 years since 1990, therefore the
interval is [2, 5]. Find the slope between the two points (2, 16.6) and (5, 17.1).
m
17.1 16.6 0.5
0.16
52
3
Example:
Joe is visiting the Eiffel Tower in Paris. He accidentally
drops his camera. The camera’s height is graphed. Use
the graph to estimate the average rate of change of the
camera from 4 to 7 seconds and interpret its meaning.
At 4 seconds the height of the camera is approximately
650 feet. At 7 seconds the height of the camera is
approximately 100 feet. Find the slope between the
points (4, 650) and (7, 100).
m
100 650 550
183.3
74
3
Height in feet
The average rate of change from 1992 to 1995 is 0.16 pounds per year. This means that
each household increased their cereal consumption an average of 0.16 pounds each year
from 1992 to 1995.
Time in seconds
The negative indicates that the camera is falling. The camera is picking up speed as it is
falling. This means that for each second the camera is falling from 4 to 7 seconds, it
increases in speed an average of 183.3 feet per second from 4 to 7 seconds.
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Secondary Mathematics 2
Practice Exercises A
Find the average rate of change for each function on the specified interval.
1. f x 3x 2 x 5 on [-1, 3]
2. f x 4 x 2 12 x 9 on [-3, 0]
3. f x x 2 4 on [-4, -2]
4. f x 2 x 2 6 x on [-1, 0]
Find the average rate of change on the specified interval and interpret its meaning.
5. Many of the elderly are placed in nursing
care facilities. The cost of these has risen
significantly since 1960. Use the table
below find the average rate of change from
2000 to 2010.
Years since
1960
0
10
20
30
40
50
6. The height of an object thrown straight up
is shown in the table below. Find the
average rate of change from 1 to 2
seconds.
Time
(seconds)
0
1
2
3
4
Nursing Care
Cost
(billions of $)
1
4
18
53
96
157
Height
(feet)
140
162
152
110
36
Fuel Consumption
Net Sales in millions of $
7. The net sales of a company are shown in the 8. The graph below shows fuel consumption
graph below. Estimate the average rate of
in billions of gallons for vans, pickups and
change for 2007 to 2009.
SUVs. Estimate the average rate of
change for 2005 to 2012.
Years since 1980
Years since 1999
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Secondary Mathematics 2
Practice Exercises B
Complete the tables.
x
f ( x) 2 x
x
g ( x) x 2
h( x) 2 x
x
-2
-2
-2
-1
-1
-1
0
0
0
1
1
1
2
2
2
3
3
3
4
4
4
5
5
5
Practice Exercises C
Find the average rate of change for functions f ( x), g ( x),and h( x) for the specified intervals.
Determine which of the three functions is increasing the fastest.
1. [0, 1]
2. [3, 5]
3. [-2, 5]
4. [0, 3]
5. [-2, 0]
6. [0, 5]
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Secondary Mathematics 2
Practice Exercises D
1. Graph the following functions on the same coordinate plane.
a. k ( x)
2.
3
x 1
2
b. p( x) 2 x 3 5
2
c. r ( x) 3x 2 7
Find the average rate of change for functions k ( x), p( x), and r ( x) for the specified
intervals. Determine which of the three functions is increasing the fastest.
a. [-4, 2]
b. [3, 5]
c. [0, 10]
You Decide
Use exercises C and D to help you answer the following questions.
1.
For each exercise, determine which function has the greatest average rate of change on
the interval 0, ?
2.
In general, what type of function will increase faster? Explain your reasoning.
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Secondary Mathematics 2
FACTORING
(To be used before F.IF.8)
VOCABULARY
Factoring is the reverse of multiplication. It means to write an equivalent expression that is a
product. Each of the items that are multiplied together in a product is a factor. An expression is
said to be factored completely when all of the factors are prime polynomials, that is they
cannot be factored any further.
The greatest common factor is the largest expression that all the terms have in common.
FACTOR OUT A COMMON TERM
Example:
2 x2 6 x 8
What is the largest factor that evenly divides
2 x2 ,6 x, and 8 ?
2 x2 : 1 2 x x
6x : 1 2 3 x
8 :1 2 2 2
The common numbers are 1and 2. Multiply
them and the product is the greatest common
factor.
2 x2 6 x 8
2
2 2
Divide each term by the greatest common
factor.
2 x 2 3x 4
Rewrite with the common term on the outside
of the parenthesis and the simplified terms
inside the parenthesis.
Example:
8w4 3w3 5w2
What is the largest factor that evenly divides
8w4 ,3w3 , and 5w2 ?
8 w4 : w w w w 1 2 2 2
3w : w w w 1 3
3
5 w2 : w w 1 5
8w4 3w3 5w2
2 2
w2
w
w
w2 8w2 3w 5
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The common numbers are w and w. Multiply
them and the product is the greatest common
factor.
Divide each term by the greatest common
factor.
Rewrite with the common term on the outside
of the parenthesis and the simplified terms
inside the parenthesis.
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Secondary Mathematics 2
Example:
What is the largest factor that evenly divides
9 z 3 ,12 z 2 , and 15z ?
9 z 3z 15z
3
2
9z : 1 3 3 z z z
3
3z 2 : 1 3 z z
15 z : 1 3 5 z
9 z 3 3z 2 15 z
3z 3z
3z
3z 3z 2 z 5
The common numbers are 3 and z. Multiply
them and the product is the greatest common
factor.
Divide each term by the greatest common
factor.
Rewrite with the common term on the outside
of the parenthesis and the simplified terms
inside the parenthesis.
Practice Exercises A
Factor out the greatest common factor.
1.
4 x2 12 x 16
2.
5x5 10 x4 15x3
3.
8x4 32 x3 16 x2
4.
2 x3 x2 3x
5.
27 x2 36 x 18
6.
14 x2 21x 49
FACTOR A TRINOMIAL WITH A LEADING COEFFICIENT OF 1
When factoring a trinomial of the form ax2 bx c ,where a = 1, and b and c are
integers, find factors of c that add to equal b.
If p and q are the factors, the factored form looks like x p x q .
Example:
x2 5x 6
Factors of 6
1
2
-1
-2
6
3
-6
-3
Sum
(adds to be)
7
5
-7
-5
x 2 x 3
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Find factors of 6 that add to be 5. The factors
are 2 and 3.
This is the factored form.
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Secondary Mathematics 2
Another way to look at factoring is with an area model like the one pictured below.
x
1
1
1
x
x2
x
x
x
1
x
1
1
1
1
x
1
1
1
The rectangular area represents the trinomial
x2 5x 6 . The width across the top is
x 1 1 1 or x 3 and the length down the
side is x 1 1 or x 2 . To obtain the area
of the rectangle, you would multiply the
length times the width or x 2 x 3 .
Notice that this result is the same as when we
found factors of the constant term that added
to the coefficient of the x term.
When factoring a trinomial of the form ax2 bx c ,where a = 1, and b and c are
integers, find factors of c that add to equal b.
If p and q are the factors, the factored form looks like x p x q .
Example:
x2 5x 6
Sum
(adds to be)
7
5
-7
-5
Factors of 6
1
2
-1
-2
6
3
-6
-3
Find factors of 6 that add to be -5. The factors
are -2 and -3.
x 2 x 3
This is the factored form.
This example can also be modeled with an area model as the picture below demonstrates.
x
1 1 1
x
x2
x x x
1
x
1
1
1
1
x
1
1
1
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The rectangular area represents the trinomial
x2 5x 6 . The width across the top is
x 1 1 1 or x 3 and the length down the
side is x 1 1 or x 2 . To obtain the area
of the rectangle, you would multiply the
length times the width or x 2 x 3 .
Notice that this result is the same as when we
found factors of the constant term that added
to the coefficient of the x term.
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Secondary Mathematics 2
When factoring a trinomial of the form ax2 bx c ,where a = 1, and b and c are
integers, find factors of c that add to equal b.
If p and q are the factors, the factored form looks like x p x q .
Example:
x2 5x 6
Factors of 6
1
2
-1
-2
-6
-3
6
3
Sum
(adds to be)
-5
-1
5
1
Find factors of -6 that add to be 5. The factors
are -1 and 6.
x 1 x 6
This is the factored form.
This example can also be modeled with an area model as the picture below demonstrates.
x
1
x
1
1
1
1
1
1
x2
x
x
x
x
x
x
x
1 1 1 1 1 1
The rectangular area represents the trinomial x2 5x 6 . The width across the top is
x 1 1 1 1 1 1 or x 6 and the length down the side is x 1 . To obtain the area of
the rectangle, you would multiply the length times the width or x 1 x 6 . Notice
that this result is the same as when we found factors of the constant term that added to the
coefficient of the x term.
When factoring a trinomial of the form ax2 bx c ,where a = 1, and b and c are
integers, find factors of c that add to equal b.
If p and q are the factors, the factored form looks like x p x q .
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Secondary Mathematics 2
Example:
x2 5x 6
Factors of 6
1
2
-1
-2
-6
-3
6
3
Sum
(adds to be)
-5
-1
5
1
Find factors of -6 that add to be -5. The factors
are 1 and -6.
x 1 x 6
This is the factored form.
This example can also be modeled with an area model as the picture below demonstrates.
x
1 1 1 1 1 1
x
x2
x x x x x x
1
x
1 1 1 1 1 1
The rectangular area represents the trinomial x2 5x 6 . The width across the top is
x 1 1 1 1 1 1 or x 6 and the length down the side is x 1 . To obtain the area of
the rectangle, you would multiply the length times the width or x 1 x 6 . Notice
that this result is the same as when we found factors of the constant term that added to the
coefficient of the x term.
Practice Exercises B
Factor each expression.
1. x2 4 x 21
2. x2 4 x 12
3. x2 x 2
4. x2 9 x 18
5. x2 x 72
6. x2 5x 36
7. x2 15xy 14 y 2
8. x 2 3xy 2 y 2
9. x2 17 xy 72 y 2
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Secondary Mathematics 2
VOCABULARY
A perfect square is a number that can be expressed as the product of two equal integers. For
example: 100 is a perfect square because 10 10 100 and x 2 is a perfect square because
x x x2 .
FACTOR USING THE DIFFERENCE OF TWO SQUARES
When something is in the form a 2 b2 , where a and b are perfect square expressions, the
factored form looks like a b a b .
Example:
x 2 49
x 2 and 49 are both perfect squares and you are
x2 x x
49 7 7
x 7 x 7
finding the difference between them, so you
can use the difference of two squares to factor.
Therefore:
a x and b 7
This is the factored form.
Example:
25x 2 36 y 2
25 x 2 5 x 5 x
36 y 2 6 y 6 y
25x 2 and 36 y 2 are both perfect squares and
you are finding the difference between them,
so you can use the difference of two squares to
factor.
Therefore:
a 5x and b 6 y
5x 6 y 5x 6 y
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This is the factored form.
Page 53
Secondary Mathematics 2
Practice Exercises C
Factor each expression.
1. 49 x2 25
3. 9 x 2 4
5. 36 x2 121
2. x 2 64 y 2
4. 16 x 2 81y 2
6. 100 x 2 64 y 2
FACTOR BY GROUPING
When factoring a trinomial of the form ax2 bx c ,where a, b, and c are integers,
you will need to use the technique of factoring by grouping.
Example:
6 x 2 x 15
Multiply the leading coefficient and
the constant.
Choose the combination that will
either give the sum or difference
needed to result in the coefficient of
the x term.
(6)(15) = 90
Factors of 90
1
90
2
45
3
30
5
18
6
15
9
10
In this case the difference should be -1,
so 9 and -10 will give you the desired
result. Or in other words, when you
combine 9 x and 10 x you will end
up with x .
Rewrite the equation using the
combination in place of the middle
term.
Group the first two terms and the last
two terms together in order to factor.
Factor the greatest common factor out
of each group.
Write down what is in the parenthesis
(they should be identical). This is one
of the factors.
Add the “left-overs” to obtain the
second factor.
6 x 9 x 10 x 15
2
6x
2
9 x 10 x 15
3 x 2 x 3 5 2 x 3
2 x 3
2x 33x 5
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Secondary Mathematics 2
Example:
12 x2 7 x 10
Multiply the leading coefficient and
the constant.
Choose the combination that will
either give the sum or difference
needed to result in the coefficient of
the x term.
(12)(10) = 120
Factors of 90
1
120
2
60
3
40
4
30
5
24
6
20
8
15
10
12
In this case the difference should be 7,
so -8 and 15 will give you the desired
result. Or in other words, when you
combine 8x and 15x you will end up
with 7x .
Rewrite the equation using the
combination in place of the middle
term.
Group the first two terms and the last
two terms together in order to factor.
Factor the greatest common factor out
of each group.
Write down what is in the parenthesis
(they should be identical). This is one
of the factors.
Add the “left-overs” to obtain the
second factor.
12 x 8x 15x 10
2
12x
2
8x 15x 10
4 x 3x 2 5 3x 2
3x 2
3x 2 4 x 5
Example:
4 x 2 25
Multiply the leading coefficient and
the constant.
Choose the combination that will
either give the sum or difference
needed to result in the coefficient of
the x term.
(4)(25) = 100
Factors of 100
1
100
2
50
4
25
5
20
10
10
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In this case the difference should be 0,
so -10 and 10 will give you the desired
result. Or in other words, when you
combine 10 x and 10 x you will end
up with 0.
Page 55
Secondary Mathematics 2
Rewrite the equation using the
combination in place of the middle
term.
Group the first two terms and the last
two terms together in order to factor.
Factor the greatest common factor out
of each group.
Write down what is in the parenthesis
(they should be identical). This is one
of the factors.
Add the “left-overs” to obtain the
second factor.
4 x 10 x 10 x 25
2
4x
2
10 x 10 x 25
2 x 2 x 5 5 2 x 5
2 x 5
2 x 5 2 x 5
Practice Exercises D
Factor the expression.
1. 2 x2 13x 6
2. 4 x2 3x 1
3. 3x2 2 x 8
4. 2 x2 11x 6
5. 2 x2 4 x 2
6. 3x2 6 x 3
7. 10 x2 x 6
8. 6 x2 7 x 20
9. 12 x2 17 x 6
FACTORING GUIDELINES
#1: Always look for a greatest common factor. Then factor it out if there is one.
#2: Count the number of terms. If there are two terms, determine if you can use the
difference of two squares. If you can, factor. If not, proceed to #3.
#3: If there are three terms, check the leading coefficient. If it is “1”, then find factors of
the constant term that add to the coefficient of the x-term. If not, proceed to #4.
#4: If the leading coefficient is not “1”, factor by grouping.
Mixed practices E
Factor the expression.
1. 2 x 2 50
2. 2 x2 16 xy 32 y 2
3. 3x2 5x 12
4. 5x2 10 x 5
5. 25x 2 64 y 2
6. 3x 2 27
7. 4 y 2 4 y
8. x2 13x 42
9. 4 x2 12 x 9
10. x 2 6 x
11. 9 x2 12 x 4
12. 8x2 2 x 3
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Secondary Mathematics 2
Unit 2 Cluster 2 (F.IF.8) , Unit 3 Cluster 1 (A.SSE.1a) and Unit 3
Cluster 2 (A.SSE.3a,b)
Forms of Quadratic Functions
Cluster 2: Analyzing functions using different representations
2.2.2 Writing functions in different but equivalent forms (quadratics: standard,
vertex, factored) using the processes of factoring or completing the square to
reveal and explain different properties of functions. Interpret these in terms of
a context.
Cluster 1: Interpret the structure of expressions
3.1.1a Interpret parts of an expression, such as terms, factors, and coefficients
Cluster 2: Writing expressions in equivalent forms and solving
3.2.1 Choose an appropriate from of an equation to solve problems (factor to find
zeros, complete the square to find maximums and minimums
VOCABULARY
Forms of Quadratic Functions
Standard Form: f x ax 2 bx c , where a 0 . Example: f ( x) 4 x 2 6 x 3
Vertex Form: f ( x) a( x h)2 k , where a 0 . Example: f ( x) 2( x 3)2 5
Factored Form: f ( x) a( x p)( x q) , where a 0 . Example: f ( x) ( x 4)( x 7)
A zero of a function is a value of the input x that makes the output f x equal zero. The
zeros of a function are also known as roots, x-intercepts, and solutions of ax2 bx c 0 .
The Zero Product Property states that if the product of two quantities equals zero, at least
one of the quantities equals zero. If ab 0 then a 0 or b 0 .
Finding Zeros (Intercepts) of a Quadratic Function
When a function is in factored form, the Zero Product Property can be used to find the zeros
of the function.
If f x ax x p then ax x p 0 can be used to find the zeros of f x .
If 0 ax x p then either ax 0 or x p 0 .
Therefore, either x 0 or x p .
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Secondary Mathematics 2
Example: Find the zeros of f x 2 x x 7
f x 2x x 7
2x x 7 0
x 0 or x 7
Substitute zero in for f(x).
Use the zero product property to set each
factor equal to zero.
Solve each equation.
The zeros are 0, 0 and 7, 0
Write them as ordered pairs.
2 x 0 or x 7 0
If f x x p x q then x p x q 0 can be used to find the zeros of
f x .
If x p x q 0 then either x p 0 or x q 0 .
Therefore, either x p or x q .
Example: Find the zeros of f x x 5 x 9
f x x 5 x 9
x 5 x 9 0
Substitute zero in for f(x).
x 5 0 or x 9 0
x 5 or x 9
Use the zero product property to set each
factor equal to zero.
Solve each equation.
The zeros are 5, 0 and 9,0
Write them as ordered pairs.
NOTE: If a quadratic function is given in standard form, factor first then
apply the Zero Product Property.
2
Example: Find the zeros of f x x 11x 24
f x x 2 11x 24
x2 11x 24 0
Substitute zero in for f(x).
x 8 x 3 0
x 8 or x 3
Factor the trinomial. (See factoring
lesson in Unit 2 for extra help.)
Use the zero product property to set each
factor equal to zero.
Solve each equation.
The zeros are 8, 0 and 3, 0
Write them as ordered pairs.
x 8 0 or x 3 0
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Secondary Mathematics 2
Example: Find the zeros of f x 4 x 2 4 x 15
f x 4 x 2 4 x 15
4 x2 4 x 15 0
Substitute zero in for f(x).
2 x 5 2 x 3 0
Factor the trinomial.
Use the zero product property to set each
factor equal to zero.
2 x 5 0 or 2 x 3 0
2 x 3
2x 5
5 or
3
x
x
2
2
3
5
The zeros are , 0 and , 0
2
2
Solve each equation.
Write them as ordered pairs.
Practice Exercises A
Find the zeros of each function.
1. f x x x 7
2.
f x 2 x x 6
3.
f x x 13 x 4
4.
f x x 21 x 3
5.
f x x2 7 x 6
6.
f x x2 x 2
7.
f x x 2 8x 12
8.
f x x 2 10 x 24
9.
f x 4 x 2 12 x
2
10. f x 9 x 25
2
11. f x 5x 4 x 12
2
12. f x 3x 17 x 10
COMPLETING THE SQUARE
x 2 bx
To complete the square of
2
b
x bx, add . In other
2
words, divide the x
coefficient by two and square
the result.
2
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x2 6x
b
x bx
2
2
2
b
b
x x
2
2
b
x
2
2
Page 59
6
x 6x
2
2
2
x 2 6 x 32
x2 6x 9
x 3 x 3
2
x 3
Secondary Mathematics 2
An area model can be used to represent the process of completing the square for the
expression x 2 6 x ___ .
x
1
1
1
x
x2
1
x
1
x
1
1
1
1
x
1
1
1
1
1
1
x
x
x
The goal is to arrange the pieces into a square.
The x pieces are divided evenly between the
two sides so that each side is x 3 long.
However, there is a large piece of the square
that is missing. In order to complete the
square you need to add 9 ones pieces.
x2 5x
x 2 bx
To complete the square of
2
b
x bx, add . In other
2
words, divide the x
coefficient by two and square
the result.
2
b
x 2 bx
2
2
b
b
x x
2
2
b
x
2
2
2
b
, which is the square
2a
of the coefficient of x divided
by two.
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25
x2 5x
4
5
x
2
3x 2 6 x
b
a x2 x
a
2
2 b
b
a x x
a
2a
b
b
a x
x
2a
2a
b
a x
2a
2
5
5
x x
2
2
ax 2 bx
To complete the square of
ax 2 bx, factor out the
leading coefficient, a, giving
b
you a x 2 x . Now add
a
5
x2 5x
2
2
Page 60
2
6
3 x2 x
3
2
2 6
6
3 x x
3
2 3
3 x 2 2 x 1
2
3 x 1 x 1
3 x 1
2
Secondary Mathematics 2
Practice Exercises B
For each expression complete the square.
1. x2 10 x ___
2. x2 7 x ___
3. x2 22 x ___
4. 4 x2 16 x ___
5. 2 x2 12 x ___
6. 5x2 20 x ___
Finding Maximum/Minimum (the vertex) Points of a Quadratic Function
VOCABULARY
Remember when a quadratic function is in vertex form f ( x) a( x h)2 k the point h, k
is the vertex of the parabola. The value of a determines whether the parabola opens up or
down.
The vertex of a parabola that opens up, when a 0 , is the minimum point of a quadratic
function.
The vertex of a parabola that opens up, when a 0 , is the maximum point of a quadratic
function.
Example:
2
Find the vertex of f ( x) x 2 3 , then determine whether it is a maximum
or minimum point.
f ( x) x 2 3
2
f ( x) x 2 3
2
Rewrite the equation so it is in the general
vertex form f ( x) a( x h)2 k .
Vertex: (-2, -3)
h 2 and k 3
The vertex is a minimum.
The leading coefficient is 1, which makes
a0
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Page 61
Secondary Mathematics 2
Example:
2
Find the vertex of f ( x) 5 x 8 4 , then determine whether it is a maximum
or minimum point.
f ( x) 5 x 8 4
2
This equation is already in the general vertex
form f ( x) a( x h)2 k .
f ( x) 5 x 8 4
2
Vertex: (8, 4)
h 8 and k 4
The vertex is a maximum.
The leading coefficient is -5, which makes
a 0.
Practice Exercises C
Find the vertex and determine whether it is a maximum or minimum point.
1.
f x 4 x 5 3
2.
f x x 3 7
3.
f x 6x2 5
4.
f x 2 x 6
5.
f x 5 x 2 3
6.
f x 7 x 1 2
2
2
2
2
2
NOTE: If a quadratic function is given in standard form, complete the
square to rewrite the equation in vertex form.
Example:
Find the vertex of f ( x) x 2 12 x 7 , then determine whether it is a maximum
or minimum point.
f ( x) x 2 12 x 7
f ( x) x 12 x
2
Collect variable terms together inside
parenthesis with constant term outside
the parenthesis.
7
2
2
2
12
12
f ( x) x 12 x 7
2
2
f ( x) x 2 12 x 6 7 6
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2
2
Page 62
b
Complete the square by adding
2
inside the parenthesis. Now subtract
2
2
b
outside the parenthesis to
2
maintain equality. In other words you
are really adding zero to the equation.
Secondary Mathematics 2
f ( x) x 2 12 x 36 7 36
Simplify
f ( x) x 6 29
Factor and combine like terms.
2
Vertex: (-6, -29)
h 6 and k 29
The vertex is a minimum.
The leading coefficient is 1, which
makes a 0
Example: Find the vertex of f ( x) 3x2 18x 2 , then determine whether it is a
maximum or minimum point.
f ( x) 3x2 18x 2
Collect variable terms together inside
parenthesis with constant term outside
the parenthesis.
Factor out the leading coefficient. In
this case 3.
2
f ( x) 3x 2 18x
2
f ( x) 3 x 2 6 x
2
2
2
2
6
6
f ( x) 3 x 6 x 2 3
2
2
f ( x ) 3 x 2 6 x 3 2 3 3
2
2
b
Complete the square by adding
2
inside the parenthesis. Notice that
everything in the parenthesis is
multiplied by 3 so we need to subtract
2
b
3 outside the parenthesis to
2
maintain equality. In other words you
are really adding zero to the equation.
f ( x) 3 x 2 6 x 9 2 27
Simplify
f ( x) 3 x 3 29
Factor and combine like terms.
2
Vertex: (-3, -29)
h 3 and k 29
The vertex is a minimum.
The leading coefficient is 3, which
makes a 0
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Page 63
Secondary Mathematics 2
Example: Find the vertex of f ( x) 4 x 2 8x 3 , then determine whether it is a
maximum or minimum point.
f ( x) 4 x 2 8x 3
f ( x) 4 x 2 8 x
3
3
f ( x) 4 x 2 2 x
Collect variable terms together inside
parenthesis with constant term outside
the parenthesis.
Factor out the leading coefficient. In
this case -4.
2
2
2
2
2
2
f ( x) 4 x 2 x 3 4
2
2
f ( x) 4 x 2 2 x 1 3 4 1
2
2
b
Complete the square by adding
2
inside the parenthesis. Notice that
everything in the parenthesis is
multiplied by -4 so we need to subtract
2
b
4 outside the parenthesis to
2
maintain equality. In other words you
are really adding zero to the equation.
f ( x) 4 x 2 2 x 1 3 4
Simplify
f ( x) 4 x 1 7
Factor and combine like terms.
2
Vertex: (-1, 7)
h 1 and k 7
The vertex is a maximum.
The leading coefficient is -4, which
makes a 0
Practice Exercises D
Find the vertex of each equation by completing the square. Determine if the vertex is a
maximum or minimum.
1.
f ( x) x 2 10 x 20
3.
f ( x) 5x 2 20 x 9
5.
f ( x) x2 8x 10
2.
f ( x) x2 24 x 1
4.
f ( x) 2 x 2 16 x 26
6.
f ( x) x 2 2 x 9
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Page 64
Secondary Mathematics 2
The axis of symmetry is the vertical line that
divides a parabola in half. The zeros will
always be the same distance from the axis of
symmetry.
The vertex always lies on the axis of
symmetry.
Axis of symmetry
When completing the square we end up with
Example:
f ( x) 3x 2 2 x 1
2
b
f ( x) a x
k
2a
h
2
b
f ( x) a x k
2a
f ( x) a x h k
2
12
12
2
2(3) 6
k f (2) 3(2)2 12(2) 1 11
b
.
2a
The y-coordinate can be found by evaluating
b
the function at
.
2a
Notice the x-coordinate of the vertex is
The point (2, 11) is the vertex. Since 3 0 ,
(2, 11) is the maximum point of the function.
Therefore, another method for finding the
vertex (h, k) from a standard form equation is
b
b
to use h
and k f .
2a
2a
Practice Exercises E
Identify the vertex of each function. Then tell if it is a maximum or minimum point.
1.
f x 4 x2 8x 7
2.
f x x2 12 x 30
3.
f x 2 x2 12 x 3
4.
f x x2 14 x 1
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Page 65
Secondary Mathematics 2
YOU DECIDE
A model rocket is launched from ground level. The function h(t ) 16t 2 160t models the
height h (measured in feet) of the rocket after time t (measured in seconds).
Find the zeros and the vertex of the function. Explain what each means in context of the
problem.
Practice Exercises F
Solve
1. The height h(t), in feet, of a “weeping willow” firework display, t seconds after having been
launched from an 80-ft high rooftop, is given by h t 16t 2 64t 80 . When will it reach
its maximum height? What is its maximum height?
2. The value of some stock can be represented by V x 2 x 2 8x 10 , where x is the number
of months after January 2012. What is the lowest value V(x) will reach, and when did that
occur?
3. Suppose that a flare is launched upward with an initial velocity of 80 ft/sec from a height of
224 ft. Its height in feet, h(t), after t seconds is given by h t 16t 2 80t 224 . How long
will it take the flare to reach the ground?
4. A company’s profit can be modeled by the equation p( x) x2 980 x 3000 where x is the
number of units sold. Find the maximum profit of the company.
5. The Rainbow Bridge Arch at Lake Powell is the world’s highest natural arch. The height of
an object that has been dropped from the top of the arch can be modeled by the equation
h(t ) 16t 2 256 , where t is the time in seconds and h is the height in feet. How long does
it take for the object to reach the ground?
6. The amount spent by U.S. companies for online advertising can be approximated by
1
a t t 2 2t 8 , where a(t) is in billions of dollars and t is the number of years after 2010.
2
In what year after 2010 did U.S. companies spend the least amount of money?
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Page 66
Secondary Mathematics 2
Unit 6 Cluster 3 (G.GPE.2): Parabolas as Conics
Cluster 3: Translating between descriptions and equations for a conic section
6.3.2 Find the equation of a parabola given the focus and directrix parallel to a
coordinate axis.
VOCABULARY
A parabola is the set of all points , P x, y , in a plane that are an equal distance from both a fixed
point, the focus, and a fixed line, the directrix.
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Page 67
Secondary Mathematics 2
Standard Form for the Equation of a Parabola
Vertex at (0, 0)
1 2
y
x
4p
Opens upward if p 0
Opens downward if p 0
Vertex at (h, k)
1
2
yk
x h
4p
Opens upward if p 0
Opens downward if p 0
Focus
(0, p)
(h, k + p)
Directrix
y p
ykp
Equation
Direction
Graph
Example 1:
Use the Distance Formula to find the equation of a parabola with focus 0,3 and
directrix y 3 .
A point P x, y on the graph of a parabola is
PF PD
( x x1 ) ( y y1 ) ( x x2 ) ( y y2 )
2
x 0
2
2
( y 3)2 ( x x)2 ( y 3)2
x 2 ( y 3)2
2
y 3
2
2
x ( y 3)
y 3
2
2
2
x ( y 3) ( y 3)
2
2
2
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the same distance from the focus F 0,3 and
2
a point on the directrix D x, 3 .
Substitute in known values.
Simplify.
2
Square both sides of the equation and use the
properties of exponents to simplify.
Page 68
Secondary Mathematics 2
x 2 ( y 3) 2 ( y 3) 2
Solve for y.
x2 y 2 6 y 9 y 2 6 y 9
x 12 y
1
y x2
12
2
Example:
Use the Distance Formula to find the equation of a parabola with focus (-5, 3) and
directrix y 9 .
A point P x, y on the graph of a parabola is
PF PD
( x x1 ) ( y y1 ) ( x x2 ) ( y y2 )
2
2
x 5
x 5
2
2
2
( y 3)2 ( x x)2 ( y 9) 2
y 9
( y 3)2
2
x 5 ( y 3)
2
2
x 5 12 y 72
2
x 5 12 y 6
2
a point on the directrix D x,9 .
Substitute in known values.
2
Simplify.
y 9
2
2
x 5 ( y 3)2 ( y 9)2
2
x 5 ( y 9)2 ( y 3) 2
2
x 5 y 2 18 y 81 y 2 6 y 9
2
the same distance from the focus F 5,3 and
2
Square both sides of the equation and use the
properties of exponents to simplify.
Combine the x terms on one side of the
equation and the y terms on the other side of
the equation.
1
2
x 5 y 6
12
Practice Exercises A
Use the distance formula to find the equation of parabola with the given information.
1. focus 0, 5
directrix y 5
4. focus 2, 6
directrix y 8
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2. focus 0, 7
directrix y 7
5. focus 3, 4
directrix y 1
Page 69
3. focus 0, 3
directrix y 6
6. focus 3,3
directrix y 7
Secondary Mathematics 2
Standard Form for the Equation of a Parabola
Vertex at (0, 0)
1 2
x
y
4p
Opens to the right if p 0
Opens to the left if p 0
Vertex at (h, k)
1
2
xh
y k
4p
Opens to the right if p 0
Opens to the left if p 0
Focus
p, 0
h p, k
Directrix
x p
x h p
Equation
Direction
Graph
Example:
Use the Distance Formula to find the equation of a parabola with focus 2, 0 and
directrix x 2 .
A point x, y on the graph of a parabola is the
PF PD
( x x1 ) ( y y1 ) ( x x2 ) ( y y2 )
2
2
2
x 2
2
( y 0)2 ( x 2) 2 ( y y) 2
x 2
2
y2
x 2
2
2
x 2 y 2
x 2
2
2
2
x 2
y 2 ( x 2)2
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same distance from the focus 2, 0 and a point
2
on the directrix 2, y .
Substitute in known values.
Simplify.
2
Square both sides of the equation and use the
properties of exponents to simplify.
Page 70
Secondary Mathematics 2
y2 x 2 x 2
2
Solve for x.
2
y 2 x2 4 x 4 x2 4 x 4
y2 8x
1 2
y x
8
Example:
Use the Distance Formula to find the equation of a parabola with focus 4,3 and
directrix x 6 .
A point x, y on the graph of a parabola is the
PF PD
( x x1 ) ( y y1 ) ( x x2 ) ( y y2 )
2
2
2
x 4
2
y 3
x 6
x 4
2
y 3
x 6
2
2
2
2
2
x 4 y 3
2
y y
2
2
2
y 3 4 x 20
2
y 3 4 x 5
2
on the directrix 6, y .
Substitute in known values.
Simplify.
2
x 6
2
x 4 y 3 x 6
2
2
2
y 3 x 6 x 4
2
y 3 x 2 12 x 36 x 2 8 x 16
2
same distance from the focus 4,3 and a point
2
2
Square both sides of the equation and use the
properties of exponents to simplify.
Combine the x terms on one side of the
equation and the y terms on the other side of
the equation.
1
2
y 3 x 5
4
Practice Exercises B
Use the distance formula to find the equation of parabola with the given information.
1. focus 4, 0
directrix x 4
4. focus 2, 3
directrix x 5
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2. focus 5, 0
directrix x 5
5. focus 2, 4
directrix x 6
Page 71
3. focus 3, 0
directrix x 3
6. focus 1,1
directrix x 5
Secondary Mathematics 2
Practice Exercises C
Determine the vertex, focus, directrix and the direction for each of the following parabolas.
1. 12 y 1 x 3
4. 6 y 3 x 1
2
2
2. x 4 6 y 1
2
5.
y 3
2
12 x 2
3.
y 1
2
4 x 5
6.
y 6
16 x 4
2
You Decide
A parabola has focus (-2,1) and directrix y = -3. Determine whether or not the point (2,1) is part
of the parabola. Justify your response.
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Page 72
Secondary Mathematics 2
Unit 6 Cluster 3 Honors (G.GPE.3)
Deriving Equations of Ellipses and Hyperbolas
Cluster 3: Translate between the geometric description and the equation for a conic section
H.5.1 Derive the equations of ellipses and hyperbolas given the foci, using the fact that
the sum or difference of distances from the foci is constant.
VOCABULARY
An ellipse is the set of all points in a plane the sum of whose distances from two fixed points
(called foci), F1 and F2 , is constant. The midpoint of the segments connecting the foci is the
center of the ellipse.
An ellipse can be elongated horizontally or vertically. The line through the foci intersects the
ellipse at its vertices. The segment whose endpoints are the vertices is called the major axis.
The minor axis is a segment that is perpendicular to the major axis and its endpoints intersect
the ellipse.
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Page 73
Secondary Mathematics 2
Deriving the Standard Equation of an Ellipse
The sum of the
distance from a point
P x, y on the
ellipse to each foci,
c,0 and c,0 , is
equal to 2a.
Use the distance
formula and
substitute in known
values.
Isolate one of the
radicals.
PF1 PF2 2a
x c y 0
2
x c y 0
2
x c y 0
2
2
2
2
x c y 0
2a
x c y 0
2
2
2a
2
x c y 0
2a
2
x c y 0
2
2
2
2
4a 2 4a
x c y 0
2
x c y 0
x 2 2 xc c 2 y 2 4a 2 4a
x c y 0
2
x 2 2 xc c 2 y 2
4 xc 4a 2 4a
x c y 0
2
2
2
2
2
2
4cx 4a 2 4a
x c y 0
2
4cx 4a 2
a
4
x c y 0
2
cx a 2 a
x c y 0
2
a 2 cx a
x c y 0
2
a
2
2
2
2
cx a
2
2
a 2a cx c x a
2
2
2
2
Isolate the radical
again.
x c y 0
2
x
2
2cx c y
2
Square each side
then simplify.
2
2
2
2
2
a 4 2a 2 cx c 2 x 2 a 2 x c y 0
4
2
2
Square each side
then simplify.
a 2a cx c x a x 2a cx a c a 2 y 2
4
2
2
2
2
2
2
2 2
a 4 c 2 x 2 a 2 x 2 a 2c 2 a 2 y 2
a 4 a 2c 2 a 2 x 2 c 2 x 2 a 2 y 2
a2 a2 c2 x2 a2 c2 a2 y 2
a 2b 2 x 2b 2 a 2 y 2
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Page 74
Combine all the
terms containing x
and y on one side.
Let b2 a 2 c2 .
Secondary Mathematics 2
a 2b 2 x 2b 2 a 2 y 2
a 2b 2
a 2b 2
x2 y 2
1 2 2
a
b
Divide by a 2b 2 .
Standard Form for the Equation of an Ellipse Centered at (0, 0)
Horizontal Ellipse
Vertical Ellipse
x2 y 2
2 1, a 2 b2
2
a
b
Along the x-axis
Length: 2a
Along the y-axis
Length: 2b
x2 y 2
2 1, a 2 b2
2
b
a
Along the y-axis
Length: 2a
Along the x-axis
Length: 2b
Foci
c,0 and c,0
0, c and 0, c
Vertices
a,0 and a,0
0, a and 0, a
a 2 b2 c 2
a 2 b2 c 2
Equation
Major Axis
Minor Axis
Pythagorean
Relation
Basic Graph
Example:
Locate the vertices and foci for the ellipse 25x2 4 y 2 100 . Graph the ellipse.
25x2 4 y 2 100
25 x 2 4 y 2 100
100 100 100
x2 y 2
1
4 25
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The standard equation of an ellipse is equal
to 1. Divide each side of the equation by
100 and simplify.
Page 75
Secondary Mathematics 2
Identify a and b. Remember a 2 b2 .
x2 y 2
1
4 25
a 25 5
Note: a and b are lengths therefore the
positive square root will ALWAYS be
used.
b 42
25 4 c 2
Use a and b to find c.
Remember a 2 b2 c2 .
21 c 2
21 c
Vertices: 0, 5 and 0,5
Foci: 0, 21 and 0, 21
The vertices are 0, a and 0, a and the
foci are 0, c and 0, c because the
ellipse is vertical.
Begin graphing the ellipse by plotting the
center which is at (0, 0). Then plot the
vertices which are at 0, 5 and 0,5 .
Use the length of b to plot the endpoints
of the minor axis. b 2 so the endpoints
are 2 units to the left and right of the
center (0, 0). They are at 2, 0 and
2, 0 .
Connect your points with a curve.
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Page 76
Secondary Mathematics 2
Example:
Write an equation in standard form for an ellipse with foci located at 2,0 and 2,0
and vertices located at 6,0 and 6,0 .
The ellipse is horizontal because the foci
and vertices are along the x-axis. Use the
standard equation for a horizontal ellipse.
a6
c2
Find b 2 using a 2 b2 c2 .
x2 y 2
1
a 2 b2
62 b 2 22
36 b 2 4
32 b 2
x2 y 2
1
36 32
Substitute in known values.
Practice Exercises A
Locate the vertices and foci of the ellipse, then graph.
1.
x2 y 2
1
16 7
4.
3x 2 4 y 2 12
2.
x2 y 2
1
21 25
3.
x2 y 2
1
27 36
5.
9 x2 4 y 2 36
6.
x2 1 4 y 2
Write an equation in standard form for the ellipse that satisfies the given conditions.
7.
Foci: 5,0 and 5,0
8.
Vertices: 0, 7 and 0,7
Vertices: 8,0 and 8,0
9.
Foci: 0, 3 and 0,3
10.
Vertices: 0, 4 and 0, 4
11.
Major axis endpoints: 0, 6
Minor axis length 8
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Foci: 0, 4 and 0, 4
Foci: 6,0 and 6,0
Vertices: 10,0 and 10,0
12.
Endpoints of axes are 5, 0 and
0, 4
Page 77
Secondary Mathematics 2
Ellipses Centered at (h, k)
Standard Form for the Equation of an Ellipse Centered at (h, k)
x h
Equation
a
2
2
y k
b
2
1, a b
2
2
2
x h
b
2
2
y k
a
2
2
1, a 2 b2
h, k
h, k
Parallel to the x-axis
Length: 2a
Parallel to the y-axis
Length: 2b
Parallel to the y-axis
Length: 2a
Parallel to the x-axis
Length: 2b
Foci
h c, k and h c, k
h, k c and h, k c
Vertices
h a, k and h a, k
h, k a and h, k a
a 2 b2 c 2
a 2 b2 c 2
Center
Major Axis
Minor Axis
Pythagorean
Relation
Example:
2
2
Locate the center, the vertices and the foci of the ellipse x 3 4 y 2 16 . Graph
the ellipse.
x 3 4 y 2 16
2
2
x 3 4 y 2 16
2
2
16
16
x 3
2
16
y 2
2
y 2
2
1
4
x 3
2
The standard equation of an ellipse is
equal to 1. Divide each side of the
equation by 16 and simplify.
16
1
16
4
a 16 4
Identify a and b. Remember a 2 b2 .
b 4 2
16 4 c 2
12 c 2
Use a and b to find c.
Remember a 2 b2 c2 .
12 c
2 3c
Center: 3, 2
Vertices:
3 4, 2 and 3 4, 2
7, 2 and 1, 2
Foci: 3 2 3, 2 and 3 2 3, 2
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h 3 and k 2
The ellipse is horizontal, therefore the
vertices are h a, k and h a, k and
Page 78
the foci are h c, k and h c, k .
Secondary Mathematics 2
Begin graphing the ellipse by plotting the
center of the ellipse 3, 2 . Then plot the
vertices 7, 2 and 1, 2 .
Use the length of b to plot the endpoints of
the minor axis. b 2 so the endpoints are
2 units above and below the center
3, 2 . They are at 3, 0 and 3, 4 .
Connect your points with a curve.
Example:
Write an equation in standard form for an ellipse with foci at 2,1 and 2,5 and
vertices at 2, 1 and 2,7 .
x h
2
b2
y k
The ellipse is vertical because the foci and
vertices are parallel to the y-axis. Use the
standard equation for a horizontal ellipse.
2a 7 (1) 2c 5 1
2
a2
1
42 b 2 22
2a 8
16 b 2 4
2c 4
12 b 2
a4
c2
2
Find b using a 2 b2 c2 .
2 2 1 7
Center:
,
2,3
2
2
The center is the midpoint of the vertices.
x 2
12
2
y 3
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16
2
1
Substitute in known values.
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Secondary Mathematics 2
Practice Exercises B
Locate the center, vertices and foci of the ellipse, then graph.
1.
x 2
9
2
y 1
2
4
1
4. x 3 9 y 2 18
2
2
2.
x 4
9
2
y 2
2
25
1
5. 9 x 1 4 y 3 36
2
2
3.
x 3
2
9
y 1
2
1
16
6. 2 x 4 4 y 3 24
2
2
Write an equation in standard form for the ellipse that satisfies the given conditions.
7.
Foci: 1, 4 and 5, 4
8.
Vertices: 0, 4 and 6, 4
9.
Foci: 4, 2 and 6, 2
Vertices: 3, 7 and 3,3
10.
Vertices: 5, 2 and 3, 2
Minor axis length is 6.
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Foci: 1,0 and 1, 4
Vertices: 1,1 and 1, 5
Vertices: 2, 2 and 8, 2
11.
Foci: 3, 6 and 3, 2
12.
Page 80
Vertices: 0, 2 and 6, 2
Minor axis length is 2.
Secondary Mathematics 2
VOCABULARY
A hyperbola is the set of all points in a plane whose distances from two fixed points in the plane
have a constant difference. The fixed points are the foci of the hyperbola.
The line through the foci intersects the hyperbola at its vertices. The segment connecting the
vertices is called the transverse axis. The center of the hyperbola is the midpoint of the
transverse axis. Hyperbolas have two oblique asymptotes that intersect at the center.
Deriving the Standard Equation of a Hyperbola
The difference of the
distance from a point
P x, y on the
hyperbola to each foci,
c,0 and c,0 , is
PF1 PF2 2a
equal to 2a .
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Secondary Mathematics 2
x c y 0
2
2
x c y 0
2
x c
2
y2
x c
2
y 2 2a
x c
x c
2
y2
2
2
x c
2a
2
x c
x c y 2 x c y 2 4a 2 4a
2
Use the distance formula
and substitute in known
values.
y2
2
2
2
x 2 2cx c 2 x 2 2cx c 2 4a 2 4a
x c
2
y2
4cx 4a 2 4a
x c
2
y2
x c
2
x c
2
y2
cx a 2 a
x c
2
y2
2
Square each side then
simplify.
y2
4cx 4a 2
a
4
cx a 2 a
2
y2 x c y2
x c y2
2
4cx 4a 2 4a
Isolate one of the
radicals.
y2
y 2 4a 2 4a
2
2a
y 2 2a
2
x c
x c
2
x c
Isolate the radical again.
2
y2
2
c 2 x 2 2a 2 cx a 4 a 2 x c y 2
2
c x 2a cx a a
2
2
2
4
2
x
2
2cx c y
2
2
Square each side then
simplify.
c 2 x 2 2a 2 cx a 4 a 2 x 2 2a 2cx a 2c 2 a 2 y 2
c 2 x 2 a 4 a 2 x 2 a 2c 2 a 2 y 2
c 2 x 2 a 2 x 2 a 2 y 2 a 2c 2 a 4
x2 c2 a2 a2 y 2 a2 c2 a2
Combine all the terms
containing x and y on
one side.
x 2b 2 a 2 y 2 a 2b 2
Let b2 c2 a 2 .
x 2b 2 a 2 y 2 a 2b 2
2 2
a 2b 2
ab
2
2
x
y
2 1
2
a
b
Divide by a 2b 2 .
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Secondary Mathematics 2
Standard Form for the Equation of a Hyperbola Centered at (0, 0)
Opens Left and Right
Opens Up and Down
x2 y 2
1
a 2 b2
x-axis
Length: 2a
y-axis
Length:2b
y 2 x2
1
a 2 b2
y-axis
Length: 2a
x-axis
Length:2b
Foci
c,0 and c,0
0, c and 0, c
Vertices
a,0 and a,0
0, a and 0, a
Pythagorean
Relation
c 2 a 2 b2
c 2 a 2 b2
Asymptotes
b
y x
a
a
y x
b
Equation
Transverse
Axis
Conjugate
Axis
Basic Graph
Example:
Find the vertices, foci and asymptotes of the hyperbola 4 x 2 9 y 2 36 . Then graph the
hyperbola.
4 x 2 9 y 2 36
4 x 2 9 y 2 36
36 36 36
x2 y 2
1
9
4
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The standard equation of an ellipse is
equal to 1. Divide each side of the
equation by 36 and simplify.
Page 83
Secondary Mathematics 2
x2 y 2
1
9
4
a 9 3
Identify a and b.
b 42
c2 9 4
Use a and b to find c.
Remember c2 a 2 b2 .
c 2 13
c 13
Vertices: 3,0 and 3,0
Foci: 13,0 and 13,0
2
2
Asymptotes: y x and y x
3
3
This hyperbola opens left and right so the
vertices are a, 0 and a, 0 and the foci
are c,0 and c,0 .
The asymptotes are y
b
x.
a
Begin graphing the hyperbola by plotting
the center at (0, 0). Then plot the vertices
at 3,0 and 3,0 .
Use the length of b to plot the endpoints of
the conjugate axis. b 2 so the endpoints
are 2 units above and below the center
0, 0 . They are at 0, 2 and 0, 2 .
Construct a rectangle using the points.
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Secondary Mathematics 2
Draw the asymptotes by drawing a line
that connects the diagonal corners of the
rectangle and the center.
Use the asymptotes to help you draw the
hyperbola. The hyperbola will open left
and right and pass through each vertex.
Example:
Write an equation in standard form for the hyperbola with foci 0, 3 and 0,3 whose
conjugate axis has length 4.
y 2 x2
1
a 2 b2
2b 4
The foci are along the y-axis so the
hyperbola’s branches open up and down.
The conjugate axis is length 4. Use it to
solve for b.
b2
32 a 2 22
Use b 2 and c 3 to solve for a 2 .
Remember c2 a 2 b2 .
9 a2 4
5 a2
y 2 x2
1
5
4
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Substitute in known values.
Page 85
Secondary Mathematics 2
Practice Exercises C
Locate the center, vertices, foci and asymptotes of the hyperbola, then graph.
1.
x2 y 2
1
4 16
2.
y 2 x2
1
25 36
5. 4 y 2 16 x2 64
4. 20 y 2 25x2 100
x2 y 2
1
1
9
3.
6. 2 x2 4 y 2 16
Write an equation in standard form for the hyperbola that satisfies the given conditions.
7.
Foci: 0, 2 and 0, 2
8.
Vertices: 3,0 and 3,0
Vertices: 0, 1 and 0,1
9.
Foci: 0, 7 and 0,7
10.
Foci: 10,0 and 10,0
Vertices: 6,0 and 6,0
Vertices: 0, 5 and 0,5
11.
Foci: 5,0 and 5,0
Vertices: 4,0 and 4,0
Conjugate axis length is 10.
12.
Vertices: 0, 3 and 0,3
Conjugate axis length is 6.
Standard Form for the Equation of a Hyperbola Centered at (h, k)
Equation
Opens Left and Right
Opens Up and Down
x h
y k
2
y k
2
1
a2
b2
Parallel to x-axis
Length: 2a
y-axis
Length:2b
Transverse
Axis
Conjugate
Axis
2
x h
2
1
a2
b2
Parallel to y-axis
Length: 2a
x-axis
Length:2b
Foci
h c, k and h c, k
h, k c and h, k c
Vertices
h a, k and h a, k
h, k a and h, k a
c 2 a 2 b2
c 2 a 2 b2
Pythagorean
Relation
Asymptotes
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yk
b
x h
a
yk
Page 86
a
x h
b
Secondary Mathematics 2
Example:
Find the center, vertices, foci and asymptotes of the hyperbola
x 2
9
2
y 5
49
2
1.
Then graph the hyperbola.
x 2
2
9
y 5
2
y 5
2
1
49
x 2
2
9
1
49
Identify a and b.
a 9 3
b 49 7
c 2 9 49
Use a and b to find c 2 . Remember that
c 2 a 2 b2 .
c 2 58
c 58
Center: 2,5
Vertices:
2 3,5 and 2 3,5
5,5 and 1,5
Foci: 2 58,5 and 2 58,5
7
Asymptotes: y 5 x 2 and
3
7
y 5 x 2
3
The hyperbola’s branches open left and
right so the vertices are h a, k and
h a, k .
h c, k .
The foci are h c, k and
Begin graphing the hyperbola by plotting
the center at (-2, 5). Then plot the
vertices at 5,5 and 1,5 .
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Secondary Mathematics 2
Use the length of b to plot the endpoints
of the conjugate axis. b 7 so the
endpoints are 7 units above and below the
center 2,5 . They are at 2, 2 and
2,12 .
Construct a rectangle using the points.
Draw the asymptotes by drawing a line
that connects the diagonal corners of the
rectangle and the center.
Use the asymptotes to help you draw the
hyperbola. The hyperbola will open left
and right and pass through each vertex.
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Secondary Mathematics 2
Example:
Write an equation in standard form for the hyperbola whose vertices are 2, 1 and
8, 1 and whose conjugate axis has length 8.
y k
2
a2
x h
2
The foci are parallel to the x-axis so the
hyperbola’s branches open left and right.
1
b2
2 8 1 1
Center:
,
3, 1
2
2
2a 8 (2)
The midpoint of the vertices is the center
of the hyperbola.
The vertices are at 2, 1 and 8, 1 .
Use the distance between them to find a.
2a 10
a5
2b 8
b4
x 3
The conjugate axis is length 8. Use it to
solve for b.
2
25
y 1
2
16
1
Substitute in known values.
Practice Exercises D
Locate the center, vertices, foci and asymptotes of the hyperbola, then graph.
1.
y 5
2
25
x 6
2
16
1
4. 4 y 2 x 6 16
2
2
2.
x 5
4
2
y2
1
36
3.
5. y 6 5 x 4 100
2
2
x 1
2
49
y 3
2
1
16
6. 7 x 4 4 y 2 28
2
2
Write an equation in standard form for the hyperbola that satisfies the given conditions.
7.
Foci: 1,9 and 1,1
8.
Vertices: 0, 5 and 6, 5
Vertices: 1,7 and 1,3
9.
Foci: 8, 4 and 4, 4
10.
Vertices: 3,6 and 3, 2
Minor axis length is 8.
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Foci: 3,5 and 3, 11
Vertices: 3,1 and 3, 7
Vertices: 7, 4 and 3, 4
11.
Foci: 2, 5 and 8, 5
12.
Page 89
Vertices: 7, 2 and 3, 2
Minor axis length is 6.
Secondary Mathematics 2
Unit 2 Cluster 3 (F.BF.1)
Building Functions That Model Relationships Between Two
Quantities
Cluster 3: Building functions that model relationships between two quantities
2.3.1 Focus on quadratics and exponentials to write a function that describes a
relationship between 2 quantities (2nd difference for quadratics)
2.3.1 Determine an explicit expression or steps for calculation from context.
2.3.1 Combine functions using arithmetic operations.
Vocabulary
A function is a relation for which each input has exactly one output. In an ordered pair the
first number is considered the input and the second number is considered the output. If any
input has more than one output, then the relation is not a function.
For example the set of ordered pairs {(1,2), (3,5),(8,11)} is a function because each input
value has an output value. The set {(1, 2) (1, 3), (6, 7)} does not represent a function because
the input 1 has two different outputs 2 and 3.
Linear Function- a function that can be written in the
form y mx b ,where m and b are constants. The graph
of a linear function is a line.
A linear function can be expressed in two different ways:
Linear notation: y mx b
Function notation: f x mx b
f x 2x 1
Linear functions can model arithmetic sequences, where
the domain is the set of positive integers, because there is
a common difference between each successive term. The
common difference can also be called the first difference.
Linear functions can model any pattern where the first
difference is the same number.
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1, 3, 5, 7, ...
+2 +2 +2 1st difference
Secondary Mathematics 2
Exponential Function- a function of the form f ( x) ab x
where a and b are constants and a 0, b 0 , and b 1.
Exponential functions are most easily recognized by the
variable in the exponent. The values of f(x) are either
increasing (exponential growth) if a 0 and b 1 or
decreasing (exponential decay) if a 0 and 0 b 1 .
f x 2x
Exponential functions can model geometric sequences,
where the domain is the set of positive integers, because
each successive term is multiplied by the same number
called the common ratio. Exponential functions can
model any pattern where the next term is obtained by
multiplying each successive term by the same number.
1, 3, 9, 27, ...
3 3 3 common ratio
Quadratic Function- a function that can be written in the
form f ( x) ax2 bx c where a 0 .
Quadratic functions are most easily recognized by the x 2
term. The graph is a parabola. A quadratic function can
be formed by multiplying two linear functions. The
quadratic function to the right can also be written as
f x ( x 3)(2 x 1) .
f x 2 x2 5x 3
To determine if a pattern or a sequence can be modeled by
a quadratic function, you have to look at the first and
second difference. The second difference is the difference
between the numbers in the first difference. If the first
difference is not the same number but the second
difference is, then the pattern or sequence can be modeled
by a quadratic function.
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1, 4, 9, 16, ...
+3 +5 +7 1st difference
+2 +2 2nd difference
Secondary Mathematics 2
Example:
Determine if the pattern 1, 3, 9, 19, … would be modeled by a linear function, an
exponential function, or a quadratic function.
Answer:
Check the first difference to see if it is the same
number each time. For this pattern, it is not the
same, so it will not be modeled by a linear
function.
1, 3, 9, 19, ...
+2 +6 +10
Check to see if each term is being multiplied by
the same factor. For this pattern, it is not the
same, so it will not be modeled by an exponential
function.
1, 3, 9, 19, ...
3 3 2.1
Check the second difference to see if it is the
same number each time. For this pattern, it is the
same, so the pattern can be modeled by a
quadratic function.
1, 3, 9, 19, ...
+2 +6 +10
+4 +4
Conclusion: The pattern can be modeled by a quadratic function.
Practice Exercises A
Determine if the pattern would be modeled by a linear function, an exponential function, or a
quadratic function.
1.
2.
1
2
3
1
4
2
4. 10, 18, 28, …
3.
1
2
3
5. 81, 27, 9, …
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3
4
6. 8, 16, 24, …
Page 92
Secondary Mathematics 2
Example:
Using a graphing calculator determine the quadratic function modeled by the given
data
x
f(x)
1
1
2
9
3
23
4
43
5
69
6
101
Input the data into a TI-83 or TI-84 calculator list
Enter the information into your lists by pushing STAT
followed by Edit.
If you have values in your lists already, you can clear the
information by highlighting the name of the list then
pushing CLEAR and ENTER. Do not push DEL or it will
delete the entire list.
Enter the x values into L1 and the f(x) values into L2.
Push 2nd MODE to get back to the home screen.
Make a scatter plot
Push 2nd Y= to bring up the STAT PLOT menu.
Select Plot1 by pushing ENTER or 1.
Turn Plot1 on by pushing ENTER when ON is highlighted.
Make sure that the scatter plot option is highlighted. If it
isn’t, select it by pushing ENTER when the scatter plot
graphic is highlighted.
The Xlist should say L1 and the Ylist should say L2. If it
doesn’t, L1 can be entered by pushing 2nd 1 and L2 by 2nd
2.
To view the graph you can push GRAPH. If you want a
nice viewing window, first push ZOOM arrow down to
option 9 ZOOMSTAT and either push ENTER or push 9.
Creating a quadratic regression equation
You do not have to graph a function to create a regression,
but it is recommended that you compare your regression to
the data points to determine visually if it is a good model or
not.
From the home screen push STAT, arrow right to CALC
and either push 5 for QuadReg or arrow down to 5 and push
ENTER. (To do an exponential regression, push 0 for
ExpReg or arrow down to 0 and push ENTER.)
Type 2nd 1, (the comma is located above 7) 2nd 2, VARS
arrow right to Y-VARS select FUNCTION and Y1 then
push ENTER.
The quadratic regression is f x 3x 2 x 1 . It has been
pasted into Y1 so that you can push GRAPH again and
compare your regression to the data.
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Secondary Mathematics 2
Practice Exercises B
Find the regression equation. Round to three decimals when necessary.
1.
Given the table of values use a graphing calculator to find the quadratic function.
x
f(x)
2.
1
3
2
-40
3
-57
4
-66
5
-61
2
1
3
1
4
3
5
7
6
13
From 1972 to 1998 the U.S. Fish and Wildlife Service has kept a list of endangered
species in the United States. The table below shows the number of endangered
species. Find an appropriate exponential equation to model the data.
Year
Number of
species
4.
1
-21
Use a graphing calculator to find a quadratic model for the data.
x
f(x)
3.
0
-6
1972
1975
1978
1981
1984
1987
1990
1993
1996
119.6
157.5
207.3
273
359.4
473.3
623.1
820.5 1080.3
The cell phone subscribers of the small town of Herriman are shown below. Find an
exponential equation to model the data.
Year
Subscribers
1990
285
1995
802
2000
2,259
2005
6,360
2010
17,904
Example:
When doctors prescribe medicine, they must consider how much the effectiveness of
the drug will decrease as time passes. The table below provides information on how
much of the drug remains in a person’s system after t hours. Find a model for the
data.
t (hours)
Amount (mg)
0
250
2
4
6
8
10
225.6 203.6 183.8 165.9 149.7
Answer:
Sometimes it is helpful to look at the graph of the points.
For this particular example, it is difficult to determine if this
should be modeled by an exponential or a quadratic function
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Secondary Mathematics 2
from the graph. Therefore, consider the context of the
example. The amount of the drug will continue to decrease
unless more is given to the patient. If the patient does not
receive more medication, at some point there will only be trace
amounts of the drug left in the patient’s system. This would
suggest a function that continues to decrease until it reaches a
leveling off point. An exponential model would be better suited for this situation.
Use the regression capabilities of your graphing calculator to find an exponential
model for the data. Follow the instructions for the previous example but make sure
that you select option 0: ExpReg. The function that models the data is:
x
f x 249.977 0.950 .
Practice Exercises C
Determine if the data is best modeled by an exponential or quadratic function. Then find the
appropriate regression equation.
1.
The pesticide DDT was widely used in the United States until its ban in 1972. DDT
is toxic to a wide range of animals and aquatic life, and is suspected to cause cancer
in humans. The half-life of DDT can be 15 or more years. Half-life is the amount of
time it takes for half of the amount of a substance to decay. Scientists and
environmentalists worry about such substances because these hazardous materials
continue to be dangerous for many years after their disposal. Write an equation to
model the data below.
Year
Amount of DDT (in grams)
2.
1972
50
1982
9.8
1992
1.9
2012
0.4
Use a graphing calculator to find a model for the data.
x
f(x)
1
0
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2
-7
3
-4
4
21
5
80
6
185
Page 95
Secondary Mathematics 2
3.
The table shows the average movie ticket price in dollars for various years from 1983
to 2003. Find the model for the data.
Years since 1983, t
Movie ticket price, m
4.
0
4.75
4
4.07
8
3.65
12
4.10
16
5.08
20
6.03
The table below shows the value of car each year after it was purchased. Find a
model for the data.
Years after purchase
Value of car
0
24,000
1
20,160
2
16,934
3
14,225
4
11,949
5
10,037
Combining functions using arithmetic operations
Let f and g be any two functions. A new function h can be created by performing any of the
four basic operations on f and g.
Operation
Definition
Example: f ( x) 5x 2 2 x , g ( x) 3x 2
Addition
h x f x g x
h( x) 5x2 2 x (3x2 ) 2 x 2 2 x
Subtraction
h x f x g x
h( x) 5x2 2 x (3x 2 ) 8x 2 2 x
Multiplication
h x f x g x
h( x) (5x2 2 x) (3x 2 ) 15x 4 6 x3
Division
h( x )
f ( x)
g ( x)
h( x )
5x2 2 x x 5x 2 5x 2
3x 2
x 3x
3x
The domain of h consists of the x-values that are in the domains of both f and g. Additionally,
the domain of a quotient does not include x-values for which g(x) = 0.
Adding and Subtracting Functions
Example:
Let f x 2 x 1 and g x x 2 3x 4 . Perform the indicated operation and state
the domain of the new function.
a. h x f x g x
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b. h x g x f x
Page 96
c. h x 2 f x g x
Secondary Mathematics 2
Answer:
a.
h x f x g x
h x 2 x 1 x 2 3x 4
Replace f x with 2 x 1 and g x
with x2 3x 4 .
Remove the parentheses because we
can add in any order.
Combine the like terms. (2x + 3x),
(1 + – 4)
The domain for both f x and g x
h x 2 x 1 x 2 3x 4
h x x 2 5x 3
The domain is , .
is , . h x a quadratic function
just like g x so it has the same
domain.
b.
h x g x f x
h x x 2 3x 4 2 x 1
Replace g x with x2 3x 4 and
h x x2 x 5
f x with 2 x 1 .
Distribute the negative throughout the
second term and remove the
parentheses.
Combine the like terms.
The domain is , .
The domain for both f x and g x
h x x 2 3x 4 2 x 1
is , . h x a quadratic function,
just like g x , so it has the same
domain.
c.
h x 2 f x g x
h x 2 2 x 1 x 2 3x 4
Replace f x with 2 x 1 and g x
h x x2 x 6
with x2 3x 4 .
Distribute the two through the first
term and distribute the negative
through the second term.
Combine like terms.
The domain is , .
The domain for both f x and g x
h x 4 x 2 x 2 3x 4
is , . h x a quadratic function
just like g x so it has the same
domain.
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Secondary Mathematics 2
Multiplying Functions
Example:
Let f x 2 x 1 and g x x 2 3x 4 . Perform the indicated operation and state
the domain of the new function.
a. h x f x g x
Answer:
a.
b. h x g x g x
C. h x f x f x
h x f x g x
h x 2 x 1 x 2 3x 4
Replace f x with 2 x 1 and g x
with x2 3x 4
Multiply using your method of choice.
(See Unit 1 Cluster 4 lesson)
The domain for both f x and g x
h x 2 x3 7 x 2 5 x 4
The domain is , .
is , . h x a polynomial
function just like g x so it has the
same domain.
b.
h x g x g x
h x x 2 3x 4 x 2 3x 4
Replace g x with x2 3x 4
h x x 4 6 x3 x 2 24 x 16
Multiply using your method of
choice. (See Unit 1 Cluster 4 lesson)
The domain for g x is , .
The domain is , .
h x a polynomial function just like
g x so it has the same domain.
c.
h x f x f x
Replace f x with 2 x 1 .
Multiply using your method of choice.
(See Unit 1 Cluster 4 lesson)
The domain for f x is , .
h x 2 x 1 2 x 1
h x 4x 4x 1
2
The domain is , .
h x a quadratic function so it has the
same domain.
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Secondary Mathematics 2
Dividing Functions
VOCABULARY
A rational function is a function of the form r x
p x
where p x and q x are
q x
polynomials and q x 0 . The domain of a rational function includes all real numbers
except for those that would make q x 0 .
A rational expression is in simplified form if the numerator and the denominator have no
common factors other than 1 or -1.
Example:
Let f x 2 x 1 and g x x 2 3x 4 . Perform the indicated operation and state
the domain of the new function.
2 f x
g x
f x
a. h x
b. h x
C. h x
g x
f x
f x
Answer:
f x
a.
h x
g x
h x
2x 1
x 3x 4
h x
2x 1
( x 1)( x 4)
Replace f x with 2 x 1 and g x
2
with x2 3x 4
Factor the numerator and the
denominator to see if the function can
be simplified. (See Unit 2 Cluster 2
(F.IF.8) for help with factoring)
The new function h x is a rational
function. The domain cannot include
any numbers for which the denominator
is zero. The denominator is zero when
x 4 and x 1.
The domain is
, 4 4,1 1, .
“and”
b.
h x
g x
f x
h x
x 2 3x 4
2x 1
h x
( x 1)( x 4)
2x 1
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Replace f x with 2 x 1 and g x
with x2 3x 4
Factor the numerator and the
denominator to see if the function can
be simplified. (See Unit 2 Cluster 2
Page 99
Secondary Mathematics 2
(F.IF.8) for help with factoring)
The new function h x is a rational
function. The domain cannot include
any numbers for which the denominator
is zero. The denominator is zero when
1
x .
2
The domain is
1 1
, , .
2 2
c.
h x
2 f x
f x
2 2 x 1
2x 1
2(2 x 1)
h x
2x 1
Replace f x with 2 x 1 .
h x
Factor the numerator and the
denominator to see if the function can
be simplified. (See Unit 2 Cluster 2
(F.IF.8) for help with factoring)
Divide out the factors and simplify the
expression.
Although the simplified form of h x
is not a rational function, it started out
as a rational function and the same
restrictions apply on the simplified
form. The denominator is zero when
1
x .
2
h x 2
The domain is
1 1
, , .
2 2
Practice Exercises E
If f x 4 x 3 , g x 3x 2 and h x 12 x 2 x 6 , find the following. State the
domain of the new function.
1. f x h x
2. f x 2 g x
3. h x 3g x
4. g x 3 f x 5
5. g x h x
6. h x 4 g x 2
7. g x h x
8. f x h x
9. h x h x
10.
f x
h x
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11.
g x
h x
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12.
f x
g x
Secondary Mathematics 2
Evaluating Combined Functions
Example:
Let f x x 5 and g x 2 x 2 8x 5 . Evaluate each expression.
b. f 3 g 0
a. f 2 g 1
c.
g 1
f 1
Answer:
a.
This expression tells you to find the value of f
at x = 2 and the value of g at x = 1 and add the
results.
f 2 g 1
f 2 2 5
Find the value of f at x = 2.
f 2 7
g 1 2(1) 2 8(1) 5
g 1 2 1 8 5
g 1 2 8 5
g 1 15
7 15 22
f 2 g 1 22
b.
f 3 g 0
f 3 3 5
f 3 2
Find the value of g at x = 1.
Add the results.
This expression tells you to find the value of f
at x 3 and the value of g at x = 0 and
multiply the results.
Find the value of f at x 3 .
g 0 2 0 8 0 5
2
g 0 2 0 0 5
g 0 0 0 5
g 0 5
2 5 10
f 3 g 0 10
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Find the value of g at x = 0.
Multiply the results.
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Secondary Mathematics 2
c.
This expression tells you to find the value of g
at x 1 and the value of f at x 1 and
divide the results.
g 1
f 1
g 1 2 1 8 1 5
2
g 1 2 1 8 5
Find the value of g at x 1 .
g 1 2 8 5
g 1 1
f 1 1 5
Find the value of f at x 1 .
f 1 4
1
4
g 1
1
or -0.25
f 1
4
Divide the results.
Practice Exercises F
If f x 3x 1 and g x 3x 2 5x 2 , find the value of each expression.
1. 5 f 2
4. 2 f 1 g 5
7. f 2 g 4
2. g 1 f 3
5. f 2 g 4
8. f 1 3g 1
3.
f 3
f 3
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6.
g 1
f 4
9.
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g 0
2 f 0
Secondary Mathematics 2
Practice Exercises G
1. A company estimates that its cost and revenue can be modeled by the functions
C x 0.75x 2 100 x 20,000 and R x 150 x 100 where x is the number of units
produced. The company’s profit, P, is modeled by R x C x . Find the profit equation
and determine the profit when 1,000,000 units are produced.
1
x 30; 0 x 450 where p represents the price
15
and x the number of units sold. Write an equation for the revenue, R, if the revenue is the
price times the number of units sold. What price should the company charge to have
maximum revenue?
2. Consider the demand equation p x
3. The average Cost C of manufacturing x computers per day is obtained by dividing the cost
function by the number of computers produced that day, x. If the cost function is
C x 0.5x3 34 x 2 1213x , find an equation for the average cost of manufacturing. What
is the average cost of producing 100 computers per day?
4. The service committee wants to organize a fund-raising dinner. The cost of renting a facility
is $300 plus $5 per chair or C x 5x 300 , where x represents the number of people
attending the fund-raiser. The committee wants to charge attendees $30 each or R x 30 x .
How many people need to attend the fund-raiser for the event to raise $1,000?
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Secondary Mathematics 2
Unit 2 Cluster 4 (F.BF.3 and F.BF.4): Transformations and Inverses
Cluster 4: Building New Functions from Existing Functions
2.4.1 Transformations, odd and even graphically and algebraically
2.4.2 Find inverse functions (simple) focus on linear and basic restrictions for
quadratics, introduce one-to-one and horizontal line test
VOCABULARY
There are several types of functions (linear, exponential, quadratic, absolute value, etc.). Each of
these could be considered a family. Each family has their own unique characteristics that are
shared among the members. The parent function is the basic function that is used to create
more complicated functions.
The graph of a quadratic function is in the shape of a parabola. This is generally described as
being “u” shaped.
The maximum or minimum point of a quadratic function is the vertex. When a quadratic
2
function is written in vertex form, f ( x) a x h k , then the vertex, (h, k ) , is highlighted.
The axis of symmetry is the vertical line that divides the graph in half, with each half being a
reflection of the other. The equation for the axis of symmetry is x h .
Quadratic parent function f x x 2
x
f x x2
-3
(3)2 9
-2
(2)2 4
-1
(1)2 1
0
(0)2 0
1
(1)2 1
2
(2)2 4
3
(3)2 9
The axis of symmetry is the line x 0 . The vertex is the point (0, 0). The domain is the set of
all real numbers , . The range is the set of positive real numbers including zero 0, .
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Secondary Mathematics 2
Vertical Shift: f x x 2 2
x
f x x2 2
-3
(3)2 2 7
Axis of symmetry: x 0
-2
(2)2 2 2
Vertex: (0, -2)
-1
(1) 2 1
0
(0)2 2 2
1
(1)2 2 1
2
(2)2 2 2
2
Domain: ,
Range: 2,
3
(3)2 2 7
Effect on the graph: The parabola has been shifted down 2 units.
Vertical Shift: f x x 2 1
x
f x x2 1
-3
(3)2 1 10
Axis of symmetry: x 0
-2
(2)2 1 5
Vertex: (0, 1)
-1
(1) 1 2
0
(0)2 1 1
1
(1)2 1 2
2
(2)2 1 5
2
Domain: ,
Range: 1,
3
(3)2 1 10
Effect on the graph: The parabola has been shifted up 1 unit.
Horizontal Shift: f x x 2
x
f x x 2
-1
(1 2)2 9
0
(0 2)2 4
1
(1 2) 1
2
(2 2)2 0
3
(3 2)2 1
4
(4 2)2 4
2
2
Axis of symmetry: x 2
Vertex: (2, 0)
2
Domain: ,
Range: 0,
5
(5 2)2 9
Effect on the graph: the parabola has been shifted 2 units to the right.
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Secondary Mathematics 2
Horizontal Shift: f x x 3
x
f x x 3
-6
(6 3)2 9
-5
(5 3)2 4
-4
(4 3) 1
-3
(3 3)2 0
-2
(2 3)2 1
-1
(1 3)2 4
2
2
Axis of symmetry: x 3
Vertex: (-3, 0)
2
Domain: ,
Range: 0,
0
(0 3)2 9
Effect on the graph: the parabola has been shifted three units to the left.
Reflection: f x x 2
x
f x x2
-3
(3)2 9
Axis of symmetry: x 0
-2
(2)2 4
Vertex: (0, 0)
-1
(1) 1
0
(0)2 0
1
(1)2 1
2
(2)2 4
2
Domain: ,
Range: , 0
3
(3)2 9
Effect on the graph: the parabola has been reflected over the x-axis.
Vertical Stretch: f x 2 x 2
x
f x 2 x2
-3
2(3)2 18
Axis of symmetry: x 0
-2
2(2)2 8
Vertex: (0, 0)
-1
2(1) 2
0
2(0)2 0
1
2(1)2 2
2
2(2)2 4
2
Domain: ,
Range: 0,
3
2(3)2 18
Effect on the graph: the y-coordinates of the parabola have been multiplied by 2.
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Secondary Mathematics 2
Vertical Shrink: f x 12 x 2
f x 12 x 2
x
-3
1
2
(3)2
-2
1
2
(2)2 2
-1
1
2
(1)
Axis of symmetry: x 0
9
2
2
Vertex: (0, 0)
1
2
0
1
2
(0)2 0
1
1
2
(1)2 12
2
1
2
(2)2 4
Domain: ,
Range: 0,
2
9
1
3
2 (3) 2
Effect on the graph: the y-coordinates of the parabola have been divided 2.
Horizontal Shift
f x a ( x h) k
2
Vertical Stretch
or Reflection
Vertical Shift
Example:
Describe the transformations performed on f x x 2 to make it f x ( x 1)2 5 .
Then graph the function and identify the axis of symmetry, the vertex, the domain and the
range.
Transformations:
Axis of symmetry: x 1
reflected over the x-axis
Vertex: (1, 5)
shifted one unit to the right
Domain: ,
shifted up three units
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Range: ,5
Page 107
Secondary Mathematics 2
Example:
3
Describe the transformations performed on f x x 2 to make it f x ( x 2)2 4 .
2
Then graph the function and identify the axis of symmetry, the vertex, the domain and the
range.
Transformations:
Axis of symmetry: x 2
y-coordinates multiplied by
3/2
Vertex: (-2, -4)
shifted two units to the left
Domain: ,
shifted down four units
Range: 4,
Practice Exercises A
Describe the transformations performed on f x x 2 to make it the following:
1. f x x 2 6
2. f x x 5 7
2
3. f x 3 x 4
2
Graph each function and identify the axis of symmetry, the vertex, the domain and the range.
2
2
6. f x 12 x 2 2
4. f x x 2 6
5. f x 2 x 1 3
7. f x 5 x 6 4
2
8. f x 3x 2 4
9. f x
3
2
x 3
2
VOCABULARY
The absolute value function is actually a piecewise-defined function consisting of two linear
equations.
x, if x 0
f x x
x, if x 0
Absolute value is often defined as the distance from zero. Therefore, the output is positive.
The point where the two linear equations meet is called the vertex. It is also the minimum or
maximum of the function. The vertex, h, k , can easily be identified when the absolute value
function is represented in the form f x a x h k .
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Secondary Mathematics 2
Absolute Value parent function f x x
x
f x x
-3
-2
-1
0
1
2
3
| 3 | 3
| 2 | 2
| 1| 1
| 0 | 0
|1| 1
| 2 | 2
| 3 | 3
The vertex is the point (0, 0). The domain is the set of all real numbers , . The range is
the set of positive real numbers including zero 0, .
Vertical Shift: f x x 4
x
f x x 4
-3
-2
-1
0
1
2
3
| 3| 4 1
| 2 | 4 2
| 1| 4 3
| 0 | 4 4
|1| 4 3
| 2 | 4 2
| 3| 4 1
Vertex: (0, -4)
Domain: ,
Range: 4,
Effect on the graph: The function has been shifted down 4 units.
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Secondary Mathematics 2
Horizontal Shift: f x x 1
x
f x x 1
-2
-1
0
1
2
3
4
| 2 1| 3
| 1 1| 2
| 0 1| 1
|1 1| 0
| 2 1| 1
| 3 1| 2
| 4 1| 3
Vertex: (1, 0)
Domain: ,
Range: 0,
Effect on the graph: The function has been shifted right 1 unit.
Reflection: f x x
x
f x x
-3
-2
-1
0
1
2
3
| 3| 3
| 2 | 2
| 1| 1
| 0 | 0
|1| 1
| 2 | 2
| 3| 3
Vertex: (0, 0)
Domain: ,
Range: , 0
Effect on the graph: The function has been reflected over the x-axis.
Vertical Stretch: f x 3 x
x
f x 3 x
-3
-2
-1
0
1
2
3
3| 3| 9
3| 2 | 6
3 | 1| 3
3| 0 | 0
3 |1| 3
3| 2 | 6
3 | 3 | 9
Vertex: (0, 0)
Domain: ,
Range: 0,
Effect on the graph: The y-coordinates of the function have been multiplied by 3.
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Secondary Mathematics 2
Horizontal Shift
f x a x h k
Vertical Stretch
or Reflection
Vertical Shift
Example:
Describe the transformations performed on f x x to make it f x 2 x 3 5 .
Then graph the function and identify the axis of symmetry, the vertex, the domain and the
range.
Transformations:
reflected over the x-axis
y-coordinates multiplied
by 2
shifted three units to the
right
shifted up five units
Vertex: (3, 5)
Domain: ,
Range: ,5
Practice Exercises B
Describe the transformations performed on f x x to make it the following:
1. f x 2 x 5
2. f x x 3 4
3. f x 3 x 2 5
Graph each function and identify the vertex, the domain and the range.
4. f x x 6
5. f x x 2 4
6. f x 12 x 1
7. f x 2 x 5
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8. f x x 3 1
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9. f x
3
2
x4
Secondary Mathematics 2
Vocabulary
The words even and odd describe the symmetry that exists for the graph of a function.
A function is considered to be even if, for every number x in its domain, the number –x is also in the
domain and f x f x . Even functions have y-axis symmetry.
A function is considered to be odd if, for every number x in its domain, the number –x is also in the
domain and f x f x . Odd functions have origin symmetry.
Even Function
The function graphed at the left is even because (2, 1) is a
point on the graph and (-2, 1) is also a point on the graph.
Notice that -2 is the opposite of 2, but both inputs give the
same output. Therefore, f x f x , i.e. opposite inputs
generate the same output.
Odd Function
The function graphed at the left is odd because (2, 2) is a point
on the graph and (-2, -2) is also a point on the graph. Notice
that the input -2 is the opposite of 2, and gives the opposite
output from 2. Therefore, f x f x , i.e. opposite
inputs generate outputs that are opposites of each other.
Neither Even nor Odd
The function graphed at the left is neither even nor odd. It is
not even because the point (4, 2) is on the graph, but (-4, 2) is
not. Similarly, it is not odd because the point (-4, -2) is not a
point on the graph.
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Secondary Mathematics 2
Practice Exercises C
Determine if the following graphs represent functions that are even, odd or neither.
1.
2.
3.
4.
5.
6.
Determining Even and Odd Algebraically
It is possible to graph functions and visually determine whether the function is even or odd, but
there is also an algebraic test that can be applied. It was previously stated that if a function is
even, then evaluating the function at x and –x should produce the same output or f x f x .
If a function is odd, then evaluating the function at x and –x should produce outputs that are
opposite or f x f x .
Example:
Is the function f x 3x an even function, an odd function, or neither?
Original function.
f x 3x
Substitute –x in for each x in the function.
f x 3 x
Simplify.
f x 3x
Compare the output to the original function.
3x 3x
If they are the same, then the function is even.
If they are opposite, then the function is odd.
If they are anything else, then they are neither.
f x f x
f x f x
Conclusion: f x 3x is an odd function.
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Secondary Mathematics 2
Example:
Is the function g x x 2 4 x 4 an even function, an odd function or neither?
Original function.
g x x2 4x 4
Substitute –x in for each x in the function.
g x x 4 x 4
Simplify.
g x x2 4x 4
Compare the output to the original function.
x2 4 x 4 x2 4 x 4
If they are the same, then the function is even.
If they are opposite, then the function is odd.
If they are anything else, then they are neither.
g x g x
2
g x g x
Conclusion: g x x 2 4 x 4 is neither an even nor an odd function.
Example:
Is the function h x 2 x 4 an even function, an odd function, or neither?
Original function.
h x 2 x 4
Substitute –x in for each x in the function.
h x 2 x 4
Simplify.
h x 2 x 4
Compare the output to the original function.
2 x 4 2 x 4
If they are the same, then the function is even.
If they are opposite, then the function is odd.
If they are anything else, then they are neither.
h x h x
h x h x
Conclusion: h x 2 x 4 is an even function.
Practice Exercises D
Determine algebraically if the function is even, odd, or neither.
1
1. f x x 3 1
2. f x x
3
4. f x 4 x 5
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5. f x x 4
Page 114
3. f x 3x 2
6. f x
5
2
x 1
4
Secondary Mathematics 2
VOCABULARY
The graph of an inverse relation is the reflection of the graph of the original relation. The line
of reflection is y = x.
The original relation is the set of ordered pairs: {(-2, 1), (-1, 2), (0, 0), (1, -2), (2, -4)}. The
inverse relation is the set of ordered pairs: {(1, -2), (2, -1), (0, 0), (-2, 1), (-4, 2)}. Notice that for
the inverse relation the domain (x) and the range (y) reverse positions.
Original Relation
Domain: {-2, -1, 0, 1, 2}
Range: {-4, -2, 0, 1, 2}
The points are reflected over
the line y x . Notice that
each point is the same distance
away from the line, but on the
opposite side of the line.
Inverse Relation
Domain: {-4, -2, 0, 1, 2}
Range: {-2, -1, 0, 1, 2}
Practice Exercises E
Find the inverse relation.
1. {(1, -1), (2, -2), (3, -3), (4, -4), (5, -5)}
3. {(-10, 6), (3, -9), (-1, 4), (-7, 1), (6, 8)}
2. {(-4, 2), (-2, 1), (0, 0), (2, 1), (4, 2)}
4. {(7, 6), (2, 9), (-3, -2), (-7, 1), (8, 10)}
VOCABULARY
If no vertical line intersects the graph of a function f more than once, then f is a function. This is
called the vertical line test.
If no horizontal line intersects the graph of a function f more than once, then the inverse of f is
itself a function. This is called the horizontal line test.
The inverse of a function is formed when the independent variable is exchanged with the
dependent variable in a given relation. (Switch the x and y with each other.) A function takes a
starting value, performs some operation on this value, and creates an output answer. The inverse
of a function takes the output answer, performs some operation on it, and arrives back at the
original function's starting value. Inverses are indicated by the notation f 1 .
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Secondary Mathematics 2
This example is not one-to-one. It is a function
because the vertical line intersects the graph
only once. However, the horizontal line
intersects the graph twice. There is an inverse
to this example, but the inverse will not be a
function.
This example is one-to-one. It is a function
because the vertical and horizontal lines
intersect the graph only once. The inverse will
be a function.
Example:
Find the inverse of f x 3x 1 .
Original function
f x 3x 1
Replace f ( x) with y.
y 3x 1
Replace x with y and y with x.
x 3 y 1
Isolate y.
x 1 3y
x 1
y
3
1
1
x y
3
3
1
1
The inverse of f x 3x 1 is f 1 x x .
3
3
Example:
Find the inverse of f x
1
x 2.
4
Original function
f x
Replace f ( x) with y.
y
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1
x2
4
1
x2
4
Secondary Mathematics 2
Replace x with y and y with x.
x
Isolate y.
1
y
4
4 x 2 y
1
y2
4
x2
4x 8 y
The inverse of f x
1
x 2 is f 1 x 4 x 8 .
4
Example:
Find the inverse of the function f x
Original function
Replace f ( x) with y.
1 2
x ; Domain , 0 and Range 0,
4
1
f x x2 , x 0
4
1
y x2
4
Replace x with y and y with x.
x
Isolate y.
4x y2
1 2
y
4
4x y
Simplify the radical
(Unit 1 Cluster1:N.RN.2)
2 x y
Determine whether to use the positive or
negative answer by referring back to the
restricted domain. The domain of the
original function is restricted to the
negative real numbers including zero,
therefore, the range of the inverse function
must also be the same. This leads us to
choose the negative square root.
2 x f 1 ( x)
Domain 0, and Range , 0
Notice the domain and range have
switched from the original function’s
domain and range.
Example:
2
Find the inverse of the function f x 3 x 1 5 ; Domain 0, and Range 5,
Original function
f x 3 x 1 5, x 0
Replace f ( x) with y.
y 3 x 1 5
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2
2
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Secondary Mathematics 2
Replace x with y and y with x.
x 3 y 1 5
Isolate y.
x 5 3 y 1
2
2
x5
2
y 1
3
x5
y 1
3
Simplify the radical
(Unit 1 Cluster1:N.RN.2)
x5
1 y
3
Determine whether to use the positive or
negative answer by referring back to the
restricted domain. The domain of the
original function is restricted to the
positive real numbers including zero,
therefore, the range of the inverse function
must also be the same. This leads us to
choose the positive square root.
x5
1 f 1 ( x)
3
Domain 5, and Range 0,
Notice the domain and range have
switched from the original function’s
domain and range.
Practice Exercises F
Find the inverse of the following. State the domain and range of the inverse. For problems 7 – 9
restrict the domain before finding the inverse.
1.
f ( x) 3x 2
2.
f ( x) 4 x 7
4.
4
f ( x) x 1
5
5.
f ( x)
7.
f ( x) 3 x 2 5
8.
f ( x) x 2
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2
x4
3
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2
3.
f ( x) 6 x 5
6.
f ( x)
9.
f ( x) x 7 9
1
x 3
2
2
Secondary Mathematics 2
Unit 3
Expressions and
Equations
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Secondary Mathematics 2
Unit 3 Cluster 1 (A.SSE.2):
Interpret the Structure of Expressions
Cluster 1: Interpret the structure of expressions
3.1.2 Recognize functions that are quadratic in nature such as
x 4 x 2 6 x 2 3 x 2 2
VOCABULARY
A quadratic pattern can be found in other types of expressions and equations. If this is the case,
we say these expressions, equations, or functions are quadratic in nature. Recall the standard
form of a quadratic expression is ax2 bx c , where a, b, and c are real numbers and a 0 .
The following are examples of expressions that are quadratic in nature:
Expression:
Notice:
x
3x6 5x3 12
2 x6 5x3 y 2 12 y 4
3
x
3
2
2 x 1 15
2
16 x10 25
x4 y6
2
x
4x 6 x 9
4
1/4
Rewritten:
x6
x6 and y 2
x
9 x1/2 12 x1/4 4
x 1
2
2
y4
x1/2
2
x
2
2
16 x10 and 5 25
2
x 4 and y 3
2
2
y6
2
3
2
4
2
5
3
3
1/4
x 12 x 14
4x
x
5 x 12
2 x 5 x y 12 y
9 x 12 x 4
3 x3
2
2
2
x
2
1/4
2
6
x9
2
x 12 2 x 12 15
4 x 5
x y
5
2
2
2
3
2
2
Practice exercise A
Determine if the expression is quadratic in nature.
1. x4 x2 12
3. x3 4 x 4
5. 9 x2/3 12 x1/3 4
2. 2 2 x 3 2 x 3 1
4. x1/2 x1/4 72
6. 4 x 2 1
2
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2
Secondary Mathematics 2
FACTORING REVIEW
1. COMMON TERM
a) What number will go into all of the
numbers evenly
b) Common variable – use the common
variable with the lowest power
EXAMPLE
2. DIFFERENCE OF TWO SQUARES
Looks like:
2
2
a b a ba b
EXAMPLE
3x3 12 x 2 6 x
3x x 2 4 x 2
a) Terms must be perfect squares
b) Must be subtraction
c) Powers must be even
3. PERFECT SQUARE TRINOMIALS
Looks like:
2
2
2
a 2ab b a b or a b a b
or
2
2
2
a 2ab b a b or a b a b
4 x2 25
2 x 5 2 x 5
EXAMPLE
9 x2 30 x 25
Does the middle term equal 2ab ?
a 3x and b 5 so 2 3x 5 30 x
Yes it does!
a) First and last term must be perfect
squares
b) Middle term is equal to 2ab
c) Sign in the parenthesis is the same as the
first
sign
Therefore 9 x 2 30 x 25 factors to:
4. GROUPING
EXAMPLE
15xy 21x 10 y 14
a) Group terms that have something in
common
b) Factor out common term in each
parenthesis
c) Write down what is in the parenthesis,
they
should be identical
d) Then add the “left-overs”
5. FACTOR TRINOMIALS BY
GROUPING
ax2 bx c
3x 5
2
or 3x 5 3x 5
15xy 21x 10 y 14
3x 5 y 7 2 5 y 7
5 y 7
5 y 7 3x 2
EXAMPLE
6 x 2 x 15
(6)(15) = 90
a) Multiply a and c
1 and 90
2 and 45
3 and 30
5 and 18
b) Find all the factors of the answer
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Secondary Mathematics 2
6 and 15
9 and 10
c) Choose the combination that will either
give the
sum or difference needed to result in b. In
this
case the difference
9 and -10
6 x2 9 x 10 x 15
6x
d) Rewrite the equation using the combination
in
place of the middle term
2
9 x 10 x 15
Notice that when factoring out a -1 it changes
the sign on c
3 x 2 x 3 5 2 x 3
e) Now group in order to factor
2 x 3
f) Factor out the common term in each
parenthesis
2x 33x 5
g) Write down what is in the parenthesis, they
should be identical
h) Add the “left-overs” to complete the answer
The same strategies used to factor quadratic expressions can be used to factor anything that is
quadratic in nature. (For more information on factoring, see the factoring lesson in Unit 2.)
Expression:
3x6 5x3 12
2 x6 5x3 y 2 12 y 4
Rewritten:
5 x 12
2 x 5 x y 12 y
2
3 x3
2
3
3x
3
3
2
2
2
Factor:
2x 3 y x 4 y
3x 23x 2
3
4 x3 3
3
2
3
1/4
9 x1/2 12 x1/4 4
9 x1/4
2
12 x1/4 4
1/4
or
3x
1/4
4 x 12 x 9
16 x10 25
x4 y6
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4
x
2
12
x9
4 x 5
x y
5
2
2
2
3
Page 122
2
2
2
2
2
2
x 3 2 x 3
or
2
x 3
2
4x 5 4 x 5
x y x y
5
2
5
3
2
3
Secondary Mathematics 2
Practice set B
Factor each quadratic in nature expression.
1. 144 x 2 49 y 4
2. 8x6 2 x3 15
3. 100 x8 121y 6
4. 81x6 4
5. 2 x x 1
6. 4 x4 20 x2 25
7. 9 x10 6 x5 y y 2
8. 12 x2/5 17 x1/5 6
9. 3x2/3 10 x1/3 8
Sometimes rewriting an expression makes it easier to recognize the quadratic pattern.
Example:
x 1
4
2 x 1 15
2
You can use a new variable to replace x 1 .
u x 1
Now replace every x 1 in the expression
with u. Notice how this is quadratic in nature.
We can use quadratic factoring techniques to
factor this expression
You must remember to replace the u with
x 1 .
u 4 2u 2 15
u
5 u2 3
2
x 12 5 x 12 3
Example:
3x1/3 8x1/6 4
u x1/6
3u 2 8u 4
You can use a new variable to replace x1/6 .
Now replace every x1/6 in the expression with
u. Notice how this is quadratic in nature.
We can use quadratic factoring techniques to
factor this expression
3u 2u 2
3x
1/6
2 x1/6 2
You must remember to replace the u with x1/6 .
Practice Exercises C
Identify the “u” in each expression, then factor using “u” substitution. Write the factored form in
terms of x.
1.
2x 2x
4. 3
4
3
2
6
6
x3
3
2. 4 y 3 15 y 3 9
3. 5 2 x 5 21 2 x 5 20
5. x1/2 7 x1/4 10
6. x 2 11 x 2 12
2
3
x3 2
2
7. 4 x 25
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1 1
8. 2
x x
Page 123
8
4
2
4
9. 4 x 2 1 9
Secondary Mathematics 2
Unit 1 Cluster 3 (N.CN.1 and N.CN.2)
Performing Arithmetic Operations With Complex Numbers
Cluster 3: Performing arithmetic operations with complex numbers
1.3.1 i 2 1 , complex number form a + bi
1.3.2 Add, subtract, and multiply with complex numbers
VOCABULARY
The imaginary unit, i , is defined to be i 1 . Using this definition, it would follow that
i 2 1 because i 2 i i 1 1 1 .
The number system can be extended to include the set of complex numbers. A complex number
written in standard form is a number a + bi, where a and b are real numbers. If a = 0, then the
number is called imaginary. If b = 0 then the number is called real.
Simplifying Radicals with i
Extending the number system to include the set of complex numbers allows us to take the square
root of negative numbers.
Example:
Simplify
9
9
1 9
Rewrite the expression using the properties
of radicals.
1 9
i 3
Remember that i 1 .
3i
Example:
Simplify
24
24
1 4 6
Rewrite the expression using the properties
of radicals.
1 4 6
i 2 6
Remember that i 1 .
2i 6
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Secondary Mathematics 2
Practice Exercises A
Simplify each radical
1.
25
2.
36
3.
144
4.
98
5.
52
6.
22
Performing Arithmetic Operations with Complex Numbers
You can add, subtract, and multiply complex numbers. Similar to the set of real numbers,
addition and multiplication of complex numbers is associative and commutative.
Adding and Subtracting Complex Numbers
Example:
Add 3 2i 5 4i
3 2i 5 4i
3 2i 5 4i
3 5 2i 4i
Remove the parentheses.
Group like terms together.
8 2i
Combine like terms.
Example:
Subtract 7 5i 2 6i
7 5i 2 6i
7 5i 2 6i
7 2 5i 6i
Distribute the negative and remove the
parentheses.
Group like terms together.
9 11i
Combine like terms.
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Secondary Mathematics 2
Example:
Simplify 8 4 5i 2 5i
8 4 5i 2 5i
8 4 5i 2 5i
8 4 2 5i 5i
Distribute the negative and remove the
parentheses.
Group like terms together.
6 10i
Combine like terms.
Multiplying Complex Numbers
Example:
Multiply 3 7 6i
3 7 6i
Distribute the negative three to each term
in the parentheses.
21 18i
Example:
Multiply 4i 2 9i
4i 2 9i
Distribute the 4i to each term in the
parentheses.
By definition i 2 1 so substitute -1 in for
i2 .
Write the complex number in standard
form.
8i 36i 2
8i 36 1
8i 36
36 8i
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Secondary Mathematics 2
Example:
Multiply 2 9i 3 10i
Distributive (FOIL) Method
2 9i 3 10i
2 3 10i 9i 3 10i
6 20i 27i 90i 2
*combine like terms
6 7i 90i 2
*remember that i 2 1
Box Method
2
3
10i
6
20i
Vertical Method
9i
27i
90i 2
2 9i
3 10i
20i 90i 2
6 27i
*combine terms on the
diagonals of the unshaded
boxes(top right to lower left)
6 7i 90i 2
*remember that i 2 1
6 7i 90 1
6 7i 90i
6 7i 90
96 7i
*remember that i 2 1
2
6 7i 90 1
6 7i 90
96 7i
6 7i 90 1
6 7i 90
96 7i
Example:
Multiply 5 2i 5 2i
Distributive (FOIL) Method
5 2i 5 2i
5 5 2i 2i 5 2i
25 10i 10i 4i 2
*combine like terms
25 4i 2
*remember that i 2 1
Box Method
5
25
10i
5
2i
Vertical Method
2i
10i
4i 2
*combine terms on the
diagonals of the unshaded
boxes(top right to lower left)
25 4 1
25 4i
25 4
29
*remember that i 2 1
2
25 4 1
25 4
29
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5 2i
5 2i
10i 4i 2
25 10i
25 0i 4i 2
*remember that i 2 1
25 4 1
25 4
29
Secondary Mathematics 2
Practice Exercises B
Simplify each expression.
9 6i 4 4i
1. 10 i 3 6i
2.
4. 7 4 8i 10 2i
5. 5 8 2i
7. i 9 15i
8.
10 1 4i 12 11i
11.
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6 7i 4 12i
10 4i 7 5i
Page 128
3. 4i 7 5i 10 7i
6. 11i 2 9i
9.
12.
2 i 7 5i
10 2i 10 2i
Secondary Mathematics 2
Unit 1 Cluster 3 Honors (N.CN.3)
H2.1
Find the conjugate of a complex number; use conjugates to find quotients of
complex numbers.
VOCABULARY
The conjugate of a complex number is a number in the standard complex form a bi , where the
imaginary part bi has the opposite sign of the original, for example a bi has the opposite sign
of a bi . Conjugate pairs are any pair of complex numbers that are conjugates of each other
such as 3 4i and 3 4i .
The product of conjugate pairs is a positive real number.
a bi a bi
3 4i 3 4i
a 2 abi abi b 2i 2
9 12i 12i 16i 2
9 16
25
a 2 b 2 1
a 2 b2
This property will be used to divide complex numbers.
Example:
Find the conjugate of the following complex numbers.
a. 4i
b. 2 5i
d. 7 2i
c. 3 i
a.
b.
The opposite of 5i is 5i . The
conjugate of 2 5i is 2 5i .
The opposite of 4i is 4i . The
conjugate of 4i is 4i .
c.
d.
The opposite of i is i . The conjugate
of 3 i is 3 i .
The opposite of 2i is 2i . The
conjugate of 7 2i is 7 2i .
Practice Exercises A
Find the conjugate of the following complex numbers.
1. 6 6i
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2. 8 9i
3. 2 3i
Page 129
4. 1 7i
Secondary Mathematics 2
Divide by an imaginary number bi
If there is an imaginary number in the denominator of a fraction, then the complex number is not
in standard complex form. In order to write it in standard complex form, you must multiply the
numerator and the denominator by the conjugate of the denominator. This process removes the
imaginary unit from the denominator and replaces it with a real number (the product of conjugate
pairs is a positive real number) without changing the value of the complex number. Once this is
done, you can write the number in standard complex form by simplifying the fraction.
Example:
Write in standard complex form:
2
8i
2 8i
8i 8i
16i
64i 2
16i
64 1
The conjugate of 8i is 8i . Multiply the
numerator and the denominator by the
conjugate.
Remember i 2 1 .
16i
64
16 i i
1
i
16 4
4
4
Example:
Write in standard complex form:
Simplify.
6 8i
9i
6 8i 9i
9i 9i
54i 72i 2
81i 2
54i 72 1
81 1
The conjugate of 9i is 9i . Multiply the
numerator and the denominator by the
conjugate.
72 54i
81
72 54i
81 81
8 2
i
9 3
Rewrite numerator in standard complex
form a bi .
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Remember i 2 1 .
Rewrite whole solution in complex form
a bi , reducing as needed.
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Secondary Mathematics 2
Practice Exercises B
Write in standard complex form.
1.
3
5i
2.
6
4i
3.
5
5i
4.
3 10i
6i
5.
10 10i
5i
6.
2 3i
4i
Dividing Complex Numbers in Standard Form a bi
To divide complex numbers, find the complex conjugate of the denominator, multiply the
numerator and denominator by that conjugate, and simplify.
Example:
Divide
10
2i
10
2i
10 2 i
2i 2i
Multiply the numerator and denominator
by the conjugate of 2+i, which is 2 – i
10 2 i
2 i 2 i
20 10i
4 2i 2i i 2
20 10i
4 i2
20 10i
4 1
Distribute
Simplify
Note that i 2 1
20 10i
5
20 10
i 4 2i
5 5
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Simplify and write in standard complex
form.
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Secondary Mathematics 2
Example:
Divide
22 7i
4 5i
22 7i
4 5i
22 7i 4 5i
4 5i 4 5i
Multiply by the conjugate of the
denominator.
22 7i 4 5i
4 5i 4 5i
88 110i 28i 35i 2
16 20i 20i 25i 2
Distribute
88 82i 35i 2
16 25i 2
88 82i 35 1
16 25 1
Combine like terms.
88 82i 35
16 25
Combine like terms again.
Remember that i 2 1 .
123 82i
41
123 82
i
41 41
3 2i
Example:
Divide
Simplify and write in standard complex
form.
6 2i
1 2i
6 2i
1 2i
6 2i 1 2i
1 2i 1 2i
Multiply by the conjugate of the
denominator.
6 2i 1 2i
1 2i 1 2i
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Secondary Mathematics 2
6 12i 2i 4i 2
1 2i 2i 4i 2
6 14i 4i 2
1 4i 2
6 14i 4 1
Distribute.
Combine like terms.
1 4 1
Remember i 2 1
6 14i 4
1 4
2 14i
5
2 14
i
5 5
Combine like terms again.
Put in standard complex form a bi .
Practice Exercises C
Divide each complex rational expression and write in standard complex form.
1.
5i
2 6i
2.
8i
1 3i
3.
10
3 i
4.
26 18i
3 4i
5.
10 5i
6 6i
6.
3 7i
7 10i
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Secondary Mathematics 2
Unit 3 Cluster 4 (A.REI.4) and Unit 3 Cluster 5 (N.CN.7):
Solve Equations and Inequalities in One Variable
Cluster 4: Solving equations in one variable
3.4.1a Derive the quadratic formula by completing the square.
3.4.1b Solve equations by taking the square root, completing the square, using the
quadratic formula and by factoring (recognize when the quadratic formula gives
complex solutions and write them as a bi )
3.5.1 Solve with complex numbers
VOCABULARY
The square root of a number is a value that, when multiplied by itself, gives the number. For
example if r 2 a , then r is the square root of a. There are two possible values for r; one
positive and one negative. For instance, the square root of 9 could be 3 because 32 9 but it
2
could also be 3 because 3 9 .
A perfect square is a number that can be expressed as the product of two equal integers. For
example: 100 is a perfect square because 10 10 100 and x 2 is a perfect square because
x x x2 .
Solving Equations by Taking the Square Root
When solving a quadratic equation by taking the square root, you want to isolate the squared
term so that you can take the square root of both sides of the equation.
Example:
Solve the quadratic equation x 2 4 .
In this example the squared term is x 2 and it is
already isolated.
x2 4
x2 4
x
2 1/2
x
4
2
x 2
2/2
x 2 or x 2
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Take the square root of each side of the equation.
Using the properties of rational exponents you can
simplify the left side of the equation to x.
The number 16 is a perfect square because
2 2 4 .
Page 134
Secondary Mathematics 2
Example:
Solve the quadratic equation
2 2
x 4 16 .
3
2 2
x 4 16
3
2 2
x 12
3
x 2 18
In this example the squared term is x 2 and it needs
to be isolated. Use a reverse order of operations to
isolate x 2 .
x 2 18
x
2 1/2
Take the square root of each side of the equation.
92
x 2/2 9 2
x 3 2
x 3 2 or x 3 2
Using the properties of rational exponents you can
simplify the left side of the equation to x.
Using the properties of radical expressions you can
simplify the right side of the equation. (See Unit 1
Cluster 2 for help with simplifying)
Example:
2
Solve the quadratic equation 3 x 2 4 52 .
3 x 2 4 52
2
3 x 2 48
2
x 2
2
16
x 2
16
2
1/2
x 2 2
x 2
2/2
16
4
x 2 4
x 2 4 or x 2 4
x 2 4 or x 2 4
x 6 or x 2
Jordan School District
In this example the squared term is x 2 and it
needs to be isolated. Use a reverse order of
2
operations to isolate x 2 .
2
Take the square root of each side of the equation.
Using the properties of rational exponents you can
simplify the left side of the equation to x 2 .
The number 16 is a perfect square because
4 4 16 .
You still need to solve each equation for x.
Page 135
Secondary Mathematics 2
Practice Exercises A
Solve each quadratic equation.
1. x 2 25
2. x 2 8
3. 4 x2 36
4. 4 x2 5 1
5. 9 x2 3 33
6. 16 x2 9
7. 2 x 3 4 8
8. 3 x 3 1 2
9.
2
2
1
3
x 1
2
5
Solving Quadratic Equations by Completing the Square
Sometimes you have to rewrite a quadratic equation, using the method of completing the square,
so that it can be solved by taking the square root.
Example:
Solve x2 10 x 23 .
x2 10 x 23
2
10
10
x 2 10 x 23
2
2
2
x 10 x 25 23 25
2
Complete the square on the left side of the
equation.
x 2 10 x 25 2
x 5 x 5 2
x 5
2
Factor the expression on the left side.
2
x 5
2
2
Take the square root of each side.
x 5 2
Simplify.
x 5 2 or x 5 2
Solve for x.
x 5 2 or x 5 2
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Secondary Mathematics 2
Example:
Solve x2 4 x 6 .
x2 4 x 5
2
4
4
x 2 4 x 5
2
2
2
x 4 x 4 5 4
2
Complete the square on the left side of the
equation.
x 2 4 x 4 1
x 2 x 2 1
x 2
2
Factor the expression on the left side.
1
x 2
2
1
Take the square root of each side.
x 2 i
Simplify. Remember that the
x 2 i or x 2 i
x 2 i or x 2 i
Solve for x.
1 i
Example:
Solve 3x2 5x 2 0 .
Collect the terms with variables on one
side of the equation and the constant
term on the other side.
3x 2 5 x 2 0
3x 2 5 x 2
5
3 x2 x 2
3
2
2
5 5
5
3 x 2 x
2
3
3 2 3
23
5
25
25
3 x2 x 2 3
3
36
36
5
25
75
3 x2 x 2
3
36
36
Complete the square on the left side of
the equation.
5
25 49
3 x2 x
3
36 12
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Secondary Mathematics 2
5
5 49
3 x x
6
6 12
Factor the expression on the left side of
the equation.
2
5
49
3 x
6 12
Isolate the squared term.
2
5
49
x
6 36
2
5
49
x
6
36
5
7
x
6
6
5 7
5
7
x or x
6 6
6
6
5 7
5 7
x or x
6 6
6 6
1
x 2 or x
3
Take the square root of each side.
Simplify.
Solve for x.
Practice Exercises B
Solve the quadratic equations by completing the square.
1.
x2 4 x 1
2.
x2 12 x 32
3.
x2 16 x 15 0
4.
x2 8x 3 0
5.
x2 8 6 x
6.
2 x2 4 x 5 0
VOCABULARY
b b 2 4ac
The quadratic formula, x
, can be used to find the solutions of the quadratic
2a
equation ax2 bx c 0 , when a 0 . The portion of the quadratic equation that is under the
radical, b2 4ac , is called the discriminant. It can be used to determine the number and type of
solutions to the quadratic equation ax2 bx c 0 .
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Secondary Mathematics 2
Using the Discriminant to Determine Number and Type of Solutions
If b2 4ac 0 , then there are two real solutions to the quadratic equation.
Example:
Determine the number and type of solutions for the equation x2 x 6 0 .
x2 x 6 0
a 1, b 1, and c 6
b 2 4ac
12 4(1)(6)
1 4(6)
1 (24)
1 24
25
Identify a, b, and c.
Substitute the values of a, b,
and c into the discriminant
formula.
Simplify using order of
operations.
Determine if the result is
greater than zero, equal to
25 0
zero, or less than zero.
2
The quadratic equation x x 6 0 has two real solutions. You can see from the graph
that the function crosses the x-axis twice.
If b2 4ac 0 , then there is one real solution to the quadratic equation.
Example:
Determine the number and type of solutions for the equation x2 4 x 4 0 .
x2 4 x 4 0
a 1, b 4, and c 4
b 2 4ac
42 4(1)(4)
16 4(4)
16 16
0
Identify a, b, and c.
Substitute the values of a, b,
and c into the discriminant
formula.
Simplify using order of
operations.
Determine if the result is
greater than zero, equal to
zero, or less than zero.
2
The quadratic equation x 4 x 4 0 has one real solution. You can see from the graph
that the function touches the x-axis only once.
00
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Page 139
Secondary Mathematics 2
If b2 4ac 0 , then there no real, but two imaginary solutions to the quadratic
equation.
Example:
Determine the number and type of solutions for the equation x2 1 0 .
x2 1 0
a 1, b 0, and c 1
Identify a, b, and c.
Substitute the values of a, b,
and c into the discriminant
formula.
b 2 4ac
02 4(1)(1)
0 4(1)
04
4
Simplify using order of
operations.
Determine if the result is
greater than zero, equal to
zero, or less than zero.
2
The quadratic equation x 1 0 has no real solutions, but it has two imaginary
solutions. You can see from the graph that the function never crosses the x-axis.
4 0
Practice Exercises C
Determine the number and type of solutions that each quadratic equation has.
1.
4. x2 2 x 5 0
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2.
3.
5. 4 x2 12 x 9 0
Page 140
6. 2 x2 4 x 2
Secondary Mathematics 2
Solving Quadratic Equations by Using the Quadratic Formula
Example:
Solve 3x2 5x 4 0 using the quadratic formula.
3x 2 5 x 4 0
a 3, b 5, and c 4
x
x
x
5 52 4(3)(4)
2(3)
Make sure all the terms are on the same side
and that the equation equals 0.
Identify a, b, and c.
Substitute the values for a, b, and c into the
quadratic formula.
5 25 12 4
6
5 25 48
6
5 25 48
x
6
5 73
x
6
5 73
5 73
and x
x
6
6
Use order of operations to simplify.
These are actually two different solutions.
Example:
Solve 2 x2 x 3 0 using the quadratic formula.
2 x2 x 3 0
a 2, b 1, and c 3
x
1 (1)2 4(2)(3)
2(2)
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Make sure all the terms are on the same side
and that the equation equals 0.
Identify a, b, and c.
Substitute the values for a, b, and c into the
quadratic formula.
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Secondary Mathematics 2
1 1 8(3)
4
1 1 (24)
x
4
1 1 24
x
4
1 25
x
4
1 5
x
4
1 5 6 3
x
and
4
4 2
1 5 4
x
1
4
4
x
Use order of operations to simplify.
Simplify each answer.
Example:
Solve 25x2 10 x 1 using the quadratic formula.
25 x 2 10 x 1
25 x 10 x 1 0
Make sure all the terms are on the same side
and that the equation equals 0.
a 25, b 10, and c 1
Identify a, b, and c.
2
x
x
x
x
x
x
10 102 4(25)(1)
2(25)
Substitute the values for a, b, and c into the
quadratic formula.
10 100 100(1)
50
10 100 100
50
10 0
50
10 0
50
10
50
Use order of operations to simplify.
1
x
5
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Simplify the answer. Notice that we only got
one answer this time because the discriminant
was 0.
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Secondary Mathematics 2
Example:
Solve 2 x2 12 x 20 using the quadratic formula.
2 x 2 12 x 20
2 x 12 x 20 0
Make sure all the terms are on the same side
and that the equation equals 0.
a 2, b 12, and c 20
Identify a, b, and c.
2
x
12 122 4(2)(20)
2(2)
Substitute the values for a, b, and c into the
quadratic formula.
12 144 8(20)
4
12 144 160
x
4
12 16
x
4
12 4i
x
4
x 3 i
Use order of operations to simplify.
x 3 i or x 3 i
Simplify the answer. Notice that we got two
imaginary answers this time because the
discriminant was less than 0.
x
Practice Exercises D
Solve the quadratic equation using the quadratic formula.
1. 3x2 5x 7 0
2. 4 x2 12 x 9 0
3. 6 x2 11x 7 0
4. x2 8x 12 0
5. x2 3 6 x
6. x2 2 x 2
7. x2 x 1 0
8. 2 x2 5x 4 0
9. 3x2 4 2 x
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Practice Exercises E
Solve each of the following equations using the method of your choice.
1. 2 x2 5x 4 0
2. x2 10 6 x
3. x2 4 x 6 0
4. x( x 3) x 9
5. x( x 1) 2 x 7
6. x2 10 x 26 0
7. 4 x2 81 0
8.
10. x2 x 4 0
11. x2 4 0
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x 1
2
9
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9.
x 1
2
5 44
12. 2 x 3 5 23
2
Secondary Mathematics 2
Unit 3 Cluster 5 (N.CN.8, N.CN.9-Honors):
Use complex numbers in polynomial identities and equations
3.5.2
3.5.3
Extend polynomial identities to the complex numbers.
Know the Fundamental Theorem of algebra; show that it is true for quadratic
polynomials.
The Fundamental Theorem of Algebra states that every polynomial of degree n with complex
coefficients has exactly n roots in the complex numbers.
Note: Remember that every root can be written as a complex number in the form of a bi . For
instance x 3 can be written as x 3 0i . In addition, all complex numbers come in conjugate
pairs, a bi and a bi .
Example:
f ( x) x 2 x 3
f ( x) 5 x 3 2 x 2 5 x 4
Degree: 2
Complex Roots: 2
Degree: 3
Complex Roots: 3
Polynomial Identities
1.
a b
a 2 2ab b2
2.
a b c d ac ad bc bd
3.
a 2 b2 a b a b
4.
x2 a b x AB x a x b
5.
b b 2 4ac
If ax bx c 0 then x
2a
2
2
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Secondary Mathematics 2
Example:
Find the complex roots of f ( x) x 2 64 and write in factored form.
0 x 2 64
64 x 2
1.
Set equal to zero to find the roots of the
function. Solve.
2.
Recall factored form is
x p x q f ( x) .
Substitute the zeros in for p and q.
64 x 2
8i x
x 8i x 8i f ( x)
x 8i x 8i f ( x)
Example:
Find the complex roots of f ( x) x 2 16 x 65 and write in factored form.
x
16
16 4 1 65
2 1
2
16 256 260
2
16 4
x
2
16 2i
x
2
x 8 i
x
x 8 i x 8 i f ( x)
x 8 i x 8 i
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1.
Use the quadratic formula to find the roots
of the function.
2.
Recall factored form is
x p x q f ( x) .
Substitute the zeros in for p and q.
f ( x)
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Secondary Mathematics 2
Example:
Find the complex roots of f ( x) x 4 10 x 2 24 and write in factored form.
f ( x) x 2 4 x 2 6
x2 4 0
x2 6 0
x 2 4
x 2 6
x 2i
x i 6
1. Factor the quadratic in nature function.
2. Set each factor equal to zero to find the
roots.
x i 6
6 x i 6
f ( x) ( x 2i )( x (2i)) x i 6
f ( x) ( x 2i )( x 2i) x i
3. Recall factored form is
x p x q f ( x) .
Substitute the zeros in for p and q.
Practice Exercises A
Find the complex roots. Write in factored form.
1.
x2 9
2.
x2 x 1
3.
x2 2 x 2
4.
x2 6 x 10
5.
x2 4 x 5
6.
x2 2 x 5
7.
x4 5x2 4
8.
x4 13x2 36
9.
x4 1
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Secondary Mathematics 2
Unit 3 Cluster 3 (A.CED.1, A.CED.4)
Writing and Solving Equations and Inequalities
Cluster 3: Creating equations that describe numbers or relationships
3.3.1 Write and solve equations and inequalities in one variable (including linear,
simple exponential, and quadratic functions)
3.3.3 Solve formulas for a variable including those involving squared variables
Writing and Solving Quadratic Equations in One Variable
When solving contextual type problems it is important to:
Identify what you know.
Determine what you are trying to find.
Draw a picture to help you visualize the situation when possible. Remember to label all
parts of your drawing.
Use familiar formulas to help you write equations.
Check your answer for reasonableness and accuracy.
Make sure you answered the entire question.
Use appropriate units.
Example:
Find three consecutive integers such that the product of the first two plus the square of
the third is equal to 137.
The first number is x. Since they are
consecutive numbers, the second term is
one more than the first or x + 1. The third
term is one more than the second term or
x 11 x 2 .
Multiply the first two together and add the
result to the third term squared. This is
equal to 137.
First term: x
Second term: x + 1
Third term: x + 2
x( x 1) ( x 2)2 137
x( x 1) ( x 2)( x 2) 137
x 2 x x 2 4 x 4 137
Multiply and combine like terms.
2 x 2 5 x 4 137
2 x2 5x 133 0
Make sure the equation is equal to 0.
2 x 2 14 x 19 x 133 0
2x
2
14 x 19 x 133 0
Factor.
2 x x 7 19 x 7 0
x 7 2 x 19 0
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Secondary Mathematics 2
x7 0
x7
2 x 19 0
2 x 19
19
x
2
First term: 7
Second term: 8
Third term: 9
Use the Zero Product Property to solve for
x.
The numbers are integers so x has to be 7.
Example:
A photo is 6 in longer than it is wide. Find the length and width if the area is 187 in 2 .
width x
length x 6
187 x x 6
The width is the basic unit, so let it equal
x. The length is 6 inches longer than width
or x 6 . A photo is rectangular so the
area is equal to the width times the length.
The area is 187 square inches.
187 x2 6 x
Multiply the right side.
0 x2 6 x 187
Make sure the equation equals 0.
0 x 11 x 17
Factor the expression on the right side of
the equation.
Use the Zero Product Property to solve for
x.
The length of a photo cannot be negative.
Therefore, x must be 11. The length is
x 6 11 6 17 .
A x wl x x 6
x 11 0
x 11
x 17 0
x 17
The width is 11 inches and the length is
17 inches.
Note: Often problems will require information from more than one equation to solve. For
example, you might need the perimeter equation to help you write the area equation or vice
versa. The primary equation is the equation you solve to find the answer you are looking for.
The secondary equation is the equation you use to help set up your primary equation.
Example:
Find two numbers that add to 150 and have a maximum product. What is the maximum
product?
One number is x. The other
number is y.
The sum is 150. Write an
equation for this.
Secondary equation:
x y 150
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Secondary Mathematics 2
y 150 x
Use the sum equation to solve for
y.
Primary Equation:
P x x 150 x
Write an equation for the product
in terms of x.
Simplify the right side of the
equation.
P x 150 x x 2
P x x 2 150 x
P x ( x 2 150 x)
P x ( x 2 150 x 5625) 5625
P x ( x 75) 2 5625
Find the maximum. Remember
the maximum is the vertex. Using
the method of your choice from
Unit 2 Lesson F.IF.8.
P x 150 x x 2
150
x
2(1)
x 75
P 75 150 75 75
2
P 75 11250 5625
P 75 5625
x 75
y 75
The two numbers are 75 and 75
75 75 5625
The maximum product is 5625.
The second number is 150 x or
75.
Find the maximum product.
Example:
Jason wants to fence in a rectangular garden in his backyard. If one side of the garden is
against the house and Jason has 48 feet of fencing, what dimensions will maximize the
garden area while utilizing all of the fencing?
x
Garden
x
First draw a picture of the house
and garden. Label the sides of
your garden. The amount of fence
used is the distance around the
garden excluding the side next to
the house. This is the same as the
perimeter.
y
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Secondary Mathematics 2
The length of the garden is x and
the width is y.
The perimeter is 48. Write an
equation for this.
Use the sum equation to solve for
y.
2 x y 48
y 48 2 x
A x lw
Write an equation for the product
in terms of x.
A x x 48 2 x
P x 48x 2 x 2
Simplify the right side of the
equation.
A x 2 x 2 48 x
A x 2( x 2 24 x)
A x 2( x 2 24 x 144) 288
A x 2( x 12) 2 288
48
2(2)
x 12
x
Find the maximum area.
Remember the maximum is the
vertex. Using the method of your
choice from Unit 2 Lesson F.IF.8.
A x 48 x 2 x 2
P 12 48 12 2 12
2
P 12 576 288
P 12 288
x 12
y 24
The length is 12 feet and the width is 24 feet.
12 24 288
The maximum area is 288 ft 2 .
The second number is 48 2 12
or 24.
Find the maximum area.
Practice Problems A
Solve
1.
The maximum size envelope that can
be mailed with a large envelope rate is
3 inches longer than it is wide. The
area is 180 in 2 . Find the length and the
width.
2.
A rectangular garden is 30 ft. by 40 ft.
Part of the garden is removed in order
to install a walkway of uniform width
around it. The area of the new garden
is one-half the area of the old garden.
How wide is the walkway?
3.
The base of a triangular tabletop is 20
inches longer than the height. The area
is 750 in 2 . Find the height and the
base.
4.
Find two numbers that differ by 8 and
have a minimum product.
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Secondary Mathematics 2
5.
Alec has written an award winning
short story. His mother wants to frame
it with a uniform border. She wants the
finished product to have an area of
315 in 2 . The writing portion occupies
an area that is 11 inches wide and 17
inches long. How wide is the border?
6.
Britton wants to build a pen for his
teacup pig. He has 36 feet of fencing
and he wants to use all of it. What
should the dimensions of the pen be to
maximize the area for his pig? What is
the maximum area?
7.
The product of 2 numbers is 476. One
number is 6 more than twice the first
number. Find the two numbers.
8.
Find three consecutive integers such
that the square of the second number
plus the product of the first and third
numbers is a minimum.
VOCABULARY
Objects that are shot, thrown, or dropped into the air are called projectiles. Their height,
measured from the ground, can be modeled by a projectile motion equation. The object is
always affected by gravity. The gravitational constant is different depending on the units of
measurement. For example, the gravitational constant in feet is 32 ft/sec2 and in meters it is
9.8 m/sec2 . Similarly, the projectile motion equation for an object shot or thrown straight up or
down is different depending on the units of measurement.
Feet: h(t ) 16t 2 v0t h0
Meters: h(t ) 4.9t 2 v0t h0
h(t ) represents the height at any time t. The time is measured in seconds. The initial velocity,
v0 , is the speed at which the object is thrown or shot. It is measured in ft/sec or m/sec. The
initial height, h0 , is the height that the object is shot or thrown from. It is measured in feet or
meters.
Example:
The Willis Tower (formerly Sears Tower) in Chicago, Illinois is the tallest building in the
United States. It is 108 stories or about 1,451 feet high. (Assume that each floor is 13
feet high.)
a.
A window washer is 28 floors from the top and he drops a piece of equipment,
how long will it take for the equipment to reach the ground?
b.
How far from the ground is the piece of equipment after 5 seconds?
c.
When does the equipment pass the 16th floor?
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Secondary Mathematics 2
a.
h(t ) 16t 2 v0t h0
h(t ) 16t 2 (0)t 108 28 13
h(t ) 16t 2 80 13
h(t ) 16t 2 1040
0 16t 2 1040
The building is measured in feet so use
the projectile motion equation for feet.
The equipment was dropped making the
initial velocity 0 ft./sec. The building has
108 floors, but he stopped 28 short of the
top floor. Each floor is 13 feet high;
multiply the number of floors by the
height of each floor to get the initial
height.
We want to know when the equipment
hits the ground making the final height
zero.
0 16t 2 1040
16t 2 1040
Solve for t.
t 2 65
t 65
Negative time means you are going back
in time. Therefore, time is positive.
t 65 sec. or 8.062 sec.
b.
h t 16t 2 1040
h 5 16 5 1040
We want to know when the height of the
equipment at 5 seconds.
2
h 5 400 1040
h 5 640
The equipment is 640 feet from the ground after 5 seconds.
c.
We want to know when the equipment
passes the 16th floor. The equation is the
same equation written in part a.
h t 16t 2 1040
16 13 16t 2 1040
The 16th floor is 208 feet above the
ground.
208 16t 2 1040
0 16t 2 832
832 16t 2
52 t 2
Solve for t.
52 t 2
2 13 t
Negative time means you are going back
in time. Therefore, time is positive.
t 2 13 sec. or 7.211 sec.
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Secondary Mathematics 2
Example:
The Salt Lake Bees are planning to have a fireworks display after their game with the
Tacoma Rainiers. Their launch platform is 5 feet off the ground and the fireworks will be
launched with an initial of 32 feet per second. How long will it take each firework to
reach their maximum height?
The height of the fireworks is measured in
feet so use the projectile motion equation
for feet.
The fireworks were launched with an
initial velocity 32 ft./sec. The launch
platform is 5 feet off the ground.
a.
h(t ) 16t 2 v0t h0
h(t ) 16t 2 (32)t 5
h(t ) 16t 2 32t 5
h(t ) 16t 2 32t 5
h t 16 t 2 2t 5
Using the method of your choice from
Unit 2 Lesson F.IF.8. Find the amount of
time it will take to reach the maximum
height. The t coordinate of the vertex
indicates WHEN the firework will reach
its maximum height.
h t 16 t 2 2t 1 5 16
h t 16 t 1 21
2
b
2a
32
t
1
2 16
t
The firework will reach its maximum height after 1 second.
Practice Exercises B
Solve.
1.
A bolt falls off an airplane at an altitude of
500 m. How long will it take the bolt to
reach the ground?
2.
A ball is thrown upward at a speed of 30
m/sec from an altitude of 20 m. What is
the maximum height of the ball?
3.
How far will an object fall in 5 seconds if
it is thrown downward at an initial
velocity of 30 m/sec from a height of 200
m?
A coin is tossed upward with an initial
velocity of 30 ft/sec from an altitude of 8
feet. What is the maximum height of the
coin?
A water balloon is dropped from a height
of 26 feet. How long before it lands on
someone who is 6 feet tall?
4.
A ring is dropped from a helicopter at an
altitude of 246 feet. How long does it take
the ring to reach the ground?
6.
What is the height of an object after two
seconds, if thrown downward at an initial
velocity of 20 ft/sec from a height of 175
feet?
A potato is launched from the ground with
an initial velocity of 15 m/sec. What is its
maximum height?
5.
7.
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8.
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Secondary Mathematics 2
Solving Quadratic Inequalities in One Variable
Example:
Solve x2 2 x 3 0 .
Find where the expression on the left side of the inequality equals zero.
x2 2 x 3 0
x 3 x 1 0
x 3 0 or x 1 0
x 3 or x 1
We are asked to find where the expression is greater than zero, in other words, where the
expression is positive. Determine if the expression is positive or negative around each
zero. Select a value in the interval and evaluate the expression at that value, then decide
if the result is positive or negative.
1 x 3
x0
x 1
x 2
2
2
2 2 3
4 4 3
4 43
5
Positive
0
2
2 0 3
003
03
3
Negative
x3
x4
4
2
2 4 3
16 8 3
83
5
Positive
There are two intervals where the expression is positive: when x 1 and when x 3 .
Therefore, the answer to the inequality is , 1 3, . The answer could be
represented on a number line as follows:
Look at the graph of the function f x x 2 2 x 3 . Determine the intervals where the
function is positive.
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Secondary Mathematics 2
The function is positive , 1 3, . Notice that this interval is the same interval
we obtained when we tested values around the zeros of the expression.
Example:
Solve 3x2 5x 2 0 .
Find where the expression on the left side of the inequality equals zero.
3x 2 5 x 2 0
3x 2 5 x 2 0
3x 1 x 2 0
3x 1 0 or x 2 0
3x 1 or x 2
1
x or x 2
3
We are asked to find where the expression is less than or equal to zero, in other words,
where the expression is negative or zero. Determine if the expression is positive or
negative around each zero. Select a value in the interval and evaluate the expression at
that value, then decide if the result is positive or negative.
x 13
13 x 2
x2
x 1
x0
3 0 5 0 2
x3
3 3 5 3 2
3 0 0 2
002
2
Positive
3 9 15 2
27 15 2
10
Negative
3 1 5 1 2
2
3 1 5 2
3 5 2
6
Negative
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2
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2
Secondary Mathematics 2
There are two intervals where the expression is negative: when x 13 and when x 2 .
It equals zero at x 13 and x 2 . Therefore, the answer to the inequality is
, 13 2, .
The answer could be represented on a number line as follows:
Example:
Solve
2
0.
x4
Find where the denominator is equal to zero.
x40
x 4
We are asked to find where the expression is less than zero, in other words, where the
expression is negative. Determine if the expression is positive or negative around where
the denominator is equal to zero. Select a value in the interval and evaluate the
expression at that value, then decide if the result is positive or negative.
x 4
x 5
2
5 4
2
1
2
Negative
x 4
x 3
2
3 4
2
1
2
Positive
The function is negative when x 4 . Therefore, the answer to the inequality is
, 4 . The answer could be represented on a number line as follows:
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Secondary Mathematics 2
Example:
Solve
x3
0.
2x 5
Find where both the numerator and the denominator are equal to zero.
2x 5 0
x3 0
5
x
x 3
2
We are asked to find where the expression is greater than or equal to zero, in other words,
where the expression is positive. Determine if the expression is positive or negative
around where the numerator and denominator are equal to zero. Select a value in the
interval and evaluate the expression at that value, then decide if the result is positive or
negative.
5
2
5
2
x3
x 3
3 x
x 4
x0
x3
2x 5
4 3
2 4 5
x3
2x 5
03
2 0 5
x3
2x 5
33
2 3 5
1
13
1
13
3
5
3
5
6
1
6
Positive
Negative
x
Positive
There are two intervals where the expression is positive: when x 3 and when x 52 .
The function is only equal to zero when x 3 because the denominator cannot equal
5
zero. Therefore, the answer to the inequality is , 3 , . The answer could be
2
represented on a number line as follows:
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Secondary Mathematics 2
Example:
A rocket is launched with an initial velocity of 160 ft/sec from a 4 foot high platform.
How long is the rocket at least 260 feet?
Write an inequality to represent the situation. The initial velocity, v0 , is 160 ft/sec and
the initial height, h0 , is 4 feet. The height will be at least (greater than or equal to) 260 ft.
16t 2 v0t h0 h
16t 2 160t 4 260
You need to determine the real world domain for this situation. Time is the independent
variable. It starts at 0 seconds and ends when the rocket hits the ground at 10 seconds.
Move all the terms to one side of the inequality so the expression is compared to zero.
16t 2 160t 256 0
Find where the expression is equal to zero.
16t 2 160t 256 0
16 t 2 10t 16 0
16 t 2 t 8 0
t 2 0 or t 8 0
t 2 or t 8
Determine if the expression is positive or negative around each zero. Select a value in
the interval and evaluate the expression at that value, then decide if the result is positive
or negative.
0t 2
t 1
2
16t 160t 256
2t 8
t 3
2
16t 160t 256
8 t 10
t 9
2
16t 160t 256
16 1 160 1 256
16 3 160 3 256
16 9 160 9 256
16 1 160 1 256
16 9 480 256
144 480 256
80
Positive
16 81 1440 256
2
16 160 256
112
negative
2
2
1296 1440 256
112
negative
The rocket is at or above 260 feet from t 2 seconds to t 8 seconds. The difference is
6, so the rocket is at least 260 feet for 6 seconds.
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Secondary Mathematics 2
Practice Exercises C
Solve.
1.
x2 18x 80 0
2.
x2 11x 30 0
3.
x2 6 x 8
4.
x2 4 x 3
5.
3x 2 6 x 0
6.
3x2 17 x 10 0
7.
1
0
x 5
8.
x2
0
x4
9.
2x 1
0
x 3
10. A bottle of water is thrown upward with an initial velocity of 32 ft/sec from a cliff that is
1920 feet high. For what time does the height exceed 1920 feet?
11. A company determines that its total profit function can be modeled by
P x 2 x 2 480 x 16,000. Find all values of x for which it makes a profit.
12. A rocket is launched with an initial velocity of 24 m/sec from a platform that is 3 meters
high. The rocket will burst into flames unless it stays below 25 meters. Find the interval of
time before the rocket bursts into flames.
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Secondary Mathematics 2
Solving for a Specified Variable
Sometimes it is necessary to use algebraic rules to manipulate formulas in order to work with a
variable imbedded within the formula.
Given the area of a circle, solve for the radius.
A
r2
Solve for r:
Given: A r 2
A
r
Surface area of a right cylindrical solid with
radius r and height h
A 2 r 2 2 rh
You may have to use the quadratic formula.
Solve for r:
A 2 r 2 2 rh
0 2 r 2 2 rh A
a 2 b 2 h c A
r
2 h
2 h 4 2 A
2 2
2
2 h 4 2 h 2 8 A
r
4
2 h 2 2 h 2 2 A
r
4
r
h 2 h 2 2 A
2
Practice Exercises D
Solve for the indicated variable.
1.
a 2 b2 c2 ; solve for b
2.
S 2hl 2hw 2lw;solve for h
3.
A 6s 2 ; solve for s
4.
A A0 1 r ; solve for r
5.
N
k 2 3k
; solve for k
2
6.
F
7.
N
1 2
n n ; solve for n
2
8.
x 1 y 3
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2
Gm1m2
; solve for r
r2
2
2
r 2 ; solve for y
Secondary Mathematics 2
Unit 3 Cluster 3 Honors: Polynomial and Rational Inequalities
Cluster 3: Creating equations that describe numbers or relationships
H.1.2 Solve polynomial and rational inequalities in one variable.
Example:
Solve x4 13x2 30 0 .
Find where x4 13x2 30 0 .
x4 13x2 30 0
x 4 13x 2 30 0
x
2
10 x 2 3 0
x 2 10 0
x 2 10
This expression is quadratic in nature.
Factor the expression using that technique
and use the zero product property to solve
for each factor.
x2 3 0
or
x2 3
x 3
x 10
Test around each zero to determine if the expression is positive or negative on the interval.
Test
Expression evaluated
Interval
Positive/Negative
Point
at point
4
2
4 13 4 30
Positive
x 4
x 10
78
4
2
2 13 2 30
Negative
x
2
10 x 3
6
4
2
0 13 0 30
Positive
x
0
3x 3
30
2
24 13 2 30
Negative
x2
3 x 10
6
4
2
4 13 4 30
x4
Positive
x 10
78
The expression is less than zero when it is negative.
The expression is negative on the intervals 10 x 3 and
The answer can also be written as 10, 3
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3 x 10 .
10, 3 .
Secondary Mathematics 2
The answer could also be represented on a number line.
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
below the x-axis. Notice the intervals are the same.
Example:
Solve x3 6 x 2 0
x3 6 x 2 0
x3 6 x 2 0
Find where x3 6 x 2 0
x2 x 6 0
Factor the expression and use the zero product
property to solve for each factor.
x6 0
x 0
or
x 6
x0
Test around each zero to determine if the expression is positive or negative on the interval.
Test
Expression evaluated
Interval
Positive/Negative
Point
at point
2
10
3
6 10
x 6
x 10
6 x 0
x 1
400
3
2
1 6 1
x5
5
3
2
5 6 5
x0
2
Negative
Positive
Positive
275
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Secondary Mathematics 2
The expression is greater than zero when it is positive.
The expression is positive on the intervals 6 x 0 and x 0 . Keep in mind it is also
equal to zero so the endpoints are also included. The intervals that satisfy the inequality
are 6 x 0 and x 0. The interval would be written x 6 .
The answer can also be written as 6, .
The answer could also be represented on a number line.
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
above the x-axis, including the values on the x-axis. Notice the intervals are the same.
Example:
x 1 x 2 x 4 0
Find where x 1 x 2 x 4 0
x 1 x 2 x 4 0
x 1 x 2 x 4 0
x 1 0
x20
x4 0
or
or
x 1
x 2
x4
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Use the zero product property to solve for each
factor.
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Secondary Mathematics 2
Test around each zero to determine if the expression is positive or negative on the interval.
Interval
Test Point
x 2
x 5
2 x 1
x0
1 x 4
x2
x4
x6
Expression evaluated at
point
5 1 5 2 5 4
162
0 1 0 2 0 4
8
2
1
2
2 2 4
8
6 1 6 2 6 4
80
Positive/Negative
Negative
Positive
Negative
Positive
The expression is less than zero when it is negative.
The expression is negative on the intervals x 2 and 1 x 4 . Keep in mind it is also
equal to zero so the endpoints are also included. The intervals that satisfy the inequality
are x 2 and 1 x 4.
The answer can also be written , 2 1, 4 .
The answer could also be represented on a number line.
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
below the x-axis, including the values on the x-axis. Notice the intervals are the same.
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Page 165
Secondary Mathematics 2
Example:
2x 5 x 1
x 1 x 1
2x 5 x 1
x 1 x 1
Compare the inequality to zero.
2x 5 x 1
0
x 1 x 1
Make sure the denominators (the bottoms) are
2 x 5 x 1
the same.
0
x 1
2x 5 x 1
Combine the numerators (the tops).
0
x 1
x6
0
x 1
x6 0
x 1 0
Set all of the factors in the numerator and
or
denominator equal to zero, and then solve.
x 1
x 6
Test around each zero to determine if the expression is positive or negative on the interval.
Expression
Positive/Negative
evaluated at point
10 6 4
10 1 9
Positive
x 10
x 6
4
9
3 6 3
3 1 2
Negative
6 x 1
x 3
3
2
06 6
x 1
x0
Positive
0 1 1
6
The expression is greater than zero when it is positive.
The expression is positive on the intervals x 6 and x 1 . Keep in mind it is also equal
to zero so the endpoints are also included except for the denominator which cannot be
equal to zero. The intervals that satisfy the inequality are x 6 and x 1.
The answer can also be written as , 6 1, .
Interval
Test Point
The answer could also be represented on a number line.
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Page 166
Secondary Mathematics 2
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
above the x-axis, including the values on the x-axis. Notice the intervals are the same.
Example:
x3
1
x 3
x3
1
x3
x3
1 0
x3
x3
x 3
1
0
x3
x3
x 3 x 3
0
x3 x3
2x
0
x3
2x 0
x 3 0
or
x0
x3
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Compare the inequality to zero.
Make sure the denominators (the bottoms) are
the same.
Combine the numerators (the tops).
Set all of the factors in the numerator and
denominator equal to zero, and then solve.
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Secondary Mathematics 2
Test around each zero to determine if the expression is positive or negative on the interval.
Interval
Test Point
x0
x 5
Expression
evaluated at point
2 5 10
5 3 8
Positive/Negative
Positive
5
4
0 x3
x 1
x3
x5
2 1
2
2
1 3
1
2 5 10
5 3 2
Negative
Positive
5
The expression is greater than zero when it is positive.
The expression is positive on the intervals x 0 and x 3 . Keep in mind the denominator
cannot be equal to zero.
The answer can also be written as ,0 3, .
The answer could also be represented on a number line.
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
above the x-axis, including the values on the x-axis. Notice the intervals are the same.
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Page 168
Secondary Mathematics 2
Example:
4
3
x 2 x 1
4
3
x 2 x 1
4
3
0
x 2 x 1
4 x 1
3 x2
0
x 2 x 1 x 1 x 2
4 x 1
3 x 2
0
x 2 x 1 x 2 x 1
Compare the inequality to zero.
Make sure the denominators (the bottoms)
are the same.
Combine the numerators (the tops).
4 x 4 3x 6
0
x 2 x 1
x 10
0
x 2 x 1
x 10 0
x 1 0
x2 0
Set all of the factors in the numerator and
or
or
denominator equal to zero, and then solve.
x 1
x 10
x2
Test around each zero to determine if the expression is positive or negative on the interval.
Interval
Test Point
x 10
x 15
10 x 1
x 5
1 x 2
x0
x2
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Expression
evaluated at point
15 10
15 2 15 1
5
238
5 10
5 2 5 1
5
28
0 10
10
0 2 0 1 2
5
5 10
15
5 2 5 1 18
x5
5
6
Page 169
Positive/Negative
Negative
Positive
Negative
Positive
Secondary Mathematics 2
The expression is less than zero when it is negative.
The expression is negative on the intervals x 10 and 1 x 2 . Keep in mind the
denominator cannot be equal to zero.
The answer can also be written as , 10 1, 2 .
The answer could also be represented on a number line.
Looking at the graph, the intervals that satisfy this inequality are the parts of the function
below the x-axis. Notice the intervals are the same.
Practice Exercises A
Solve.
1.
x 3 x 4 x 1 0
2.
x 5 x 1 x 2 0
3. x3 25x 0
4. x 4 x2 2
5. x3 x 2
6. x 4 4 x 2
7.
10.
x
6
x2 x2
5x 7
8
x2 x2
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8.
11.
4x 5
3
x2
1
1
x2 x3
Page 170
9.
x 3 x 2 0
2
x 1
12.
5
3
x5
2x 1
Secondary Mathematics 2
You Decide
Carter’s spaceship is trapped in a gravitational field of a newly discovered Class M planet. Carter
will be in danger if his spaceship’s acceleration exceeds 500 m/h/h. If his acceleration can be
2500
modeled by the equation A(t )
m/h/h , for what range of time is Carter’s spaceship
2
5 t
below the danger zone?
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Secondary Mathematics 2
Unit 3 Cluster 3 (A.CED.2)
Writing and Graphing Equations in Two Variables
Cluster 3: Creating equations that describe numbers or relationships
3.3.2 Write and graph equations in 2 or more variables with labels and scales
Writing Quadratic Functions Given Key Features
A quadratic function can be expressed in several ways to highlight key features.
f x a x h k highlights the vertex h, k .
2
Vertex form:
Factored from: f x a x p x q highlights the x-intercepts p, 0 and q, 0 .
It is possible to write a quadratic function when given key features such as the vertex or the
x-intercepts and another point on the graph of the parabola.
Example:
Write a quadratic equation for a parabola that has its vertex at 2, 4 and passes through
the point 1, 6 .
Answer:
f x a x h k
2
f x a x 2 4
2
6 a 1 2 4
2
You are given the vertex which is a key feature
that is highlighted by the vertex form of a
quadratic function. Use this form to help you
write the equation for the parabola graphed.
The vertex is 2, 4 . h 2 and k 4 . Substitute
these values into the equation and simplify if
necessary.
Use the point 1, 6 to help you find the value of
a. The value of the function is 6 when x = 1 so
substitute 1 in for x and 6 in for f(x).
6 a 1 2 4
2
6 a 1 4
2
Use order of operations to simplify the expression
on the right side of the equation then solve for a.
6 a 1 4
6a4
2a
f x 2 x 2 4
2
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Rewrite the expression substituting in the value
for a.
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Secondary Mathematics 2
Example: Write an equation for the parabola graphed below.
vertex
3, 1
5, 5
Answer:
f x a x h k
2
f x a x (3) (1)
2
f x a x 3 1
2
5 a 5 3 1
2
You are given the vertex which is a key feature
that is highlighted by the vertex form of a
quadratic function. Use this form to help you
write the equation for the parabola graphed.
The vertex is 3, 1 . h 3 and k 1 .
Substitute these values into the equation and
simplify if necessary.
Use the point 5, 5 to help you find the value
of a. The value of the function is -5 when x = -5
so substitute -5 in for x and -5 in for f(x).
5 a 2 1
2
5 a 4 1
Use order of operations to simplify the expression
on the right side of the equation then solve for a.
5 4 a 1
4 4 a
1 a
f x x 3 1
2
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Rewrite the expression substituting in the value
for a.
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Secondary Mathematics 2
Example:
Write an equation for a parabola with x-intercepts 3, 0 and 5, 0 and passes through
the point 3, 3 .
Answer:
f x a x p x q
f x a x (3) x 5
f x a x 3 x 5
3 a 3 3 3 5
3 a 3 3 3 5
You are given the x-intercepts which are a key
feature that is highlighted by the factored form of
a quadratic function. Use this form to help you
write the equation for the parabola graphed.
One x-intercept is 3, 0 so p 3 . The other
x-intercept is 5, 0 q 5 . Substitute these
values into the equation and simplify if
necessary.
Use the point 3, 3 to help you find the value
of a. The value of the function is -3 when x = 3
so substitute 3 in for each x and -3 in for f(x).
3 a 6 2
3 12a
3
a
12
1
a
4
1
f x x 3 x 5
4
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Use order of operations to simplify the
expression on the right side of the equation then
solve for a.
Rewrite the expression substituting in the value
for a.
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Secondary Mathematics 2
Example:
Write an equation for the parabola graphed below.
Answer:
f x a x p x q
f x a x (2) x 2
f x a x 2 x 2
6 a 0 2 0 2
6 a 0 2 0 2
You are given the x-intercepts which are a key
feature that is highlighted by the factored form of
a quadratic function. Use this form to help you
write the equation for the parabola graphed.
One x-intercept is 2, 0 so p 2 . The other
x-intercept is 2, 0 q 2 . Substitute these
values into the equation and simplify if
necessary.
Use the point 0, 6 to help you find the value of
a. The value of the function is 6 when x = 0 so
substitute 0 in for each x and 6 in for f(x).
6 a 2 2
6 4a
6
a
4
3
a
2
f x
Use order of operations to simplify the
expression on the right side of the equation then
solve for a.
3
x 2 x 2
2
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Rewrite the expression substituting in the value
for a.
Page 175
Secondary Mathematics 2
Practice Exercises A
Write a quadratic equation for the parabola described.
1. Vertex: 2,3
2. Vertex: 1, 4
3. Vertex: 3, 1
Point: 1,8
Point: 2, 0
Point: 0, 7
4. Intercepts: 2, 0 4, 0
6. Intercepts: 5,0 4, 0
5. Intercepts: 1,0 7, 0
Point: 1,3
Point: 3,8
Point: 5, 12
Write a quadratic equation for the parabola graphed.
7.
8.
9.
10.
12.
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11.
Page 176
Secondary Mathematics 2
Graphing Quadratic Equations
Graphing from Standard Form f x ax 2 bx c
Example:
f x x2 2x 2
x
x
b
2a
2
2 1
f x x2 2x 2
f 1 1 2 1 2
2
1
f 0 0 2 0 2
2
f (0) 2
f x 3
Find the vertex.
The vertex is (1, -3)
Plot the vertex.
Find the y-intercept.
The y-intercept is (0, -2)
Plot the y-intercept.
Use the axis of symmetry to find another point
that is the reflection of the y-intercept.
Connect the points, drawing a smooth curve.
Remember quadratic functions are “U”
shaped.
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Secondary Mathematics 2
Graphing from Vertex Form f x a( x h)2 k
Example:
1
2
f x x 5 2
5
f x a ( x h) 2 k
f x
Find the vertex.
The vertex is (-5, -2)
Plot the vertex.
1
2
0 5 2
5
f (0) 3
Find the y-intercept.
The y-intercept is (0, 3)
Plot the y-intercept.
1
2
x 5 2
5
The vertex is (h, k).
f 0
Use the axis of symmetry to find another point
that is the reflection of the y-intercept.
Connect the points, drawing a smooth curve.
Remember quadratic functions are “U”
shaped.
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Secondary Mathematics 2
Graphing from Factored Form f x a( x p) x q
Example:
1
f x x 4 x 3
2
f x
1
x 4 x 3
2
x40
x 4
x3 0
x3
4 3
1
2
2
1 1
1
1
f 4 3
2 2
2
2
1 7 7
1
f
2 2 2
2
1 49
f
6.125
2 8
Find the x-intercepts.
The x-intercepts are
(-4, 0) and (3, 0)
Plot the x-intercepts.
Find the x-coordinate between the two
intercepts.
Use the x-coordinate
to find the ycoordinate.
The vertex is
1 49
,
2 8
Plot the vertex.
Connect the points, drawing a smooth curve.
Remember quadratic functions are “U”
shaped.
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Page 179
Secondary Mathematics 2
Practice Exercises B
Graph the following equations.
1.
f ( x) x 2 6 x 6
2.
f ( x) 2 x 2 4 x 1
3.
1
f ( x) x 2 x 2
3
4.
f ( x)
1
2
x 1 3
2
5.
f ( x) x 2 5
6.
f ( x) x 3 8
1
x 2 x 5
2
8.
f ( x) x 1 x 5
9.
7.
f ( x)
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2
Page 180
2
f ( x) 2 x 1 x 3
Secondary Mathematics 2
Unit 3 Cluster 6 (A.REI.7): Solve Systems of Equations
3.6.1
Solve simple systems containing linear and quadratic functions algebraically and
graphically.
Recall solving systems of equations in Secondary 1. We are looking for the intersection of the
two lines. There were three methods used. Below are examples of each method.
x 2y 4
Solve:
2x 3y 1
Graphing
Graph the two equations and find the
intersection.
The intersection is (-10, -7).
Substitution
x 2y 4
2(2 y 4) 3 y 1
4 y 8 3y 1
y 8 1
y 7
1. Solve for x in the first equation.
2. Substitute the solution for x in the second
equation
3. Solve for y
x 2y 4
x 2 7 4
x 14 4
x 10
The solution is (-10, -7)
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4. Substitute y back into the first equation to
solve for x.
5. The solution is the intersection.
Page 181
Secondary Mathematics 2
Elimination
x 2y 4
2x 3y 1
2 x 2 y 4
2x 3y 1
2 x 4 y 8
2x 3y 1
y 7
x 2y 4
x 2 7 4
x 14 4
x 10
The intersection is (-10, -7)
1. In order to eliminate the x’s, multiply the top
equation by -2.
2. Combine the two equations.
3. Substitute y 7 into either original
equation in order to solve for x
4. The solution is the intersection.
We will use these methods to help solve systems involving quadratic equations.
Example:
Find the intersection of the following two equations:
y x2 4
x y 2
Graphing
Graph the two equations and find the
intersection(s).
The intersections are (-2, 0) and (3, 5).
Substitution
x x 2 4 2
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1. The first equation is already solved for y;
substitute x 2 4 for y in the second
equation.
Page 182
Secondary Mathematics 2
x 4 2
x 2 x 4 2
2. Simplify and write in standard polynomial
form.
x2 x 6 0
x2 x 6 0
x2 x 6 0
x 3 x 2 0
3. Solve for x using the method of your choice
x 3 or x 2
y x2 4
y x2 4
y 3 4
y 2 4
y 94
y5
y 44
y0
2
2
4. Substitute the x values back into the first
equation to solve for y.
The solutions are (3, 5) and (-2, 0)
5. The solutions are the intersections.
Elimination
x2 0x 4 y
1. Line up like variables.
x2 y
x 0x 4 y
1( x 2 y )
2
x2 0 x 4 y
x 2 y
2. Multiply the second equation by -1 then
combine the two equations.
x2 x 6 0
x2 x 6 0
x 3 x 2 0
3. Solve using the method of your choice.
x 3 or x 2
y x2 4
y x2 4
y 3 4
y 2 4
y 94
y5
y 44
y0
2
2
The solutions are (3, 5) and (-2, 0)
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4. Substitute the x values back into the first
equation to solve for y.
5. The solutions are the intersections.
Page 183
Secondary Mathematics 2
Example:
Using the method of your choice, find the intersection between the following equations:
x2 y 2 4
2 x y 1
Substitution
y 2 x 1
1. Solve for x in the second equation.
2. Substitute 2 x 1 for y in the first
equation.
x 2 2 x 1 4
2
x2 4 x2 4 x 1 4
5x2 4 x 3 0
x
4 42 4 5 3
2 5
4 16 60
10
4 76
x
10
4 2 19
x
10
2 19
x
5
2 19
2 19
x
or x
5
5
x 0.472
or x 1.272
x
3. Simplify and solve for x.
y 2 x 1
y 2 x 1
2 19
y 2
1
5
2 19
y 2
1
5
1 2 19
y
5
y 1.944
1 2 19
y
5
y 1.544
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Page 184
4. Substitute the x values back into the
first equation to solve for y.
Secondary Mathematics 2
2 19 1 2 19
,
The solutions are
and
5
5
2 19 1 2 19
,
5
5
5. The solutions are the intersections.
Or approximately (0.472, -1.944) or (-1.272, 1.544)
Practice Exercises A
Solve each of the systems of equations.
1.
4.
7.
10.
x y 6
y2 x 3
y x2
3x y 2
y x2
x y2
x 2 y 2 45
yx3
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2.
x 2 y 2 41
y x 1
5.
y2 x 3
2y x 4
8.
11.
2x y 1
y 4 x2
x 2 y 2 14
x2 y 2 4
Page 185
3.
3x y 7
4 x 2 5 y 24
6.
4 x 2 9 y 2 36
3y 2x 6
9.
x 2 y 2 89
x y 3
12.
x 12 y 32 4
yx
Secondary Mathematics 2
Unit 3 Cluster 6 Honors (A.REI.8 and A.REI.9)
Solving Systems of Equations with Vectors and Matrices
H.3.1 Represent a system of linear equations as a single matrix equation in a vector
variable.
H.3.2 Find the inverse of a matrix if it exists and use it to solve systems of linear
equations (using technology for matrices of dimension 3 × 3 or greater).
In Secondary Mathematics 1 Honors you learned to solve a system of two equations by writing
the corresponding augmented matrix and using row operations to simplify the matrix so that it
had ones down the diagonal from upper left to lower right, and zeros above and below the ones.
1 0 g
0 1 h
When a matrix is in this form it is said to be in reduced row-echelon form. The process for
simplifying a matrix to reduced row-echelon form is called Gauss-Jordan elimination after the
two mathematicians, Carl Friedrich Gauss and Wilhelm Jordan. This process, using row
operations, can be used for systems of two or more variables.
0 2 6 3
Row operations are listed below using the original matrix 1 4 7 10 .
2 6 9 1
Note: R indicates row and the number following indicates which row. For instance, R1 indicated
row one.
Interchange Rows
We want to have ones
along the diagonal. We
can switch rows so that
the one ends up in the
correct position.
Symbol: R1 R2
All of the elements of
Row 1 will switch
positions with all the
elements of Row 2.
Multiply a row by a scalar
We want positive ones
down the diagonal.
Multiplying each
element by a scalar k,
Symbol: k R1
k 0 , allows us to
change values in one
row while preserving
the overall equality.
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Page 186
1 4 7 10
0 2 6 3
2 6 9 1
3
1
R1 = 0 1 3
2
2
The matrix is now:
0 1 3 32
1 4 7 10
2 6 9 1
Secondary Mathematics 2
Combine Two Rows
You can add or subtract
two rows to replace a
single row. This enables
you to get ones along
the diagonal and zeros
elsewhere.
R 2 : 0 2 6 3 1 4 7 10
R 2 : 1 6 1 13
R2 : R1 R2
The matrix is now:
0 2 6 3
1 6 1 13
2 6 9 1
Multiply by a scalar, then combine rows
Sometimes it is
necessary to multiply a
row by a scalar before
combining with another
row to get a one or a
zero where needed.
R3 : 2 1 4 7 10 2 6 9 1
R3 : 2 8 14 20 2 6 9 1
R3 : 0 2 23 19
R3: 2 R2 R3
The matrix is now:
0 2 6 3
1 4 7 10
0 2 23 19
Practice Exercises A
Perform the row operations on the given matrix.
3 3 1 6
8 5 5 1
0 6 9 3
1.
R2 R3
2.
13 R1
3.
R2 : 1 R3 R2
Using the matrix below, write appropriate row operation(s) to get the desired results.
(Recall that amn indicates the element of the matrix is located in row m and column n.)
3 8 3 10
2 5 10 9
7 5 5 5
4.
a11 1
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5.
a21 0
Page 187
6.
a32 0
Secondary Mathematics 2
Example: Solve the following systems of equations using the Gauss-Jordan elimination method.
x 2y z 7
3x 5 y z 14
2x 2 y z 3
1 2 1 7
3 5 1 14
2 2 1 3
Rewrite the equations in matrix form.
1 2 1 7
0 1 2 7
2 2 1 3
R 2 : 3R1 R 2
1 2 1 7
0 1 2 7
0 2 3 11
R3 : 2 R1 R3
1 0 3 7
0 1 2 7
0 2 3 11
R1: 2 R 2 R1
1 0 3 7
0 1 2 7
0 0 1 3
R3 : 2 R 2 R 2
1 0 0 2
0 1 2 7
0 0 1 3
R1: 3R3 R1
1 0 0 2
0 1 0 1
0 0 1 3
R 2 : 2 R3 R 2
The solution is 2, 1,3
R 2 : 3 6 3 21 3 5 1 14
R 2 : 0 1 2 7
R3 : 2 4 2 14 2 2 1 3
R3 : 0 2 3 11
R1: 0 2 4 14 1 2 1 7
R1: 1 0 3 7
R3 : 0 2 4 14 0 2 3 11
R3 : 0 0 1 3
R1: 0 0 3 9 1 0 3 7
R1: 1 0 0 2
R 2 : 0 0 2 6 0 1 2 7
R 2 : 0 1 0 1
Rewriting the answer in equation form you end
up with:
x2
y 1
z 3
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Page 188
Secondary Mathematics 2
Practice Exercises B
Using Gauss-Jordan Elimination solve the following systems of equations.
1.
3x 2 y 19
2x y 1
2.
2 x y 3z 13
x 2 y 2z 3
3x y z 4
Technology can be used to help solve systems of equations. The following system of equations
can be solved using your graphing calculator.
2 x y z 11
x 2 y 2 z 11
x y 3z 24
2 1 1 11
1 2 2 11
1 1 3 24
Write the augmented matrix.
Enter the matrix into your calculator.
Note: [A] is the default matrix
Push 2nd x 1
Arrow over to choose EDIT
Push ENTER
Type the dimensions. After each number push
ENTER.
For this example: 3 ENTER 4 ENTER
Type each element in the first row. Push
ENTER after each number. Continue until
every row has been entered.
Push 2nd MODE to return to the home screen
Push 2nd x 1 Then ENTER ENTER
This will give you a chance to check your
matrix for accuracy.
Push 2nd x 1 arrow over to MATH.
Either arrow down to option B or push
ALPHA APPS to get to rref(
This is row reduced echelon form.
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Secondary Mathematics 2
Push 2nd x 1 ENTER )
Or select the matrix in which your equations
are stored.
Push ENTER to obtain the answer.
Rewriting the answer in equation form you end
up with:
x 1
The answer is (-1, 2, 7)
y2
z7
Practice Exercises C
Solve each of the following systems using technology.
1.
3x y 2 z 31
x y 2 z 19
x 3 y 2 z 25
2.
5 x 2 y 3z 0
x y 5
2 x 3z 4
3.
2x y 2z 2
3x 5 y z 4
x 2 y 3z 6
Not every system has a single point as the solution. The following situations may also occur.
A single point solution
Consistent and
Independent
x yz 5
4 x 2 y z 1
9 x 3 y z 13
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1 0 0 4
0 1 0 6
0 0 1 5
4, 6, 5
Page 190
Secondary Mathematics 2
A line solution defined by one
variable
Consistent and
Dependent
6x y z 4
12 x 2 y 2 z 8
5x y z 3
1 0 112 117
1
2
0 1 11 11
0 0 0 0
Rewriting the answer in
equation form you end up with:
x 112 z 117
y 111 z 112
0 0 this is always true
Note: z is an independent
variable and can take on
any real value
The solution is written as:
117 112 z, 112 111 z, z
A plane solution defined by
two variables
1 0 1 1 2
0 1 2 1 1
0 0 0 0 0
Consistent and
Dependent
x y 3z 1
xzw2
2x y 4z w 3
Rewriting the answer in
equation form you end up with:
xzw2
y 2 z w 1
0 0 this is always true
All three planes coincide.
Note: z and w are independent
variables and can take on any
real value
The solution is written as:
2 w z, 1 w 2 z, z, w
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Page 191
Secondary Mathematics 2
No solution
There are no intersections
common to all three planes or
the three planes are parallel
Inconsistent
x yz 6
2x y z 3
x 2 y 2z 0
1 0 0 0
0 1 1 0
0 0 0 1
The equations would be:
x0
yz 0
0 1 this is not true
or
No Solution
Practice Exercises D
Solve the following systems. Indicate if the system is consistent or inconsistent.
1.
4.
x 2y z 0
2 x 2 y 3 z 3
y z 1
x 4 y 2 z 13
x y 2z 2
2x 3y 6z 5
3x 4 y 4 z 12
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2.
5.
5 x 8 y 6 z 14
3x 4 y 2 z 8
x 2 y 2z 3
x y 2w 3
3x 2 y w 4
x 3 y z 3w 1
2 x 4 y z w 2
Page 192
3.
6.
5 x 12 y z 10
2 x 5 y 2 z 1
x 2 y 3z 5
2x 3y z w 7
2x 3y z 4
4 x y w 3
Secondary Mathematics 2
Find the Inverse of a Matrix
Two n n matrices are inverses of one another if their product is the n n identity matrix. Not
all matrices have an inverse. An n n matrix has an inverse if and only if the determinant is not
zero. The inverse of A is denoted by A1 . There are two ways to find the inverse both can be
done using technology.
1 1 0
1 1 0
A 1 3 4
1 3 4 4
0 4 3
0 4 3
The determinant is not zero therefore the matrix has an inverse.
Method 1:
Method 2:
Enter the matrix in your calculator as
Rewrite the matrix with the 3 3 identity
matrix A.
matrix.
1 1 0 1 0 0
From the home screen push 2nd x 1 to select
1
3
4
0
1
0
your matrix.
0 4 3 0 0 1
Enter the matrix in your calculator and find Then push x 1
reduced row echelon form.
Then push ENTER
To convert to fractions push MATH Frac
To convert to fractions push MATH Frac
The inverse matrix is:
3
1
74
4
1
A 34 34 1
1
1 1
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The inverse matrix is:
3
74
4
1
3
A 4 34
1
1
Page 193
1
1
1
Secondary Mathematics 2
Practice Exercises E
Find the inverse of the following matrices.
1.
2.
1 1 1
0 2 1
2 3 0
1 1 1
4 5 0
0 1 3
4.
3.
5.
1 2 1
2 1 3
1 0 1
2 1 3
1 2 2
0 1 1
6.
2 3 1
1 0 4
0 1 1
5 0 2
2 2 1
3 1 1
Using the Inverse to Solve a System of Linear Equations
If AX B has a unique solution, then X A1B . Where A is the coefficient matrix, X is the
column variable matrix, and B is the column solution matrix.
ax by cz d
a b c
d
x
Given: ex fy gz h then, A e f g , X y , and B h
k m n
p
z
kx my nz p
Example:
x y 3
x 3 y 4 z 3
4 y 3z 2
1 1
A 1 3
0 4
74
A1 34
1
0
x
3
4 , X y , B 3
z
2
3
3
1
4
3
4 1
1 1
74
X 34
1
3
4
1
3
4
1
1
1
3
3
2
1
The solution is: 2 Written as: (1, 2, -2)
2
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Identify the A, X, and B matrices
Find the inverse of A
Use X A1B to find the solution.
x 1
y2
z 2
Page 194
Secondary Mathematics 2
Practice Exercises F
Solve using the inverse matrix.
1.
4.
7.
2x 6 y 6z 8
2 x 7 y 6 z 10
2x 7 y 7 z 9
x 6 y 3z 11
2 x 7 y 3z 14
4 x 12 y 5 z 25
3x 2 y z 2
4 x 5 y 3z 9
2 x y 5 z 5
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2.
5.
8.
x 2 y 5z 2
2 x 3 y 8z 3
x y 2z 3
x y z 6
4x 2 y z 9
4 x 2 y z 3
x y 1
6 x y 20 z 14
y 3z 1
Page 195
3.
6.
9.
x yz 8
2 y z 7
2x 3y 1
y 2z 0
x y 1
2 x y z 1
x 3 y 4 z 3
x 2 y 3z 2
x 4 y 3z 6
Secondary Mathematics 2
Unit 2 Cluster 2b (F.IF.8b), Unit 3 Cluster 1b (A.SSE.1b), Unit 3
Cluster 2c (A.SSE.3c)
Forms and Uses of Exponential Functions
Cluster 2: Analyzing functions using different representations
2.2.2b Use properties of exponents to interpret expressions for exponential functions
Cluster 1: Interpreting the structure of expressions
3.1.1b Interpret complicated expressions by looking at one or more of their parts
separately (focus on exponential functions with rational exponents using mainly
square roots and cube roots)
Cluster 2: Writing expressions in equivalent forms to solve problems
3.2.1c Use properties of exponents to rewrite exponential functions
VOCABULARY
An exponential function is a function of the form f x ab x where a and b are constants and
a 0 , b 0 , and b 1.
Exponential functions can also be of the form A P 1 r . This is the simplified interest
formula. Each part of the formula has a specific meaning. The principal, P, is the original
amount of money that is deposited. The interest rate, r, is expressed as a decimal and represents
the growth rate of the investment. Time, t, is the number of years that the money remains in the
account. The amount, A, after t years can be calculated by using the formula.
t
Example:
Austin deposits $450 into a savings account with a 2.5% interest rate. How much money
will be in the account after 5 years?
A 450 1 0.025
P $450
r 2.5% 0.025
t 5 years
Substitute the known values into the
equation
A 450 1.025
Evaluate.
A P 1 r
t
5
5
A 509.1336958
Austin will have $509.13 in his account after 5 years.
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Page 196
Secondary Mathematics 2
VOCABULARY
Interest is commonly assessed multiple times throughout the year. This is referred to as
compound interest because the interest is compounded or applied more than once during the
nt
r
year. The formula is A P 1 where n is the number of times that the interest is
n
compounded during the year.
Interest
Rate
r
A P 1
n
Final
Amount
Principal
nt
Time
(in years)
Number of times
compounded per year
Example:
Cyndi received a lot of money for her high school graduation and she plans to invest
$1000 in a Dream CD. A CD is a certified account that pays a fixed interest rate for a
specified length of time. Cyndi chose to do a 3 year CD with a 0.896% interest rate
compounded monthly. How much money will she have in 3 years?
r
A P 1
n
P $1000
r 0.896% 0.00896
n 12 times
t 3 years
nt
123
0.00896
A 1000 1
12
0.00896
A 1000 1
12
A 1027.234223
Substitute the known values into the
equation
36
Simplify the exponent and evaluate.
Cyndi will have $1027.23 after 3 years.
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Page 197
Secondary Mathematics 2
VOCABULARY
There is another interest formula where the interest is assessed continuously. It is called
continuous interest. The formula is A Pert . It uses the same P, r, and t from the other
interest formulas but it also utilizes the Euler constant e. Similar to pi, e is an irrational number.
n
1
It is defined to be lim 1 e 2.718281828 . It is also called the natural base.
n
n
Example:
Eva invested $750 in a savings account with an interest rate of 1.2% that is compounded
continuously. How much money will be in the account after 7 years?
P $750
r 1.2% 0.012
t 7 years
Substitute the known values into the
equation
A Pert
A 750e0.0127
A 750e0.084
A 815.7216704
Simplify the exponent and evaluate.
Eva will have $815.72 after 7 years.
VOCABULARY
There are times when the value of an item decreases by a fixed percent each year. This can be
t
modeled by the formula A P 1 r where P is the initial value of the item, r is the rate at
which its value decreases, and A is the value of the item after t years.
Example:
Jeff bought a new car for $27,500. The car’s value decreases by 8% each year. How
much will the car be worth in 15 years?
A 27,500 1 0.08
P $27,500
r 8% 0.08
t 15 years
Substitute the known values into the
equation
A 27,500 0.92
Evaluate.
A P 1 r
t
15
15
A 7873.178612
Jeff should sell the car for at least $7873.18.
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Secondary Mathematics 2
Practice Exercises A
Use the compound and continuous interest formulas to solve the following. Round to the nearest
cent.
1.
Bobbi is investing $10,000 in a money market account that pays 5.5% interest quarterly.
How much money will she have after 5 years?
2.
Joshua put $5,000 in a special savings account for 10 years. The account had an interest
rate of 6.5% compounded continuously. How much money does he have?
3.
Analeigh is given the option of investing $12,000 for 3 years at 7% compounded monthly
or at 6.85% compounded continuously. Which option should she choose and why?
4.
Mallory purchased a new Road Glide Ultra motorcycle for $22,879. Its value depreciates
15% each year. How much could she sell it for 8 years later?
Example:
Emily invested $1250 after 2 years she had $1281.45. What was the interest rate, if the
interested was assessed once a year?
The interest is assessed only once a year;
use the simplified interest formula.
A $1281.45
P $1250
t 2 years
Substitute the known values into the
equation
A P 1 r
t
A P 1 r
t
1281.45 1250 1 r
2
1281.45
2
1 r
1250
1281.45
2
1 r
1250
Isolate the squared term.
Find the square root of each side.
1281.45
1 r
1250
1281.45
r
1250
1 1.0125 r
1 1.0125 r
or
2.0125 r
0.0125 r
1
Solve for r.
The interest rate is positive; therefore it is 0.0125 or 1.25%.
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Page 199
Secondary Mathematics 2
Example:
Sam invested some money in a CD with an interest rate of 1.15% that was compounded
quarterly. How much money did Sam invest if he had $1500 after 10 years?
r
A P 1
n
nt
r
A P 1
n
nt
The interest is compounded quarterly; use
the compound interest formula.
A $1500
r 1.15% 0.015
n 4 times
t 10 years
0.015
1500 P 1
4
410
1500 P 1 0.00375
1500 P 1.00375
1500
1.00375
40
Substitute the known values into the
equation
40
40
Isolate P.
P
1291.424221 P
Evaluate.
Ten years ago Sam invested $1291.42.
Example:
Suzie received $500 for her birthday. She put the money in a savings account with 4%
interest compounded monthly. When will she have $750?
r
A P 1
n
P $500
r 4% 0.04
nt
n 12 times
12 t
0.04
A 500 1
12
Substitute the known values into the
equation
A $750
Put the interest formula in Y1 and $750 in
Y2. Graph the two equations and use 2nd,
Trace, intersect to find their intersection.
Suzie will have $750 in 10.154 years.
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Page 200
Secondary Mathematics 2
Practice Exercises B
Solve.
1.
Jace invested $12,000 in a 3-year Dream CD with interest compounded annually. At the
end of the 3 years, his CD is worth $12,450. What was the interest rate for the CD?
2.
Jaron has a savings account containing $5,000 with interest compounded annually. Two
years ago, it held $4,500. What was the interest rate?
3.
Lindsey needs to have $10,000 for the first semester of college. How much does she have
to invest in an account that carries an 8.5% interest rate compounded monthly in order to
reach her goal in 4 years?
4.
If Nick has $20,000 now, how long will it take him to save $50,000 in an account that
carries an interest of 5.83% compounded continuously?
VOCABULARY
When interest is assessed more than once in the year, the effective interest rate is actually higher
than the interest rate. The effective interest rate is equivalent to the annual simple rate of
interest that would yield the same amount as compounding after 1 year.
Annual compounding
Semiannual compounding
Quarterly compounding
Monthly compounding
Daily compounding
Continuous compounding
Annual Rate
10%
10%
10%
10%
10%
10%
Effective Rate
10%
10.25%
10.381%
10.471%
10.516%
10.517%
Example:
$1000 is put into a savings account with 5% interest compounded quarterly. What is the
effective interest rate?
r
A P 1
n
0.05
A 1000 1
4
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P $1000
r 5% 0.05
nt
n 4 times
4t
Substitute the known values into the
equation
Page 201
Secondary Mathematics 2
A 1000 1 0.0125
A 1000 1.0125
4t
Simplify what is inside the parentheses.
4t
A 1000 1.01254
t
A 1000 1.050945337
Use properties of exponents to rewrite the
function so that is to the power of t.
t
Rewrite it so that it is 1 plus the interest
rate. This is the simplified interest
formula.
A 1000 1 0.050945337
t
In the new simplified interest formula r 0.0509 5.09%
Example:
If $1000 were put into a savings account that paid 5% interest compounded continuously,
what would the monthly interest rate be?
P $1000
r 5% 0.05
Substitute the known values into the
equation.
A Pert
A 1000e0.05t
A 1000 e0.05
t
A 1000 1.051271096
Use properties of exponents to rewrite the
function so that is to the power of t.
t
Remember that t is the number of years
the money is being invested. It is
necessary to multiply t by 12 to convert it
to months.
A 1000 1.0512710961/12
Using algebra rules, we must also divide
by 12 so that the equation does not change
12
since
1 . Perform the division inside
12
the parenthesis so the interest rate is
affected.
12 t
A 1000 1.004175359
12 t
A 1000 1 0.004175359
12 t
Rewrite the information in the parenthesis
so that it is 1 plus the interest rate. This is
the simplified interest formula.
In the monthly interest rate is r 0.00418 0.418%
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Page 202
Secondary Mathematics 2
Practice Exercises C
Find the monthly rate or effective interest rate.
1.
If $2,500 is invested in an account with an interest rate of 7.23% compounded semiannually, what is the effective rate?
2.
If $7,700 is invested in an account with an interest rate of 9% compounded quarterly,
what is the monthly interest rate?
3.
If $235,000 is invested in an account with an interest rate of 22.351% compounded
monthly, what is the effective rate?
4.
If $550 is invested in an account with an interest rate of 45.9% compounded annually,
what is the monthly interest rate?
Exponential Growth and Decay
VOCABULARY
Money is not the only real world phenomenon that can be modeled with an exponential function.
Other phenomena such as populations, bacteria, radioactive substances, electricity, and
temperatures can be modeled by exponential functions.
A quantity that grows by a fixed percent at regular intervals is said to have exponential growth.
The formula for uninhibited growth is A t A0ekt , k 0 , where A0 is the original amount, t is
the time and k is the growth constant. This formula is similar to the continuous interest formula
A Pert . Both formulas are continuously growing or growing without any constraints.
A t A0ekt
Final
amount
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Initial
Amount
Page 203
Time
Growth
Constant
Secondary Mathematics 2
Example:
A colony of bacteria that grows according to the law of uninhibited growth is modeled by
the function A t 100e0.045t , where A is measured in grams and t is measured in days.
(a) Determine the initial amount of bacteria.
(b) What is the growth constant of the bacteria?
(c) What is the population after 7 days?
(d) How long will it take for the population to reach 140 grams?
a.
b.
c.
A0 is the initial amount and in the
equation A0 100 , therefore the initial
amount is 100 grams.
k is the growth constant and in the
equation k 0.045 , therefore the growth
rate is 0.045/day.
A t 100e0.045t
A t 100e0.045t
A t 100e0.045t
Substitute 7 in for time, t, then evaluate.
A(7) 100e0.045(7)
A(7) 137.026
After 7 days, there will be 137.026 grams
of bacteria.
d.
A t 140
Put the exponential growth equation in
Y1 and 140 in Y2. Graph the two
equations and use 2nd, Trace, intersect to
find their intersection.
It will take 7.477 days for the bacteria to grow to 140 grams.
VOCABULARY
A quantity that decreases by a fixed percent at regular intervals is said to have exponential
decay. The formula for uninhibited decay is A t A0ekt , k 0 where A0 is the original
amount, t is the time and k is the constant rate of decay.
The decay formula is the same as the growth formula. The only difference is that when k > 0 the
amount increases over time making the function have exponential growth and when k < 0 the
amount decreases over time making the function have exponential decay.
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Page 204
Secondary Mathematics 2
Example:
A dinosaur skeleton was found in Vernal, Utah. Scientists can use the equation
A(t ) 1100e0.000124t , where A is measured in kilograms and t is measured in years, to
determine the amount of carbon remaining in the dinosaur. This in turn helps to
determine the age of the dinosaur bones.
(a) Determine the initial amount of carbon in the dinosaur bones.
(b) What is the growth constant of the carbon?
(c) How much carbon is left after 5,600 years?
(d) How long will it take for the carbon to reach 900 kilograms?
A0 is the initial amount and in the
a.
A(t ) 1100e 0.000124t
equation A0 1,100 , therefore the initial
amount is 1,100 kilograms.
k is the growth constant and in the
equation k 0.000124 . Since k is
negative, therefore the carbon is
decaying at a rate 0.000124/year.
Substitute 5600 in for time, t, then
evaluate.
A(5600) 1100e0.000124 5600
A(5600) 549.311
After 5600 years, there will be 549.311
kilograms of carbon remaining.
A(t ) 1100e0.000124t
b.
A(t ) 1100e0.000124t
c.
d.
A t 900
Put the exponential growth equation in
Y1 and 900 in Y2. Graph the two
equations and use 2nd, Trace, intersect to
find their intersection.
It will take 1,618.312 years for the carbon to decrease to 900 kilograms.
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Page 205
Secondary Mathematics 2
Practice Problems D
Solve.
1.
India is one of the fastest growing countries in the world. A t 574e0.026t describes the
population of India in millions t years after 1974.
a.
b.
c.
d.
2.
What was the population in 1974?
Find the growth constant.
What will the population be in 2030?
When will India’s population be 1,624 million?
The amount of carbon-14 in an artifact can be modeled by A t 16e0.000121t , where A is
measured in grams and t is measured in years.
a.
b.
c.
d.
How many grams of carbon-14 were present initially?
Find the growth constant.
How many grams of carbon-14 will be present after 5,715 years?
When will there be 4 grams of carbon-14 remaining?
VOCABULARY
An exponential function is a function of the form A(t ) A0bt A0 1 r where A0 is the initial
t
amount, b 1 r is the growth factor, and A0 0 , b 0 , and b 1.
Growth factor = 1 + r
If r < 0, is exponential decay
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If r > 0, is exponential growth
Page 206
Secondary Mathematics 2
Example:
A culture of bacteria obeys the law of uninhibited growth. Initially there were 500
bacteria present. After 1 hour there are 800 bacteria.
a.
Identify the growth rate.
b.
Write an equation to model the growth of the bacteria
c.
How many bacteria will be present after 5 hours?
d.
How long is it until there are 20,000 bacteria?
Find the common ratio.
a.
an 1
an
800
1.6
500
Rewrite b in the form of 1 r .
b 1.6 1 0.6
Identify the growth rate.
r 0.6 60%
r 0.6
b.
A(t ) A0bt
A(t ) 500(1.6)t
c.
A(t ) 500(1.6)t
A(5) 500(1.6)5
A(5) 5242.88
The bacterial is increasing at a rate of 60%
each hour.
A0 500
b 1.6
Substitute known values into the equation.
Substitute t 5 into the equation and
evaluate.
Round down to the nearest whole number.
There are 5,242 bacteria present after 5
hours.
A t 20,000
d.
Graph this equation and your equation in
Y1 and Y2.
Use 2nd , Trace, intersect to find the
intersection.
There will be 20,000 bacteria after 7.849
hours.
Jordan School District
Page 207
Secondary Mathematics 2
Example:
Michael bought a new laptop for $1,800 last year. A month after he purchased it, the
price dropped to $1,665.
a.
Identify the growth rate.
b.
Write an equation to model the value of the computer.
c.
What will the value of the computer be after 9 months?
d.
When will the value of the computer be $500?
a.
Find the common ratio.
an 1
an
1665
0.925
1800
Rewrite b in the form of 1 r .
b 0.925 1 0.075
Identify the growth rate.
r 0.075
r 0.075 7.5%
The value of the computer is decreasing at
a rate of 7.5% each month.
A0 1800
b 0.925
Substitute known values into the equation.
b.
A(t ) A0bt
A(t ) 1800 0.925
t
Substitute t 9 into the equation and
evaluate.
c.
A(t ) 1800 0.925
t
A(9) 1800 0.925
9
Round to the nearest cent.
After 9 months, the computer is worth
$892.38.
A t 500
A(9) 892.38
d.
Graph this equation and your equation in
Y1 and Y2.
Use 2nd , Trace, intersect to find the
intersection.
It will take 16.430 months for the
computer’s value to decrease to $500.
Jordan School District
Page 208
Secondary Mathematics 2
Example:
The population of West Jordan was 106,863 in 2011. The population is growing at a rate
of 5% each year.
a.
Write an equation to model the population growth.
b.
If this trend continues, what will the population be in 2020?
c.
How long before the population grows to 125,000 people?
a.
A(t ) A0 1 r
A0 106863
t
A(t ) 106863 1.05
r 5% 0.05
1 r 1 0.05 1.05
t
Substitute known values into the equation.
b.
A(9) 106863 1.05
t 2020 2011 9
Substitute t 9 into the equation and
9
A 9 165779.587
evaluate.
Round down to the nearest whole person.
After 9 years, the population of West
Jordan is 165,779 people.
A t 125000
c.
Graph this equation and your equation in
Y1 and Y2.
Use 2nd , Trace, intersect to find the
intersection.
It will take 16.430 months for the
computer’s value to decrease to $500.
Jordan School District
Page 209
Secondary Mathematics 2
Example:
A culture of 200 bacteria is put in a petri dish and the culture doubles every hour.
a.
Write an equation to model the bacteria growth.
b.
If this trend continues, how many bacteria will there be in 5 hours?
c.
How long before the bacteria population reaches 7000,000?
A0 200
a.
A(t ) A0bt
r 100% 1.00
b 1 r 11 2
A(t ) 200 2
t
Substitute known values into the equation.
b.
A(t ) 200 2
Substitute t 5 into the equation and
evaluate.
t
A(5) 200 2
5
After 5 hours, there are 6,400 bacteria.
A(5) 6, 400
A t 700,000
c.
Graph this equation and your equation in
Y1 and Y2.
Use 2nd , Trace, intersect to find the
intersection.
It will take 11.773 hours for the bacteria to
reach 700,000.
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Page 210
Secondary Mathematics 2
Practice Exercises E
Solve
1.
2.
3.
A bird species is in danger of extinction. Last year there were 1,400 birds and today only
1,308 of the birds are alive.
a.
Identify the growth rate.
b.
Write an equation to model the population.
c.
If this trend continues, what will the population be in 10 years?
d.
If the population drops below 100 then the situation will be irreversible. When
will this happen?
There is a fruit fly in your house. Fruit fly populations triple every day until the food
source runs out.
a.
Write an equation to model the fruit fly growth.
b.
If this trend continues, how many fruit flies will there be at the end of 1 week?
c.
How long before the fruit fly population reaches 50,000?
In 2003 the population of Nigeria was 124,009,000. It has a growth rate of 3.1%.
a.
Write an equation to model the population growth since 2003.
b.
If this trend continues, what will the population be in 2050?
c.
How long before the population grows to 200,000,000 people?
YOU DECIDE
Utah’s population was 2,763,885 in 2010 and 2,817,222 in 2011, find the growth rate. Can the
population in Utah continue to grow at this rate indefinitely? Why or why not? Justify your
answer.
Jordan School District
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Secondary Mathematics 2
Unit 4
Applications of
Probability
Jordan School District
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Secondary Mathematics 2
Unit 4 Cluster 1 (S.CP.1)
Applications of Probability
Cluster 1: Understand independence and conditional probability and use them to interpret data
4.1.1 Describe events as subsets of a sample space (the set of outcomes) using
characteristics (or categories) of the outcomes, or as unions, intersections, or
complements of other events (“or,” “and,” “not”).
VOCABULARY
An event is an activity or experiment which is usually represented by a capital letter. A sample
space is a set of all possible outcomes for an activity or experiment. A smaller set of outcomes
from the sample space is called a subset. The complement of a subset is all outcomes in the
sample space that are not part of the subset. A subset and its complement make up the entire
sample space. If a subset is represented by A, the complement can be represented by any of the
following: not A, ~A, or A c .
Example 1:
Event
Flip a coin
Sample Space
S={heads, tails}
Roll a die
S={1, 2, 3, 4, 5, 6}
Pick a digit 0-9
S={0,1, 2, 3, 4, 5, 6, 7, 8, 9}
VOCABULARY
Definition
The union of two events
includes all outcomes from
each event. The union can
be indicated by the word
“or” or the symbol .
The intersection of two
events includes only those
outcomes that are in both
events. The intersection
can be indicated by the
word “and” or the symbol
. If the two events do
NOT have anything in
common, the intersection
is the “empty set”,
indicated by { } or .
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Possible Subset
B={heads}
even
E={2, 4, 6}
N={2, 5, 7, 9}
Example
Complement
B tails
c
~E ={1, 3, 5}
not N= {0, 1, 3, 4, 6, 8}
Venn Diagram
A= 0, 2, 4, 6,8,10
B= 0,5,10,15, 20
A B= 0, 2, 4,5,6,8,10,15, 20
A= 0, 2, 4, 6,8,10
B= 0,5,10,15, 20
A B 0,10
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Secondary Mathematics 2
Example 2:
The sample space is S={Green, Violet,
Turquoise, Yellow, Blue, Red, White,
Brown, Peach, Black, Magenta, Orange}
The subset of primary colors is P={Yellow,
Blue, Red}
The subset of American Flag colors is:
A={Blue, Red, White}
P A= {Yellow, Blue, Red, White}
P A= {Blue, Red}
Pc {Green, Violet, Turquoise, White,
Brown, Peach, Black, Magenta, Orange}
~ P A ={Green, Violet, Turquoise,
Brown, Peach, Black, Magenta, Orange}
Practice Exercises A
1.
Choosing a letter from the alphabet.
A.
List the sample space.
B.
List a subset of the letters in your first name.
C.
List a subset of the letters in your last name.
D.
Find the union of the subsets of your first name and last name.
E.
Find the intersection of the subsets of your first name and last name.
2.
Given the sample space S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} with event A = {3, 4, 5, 6, 7} and
event B = {1, 2, 3, 4, 5}.
A.
Draw a Venn diagram representing the sample space with events A and B.
B.
List all the outcomes for A B.
C.
List all the outcomes for A B.
D.
List all the outcomes for A c .
3.
Given a standard deck of 52 cards, event A is defined as a red card and event B is defined
as the card is a diamond.
A.
List all the outcomes for A B.
B.
List all the outcomes for A B.
C.
What is ~A?
D.
What is A A c ?
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Secondary Mathematics 2
Unit 4 Clusters 1–2 (S.CP.2, S.CP.3, S.CP.4, S.CP.5, S.CP.6, S.CP.7)
Conditional Probability
Cluster 1: Understanding and using independence and conditional probability
4.1.2 Independence of 2 events (use the product of their probabilities to determine if
they are independent)
4.1.3 Understand conditional probability and interpret the independence of A and B
using conditional probability
4.1.4 Construct and interpret two-way frequency tables; Use two-way frequency tables
to determine independence and to find conditional probabilities
4.1.5 Explain conditional probability and independence
Cluster 2: Computing probabilities of compound events
4.2.1 Find conditional probabilities
4.2.2 Apply the Addition Rule
VOCABULARY
Probability is a value that represents the likelihood that an event will occur. It can be
represented as a fraction, decimal 0 probabiltiy 1 , or percent. A probability of zero (0)
means that the event is impossible and a probability of one (1) means that the event must occur.
Events are independent if the occurrence of one event does not change the probability of
another event occurring. Events are dependent if the occurrence of one changes the probability
of another event occurring. For example, drawing marbles from a bag with replacement is
independent, while drawing marbles from a bag without replacement is dependent.
Joint probability is the likelihood of two or more events occurring at the same time.
Formula
P A
number of favorable outcomes
total number of outcomes
P(A B)=P(A) P(B)
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Description
Probability of the
individual event A
occurring.
Joint probability
of independent
events.
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Example
Flipping a coin
1
P(heads)=
2
Flipping a coin AND rolling a die
P(heads and 5)=P(heads) P(5)
1 1
=
2 6
1
12
Secondary Mathematics 2
The Addition Rule
The addition rule finds the probability of event
A occurring or event B occurring.
P(A B) P(A) P(B) P(A B)
A letter in the word Algebra or a letter in the
word Geometry. Where event A is Algebra and
event B is Geometry.
P(A B)=P(A) P(B) P(A B)
6
7
3
26 26 26
10 5
26 13
Practice Exercises A
1.
You have an equally likely chance of choosing an integer from 1 to 50. Find the
probability of each of the following events.
A.
An even number
B.
A perfect square
C.
A factor of 150 is chosen
D.
A two digit number is chosen
E.
A multiple of 4 is chosen
F.
A number less than 35 is chosen
G.
A prime number is chosen
H.
A perfect cube is chosen
2.
You randomly chose two marbles, replacing the first marble before drawing again, from a
bag containing 10 black, 8 red, 4 white, and 6 blue marbles. Find the probability of each
of the following events.
A.
A white marble, then a red marble is selected.
B.
A red marble is not selected, then a blue marble is selected.
C.
A green marble, then a green marble is selected.
D.
A blue or black marble is selected, then a white marble is selected.
3.
Drawing a card from the cards on the left, determine
the probability of each of the following.
A.
B.
C.
D.
E.
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P(Even or shaded)
P(White or odd)
P(Less than four or shaded)
P(Greater than five or shaded)
P(Factor of ten or white)
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Secondary Mathematics 2
Example 1:
P(peach)=
1
12
P(color in American flag)=
3 1
12 4
P(primary color and American flag)=
P(pink)=
2 1
12 6
0
0
12
Practice Exercises B
Using the Venn diagram, answer the following questions.
1. P(girls)
2. P(sports, not girls)
3. P(not sports)
4. P(not sports, not girls)
5. P(girls and sports) 6. P(Mr. P class)
Example 2:
P(has chores)=
Chores:
Yes
Chores:
No
Total
Curfew:
Yes
Curfew:
No
Total
13
5
18
12
3
15
25
8
33
18
33
8
33
13
P(has a curfew and chores)=
33
P(doesn’t have a curfew)=
P(has chores doesn’t have a curfew)=
P(has a curfew)=
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5
33
25
33
Secondary Mathematics 2
Practice Exercises C
Find the marginal totals. Then use the table to find the probabilities below.
Male
Female
Total
Brown hair
42
47
Blonde hair
11
16
1. P(male)
4. P(blonde hair male)
7. P(female not other hair)
Red hair
3
13
Black hair
17
9
2. P(red hair)
5. P(black hair and female)
8. P(not female not male)
Other hair
27
15
Total
3. P(other hair)
6. P(brown hair not male)
9. P(red hair and black hair)
VOCABULARY
Two events are independent if P(A) P(B)=P(A B)
Example 3:
Male
Female
Total
10th
320
285
605
11th
297
238
535
12th
215
216
431
Total
832
739
1571
1. Are being a male and being in 10th independent?
2. Are being a female and being in 12th grade independent?
1.
2.
832
739
P(male)=
P(female)=
1571
1571
605
431
P(10th grade)=
P(12th grade)=
1571
1571
320
216
P(male 10th grade)=
P(female 12th grade)=
1571
1571
?
?
832 605
320
739 431
216
1571 1571 1571
1571 1571 1571
0.204 0.204
The product of the probabilities of the
individual events is equal to the probability of
the intersection of the events; therefore the
events are independent.
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0.129 0.137
The product of the probabilities of the
individual events is not equal to the probability
of the intersection of the events; therefore the
events are not independent.
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Secondary Mathematics 2
Practice Exercises D
Determine whether or not the following events are independent.
1. If P(A)=0.7, P(B)=0.3, and P(A B)=0.21, are events A and B independent? Why or why
not?
2. Jaron has a dozen cupcakes. Three are chocolate with white frosting, three are chocolate with
yellow frosting, four are vanilla with white frosting, and two are vanilla with yellow frosting.
Are cake flavor and frosting color independent?
3.
Men
Women
Total
Dance
2
16
18
Sports
10
6
16
TV
8
8
16
Total
20
30
50
The above table represents the favorite leisure activities for 50 adults. Use it to answer the
following:
A.
B.
C.
D.
Find the probability of your gender.
Find the probability of your favorite leisure activity.
Find the probability of P(your gender your favorite leisure activity).
Are your gender and your favorite leisure activity independent?
VOCABULARY
A probability that takes into account a given condition is called a conditional probability. A
given condition is when we already know the outcome of one of the events. For example, when
flipping a coin and rolling a die, the probability of “rolling a 6 given heads”, means we already
know the coin has resulted in heads. This is written P(6 | heads).
The conditional probability formula is P A | B
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P( A B)
.
P( B)
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Secondary Mathematics 2
Example 4:
P(Vanilla Blue)
P(Blue)
5
12
P(Vanilla Blue)
A bakery sells vanilla and chocolate cupcakes
with white or blue icing.
White
3
Vanilla
Chocolate 6
9
Total
Blue
5
7
12
Total
8
13
21
P(White Chocolate)
P(Chocolate)
6
13
P(White Chocolate)
Alex’s favorite cupcake is chocolate with blue
icing. What is the probability he will get his
favorite cupcake if all the vanilla cupcakes
have already been sold?
P(Blue Chocolate)
P(Chocolate)
7
13
P(Blue Chocolate)
Example 5:
P(iPod Cell Phone)
P(Cell Phone)
10 2
25 5
P(iPod Cell Phone)
P(Cell Phone No iPod)
P(No iPod)
15
15 5
15 6 21 7
P(Cell Phone No iPod)
Miss K finds an iPod after class. What is the
probability the owner has an iPod and no cell
phone?
P(iPod No Cell Phone )
P(No Cell Phone)
4
4 2
4 6 10 5
P(iPod No Cell Phone )
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Secondary Mathematics 2
Practice Exercises E
Male
Female
Total
Bus
146
154
300
Private Car
166
185
351
Walk
82
64
146
Total
394
403
797
Use the table above to answer the following questions.
1.
P(Walk | Female)
2.
P(Male | Private Car)
3.
P(Bus | Male)
4.
P(Female | Doesn’t Walk)
5.
What is the probability that Melissa
rides the bus? Write the conditional
probability equation and then find
the probability.
6.
Jordan walks to school. What is the
probability Jordan is male? Write the
conditional probability equation and
then find the probability.
Use the Venn diagram above to answer the following questions.
7.
P(After School Job | Male)
9.
P(No After School Job | Male)
10.
P(Male | After School Job)
11.
Is the probability of having an after
school job given you are male the
same as the probability of being
male given that you have an after
school job? Use the probabilities in
#7 and #10 to justify your answer.
12.
A student works at McTaco Chimes
what is the probability the student is
female?
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8.
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P(Female | No After School Job)
Secondary Mathematics 2
VOCABULARY
Events A and B are independent if and only if they satisfy the probability A given B equals the
probability of A OR the probability of B given A equals the probability of B.
P( A B) P( A) or P( B A) P( B)
Example 6:
A bakery sells vanilla and chocolate cupcakes
with white or blue icing.
Vanilla
Chocolate
Total
White
3
6
9
Blue
5
7
12
Total
8
13
21
Are color of icing and cupcake flavor
independent?
?
P( Blue Vanilla) P( Blue)
P( Blue Vanilla ) ?
P( Blue)
P(Vanilla )
5 ? 12
8 21
0.625 0.571
Therefore the color of icing and cupcake flavor
are not independent.
Note: Keep in mind the above can also be tested using any of the following options.
?
?
1.
P(Blue Chocolate) P(Blue)
3.
P(Chocolate Blue) P(Chocolate)
5.
P(White Vanilla) P(White)
7.
P(Chocolate White) P(Chocolate)
2.
P(Vanilla Blue) P(Vanilla)
4.
P(White Chocolate) P(White)
6.
P(Vanilla White) P(Vanilla)
?
?
?
?
?
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Secondary Mathematics 2
Example 7:
Are having an iPod and having a cell phone
independent?
?
P (iPod Cell Phone) P(iPod)
P(iPod Cell Phone) ?
P(iPod)
P (Cell Phone)
10 ? 14
25 35
2 2
5 5
Therefore, having an iPod and having a cell
phone are independent.
Practice Exercises F
Students were asked what their main goal for their high school years was. The reported goals
were getting good grades, being popular, or excelling at sports.
Goals
Grades
Popular
Sports
Total
117
50
60
227
Boy
130
91
30
251
Girl
247
141
90
478
Total
Use the table above to answer the following questions.
1.
Is the probability of having good grades as a goal independent of gender?
2.
Is gender independent of having popularity as a goal?
3.
Workers at Cal Q Lus Copies were polled to see if Vitmain C was a way to reduce the
likelihood of getting a cold. According to the diagram, are you less likely to catch a cold
if you are taking vitamin C? Justify your answer using conditional probability.
4.
Real estate ads suggest that 64% of homes for sale have garages, 21% have swimming
pools, and 17% have both features. Are having a garage and having a pool independent
events? Justify your answer using conditional probability.
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Secondary Mathematics 2
Unit 4 Cluster 2 & 3 Honors (S.CP.8, S.CP.9, S.MD.6, S.MD.7)
Applications of Probability
Cluster 2: Computing probabilities of compound events
4.2.3 Apply the general Multiplication Rule
4.2.4 Use permutations and combinations to compute probabilities of compound events
Cluster 3: Using probability to evaluate outcomes of decisions
4.3.1 Use probability to make fair decisions
4.3.2 Analyze decisions and strategies
VOCABULARY
A compound event consists of two or more simple events. Tossing a coin is a simple event.
Tossing two or more coins is a compound event. A compound event is shown as
P(A B) P( A) P( B)
A tree diagram is a way of illustrating compound events. Each simple event adds new branches
to the tree diagram. The end result shows all possible outcomes.
When events are independent, the probability of a compound event is the product of the
probability of the desired outcome for each simple event. This is the general Multiplication Rule.
Example 1:
1
2
1
P (T)
2
P (H)
1
2
1
2
H
H
1
2
T
Start
1
2
Dime
Result
H
T
HHH
H
HTH
T
HTT
H
THH
1
2
T
THT
1
2
H
TTH
1
2
T
TTT
Nickel
Penny
1
2
1
2
1
2
1
2
1
2
H
T
1
2
1
2
T
HHT
This is a tree diagram
representing the possible
outcomes when tossing three
different coins.
There are eight possible
outcomes.
This would be the same
representation if looking at the
outcomes of tossing one coin
three separate times.
1 1 1 1
P(HHH)
2 2 2 8
P(THH)
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1 1 1 1
2 2 2 8
Secondary Mathematics 2
Example 2:
P( H )
1
2
P(T )
1
2
1
2
"1"
1
12
H1
5
6
not "1"
5
12
H not 1
"1"
1
12
T1
not "1"
5
12
T not 1
1
6
1
2
1
6
RESULT
1
6
H
Start
P(1)
DIE
COIN
T
5
6
5
P(not 1)
6
This is a tree diagram
representing the possible
outcomes when tossing
one coin and rolling one
die.
1 1 1
P(H 1)
2 6 12
1 5 5
P(T not 1)
2 6 12
Practice Exercises A
Using a tree diagram find the following probabilities.
1. Sophomores are required to either take English 10 or English 10H. They need Secondary
Math 2, Secondary Math 2H, or Pre-Calculus. Sophomores also need either Biology or
Chemistry.
2.
A. Draw a tree diagram representing all
sophomore choices.
B. P(Eng10, Sec2H, Chem.)
C. P(Eng10H, Sec2, Bio.)
D. P(Eng10H, Sec2H, PreCalc)
You have the following objects: a spinner with five choices, a six-sided die, and a coin.
P(spinner, die, coin)
1¢
A.
Draw a tree diagram satisfying the following: a choice on the spinner, a one or two
vs. anything else on the die (Hint: P(1 2) vs P(not 1 2) ), and heads or tails.
B. P(5, 1 or 2, H)
C. P(even, not 1 or 2, T)
D. P(1, 1 or 2, H)
E. P(6, not 1 or 2, T)
F. P(number < 6, not 1 or 2, H)
G. P(odd, 1 or 2, T)
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Secondary Mathematics 2
VOCABULARY
Probability of Two Dependent Events: If two events, A and B, are dependent, then the
probability of both events occurring is P A B P A P B A . In other words, the
probability of both events occurring is the probability of event A times the probability of event B
given that A has already occurred. Likewise, P B A P( B) P A B . These formulas can be
extended to any number of independent events.
Example 3:
There are 7 dimes and 9 pennies in a wallet. Suppose two coins are to be selected at random,
without replacing the first one. Find the probability of picking a penny and then a dime.
P(penny, then dime) P penny P dime|first coin was penny
number of pennies number of dimes after a penny has been drawn
number of coins number of coins after a penny has been drawn
9 7
16 15
21
80
Example 4:
A basket contains 4 plums, 6 peaches, and 5 oranges. What is the probability of picking 2
oranges, then a peach if 3 pieces of fruit are selected at random?
P 2 oranges, then peach P orange P orange after orange P peach after 2 oranges
5 4 6
120
4
15 14 13 2730 91
Example 5:
A cereal company conducts a blind taste test. 55% of those surveyed are women and 45% are
men. 75% of the women surveyed like the cereal, and 85% of the men like the cereal. What is the
probability that a person selected at random is:
A. A woman who likes the cereal?
A. P woman like P woman P like woman
B. A man who likes the cereal?
C. A woman who doesn’t like the
cereal?
D. A man who doesn’t like the
cereal?
0.55 0.75 0.4125
B. P man like P man P like man
0.45 0.85 0.3825
C. P woman dislike P woman P dislike woman
0.55 0.25 0.1375
D. P man dislike P man P dislike man
0.45 0.15 0.0675
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Secondary Mathematics 2
Practice Exercises B
1.
A photographer has taken 8 black and white photographs and 10 color photographs for a
brochure. If 4 photographs are selected at random, what is the probability of picking first 2
black and white photographs, then 2 color photographs?
2.
There are 7 blue pens, 3 black pens, and 2 red pens in a drawer. If you select three pens at
random with no replacement, what is the probability that you will select a blue pen, then a
black pen, then another blue pen?
3.
Tammy’s mom is baking cookies for a bake sale. When Tammy comes home, there are 22
chocolate chip cookies, 18 sugar cookies, and 15 oatmeal cookies on the counter. Tammy
sneaks into the kitchen, grabs a cookie at random, and eats it. Five minutes later, she does
the same thing with another cookie. What is the probability that neither of the cookies was a
chocolate chip cookie?
4.
There are 2 Root Beers, 2 Sprites, 3 Mountain Dews, and 1 Gatorade left in the vending
machine at school. The machines buttons are broken and will randomly give you a random
drink no matter what button you push. Find the probability of each outcome?
A. P root beer, root beer
B. P root beer, mountain dew
C. P sprite, gatorade
5.
6.
D. P mountain dew, mountain dew, mountain dew
A department store employs high school students, all juniors and seniors. 60% of the
employees are juniors. 50% of the seniors are females and 75% of the juniors are males.
One student employee is chosen at random. What is the probability of choosing:
A. A female junior
B. A female senior
C. A male junior
D. A male senior
There are 400 fans at a baseball game that get popcorn and hotdogs. 75% of the fans getting
food are adults and the rest are children. 80% of the children getting food are eating hotdogs
and 40% of the adults getting food are getting popcorn. One fan is chosen at random to
receive free food. What is the probability of choosing:
A. An adult with popcorn
B. A child with popcorn
C. An adult with a hotdog
D. A child with a hotdog
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Secondary Mathematics 2
VOCABULARY
The factorial function (symbol: !) is a way to multiply a series of descending natural numbers.
n! n n 1 n 2 3 2 1 is the general formula representing a factorial function. For
instance, 5! 5 4 3 2 1 120 . By definition 0! 1 and 1! 1.
A permutation is a combination, or an arrangement of a group of objects, where order matters.
For instance a lock combination or batting orders are examples where the order matters. If we
look at the letters A, B, and C there are six ways to arrange the letters, ie: ABC, ACB, BAC,
BCA, CBA, CAB.
The number of ways to arrange n distinct objects is indicated by n !
For example, in how many orders can a person read 5 different magazines?
This is found by finding 5! or 5 4 3 2 1 120 . So, there are 120 different orders in which are
person can read 5 different magazines.
The number of ways to arrange n distinct objects taking them r at a time is indicated by
n!
, where n and r are whole numbers and n r . If n r then nPr n!.
n Pr
n r !
For example, in how many orders can a person read 5 magazines selected from a list of 9
possibilities?
9!
9 8 7 6 5 4 3 2 1
9 8 7 6 5 15,120 . So, there
4 3 2 1
9 5!
are 15,120 different orders in which a person can ready 5 magazines selected from a list of 9
possibilities.
This is found by finding 9 P5
Permutations with n objects where one or more objects repeats, requires taking into consideration
n!
each item that is repeated. Use the formula
where s1 represents the number of times
s1 ! s2 ! sk !
the first object is repeated.
For example, how many way can you arrange the letters in ? KNICKKNACK
There are 10 letters with 4 K’s, 2 C’s, and 2 N’s. The total number of arrangements is
n!
10!
37,800
s1 ! s2 ! sk ! 4! 2! 2!
Note: All of the probability functions on your graphing calculator can be found by selecting
MATH then arrow over to PRB.
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Secondary Mathematics 2
Example 6:
You have just purchased 15 new CDs and want to add them to your iPod. You don’t want to
remove any music already on your iPod and there is only room for 5 more CDs. How many ways
can you add 5 different CDs to your iPod?
n =15
15!
15 P5
r=5
15 5!
15!
360,360
10!
Use nPr
n!
n r !
There are 360,360 ways you can add 5
different CDs from your15 choices to your
iPod.
Example 7:
Find the number of distinguishable permutations for the word MISSISSIPPI.
n = 11
11!
34, 650
s1 = I = 4
4! 4! 2!
s2 = S = 4
There are 34, 650 distinguishable permutations
s3 = P = 2
for the word MISSISSIPPI
n!
Use nPr
s1 ! s2 ! sk !
Practice Exercises C
Compute.
1. 8 P3
2.
6
P6
3.
7
P0
4.
P
10 1
Find the number of distinguishable permutations for the following words.
5. MATHEMATICS
6. SALT LAKE CITY
7. CHEMISTRY
Solve the following.
8. It is time to elect sophomore class officers. There are 12 people running for four positions:
president, vice president, secretary, and historian. How many distinct ways can those
positions be filled?
9. You just received 7 new movies in the mail. You only have time to watch 3 this weekend.
How many distinct ways can you watch the movies this weekend?
10. The Discriminants are giving a short evening performance. Their latest CD has 14 songs on
it; however they only have enough time to perform 8 songs. How many distinct
performances can they give?
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VOCABULARY
When we are considering combinations we are only considering the number of groupings. For
instance selecting people to a committee or choosing pizza toppings are examples where the order
does not matter.
The number of ways to group n distinct objects taking them r at a time is indicated by
n!
, where n and r are whole numbers and n r .
n Cr
r ! n r !
Example 8:
Honors English students are required to read 8 books from a list of 25. How many combinations
could a student select?
n =25
25!
25 C8
r=8
8! 25 8 !
25!
8!17 !
Use n Cr
1, 081,575
A student has 1,081,575 different groupings of
books they could read.
n!
r ! n r !
Practice Exercises D
Compute
1. 11 C6
2.
32
C0
3.
65
C62
4.
100
C96
Solve the following.
5.
Four members from a group of 18 on the board of directors at the Fa La La School of Arts
will be selected to go to a convention (all expenses paid) in Hawaii. How many different
groups of 4 are there?
6.
You have just purchased a new video game console. With the purchase you are given the
option of obtaining three free games from a selection of ten. How many combinations of
games can you choose?
7.
You are the manager of a new clothing store. You need 5 new employees and have 20
qualified applicants. How many ways can you staff the store?
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You Decide
You are registering for your junior year in school. Your school is on a block schedule, four
periods each day. You must take the following courses: English, history, math, and science. You
can fill the other four periods with classes of your choice.
English
English 11
English 11H
Concurrent
AP
Math
Sec Math 3
Sec Math 3H
PreCalculus
Calculus
Concurrent
Intro. to Stats
AP Stats
Personal Finance
Science
Biology
Chemistry
Physics
AP Biology
AP Chem
AP Physics
Anatomy
Wild Life Bio
Bio Ag.
Genetics
Astronomy
Geology
Zoology
History
U.S.
U.S. Honors
AP U.S
Elective
Law Enforcement
PE
World Religions
Sports Medicine
European Hist. Craft
Interior Design
Psychology
Business
AP Psychology
Marketing
Sociology
Web Page Design
Language
Woods
Band
Drafting
Orchestra
Auto
Choir
Sewing
Dance
Foods
Drama
Child Development
Photo
Pre-School
Ceramics
Financial Lit
Painting
Green House
Drawing
Release Time
A. If you are not focusing on the order of your classes, how many different schedules could
you construct?
B. Now that you have chosen your classes, how many ways could you set up your schedule?
C. Now that you have chosen your classes, if your Math class must be the first period of the
day, how many different schedules can you now have?
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Using Permutations and Combinations to Determine Probabilities of Events
Recall
Probability is P
number of favorable outcomes P( E )
.
total number of outcomes
P( S )
Example 9:
A standard deck of face cards consists of 52 cards, 4 suits (red diamonds, red hearts, black
spades, black clubs), and 13 cards in each suit, (numbers 2 through 10, jack, queen, king, ace).
What is the probability that the hand consists of 5 red cards?
C
P 5 red cards 26 5
The total number of outcomes is 52 C5 .
52 C5
65, 780
2,598,960
0.025
There are 26 red cards, so 26 C5 is the number
of ways to choose 5 red cards.
Example 10:
Using a standard deck of face cards, what is the probability that the hand consists of 1 diamond?
C C
The total number of outcomes is 52 C5 .
P one diamond 39 4 13 1
C
52 5
There are 39 cards that are not diamonds, so
82, 25113
2,598,960
1, 069, 263
0.411
2,598,960
C4 is the number of ways to choose 4 cards
that are not diamonds. There are 13
diamonds so 13 C 1 is the way to choose 1
diamond.
39
Practice Exercises E
1.
There are 14 black pens and 8 blue pens in a drawer. If 3 pens are chosen at random, what is
the probability that they are all blue?
2.
Sam has 9 pairs of socks in a drawer: 5 white pairs and 4 gray pairs. If he chooses three
pairs at random to pack for a trip, find the probability that he chooses exactly two white
pairs.
3.
A bag contains 14 cherry, 15 lime, and 10 grape suckers. Find the probability of picking 3
cherry suckers and 2 grape suckers if 5 suckers are chosen at random.
4.
Barbara has a collection of 28 movies, including 12 comedies and 16 dramas. She selects 3
movies at random to lend to a friend. What is the probability of her selecting 3 comedies?
5.
Five books are chosen at random from a best-seller list that includes 12 novels and 6
biographies. Find the probability of selecting 3 novels, and 2 biographies.
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Using Probability Models to Analyze Situations and Make Decisions
VOCABULARY
A game in which all participants have an equal chance of winning is a fair game. Similarly, a
fair decision is based on choices that have the same likelihood of being chosen.
Fair Decisions and Random Numbers
In order for a decision to be fair each possible outcome must be equally likely. For example if
you are hosting a party that includes 20 people and want to randomly choose 5 people to bring
treats there are multiple ways in which to make a fair and unfair decision.
A fair decision would be to assign each person a number. Write each of these numbers on a
separate piece of paper and drop them into a hat. Shuffle the numbers in the hat and choose five
papers to match five party members. These five people would be assigned to bring treats. This is
often referred to as “drawing lots”.
An unfair decision would be to write each person’s name on a piece of paper. Then arrange the
papers in alphabetical order. Starting with the first paper, flip a coin and record the result: heads
or tails. The first five friends that have a tail flipped for them must bring treats. This is unfair
because you will probably never flip a coin for your friends that have names that begin with
letters at the end of the alphabet.
Often Random Numbers are used to help you make fair decisions. For an event to be random
there is no pre-determined bias towards any particular outcome. Often people use random
number tables, random number generators on a calculator, or simply shuffle pieces of papers that
have numbers printed on them.
Example 11:
The addition game is played by rolling two dice. Player 1 gets a point if the sum of the two dice
is even. Player 2 gets a point if the sum of the two dice is odd. Use probability to determine if
this game is fair.
There are six possible outcomes for each
dice.
Start by determining how many different
outcomes can occur when rolling two dice.
Using the general multiplication rule
6 6 36 .
Rolling Sums that are Even
1, 1
2, 2
3, 1
4, 2
1, 3
2, 4
3, 3
4, 4
1, 5
2, 6
3, 5
4, 6
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5, 1
5, 3
5, 5
6, 2
6, 4
6, 6
Determine the number of ways that you can roll
an even number and an odd number by listing all
possible outcomes
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Secondary Mathematics 2
Rolling Sums that are Odd
1, 2
2, 1
3, 2
4, 1
1, 4
2, 3
3, 4
4, 3
1, 6
2, 5
3, 6
4, 5
5, 2
5, 4
5, 6
6, 1
6, 3
6, 5
There are 18 ways to roll an even sum and 18
ways to roll an odd sum.
18
P(even)
0.50
36
18
P(odd)
0.50
36
Calculate the probability of rolling an even sum.
Calculate the probability of rolling an odd sum.
This game is fair because each player has the same probability of rolling and even or odd
number.
Example 12:
The multiplication game is played by rolling two dice. Player 1 gets a point if the product of the
two dice is even and player 2 gets a point if the product of the two dice is odd. Use probability
to determine if this game is fair.
There are six possible outcomes for each dice.
Using the general multiplication rule 6 6 36 .
Start by determining how many different
outcomes can occur when rolling two dice.
Rolling a Product that is Even
1, 2
2, 1
3, 2
4, 1
5, 2
1, 4
2, 2
3, 4
4, 2
5, 4
1, 6
2, 3
3, 6
4, 3
5, 6
2, 4
4, 4
2, 5
4, 5
2, 6
4, 6
Determine the number of ways that you
can roll and even number and an odd
number by listing all possible outcomes
6, 1
6, 2
6, 3
6, 4
6, 5
6, 6
Rolling a Product that is Even
1, 1
3, 1
5, 1
1, 3
3, 3
5, 3
1, 5
3, 5
5, 5
There are 27 ways to roll a product that is even and
9 ways to roll a product that is odd.
27
P(even)
0.75
36
9
P(odd)
0.25
36
Calculate the probability of rolling an even
product.
Calculate the probability of rolling an odd
product.
This game is not fair because the probability of rolling an even product is higher than rolling an
odd product.
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Example 13:
Suppose each player spins the spinner once. Player A using spinner A, and Player B using
spinner B. The one with the larger number wins.
1
4
9
5
8
Spinne r A
Spinne r B
To determine how many possible outcomes there
are, list all the outcomes or use the general
multiplication rule.
1, 4
5, 4
9, 4
1, 3
5, 3
9, 3
3
Start by determining how many possible
outcomes there are.
1, 8
5, 8
9, 8
3 3 9
There are 9 possible outcomes.
List the number of ways in which player A wins
5, 4
5, 3
9, 4
9, 3
9, 8
List the number of ways in which play B wins
1, 4
1, 3
1, 8
5, 8
number of ways player A wins 5
P( A)
.5
total number of outcomes
9
P( B)
number of ways player B wins 4
.4
total number of outcomes
9
Count the number of ways in which player A
wins and the number of ways in which player
B wins.
Find the probability that player A will win or
P(A) and the probability that player B will
win or P(B).
This game is not fair because player A has a higher probability of winning.
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Using Probability to Analyze Decisions
Understanding the probability or expected outcomes from probability models and experiments
can help us make good decisions.
Example 14:
Mr. Green created a frequency table to collect data about his students and how their study habits
related to performance in his class. The table is shown below.
Passed
Failed
Totals
Studied
16
2
18
Did not Study
6
12
18
Totals
22
14
36
What is the probability that a student who studies will pass the class?
What is the probability that a student that does not study will pass his class?
Based off of these probabilities, if you want to pass the class should you study or not?
Find P(studied | passed) and P(did not study | passed) .
P(studied | passed)
number of students that studied and passed 16
0.8
total number of students that studied
18
P(did not study | passed)
number of students that did not study and passed 6
.3
total number of students that studied
18
There is a higher probability that a student will pass the class if they study
Example 15:
A teacher is conducting an action research project to determine the effectiveness of an
instructional strategy. The new instructional strategy was used with two class periods of 45
students each, and the traditional teaching method was used with 2 class periods of 45 students
each. Students were given a pre- and a post-exam to determine whether or not they improved
after the instruction. The results are shown in the table below.
78
Did not
Improve
12
35
55
90
123
67
180
Improved
Received the new
strategy
Received the traditional
method
Totals
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Total
90
Secondary Mathematics 2
What is the probability that a student improved given that he was instructed with the new
strategy?
What is the probability that a student received the traditional method of instruction given that he
did not improve?
If the teacher decided that the new instructional strategy was more effective than the traditional
method of teaching, did she make a good decision?
Find P(improved | new) and P(traditional | no improvement) .
P(improved | new)
number of students that improved with the new stratgy 78
0.86
total number of students with new strategy
90
number of students that did not improve with traditional
total number of students that did not improve
55
P(traditional | no improvement)
.821
67
P(traditional | no improvement)
There is a higher probability that a student will improve from the pre- to the post-exam if the
new instructional strategy is used.
Practice Exercises F
The table below shows the number of students at a certain high school who took an ACT
preparatory class before taking the ACT exam and the number of students whose scores were at
or above the minimum requirement for college entrance.
At or above minimum
requirement
Below minimum
requirement
Totals
Prep Class
No Prep
Class
Totals
268
210
478
57
115
172
325
325
650
1.
What is the probability that a student scored at or above the minimum requirement for
college entrance given that he or she took the ACT preparatory class?
2.
What is the probability that a student scored below the minimum requirement given that he
or she did not take the ACT preparatory class?
3.
A student decides not to take the preparatory class before taking the ACT exam. Is this a
good decision? Explain your answer.
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Unit 5
Similarity, Right
Triangle Trigonometry,
and Proof
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Unit 5 Cluster 1 (G.SRT.1, G.SRT.2, G.SRT.3)
Understand Similarity in terms of similarity transformations
Cluster 1: Understanding similarity in terms of transformations
5.1.1: Dilation with a center and scale factor, with a parallel line and a line segment
5.1.2: Transformations with similarity using equality of corresponding angles and
proportionality of corresponding sides
5.1.3: Establish criterion of AA using similarity transformations
VOCABULARY
A dilation is a transformation that produces an image that is the same shape as the original
figure but the image is a different size. The dilation uses a center and a scale factor to create a
proportional figure.
The center of dilation is a fixed point in the plane about which all points are expanded or
contracted.
The scale factor is the ratio of the new image to the original image (i.e. if the original figure has
4
a length of 2 and the new figure has a length of 4, the scale factor is 2 .)
2
A ' B ' C ' (the image) is a dilation of ABC
(the pre-image) with a scale factor of 3.
The origin, point O, is the center of dilation.
A 1, 0
A ' 3, 0
B 1, 2
B ' 3, 6
C 5, 2
C ' 15, 6
The ratio of the lengths from the center of
dilation to each coordinate is equal to the scale
factor.
OA ' OB ' OC '
3
OA OB OC
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Example 1: Center at the origin
Dilate the triangle with vertices A(0, 0) , B(4, 0) and C (4,3) by a scale factor of 2 and center at
(0, 0) .
C
A
Draw the triangle and label its vertices.
B
C'
mapped to (8, 0) . This makes A ' B ' 8 which is twice the
C
A
A ' B ' will be twice as long as AB. Since the center of the
dilation is at (0, 0) , A ' is mapped to (0, 0) while B ' is
B
B'
length of AB . Similarly, A ' C ' will be twice as long as
AC , mapping C ' to (8, 6) . B ' C ' will be twice as long as
BC .
Example 2: Center at the origin
Dilate the triangle with vertices A(1,1) , B(4,1) and C (4, 4) by a scale factor of 3 and center at
(0, 0) .
C
Draw the triangle and label its vertices.
A
B
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C'
A ' B ' will be three times as long as AB . Since the center of
the dilation is at (0, 0) , A ' is mapped to (3,3) while B ' is
mapped to (12,3) . This makes A ' B ' 9 which is three
C
A'
A
B'
B
times the length of AB . Similarly, A ' C ' will be three
times as long as AC , mapping C ' to (12,12) . B ' C ' will
be three times as long as BC .
Dilations can be performed on other shapes besides triangles. The shapes can be in any of the
four quadrants or even in more than one quadrant. The dilation can also shrink the original shape
instead of enlarging it.
Example 3: Center at the origin
Dilate a parallelogram with vertices A(4, 4) , B(6, 6) , C (6, 2) and D(4, 4) by a scale factor
of
1
and center at (0, 0) .
2
B
A
Draw the parallelogram and label its vertices.
C
D
B
A
A'
D'
B'
C'
C
Measure of Side Lengths
Coordinates of Vertices
A 4, 4 A ' 2, 2
B 6, 6 B ' 3,3
CD 2 26, C ' D ' 26
C 6, 2 C ' 3, 1
AD 8, A ' D ' 4
D 4, 4 D ' 2, 2
D
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AB 2 26, A ' B ' 26
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BC 8, B ' C ' 4
Secondary Mathematics 2
If the center of the dilation is not at the origin, then you will want to use graph paper and rulers
or dynamic geometry software such as: Geogebra or Geometer’s Sketchpad. Geogebra is a free
download and can be found at http://www.geogebra.org/cms/.
Example 4: Center not at the origin
Dilate the triangle with vertices A(1,1) , B(4, 4) and C (5,1) by a scale factor of 2 and center
at (1,1) .
B'
Coordinates of Vertices
A 1,1 A ' 1,1
B
C
C'
A'
Measure of Side Lengths
AB 3 2, A ' B ' 6 2
B 4, 4 B ' 7, 7
AC 4, A ' C ' 8
C 5,1 C ' 9,1
CB 10, C ' B ' 2 10
Example 5: Center not at the origin
Dilate the square with vertices A(1,1) , B(4,1) , C (4, 4) and D(1, 4) by a scale factor of 2
and center at (4, 4) .
C
D
D'
Measure of Side Lengths
Coordinates of Vertices
A 1,1 A ' 2, 2
B
A
B'
A'
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AB 3, A ' B ' 6
B 4,1 B ' 4, 2
AD 3 A ' D ' 6
C 4, 4 C ' 4, 4
BC 3, B ' C ' 6
C 1, 4 C ' 2, 4
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CD 3, C ' D ' 6
Secondary Mathematics 2
Practice Exercises A
Draw the dilation image of each figure with given center and scale factor.
1. Center 0, 0 ; scale factor 3
2. Center 0, 0 ; scale factor 2
3. Center 0, 0 ; scale factor 12
4. Center 4, 1 ; scale factor
2
5. Center 3, 4 ; scale factor 3
6. Center 2, 5 ; scale factor
2
7.
Graph the pre-image with given vertices.
J 2, 4 , K 4, 4 , and P 3, 2 . Then
graph the image with center of dilation at
the origin and a scale factor of 2.
8.
Graph the pre-image with given vertices.
J 2, 4 , K 4, 4 , and P 3, 2 . Then
graph the image with center of dilation at
the origin and a scale factor of 12 .
Determine whether each statement is true or false.
9. A dilation with a scale factor greater than
10. For a dilation, corresponding angles of
1 will shrink the image.
the image and pre-image are congruent.
11. A dilation image cannot have any points
in common with its pre-image.
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12. A dilation preserves length.
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Secondary Mathematics 2
Similarity
VOCABULARY
Two figures are similar if and only if there is a dilation that maps one figure onto the other. In
the new figure, corresponding angles are congruent and corresponding sides are proportional to
the original figure. You can denote that two figures are similar by using the symbol . For
example, ABC
DEF .
Optional Exploration Activity
Is it sufficient to know that two angles are congruent to two corresponding angles in another
triangle in order to conclude that the two triangles are similar?
Step 1: Using graph paper, dynamic geometry software or patty paper, have students construct
any triangle and label its vertices.
Step 2: Construct a second triangle with two angles that measure the same as two angles in the
first triangle.
Step 3: Measure the lengths of the sides of both triangles and compare the ratios of the
corresponding sides.
Step 4: Compare your results to those of other students.
Example 1: State if the triangles in each pair are similar. If so, state how you know they are
similar and create the similarity statement.
V
Since VUT LKJ , VTU LJK , and
the angles of a triangle add to 180 , the third
set of angles must also be congruent. The
angles are congruent which forces the
corresponding sides to be proportional.
J
U
55
45
T
45
55
L
TUV
K
JKL because the corresponding angles are congruent.
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Example 2: State if the triangles in each pair are similar. If so, state how you know they are
similar and create the similarity statement.
Find the sets of corresponding sides and show
that they have the same ratio of proportionality.
A
AC looks similar to HF ,
AC 84
6
HF 14
BC looks similar to GF ,
BC 48
6
8
GF
H
84
72
12
C
48
B
G 8
14
F
AB 72
6
HG 12
ABC HGF because the sets of corresponding sides are proportional which will force the
angles to be congruent. This is SSS similarity.
AB looks similar to HG ,
Example 3: State if the triangles in each pair are similar. If so, state how you know they are
similar and create the similarity statement.
DE ST , therefore D T because
alternate interior angles are congruent.
EUD SUT because they are vertical
angles.
By AA similarity SUT
EUD .
You can also show that the corresponding sides
are proportional.
SU 32
SU looks similar to EU ,
2
EU 16
TU 24
2
EU 12
SUT EUD because the sets of corresponding sides are proportional and the included
angle is congruent. This is called SAS similarity.
TU looks similar to DU ,
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Practice Exercises B
Are the following triangles similar? If so, write a similarity statement.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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Practice Exercises C
Identify the similar triangles and write a similarity statement. Then find the value for x and the
measure of the indicated sides.
1.
ML and LK
2.
MN and MJ
3.
KM and OP
4.
JK and JL
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Unit 5 Cluster 2 (G.CO.9)
Prove Theorems about Lines and Angles
Cluster 2: Prove Geometric Theorems
5.2.1
Prove theorems about lines and angles: vertical angles, alternate interior
angles, corresponding angles, points on a perpendicular bisector of a segment
VOCABULARY
Two nonadjacent angles formed by two intersecting lines
are called vertical angles or opposite angles. For example,
AEB and DEC are vertical angles.
A line that intersects two or more lines in a plane at
different points is called a transversal. Transversal p
intersects lines q and r.
When two parallel lines are intersected by a transversal,
angles that are in the same position at each intersection are
called corresponding angles. For example, 1 and 5 ,
2 and 6 , 3 and 7 , 4 and 8 are all corresponding
angles.
When two parallel lines are intersected by a transversal,
angles that are inside of the two parallel lines, but on
opposite sides of the transversal are called alternate
interior angles. For example, 3 and 5 , and 4 and
6 are both alternate interior angles.
A segment, line, or ray perpendicular to a given segment
that cuts the segment into two congruent parts is called a
perpendicular bisector. For example, CD is the
perpendicular bisector of AB .
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Secondary Mathematics 2
Theorems and Postulates
This unit of the core requires students to create proofs, formal or informal, to prove the following
theorems and postulates. Example proofs will be provided; however, teachers will probably do
tasks with students to complete these proofs in class.
Vertical Angle Theorem: If two angles are vertical angles, then they are congruent.
Corresponding Angle Postulate: If two parallel lines are cut by a transversal, then each pair of
corresponding angles is congruent.
Alternate Interior Angle Theorem: If two parallel lines are cut by a transversal, then each pair
of alternate interior angles is congruent.
Perpendicular Bisector Theorem: Any point on the perpendicular bisector of a segment is
equidistant from the endpoints of the segment.
Example 1: Proof of the Vertical Angle Theorem
Given: ABC and EBD are vertical angles.
Prove: ABC EBD
ABC and CBD are vertical angles.
Given.
ABC is supplementary to CBD and
EBD is supplementary to CBD .
ABC and CBD form a straight angle.
EBD and CBD form a straight angle.
mABC mCBD 180
mEBD mCBD 180
Definition of supplementary angles.
mABC 180 mCBD
mEBD 180 mCBD
Subtraction property of equality.
mABC mEBD
Transitive property.
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Example 2: Proof of Corresponding Angle Postulate
Given: q || r , 1 and 5 are corresponding angles
Prove: 1 5
We know that angle 4 is supplementary to angle 1 from the straight angle theorem. 5 and
4 are also supplementary, because they are interior angles on the same side of transversal p
(same side interior angles theorem).
Therefore, since m4 180 m1 180 m5 , we know that m1 m5 which means that
1 5 . This can be proven for every pair of corresponding angles in the same way.
Example 3: Proof of Alternate Interior Angle Theorem
Given: q r
Prove: 3 7
1 3 and 5 7 because they are vertical angles.
If q r then 1 5 and 3 7 because they are
corresponding angles.
Therefore, using the transitive property, 3 7 .
A similar argument can be given to show 4 6 .
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Secondary Mathematics 2
Practice Exercises A
Use the figure below for problems 1–2.
1.
Identify the pairs of angles that are vertical angles, corresponding angles, and alternate
interior angles.
Vertical Angles
2.
Corresponding Angles
Alternate Interior Angles
Given m1 72, find the measure of the remaining angles.
m2
m3
m4
m6
m7
m8
m5
Use the figure below for problems 3–4.
3.
Identify the pairs of angles that are vertical angles, corresponding angles, and alternate
interior angles.
Vertical Angles
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Corresponding Angles
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Alternate Interior Angles
Secondary Mathematics 2
4.
Given m5 110 and m17 95, find the measure of the remaining angles.
m1
m2
m3
m4
m6
m7
m8
m9
m10
m11
m12
m13
m14
m15
m16
Use the figure at the right for questions 5–7.
5.
Given: l || m prove that 2 7 .
6.
Given: l || m prove that m3 m5 180 .
7.
Given: l || m prove that m2 m8 180 .
Example: Proof of the Perpendicular Bisector Theorem
Given: CD is the perpendicular bisector of AB
Prove: AC BC
Because CD is the perpendicular bisector of AB , AE BE . CEA CEB because they are
both right angles. CE is congruent to itself because of the reflexive property. This means that
CEA CEB using SAS. Since the triangles are congruent, all of their corresponding parts
must be congruent. Therefore, AC BC .
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Secondary Mathematics 2
Practice Exercises B
CD is the perpendicular bisector of AB . Solve for x.
1.
2.
Lines l, m, and n are perpendicular bisectors of
PQR and meet at T. If TQ 2 x ,
PT 3 y 1, and TR 8 , find x, y, and z.
3.
Given that PA PB and PM AB at M, prove that
PM is the perpendicular bisector of AB .
4.
Given that BD is the perpendicular bisector of AC ,
prove that ABD CBD .
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Secondary Mathematics 2
Unit 5 Cluster 2 (G.CO.10)
Prove Theorems about Triangles
Cluster 2: Prove Geometric Theorems
5.2.2
Prove theorems about triangles: sum of interior angles, base angles of
isosceles triangles, segment joining midpoints of two sides is parallel to third
side and half the length, and medians meet at a point
VOCABULARY
The angles inside of a triangle are called interior angles.
A triangle with at least two congruent sides is called an
isosceles triangle. In an isosceles triangle, the angles that
are opposite the congruent sides are called base angles.
A point that is halfway between the endpoints of a segment
is called the midpoint. Point C is the midpoint of AB .
A segment whose endpoints are the midpoints of two sides
of a triangle is called the midsegement of a tiangle.
The line connecting midpoints to the opposite vertex of a
triangle is called the median. Point S is the midpoint of
AC . Point Q is the midpoint of BC . Point R is the
midpoint of AB .
The point where all three medians of a triangle intersect is
called a centroid. Point T is the centroid of ABC .
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Secondary Mathematics 2
Theorems
This unit of the core requires students to create proofs, formal or informal, to prove the following
theorems. Example proofs will be provided; however, teachers will probably do tasks with
students to complete these proofs in class.
Angle Sum Theorem: The sum of the measures of the angles of a triangle is 180.
Isosceles Triangle Theorem: If two sides of a triangle are congruent, then the angles opposite
those sides are congruent.
Triangle Midsegment Theorem: A midsegment of a triangle is parallel to one side of the
triangle, and its length is one-half the length of that side.
Theorem: The medians of a triangle meet at a point.
Example 1: Proof of the Angle Sum Theorem
Given: ABC
Prove: mC m2 mB 180
Draw XY through A so it is parallel to CB . Because 1 and CAY form a linear pair they are
supplementary, so m1 mCAY 180 . 2 and 3 form CAY , so m2 m3 mCAY .
Using substitution, m1 m2 m3 180 . Because you drew XY parallel to CB , we know
1 C and 3 B since they are alternate interior angles. Because the angles are
congruent, their measures are equal. Therefore, using substitution again, we know
mC m2 mB 180 .
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Secondary Mathematics 2
Example 2: Proof of the Isosceles Triangle Theorem
Given: PQR, PQ RQ
Prove: P R
Let S be the midpoint of PR . Draw SQ . Since S is the midpoint, PS RS . QS is congruent
to itself. Since were were given that PQ RQ , we know that all 3 corresponding pairs of sides
are congruent and we can say PQS RQS because of SSS congruency. Therefore, P R
since they are corresponding angles of congruent triangles.
Example 3: Proof of the Triangle Midsegment Theorem
Given: Points D, E, and F are the midpoints of the sides
of the triangle.
AC
Prove: DE || AC and DE
2
Statement
Reason
1. D and E are midpoints
1
1
2. DB AB , BE BC
2
2
3. BDE ~ BAC
1. Given
2. By definition of a midpoints
3. B is the center of dilation and scale
factor is ½
4. DE || AC
5. DE
4. Dilations take lines to parallel lines
AC
2
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5. Scale factor is ½
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Secondary Mathematics 2
Example 4:
ABC has vertices A(-4, 1), B(8, -1), and C(-2, 9). DE is a midsegment of
A.
Find the coordinates of D and E.
B.
Verify that AC is parallel to DE .
C.
Verify that DE
ABC .
1
AC .
2
Answers:
A.
B.
2 8 9 (1)
6 8
D
,
D , D 3, 4
2
2
2 2
4 8 1 (1)
4 0
E
,
E , E 2, 0
2
2
2 2
Use the Midpoint Formula to find the
midpoints of AB and CB .
9 1
8
4
2 (4) 2
40 4
slope of DE =
4
3 2 1
If the slopes of AC and DE are equal,
AC || DE .
slope of AC =
Because the slopes of AC and DE are equal, AC || DE .
C.
AC (2 (4))2 (9 1)2 (2)2 (8)2 4 64 68
DE
3 2 4 0
2
2
1 4
2
2
1 16 17
Now show DE
DE
17
17
1 1
AC
68
4 2
68
Because
First, use the distance formula
to find AC and DE.
1
AC
2
DE 1
1
, DE AC .
AC 2
2
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Secondary Mathematics 2
Example 5: The medians of a triangle meet at a point.
Given a triangle with vertices at A 0,1 , B 6,1 , and C 4,5 , prove that the medians meet at a
single point.
Draw the triangle and label its vertices.
C 4,5
B 6,1
A 0,1
0 6 11 6 2
D
,
, 3,1
2 2 2
2
0 4 1 5 4 6
E
,
, 2,3
2 2 2
2
6 4 1 5 10 6
F
,
, 5,3
2 2 2
2
Find the midpoint of AB and label it D.
Find the midpoint of AC and label it E.
Find the midpoint of CB and label it F.
C 4,5
E 2,3
A 0,1
F 5,3
D 3,1
3 1 2
1
2 6 4
2
1
y 3 2 x 2
mEB
B 6,1
Write an equation for EB , DC , and FA .
1 5 4
4
3 4 1
y 1 4 x 3
mDC
3 1 2
50 5
y 3 52 x 5
mFA
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Secondary Mathematics 2
y 3 12 x 2
y 12 x 4
and
y 1 4 x 3
y 4 x 11
Using two of the equations, try to solve the
system.
12 x 4 4 x 11
4.5 x 15
10
x
3
1 10
y 4
2 3
5
y 4
3
7
y
3
y 3
2
5
x 5
7
2 10
3 5
3
5 3
2 2 5
3 5 3
2
2
3
3
10 7
See if the solution , works in the third
3 3
equation.
All three medians of a triangle intersect in a single point.
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Secondary Mathematics 2
Practice Exercises A
1.
A base angle in an isosceles triangle measures 37°. Draw and label the triangle. What is the
measure of the vertex angle?
Find the missing angle measures.
2.
3.
4.
Find the value of x.
5.
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6.
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Secondary Mathematics 2
7.
Find the values of both x and y
8.
Can the given measurements be accurate? Why or why not?
9.
Find the length of DE given that D and E are midpoints.
10. Find the length of FG given that IJ is a midsegment of the triangle.
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Secondary Mathematics 2
11. Solve for x given NO is a midsegment of the triangle.
12.
ABC has vertices A 2, 6 , B 4, 0 , and C 10, 0 . DE is a midsegment with D being
the midpoint of of AB and E being the midpoint of AC .
a. Find the coordinates of D and E.
b. Verify that BC is parallel to DE .
1
c. Verify that DE BC .
2
13. ABC has vertices at A 5, 4 , B 0, 4 , and C 13, 4 .
a. Find the coordinates of D, E, and F, the midpoints of AB , AC , and BC .
B
E
D
C
A
F
b. Find the equations for two of the medians. Use a system of equations to find the location
of the centroid (the point where the medians intersect).
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Secondary Mathematics 2
Unit 5 Cluster 2 (G.CO.11)
Prove Geometric Theorems about Parallelograms
Cluster 2:
5.2.3
Prove Geometric Theorems
Prove theorems about parallelograms: opposite sides are congruent, opposite
angles are congruent, the diagonals bisect each other, and rectangles are
parallelograms with congruent diagonals.
VOCABULARY
A polygon with four sides is called a quadrilateral.
A quadrilateral with two pairs of parallel sides is called a
parallelogram.
A segment that connects two nonconsecutive vertices is
called a diagonal.
A parallelogram with four right angles is called a
rectangle.
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Secondary Mathematics 2
Theorems
This unit of the core requires students to create proofs, formal or informal, to prove the following
theorems. Example proofs will be provided; however, teachers will probably do tasks with
students to complete these proofs in class.
If a quadrilateral is a parallelogram, then its opposite sides are congruent.
If a quadrilateral is a parallelogram, then its opposite angles are congruent.
If a quadrilateral is a parallelogram, then its diagonals bisect each other.
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle.
Exmaple 1: If a quadrilateral is a parallelogram, then its opposite sides are congruent.
Given: QUAD is a parallelogram
Prove: QU AD and DQ UA
1.
Statements
QUAD is a parallelogram.
1.
Given.
2.
QU || AD and DQ || UA
2.
Definition of parallelogram.
3.
1 3 and 2 4
3.
Alternate interior angles are congruent.
4.
QA QA
4.
Reflexive property.
5.
6.
ASA congruence.
Corresponding parts of congruent
triangles are congruent (CPCTC).
5.
6.
QUA ADQ
QU AD and DQ UA
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Reasons
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Secondary Mathematics 2
Example 2: If a quadrilateral is a parallelogram, then its opposite angles are congruent.
Given: ABCD is a parallelogram
Prove: A C and B D
Since ABCD is a parallelogram, the opposite sides must be parallel.
Draw BD as a diagonal of the parallelogram. We know AD BC and AB CD because
opposite sides of a parallelogram are congruent. BD is congruent to itself. This creates two
congruent triangles by SSS: ABD CDB . Because the triangles are congruent, the
corresponding parts will be congruent. Therefore, A C . The same logic can be used using
AC as the diagonal to show B D .
Example 3: If a quadrilateral is a parallelogram, then its diagonals bisect each other.
Given: ACDE is a parallelogram
Prove: AB BD and EB BC
It is given that ACDE is a parallelogram. Since opposite sides of a parallelogram are congruent,
EA DC . By definition of a parallelogram, EA || DC . AEB DCB and EAB CDB
because alternate interior angles are congruent. EBA CBD by ASA. Since the triangles are
congruent, their corresponding parts are also congruent. Therefore, AB BD and EB BC .
The diagonals of a parallelogram bisect each other.
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Secondary Mathematics 2
Example 4: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a
parallelogram.
Given: AC and BD bisect each other at E.
Prove: ABCD is a parallelogram
AC and BD bisect each other at E.
Given
AEB CED
Vertical Angles
AE EC
BE ED
Definition of a segment bisector
BEA DEC
SAS
BEC DEA
Vertical Angles
AED CEB
SAS
BAE DCE
Corresponding parts of congruent
triangles are congruent (CPCTC).
ECB EAD
Corresponding parts of congruent
triangles are congruent (CPCTC).
AB CD
BC AD
If alternate interior angles are
congruent, then lines are
If alternate interior angles are
congruent, then lines are
ABCD is a parallelogram.
Definition of parallelogram.
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Secondary Mathematics 2
Example 5: If the diagonals of a parallelogram are congruent, then the parallelogram is a
rectangle.
Given: WXYZ is a parallelogram with WY XZ
Prove: WXYZ is a rectangle
Statements
Reasons
1.
WY XZ
1.
Given.
2.
XY ZW
3.
WX XW
2.
3.
4.
WZX XYW
5.
ZWX YXW
6.
mZWX mYXW
6.
Definition of a parallelogram.
Segment congruent to itself (reflexive
property).
SSS congruence.
Corresponding parts of congruent
triangles are congruent.
Definition of congruent.
7.
mZWX mWXZ mXZW 180
7.
Triangle sum theorem.
8.
XZW ZXY
8.
Alternate interior angles are congruent.
9.
mXZW mZXY
9.
Definition of congruent.
4.
5.
10. mZWX mWXZ mZXY 180
10. Substitution.
11. mWXZ mZXY mYXW
11. Angle addition postulate.
12. mZWX mYXW 180
13. ZWX and YXW are right angles
12. Substitution.
13. If two angles are congruent and
supplementary, each angle is a right
angle.
14. Opposite angles of a parallelogram are
congruent.
15. Definition of a rectangle.
14. WZY and XYZ are right angles
15. WXYZ is a rectangle
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Secondary Mathematics 2
Practice Exercises A
1. Complete the flow proof by filling in the blanks for the theorem: If a quadrilateral is a
K
L
parallelogram, then its opposite angles are congruent.
Given: JKLM is a parallelogram
Prove: J L and K M
M
J
JKLM is a parallelogram
a._____________
J and K are consecutive s.
Def. of consecutive s
K and L are consecutive s.
b. ________________________
c. ___________________
Def. of consecutive s
d . ______________________
Consecutive s are supplementary
K and L are supplementary
Consecutive s are supplementary
L and M are supplementary
Consecutive s are supplementary
K M
f . ___________________
__________________
e. _______________
Supplements of the
same are
2. Complete the two column proof by filling in the blanks for the theorem: If a parallelogram is a
rectangle, then its diagonals are congruent.
Given: ABCD is a rectangle
Prove: AC BD
Statements
1. ABCD is a rectangle
1.
Reasons
a. ______________________
2.
ABCD is a
2.
b. ______________________
3.
BC CB
3.
c.______________________
4.
ABC and DCB are right s.
4.
d.______________________
5.
ABC DCB
6.
f. _______________________
5.
6.
7.
g.________________________
7.
e.________________________
Opposite sides of a parallelogram are
congruent.
SAS
8.
AC BD
8.
h._________________________
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Secondary Mathematics 2
3.
a.
Find the values(s) of the variables(s) in each parallelogram.
b.
c.
4.
a.
For what values of a and b must EFGH be a parallelogram?
b.
c.
5.
QRST is a rectangle. Find the value of x and the length of each diagonal.
a. QS x and RT 2 x - 4
6.
b. QS 7 x 2 and RT 4 x 3
c. QS 5x 8 and RT 2 x 1
Is the given information enough to prove that the quadrilateral is a parallelogram? Explain
why or why not.
a.
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b.
c.
Page 269
Secondary Mathematics 2
Unit 5 Cluster 3 (G.SRT.4, G.SRT.5)
Prove Theorems Involving Similarity
Cluster 3: Prove theorems involving similarity
5.3.1 Prove theorems about triangles: a line parallel to one side divides the other 2
proportionally, the Pythagorean Theorem
5.3.2 Use congruence and similarity criteria to solve problems and prove relationships
Theorems
This unit of the core requires students to create proofs, formal or informal, to prove the following
theorems. Example proofs will be provided; however, teachers will probably do tasks with
students to complete these proofs in class.
Triangle Proportionality Theorem: If a line is parallel to one side of a triangle, then it divides
the other two sides proportionally.
Converse of the Triangle Proportionality Theorem: If a line intersects two sides of a triangle
proportionally, then that line is parallel to one side of the triangle.
Pythagorean Theorem: If a triangle is a right triangle with hypotenuse c, then a 2 b2 c2 .
Proof of the Triangle Proportionality Theorem: A line parallel to one side of a triangle
divides the other two sides proportionally.
Given: BD || AE
BA DE
Prove:
=
CB CD
Since BD AE, 3 1 and 4 2 because they are corresponding angles. Then by AA
CA CE
Similarity, ACE BCD . By definition of similar polygons,
. From the Segment
CB CD
Addition Postulate, CA BA CB and CE DE CD . Substituting for CA and CE in the ratio,
we get the following proportion.
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Secondary Mathematics 2
BA CB DE CD
CB
CD
CB BA CD DE
CB CB CD CD
BA
DE
1
1
CB
CD
BA DE
CB CD
Rewrite as a sum.
CB
CD
1 and
1
CB
CD
Subtract 1 from each side
Proof of the Converse of the Triangle Proportionality Theorem: If a line intersects two sides
of a triangle proportionally, then that line is parallel to one side of the triangle.
BA
DE
Given:
=
CB
CD
Prove: BD || AE
Statements
1.
2.
3.
4.
5.
6.
BA
DE
=
CB
CD
BA
DE
1+
=1+
CB
CD
CB BA
CD
DE
+
=
+
CB
CB
CD
CD
CB + BA
CD + DE
=
CB
CD
CA = CB + BA and CE = CD + DE
CA
CE
=
CB
CD
Reasons
1.
Given.
2.
Addition property of equality.
3.
Substitution for 1.
4.
Common denominator.
5.
Segment addition postulate.
6.
Substitution; sides are proportional.
7.
C C
7.
Reflexive property of congruence.
8.
ACE ~ BCD
8.
SAS Similarity.
9.
CBD CAB
9.
By definition of similar triangles.
10.
CBD and CAB are corresponding
angles. Since they are congruent, the
segments BD and AE are parallel.
10. BD || AE
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Secondary Mathematics 2
Practice Exercises A
Solve for x and y given
1.
ABC ~ AED .
Solve for x.
3.
2.
4.
5.
In HKM , HM = 15, HN = 10, and HJ is twice the
length of JK . Determine whether NJ || MK .
Explain.
6.
Find TO, SP, OR, and RP.
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Secondary Mathematics 2
Proof of the Pythagorean Theorem: If a triangle is a right triangle with hypotenuse c, then
a 2 b2 c 2 .
Given: ABC is a right triangle with hypotenuse c
Prove: a 2 b2 c 2
Draw CD so it is perpendicular to AB . This creates three right triangles, ABC , ACD , and
CBD . All three of these triangles are similar by AA similarity. ABC ~ CBD because they
both have a right angle and they have B in common. ABC ~ ACD because they both have
a right angle and they have A in common. ACD ~ CBD because they are both similar to
ABC .
Since we know the triangles are similar, we also know the ratios of the sides are all the same.
Start with ABC and CBD .
BC BD
a d
a 2 cd
AB CB
c a
Now use the same logic with
ABC and ACD .
AC AD
b e
b2 ce
AB AC
c b
Next, add the two equations together and factor out c.
a 2 b2 cd ce
a 2 b2 c(d e)
Looking at the original picture we see d e c , so we can substitute.
a 2 b 2 c (c )
a 2 b2 c2
Therefore, given a right triangle with hypotenuse c, a 2 b2 c 2 .
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Secondary Mathematics 2
Practice Exercises B
Solve for x.
1.
2.
x
3
3
9
15
x
3.
4.
65
4
9
x
25
5
9
x
5.
6.
x
x
16
28
11
33
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Secondary Mathematics 2
Unit 5 Cluster 4 (G.GPE.6)
Coordinte Proofs
Cluster 4: Use coordinates to prove simple geometric theorems algebraically.
5.4.1 Find the point on a directed line segment between two given points that partitions
that segment in a given ratio
Concept
For any segment with endpoints A and B, a point C between A and B will partition the segment
into a given ratio.
Example 1: Horizontal Line Segment
Find the coordinates of C so that the ratio
AC
4.
CB
Answer:
C x, y C x, 2
3 (7) 10 10
AC AB CB
AB CB
4
CB
AB CB 4CB
AB 5CB
10 5CB
2 CB
C x, y C 1, 2
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Because AB is a horizontal line, the y-coordinate of C
must be 2.
Find the horizontal distance from point A to point B.
Use the segment addition postulate AC CB AB and
solve for AC.
AC
Substitute AC AB CB into
4.
CB
Multiply each side by CB.
Add CB to each side.
Since AB 10 , substitute 10 in for AB and solve for CB.
Since CB 2 , the x-coordinate of the point C must be
3 2 1.
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Secondary Mathematics 2
Example 2: Vertical Line Segment
Find the coordinates of C so that the ratio
AC 2
.
CB 5
Answer:
C x, y C 3, y
5 2 7 7
AC AB CB
AB CB 2
CB
5
2
AB CB CB
5
5 AB CB 2CB
5 AB 5CB 2CB
5 AB 7CB
5 7 7CB
35 7CB
5 CB
C x, y C 3,0
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Because AB is a vertical line, the x-coordinate of C must
be 3.
Find the vertical distance from point A to point B.
Use the segment addition postulate AC CB AB and
solve for AC.
AC 2
Substitute AC AB CB into
.
CB 5
Multiply each side by CB.
Multiply each side by 5 and simplify.
Add 5CB to each side.
Since AB 7 , substitute 7 in for AB and solve for CB.
Since CB 5 , the y-coordinate of the point C must be
55 0.
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Secondary Mathematics 2
Example 3: Positive Slope Line Segment
AC 2
Find the coordinates of C such that
CB 3
Answer:
AB
4 2 7 1
2
2
AB 62 82
AB 36 64
AB 100
AB 10
AC AB CB
AB CB 2
CB
3
2
AB CB CB
3
3 AB CB 2CB
3 AB 3CB 2CB
3 AB 5CB
3(10) 5CB
30 5CB
6 CB
AC CB AB
AC 6 10
AC 4
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Find the length of AB using the distance formula.
Use the segment addition postulate AC CB AB and
solve for AC.
AC 2
Substitute AC AB CB into
.
CB 3
Multiply each side by CB.
Multiply each side by 3 then simplify.
Add 3CB to each side.
Since AB 10 , substitute 10 in for AB and solve for CB.
We can find AC by using the segment addition postulate
and substituting in the known values.
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Secondary Mathematics 2
To find the coordinates of point C, you will
need to draw a right triangle with AB as the
hypotenuse. Then draw a line from point C
perpendicular to each leg of the right triangle.
The two small triangles created within the
larger triangle are similar to each other and the
larger triangle by AA similarity. Therefore the
ratio of their sides is equal.
AC AF
AB AD
4 AF
10
6
10 AF 24
AF 2.4
You need the coordinates of point F to find the x-coordinate
of point C. Use the ratios to find AF.
F 2 2.4, 1 (0.4, 1)
Point F will have the same y-coordinate as point A. The xcoordinate will be the same as A plus 2.4 because AF is 2.4.
AC DE
AB BD
4 DE
10
8
10 DE 32
DE 3.2
You need the coordinates of point E to find the y-coordinate
of point C. Use the ratios to find DE.
E 4, 1 3.2 (4, 2.2)
Point E will have the same x-coordinate as point D. The ycoordinate will be the same as D plus 3.2 because DE is 3.2.
C 0.4, 2.2
The x-coordinate of F is 0.4 and the y-coordinate of E is 2.2.
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Secondary Mathematics 2
Example 4: Negative Slope Line Segment
AC 7
Find the coordinates of C such that
CB 6
Answer:
AB
7 5 4 1
AB
12
2
2
2
52
AB 144 25
AB 169
AB 13
AC AB CB
AB CB 7
CB
6
7
AB CB CB
6
6 AB CB 7CB
6 AB 6CB 7CB
6 AB 13CB
6(13) 13CB
78 13CB
6 CB
AC CB AB
AC 6 13
AC 7
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Find the length of AB using the distance formula.
Use the segment addition postulate AC CB AB and
solve for AC.
AC 7
Substitute AC AB CB into
.
CB 6
Multiply by each side by CB.
Multiply each side by 6 then simplify.
Add 6CB to each side.
Since AB 13 , substitute 13 in for AB and solve for CB.
We can find AC by using the segment addition postulate
and substituting in the known values.
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Secondary Mathematics 2
To find the coordinates of point C,
you will need to draw a right
triangle with AB as the
hypotenuse. Then draw a line from
point C perpendicular to each leg of
the right triangle. The two small
triangles created within the larger
triangle are similar to each other
and the larger triangle by AA
similarity. Therefore the ratio of
their sides is equal.
AC AF
AB AD
7 AF
13
5
13 AF 35
35
AF
2.69
13
You need the coordinates of point F to find the ycoordinate of point C. Use the ratios to find AF.
35
17
F 7, 4 7, 7,1.31
13
13
Point F will have the same x-coordinate as point
A. The y-coordinate will be the same as A minus
35
35
13 because AF is 13 .
AC DE
AB BD
7 DE
13 12
13DE 84
84
DE
6.46
13
You need the coordinates of point E to find the xcoordinate of point C. Use the ratios to find DE.
84
7
E 7 , 1 , 1 0.54, 1
13
13
Point E will have the same y-coordinate as point
D. The x-coordinate will be the same as D plus
84
84
13 because DE is 13 .
7 17
C , C 0.54,1.31
13 13
The x-coordinate of F is 137 and the y-coordinate
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of E is
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17
13
.
Secondary Mathematics 2
Practice Exercises A
1.
2.
C is between A and B with A (3, -5) and B (3, 7). Find the coordinates of C such that
AC
3.
CB
C is between A and B with A 3, 1 and B 6, 1 . Find the coordinates of C such that
AC 5
.
CB 4
3.
C is between A and B with A 2, 4 and B 2, 4 . Find the coordinates of C such that
AC 1
.
CB 3
4.
C is between A and B with A 2,5 and B 6,5 . Find the coordinates of C such that
AC 5
.
CB 3
5.
C is between A and B with A (-5, -3) and B (3, 3). Find the coordinates of C such that
AC 3
.
CB 7
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Secondary Mathematics 2
Unit 5 Cluster 5 (G.SRT.6, G.SRT.7, G.SRT.8)
Right Triangle Trigonometry
Cluster 5: Defining trigonometric ratios and solving problems
5.5.1 Similarity of triangles leads to the trigonometric ratios of the acute angles.
5.5.2 Understand the relationship between the sine and cosine of complimentary angles.
5.5.3 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in
applied problems
Two figures that have congruent angles and proportional sides are similar. The requirements for
proving that two right triangles are similar are less than the requirements needed to prove nonright triangles similar. It is already known that both triangles have one right angle. Therefore,
two right triangles are similar if one acute angle is congruent to one acute angle in the other right
triangle. In a right triangle, the ratio between the side lengths is a function of an acute angle ( 0
to 90 ). There are six distinct trigonometric ratios that are functions of an acute angle in a right
triangle. Right triangles that are similar will have the same trigonometric ratios.
VOCABULARY
The side opposite the right angle is the called the
hypotenuse.
The side that meets the hypotenuse to form the angle
is called the adjacent side.
The side that is opposite the angle is called the
opposite side.
Trigonometric Ratios
Let be an acute angle in the right
sine sin
opposite
hypotenuse
cosine cos
adjacent
hypotenuse
tangent tan
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opposite
adjacent
ABC as shown in the figure above. Then,
cosecant csc
secant sec
1
hypotenuse
sin
opposite
1
hypotenuse
cos
adjacent
cotangent cot
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1
adjacent
tan opposite
Secondary Mathematics 2
Example: Given the figure below, identify all six trigonometric ratios of the angle .
Answer:
The opposite side is 8. The adjacent side is 15. The hypotenuse is 17. Therefore,
sin
8
17
csc
1
1 17
8
sin 17 8
cos
15
17
sec
1
1 17
15
cos 17 15
tan
8
15
cot
1
1 15
8
tan 15 8
Example: Given the figure below, identify all six trigonometric ratios of the angle .
Answer:
The opposite side is 12. The adjacent side is 5. The hypotenuse is not known. Use the
Pythagorean Theorem to find the missing hypotenuse and then calculate the six trigonometric
ratios.
52 122 c 2
25 144 c 2
169 c 2
13 c
The hypotenuse is 13.
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Secondary Mathematics 2
sin
12
13
csc
1
1 13
12
sin 13 12
cos
5
13
sec
1
1 13
5
cos 13 5
tan
12
5
cot
1
1 5
12
tan 5 12
Practice Exercises A
Given the figures below, identify all six trigonometric ratios of the angle .
1.
2.
3.
Angle in standard position
7
θ
11
4.
5.
6.
Use a calculator to find each value.
7.
sin 9
10. cos 55
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8.
cos 37
11. tan 72
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9.
tan 48
12. sin 23
Secondary Mathematics 2
When solving equations, you use the inverse operation to find the value of the variable. In
trigonometry, you can find the measure of the angle by using the inverse of sine, cosine, or
tangent.
Equation
Inverse Equation
sin
x
y
x
sin 1
y
cos
x
y
x
cos 1
y
tan
x
y
x
tan 1
y
Meaning
x
The inverse, or arcsine, of
is equal to the
y
angle .
x
The inverse, or arccosine, of
is equal to the
y
angle .
x
The inverse, or arctangent, of
is equal to
y
the angle .
Practice Exercises B
Use a calculator to find each value.
1.
sin 1 0.5
2.
cos1 0.86
3.
tan 1 6
4.
tan 1 1
5.
sin 1 0.75
6.
cos1 0.33
Given the figures below, find the measure of the angle .
7.
8.
9.
Angle in standard position
7
θ
11
10.
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11.
12.
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Secondary Mathematics 2
Relationship Between Complementary Angles (Cofunction Identities)
cos 90 1 sin 1
2
cos 2 sin 1
4 4
5 5
1
2
sin 90 1 cos 1
sin 2 cos 1
3 3
5 5
1
Practice Exercises C
Use the relationship of complementary angles to find the missing angle.
1. If sin 30
1
1
, then cos ____
2
2
2. If cos 0 1 , then sin ____ 1
3. If cos 23 0.92 , then sin ____ 0.92
4. If sin 75 0.97 , then cos ____ 0.97
VOCABULARY
An angle of elevation is the angle made with the
ground and your line of sight to an object above
you.
An angle of depression is the angle from the
horizon and your line of sight to an object below
you.
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Secondary Mathematics 2
Example 1:
You are standing 196 feet from the base of an office building in downtown Salt Lake City. The
angle of elevation to the top of the building is 65 . Find the height, h, of the building.
Draw a picture of the situation and label what
you know.
tan 65
h
196
h
196
196 tan 65 h
You know the adjacent side and want the
opposite side. Use the tangent ratio to help
you set up the problem.
tan 65
Solve for h.
420.32 h
The height of the building is approximately 420.32 feet high.
Example 2:
John is standing on the roof of a building that is 300 feet tall and sees Sarah standing on the
ground. If the angle of depression is 60 how far away is Sarah from John?
Draw a picture of the situation and label what
you know.
sin 60
300
d
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You know the opposite side and want the
hypotenuse. Use the sine ratio to help you set
up the problem.
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Secondary Mathematics 2
300
d
d sin 60 300
sin 60
Solve for d.
300
d
sin 60
d 346.41
Sarah is approximately 346.41 feet away from John.
Practice Exercises D
Solve each problem.
1.
The angle of depression from the top of a
lighthouse 150 feet above the surface of
the water to a boat is 13 . How far is the
boat from the lighthouse?
2.
A guy wire connects the top of an antenna
to a point on the level ground 7 feet from
the base of the antenna. The angle of
elevation formed by this wire is 75 .
What are the length of the wire and the
height of the antenna?
3.
A private jet is taking off from Telluride,
4.
Colorado. The runway is 46,725 feet from
the base of the mountain. The plane needs
to clear the top of Mount Sneffels, which
is 14,150 feet high, by 100 feet. What
angle should the plane maintain during
takeoff?
A person is 75 feet from the base of a
barn. The angle of elevation from the
level ground to the top of the barn is 60 .
How tall is the barn?
5.
From the top of a building 250 feet high, a
man observes a car moving toward him.
If the angle of depression of the car
changes from 18 to 37, how far does
the car travel while the man is observing
it?
6.
A rocket is launched from ground level.
A person standing 84 feet from the launch
site observes that the angle of elevation is
71 at the rocket’s highest point. How
high did the rocket reach?
7.
A hot-air balloon is 700 feet above the
ground. The angle of depression from the
balloon to an observer is 5 . How far is
the observer from the hot-air balloon?
8.
If a wheelchair access ramp has to have an
angle of elevation of no more than 4.8
and it has to rise 18 inches, how long must
the ramp be?
9.
A kite has 25 feet of string. The wind is
10. A sledding run is 400 yards long with a
blowing the kite to the west so that the
vertical drop of 40.2 yards. Find the angle
of depression of the run.
angle of elevation is 40 . How far has the
kite traveled horizontally?
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Secondary Mathematics 2
Unit 5 Cluster 5 Honors (N.CN.3, N.CN.4, N.CN.5 and N.CN.6)
Using Complex Numbers in Rectangular and Polar Form
H.5.1 Find moduli of complex numbers.
H.5.2 Represent complex numbers on the complex plane in rectangular and polar form
and explain why the rectangular and polar forms of a given complex number
represent the same number.
H.5.3 Represent addition, subtraction, multiplication, and conjugation of complex
number geometrically on the complex plane; use properties of this representation
for computation.
H.5.4 Calculate the distance between numbers in the complex plane as the modulus of
the difference
H.5.4 Calculate the midpoint of a segment as the average of the numbers at its
endpoints.
VOCABULARY
The complex plane is where the horizontal axis represents the
real component, a, and the vertical axis represents the imaginary
component, bi, of a complex number.
The rectangular form of a complex number, a bi , is written
as a, b . Traditionally known as x, y . To graph a point in
the complex plane, graph a units horizontally and b units
vertically.
The modulus (plural form is moduli) of a complex number is
defined by z a 2 b2 and is the length, or magnitude, of the
vector in component form created by the complex number in the
complex plane.
Example 1:
Write the following complex numbers in rectangular form, then graph them on the complex
plane.
a. 3 2i
b. 4 5i
c. 2 i
d. 5 3i
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Secondary Mathematics 2
Answer:
a. 3 2i 3, 2
b.
4 5i 4,5
c.
2 i 2, 1
d.
5 3i 5,3
The rectangular form of complex number
a bi is written as a, b .
Since a is the real component it is graphed
horizontally. Similarly, bi is the complex
component and is graphed vertically.
a.
b.
c.
d.
3 units right and 2 units down
4 units right and 5 units up
2 units left and 1 unit down
5 units left and 3 units up
Example 2:
Find the modulus of z 3 4i .
z 3 4i
a 3 and b 4
z 32 4
2
Modulus formula: z a 2 b2
z 9 16
z 25
Simplify.
z 5
Example 3:
Find the modulus of the complex number in rectangular form 3, 6 .
3, 6
z 32 6
a 3 and b 6
2
Modulus formula: z a 2 b2
z 9 36
z 45 6.71
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Simplify.
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Secondary Mathematics 2
Practice Exercises A
Write the following complex numbers in rectangular form then graph them on the complex
plane.
1. 2 3i
2. 5 4i
3. 1 2i
4. 3 5i
5. 1 i
6. 5 6i
Find the moduli of the following complex numbers.
7. 5 12i
8. 5 4i
9. 1, 2
10. 8, 6
VOCABULARY
The polar form of a complex number, z a bi , is
represented by z r cos i sin (sometimes called
trigonometric form) where a r cos , b r sin , r is the
modulus, r z a 2 b2 , of a complex number and is
an argument of z. An argument of a complex number,
z a bi , is the direction angle of the vector a, b .
To find , when given a complex number z a bi , use
b
tan 1 .
a
To graph a complex number in polar form, plot a point r
units from the origin on the positive x-axis, then rotate the
point the measure of .
To convert from polar to rectangular form use the fact that a r cos and b r sin .
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Secondary Mathematics 2
Example 4:
Write 3 3i in polar form then graph it.
Answer
3 3i
a 3 and b 3
z 32 32
z 99
Compute the modulus z a 2 b2 .
z 18
z 3 2
3
tan 1
3
Find . Make sure that your calculator is in
degrees.
45
3 2 cos 45 i sin 45
Remember that a complex number in polar
form is z r cos i sin . The modulus is r
so r 3 2 .
Plot the point 3 2 units from the origin on the
positive x-axis. Rotate it 45 .
Example 5:
Write 3,1 in polar form then graph it.
Answer:
3 i
z
a 3 and b 1
3
2
12
z 3 1
Compute the modulus z a 2 b2 .
z 4
z 2
1
3
tan 1
30
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Find . Make sure that your calculator is in
degrees.
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Secondary Mathematics 2
30 180 150
z 2 cos150 i sin150
The calculator only returns values between
90 and 90 . In order to get the correct
angle, so that the point is in quadrant II, add
180 to the angle. (If the point is in quadrant
III, you will also add 180 , but if it is in
quadrant IV, you will need to add 360 .)
Remember that a complex number in polar
form is z r cos i sin . The modulus is r
so r 2 .
Plot the point 2 units from the origin on the
positive x-axis. Rotate the point from 150 the
positive x-axis.
Example 6:
Write 2 cos135 i sin135 in rectangular form.
Answer:
2 cos135 i sin135
a 2 cos135 1
b 2 sin135 1
The rectangular form is 1,1 .
r 2 and 135
To convert from polar to rectangular form use
a r cos and b r sin . Make sure that
your calculator is in degree mode.
Rectangular form is a, b .
Example 7:
Write 3 cos 210 i sin 210 in rectangular form.
Answer:
3 cos 210 i sin 210
r 3 and 210
3
3
a 3 cos 210 3
1.5
2
2
3
1
b 3 sin 210 3
0.866
2
2
To convert from polar to rectangular form use
a r cos and b r sin . Make sure that
your calculator is in degree mode.
3
3
The rectangular form is ,
.
2
2
Rectangular form is x, y .
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Secondary Mathematics 2
Practice Exercises B
Write the following complex numbers in polar form.
2. 5 4i
1. 5 12i
4. 8, 6
3. 0, 2
Write the following complex numbers in rectangular form.
5. 3 cos 270 i sin 270
6.
7. 3 cos135 i sin135
8. 4 cos 30 i sin 30
9.
a.
b.
c.
d.
6 cos 60 i sin 60
Write 10 24i in rectangular form.
Write your answer to part a) in polar form.
Write your answer to part b) in rectangular form.
Compare your answers form part a) and part c).
Representing Addition, Subtraction and Multiplication of Complex Numbers
A Complex Number and its
Conjugate
The conjugate is a reflection
through the x-axis.
Adding Complex Numbers
Adding complex numbers can
be demonstrated geometrically
by showing the addition of the
vector associated with each
complex number z. The sum
of 3 2i 1 3i is the
resultant vector 2,5 or
2 5i .
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Subtracting a Complex
Number
Subtracting complex numbers
can be demonstrated
geometrically by showing the
subtraction of the vector
associated with each complex
number z. The difference of
3 2i 1 3i is the
resultant vector 4, 1 or
4 i .
Secondary Mathematics 2
Multiplying a Complex Number
Multiplication of complex numbers is
geometrically represented by adding the angles
and multiplying the moduli of the two vectors.
2 i 1 3i 2 7i 3 1 7i
7
81.870 180 98.13
1
tan 1
z
1
2
72 1 49 50 7.071
z1 2 i
1
z2 1 3i
3
1 tan 1 26.6
2
2 tan 1 71.6
1
z1 22 12 4 1 5
z2 12 32 1 9 10
Sum of angles 1 2 26.565 71.565 98.13
Product of moduli: z z1 z2 5 10 50 7.071
Example 8:
Add 5 2i 2 4i and represent the sum by graphing it on the complex plane.
Answer:
5 2i 2 4i
5 2i 2 4i
5 2 2i 4i
3 2i
Remove the parenthesis.
Combine like terms.
Represent the two complex numbers with
vectors.
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Secondary Mathematics 2
Example 9:
Subtract 4 2i 2 3i and represent the difference by graphing it on the complex plane.
Answer:
4 2i 2 3i
4 2i 2 3i
4 2 2i 3i
6 i
Remove the parenthesis.
Combine like terms.
Represent the two complex numbers with
vectors.
Example 10:
Multiply 2 3i 1 4i and represent the product by graphing it on the complex plane.
Answer:
2 3i 1 4i
2 3i 8i 12i 2
2 11i 12
10 11i
Use your preferred method to multiply.
Simplify and remember that i 2 1 .
3
4
2 tan 1 75.964
1
Sum of angles: 56.310 75.964 132.27
1 tan 1 56.310
2
z1 22 32 4 9 13
z2 12 42 1 16 17
Product of moduli: 13 17 221 14.866
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Secondary Mathematics 2
Practice Exercises C
Find the conjugate of the complex numbers.
1.
9i
2.
4 i
3.
9 6i
4.
3 7i
Add or subtract and represent the sum or difference on the complex plane.
5.
8 5i 7 6i
6.
4 7i 4i
7.
3 6i 4 6i
8.
1 3i 6 4i
9.
1 7i 5 8i
10.
5 2i 9 10i
Multiply and represent the product on the complex plane.
11. 1 i 2 i
12. 3 2i 4 3i
13. 3 i 5 2i
Powers of Complex Numbers
De Moivre’s Theorem
If z r cos i sin is a complex number and n is any positive integer, then
z n r n cos n i sin n .
Example 11:
Use De Moivre’s Theorem to simplify 3 i
Answer:
3 i
z
You need the radius and the angle. To get the
radius find the modulus.
8
3
8
2
12
z 3 1
a 3 , b 1 , and r z a 2 b2 .
z 4
z 2
1
3
tan 1
30
30 180 150
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b
To get the angle use tan 1 .
a
The point should be in the second quadrant so
add 180 to the angle.
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Secondary Mathematics 2
3 i
3 i
3 i
8
28 cos 8 150 i sin 8 150
8
256 cos 1200 i sin 1200
1
3
256
i
2
2
8
Substitute r 2 , 150 , and n 8 into De
Moivre’s Theorem.
128 128 3i
3 i 128 128 3i
3 i
8
8
Example 12:
Use De Moivre’s Theorem to simplify 1 3i
3
.
Answer:
1 3i
You need the radius and the angle. To get the
radius find the modulus.
3
z
1
2
3
2
z 1 3
a 1, b 3, and r z a 2 b2 .
z 4
z 2
3
1
tan 1
b
To get the angle use tan 1 .
a
60
The point should be in the second quadrant so
add 180 to the angle.
60 180 120
1
1
1
1
1
3i
3i
3i
3i
3
23 cos 3 120 i sin 3 120
3
8 cos 360 i sin 360
3
8 1 0i
3
8 0
3
8
3i
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Substitute r 2 , 120 , and n 3 into De
Moivre’s Theorem.
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Secondary Mathematics 2
Practice Exercises D
Use De Moivre’s Theorem to simplify the following.
1. 2 2i
4. 2 i
2. 1 i
3
4
5.
3. 2 3i
3
3 i
4
4
6. 3 3i 3
5
VOCABULARY
The distance between numbers in the complex plane is the modulus, z a 2 b2 , of the
difference of the complex numbers.
The midpoint of a segment in the complex plane is the average of the endpoints.
Example 13:
Find the distance between 6 2i and 2 3i .
Answer:
6 2i 2 3i
6 2i 2 3i
8 i
z
8 1
2
Find the difference between the two points.
2
Compute the modulus, z a 2 b2 , of the
z 64 1
difference. a 8 and b 1
z 65 8.06
Example 14:
Find the midpoint between 8 2i and 1 4i .
Answer:
8 2i
The endpoints of the first complex number are
a1 8 and b1 2 . The endpoints of the
and 1 4i
8 1 2 4
,
2
2
9 2
,
2 2
9
,1 is the midpoint or
2
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second complex number are a2 1 and b2 4 .
9
i
2
Compute the midpoint by averaging the a
endpoints and the b endpoints.
a1 a2 b1 b2
,
2
2
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Secondary Mathematics 2
Practice Exercises E
Find the distance and midpoint between the complex numbers.
1. 3 2i and 1 3i
2. 7 4i and 2 2i
3. 5 12i and 10 7i
4. 3 8i and 2 12i
5. 2 7i and 7 9i
6. 11 3i and 3 3i
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Secondary Mathematics 2
Unit 5 Cluster 6 (F.TF.8)
Using the Pyathagorean Identity
Cluster 6: Prove and apply trigionometric identities
5.6.1 Prove the Pythagorean identity sin 2 cos2 1, then use it to find sin(θ),
cos(θ), or tan(θ), given sin(θ), cos(θ), or tan(θ) (limit angles between 0 and 90
degrees).
Proof of the Pythagorean Identity sin 2 cos2 1
a 2 b2 c 2
Given a right triangle with side lengths a, b, and c, use the
Pythagorean Theorem to relate the sides of the triangle.
a 2 b2 c2
c2 c2 c2
a 2 b2
1
c2 c2
Divide each side by c 2 so that the expression on the left is
equal to 1 on the right.
2
2
a b
1
c c
Use the properties of exponents to rewrite the expression on
the left.
sin cos 1
sin 2 cos 2 1
2
2
Substitute sin
a
b
and cos .
c
c
VOCABULARY
An angle is in standard position when the vertex is at the origin,
one ray is on the positive x-axis, and the other ray extends into the
first quadrant.
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Secondary Mathematics 2
Example 1:
4
If sin is in the first quadrant, find cos and tan .
5
Draw a triangle with the angle in standard position.
Then label the information you know.
4 2 b 2 52
16 b 2 25
b 2 25 16
Use the Pythagorean Theorem to find the missing side.
b2 9
b3
adjacent
3
hypotenuse 5
opposite 4
tan
adjacent 3
cos
Use the definitions of the trigonometric ratios to find
cos and tan .
Example 2:
If tan
8
is in the first quadrant, find sin and cos .
15
Draw a triangle with the angle in standard position.
Then label the information you know.
82 152 c 2
64 225 c 2
289 c 2
c 17
opposite
8
hypotenuse 17
adjacent
15
cos
hypotenuse 17
sin
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Use the Pythagorean Theorem to find the missing side.
Use the definitions of the trigonometric ratios to find
sin and cos .
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Secondary Mathematics 2
Practice Exercises A
1.
Find sin and cos if tan
3
.
4
2.
3.
Find sin and tan if cos
1
.
4
4.
Find sin and tan if cos
5.
Find cos and tan if sin
5
.
13
6.
Find cos and tan if sin
7.
Find sin and cos if tan
8
.
5
8.
Find sin and tan if cos
9.
Find cos and tan if sin
3
2
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Find sin and cos if tan 3 .
4
.
5
1
.
2
1
.
2
10. Find sin and cos if tan 3 .
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Secondary Mathematics 2
Unit 5 Honors Unit Circle
Defining Trigonometric Ratios on the Unit Circle
H.5.6 Define trigonometric ratios and write trigonometric expressions in equivalent
forms.
There are special right triangles. These triangles have special relationships between the lengths
of their sides and their angles that can be used to simplify calculations when finding missing
angles and sides.
45 45 90 Triangle
The Pythagorean Theorem allows us to derive the relationships that
exist for these triangles. Consider a right isosceles triangle with leg
lengths x and hypotenuse h. Since this is a right isosceles triangle, the
measures of the angles are 45 45 90 .
Using the Pythagorean Theorem, we know that x2 x 2 h2 . Solving the equation for h, we get:
h2 x2 x2
h2 2 x2
h 2 x2
hx 2
45 45 90 Triangle
In any 45 45 90 triangle, the length of the hypotenuse is
2 times the length of its leg.
x 2
Therefore, in a 45 45 90 , the measures of the side lengths are x, x, and x 2 .
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Secondary Mathematics 2
30 60 90 Triangles
There is also a special relationship for triangles with angles of
30 60 90 . When an altitude, a, is drawn from the vertex of an
equilateral triangle, it bisects the base of the triangle and two
congruent 30 60 90 triangles are formed. If each triangle has a
base of length x, then the entire length of the base of the equilateral
triangle is 2x.
Using one of the right triangles and the Pythagorean Theorem, we know that a 2 x 2 2 x 2 .
Solving for a we get:
a2 x2 2x 2
a2 4 x2 x2
a 2 3x 2
a 3x 2
ax 3
Therefore, in a 30 60 90 triangle, the measures of the side lengths are x, x 3, and 2 x .
30 60 90 Triangles
In any 30 60 90 triangle, the length of the hypotenuse is twice the length of the shorter leg,
and the length of the longer leg is 3 times the length of the shorter leg.
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Trigonometric Ratios for a Circle of Radius 1
opposite
hypotenuse
y
sin
1
sin y
sin
adjacent
hypotenuse
x
cos
1
cos x
cos
For any point x, y on a circle of radius 1, the x-coordinate is the cosine of the angle and the ycoordinate is the sine of the angle. A circle of radius 1 is called a unit circle. Special values on
the unit circle are derived from the special right triangles, 45 45 90 triangles and
30 60 90 triangles, and their trigonometric ratios.
Special Right Triangles and the Unit Circle
Recall that if you reflect any point x, y , on the coordinate plane over the x axis, y axis, or the
origin, then the following relationships exist:
Reflection over the y axis ( x, y) ( x, y)
Reflection over the x axis ( x, y) ( x, y)
Reflection over the origin ( x, y) ( x, y)
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We will use this reasoning to show relationships of special right triangles on the unit circle.
To illustrate a 45 45 90 triangle on the unit circle, we are going to rotate a point 45 from
the positive x-axis. The right triangle formed has a hypotenuse of 1 and leg lengths of x and y.
Since two of the angles are congruent this is an isosceles right triangle and the lengths of the legs
are the same, x y . Finding the value of x will help us identify the numerical coordinates of the
point x, y .
The length of the hypotenuse in a 45 45 90 is equal to the length of a leg times
Using this property we can find the length of a leg.
1 x 2
1
x
2
1
2
x
2 2
The hypotenuse, which is length 1, is
times the length of leg x.
2.
2
Solve for x.
Rationalize the denominator.
2
x
2
Since both legs are equal, x
2
2
.
and y
2
2
If we relate this to the unit circle where 45 , then the following is true:
2
2
adjacent
2
opposite
2
x cos
cos 45 2
y sin
sin 45 2
hypotenuse
1
2
hypotenuse
1
2
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Secondary Mathematics 2
2 2
Thus, the point ( x, y )
,
. If we reflect this point so that it appears in Quadrants II, III,
2
2
and IV, then we can derive the following:
The reflection of the point over the y-axis is equivalent to a rotation of 135 from the
positive x-axis. The coordinates of the point are
2 2
( x, y) cos 45,sin 45 cos135,sin135
,
.
2
2
The reflection of the point through the origin is equivalent to a rotation of 225 from the
positive x-axis. The coordinates of the new point are
2
2
( x, y) cos 45, sin 45 cos 225,sin 225
,
.
2
2
The reflection of the point over the x-axis is equivalent to a rotation of 315 from the
positive x-axis. The coordinates of the new point are
2
2
( x, y ) cos 45, sin 45 cos 315,sin 315
,
.
2
2
To illustrate a 30 60 90 triangle on the unit circle, we are going to rotate a point 60 from
the positive x-axis and drop a perpendicular line to the positive x-axis. The right triangle formed
has a hypotenuse of 1 and leg lengths of x and y.
The length of the hypotenuse in a 30 60 90 is twice the length of the shorter side. Thus,
1
1 2x . Upon solving the equation for x we find that x . We can use this value of x to find
2
the length of the longer leg, y. The longer leg is 3 times the length of the shorter leg.
1
3
1
Therefore, y x 3 and since x then y 3 or y
.
2
2
2
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Secondary Mathematics 2
If we relate this to the unit circle where 60 then the following is true:
1
3
adjacent
1
opposite
3
2
x cos
cos 60
y sin
sin 60 2
hypotenuse
1 2
hypotenuse
1
2
1 3
Thus, the point ( x, y ) ,
. If we reflect this point so that it appears in Quadrants II, III,
2 2
and IV, then we can derive the following:
The reflection of the point over the y-axis is equivalent to a rotation of 120 from the
positive x-axis. The coordinates of the point are
1 3
( x, y) cos 60,sin 60 cos120,sin120 ,
.
2 2
The reflection of the point through the origin is equivalent to a rotation of 240 from the
positive x-axis. The coordinates of the new point are
1
3
( x, y) cos 60, sin 60 cos 240,sin 240 ,
.
2
2
The reflection of the point over the x-axis is equivalent to a rotation of 300 from the
positive x-axis. The coordinates of the new point are
1
3
( x, y ) cos 60, sin 60 cos 300,sin 300 ,
.
2
2
We can also consider the case where 30 and use the 30 60 90 triangle to find the
values of x and y for this value of .
3
adjacent
3
x cos
cos 30 2
hypotenuse
1
2
1
opposite
1
y sin
sin 30 2
hypotenuse
1 2
3 1
, . If we reflect this point so that it appears in Quadrants II, III,
Thus, the point ( x, y )
2
2
and IV, then we can derive the following:
The reflection of the point over the y-axis is equivalent to a rotation of 150 from the
positive x-axis. The coordinates of the point are
3 1
( x, y) cos 30,sin 30 cos150,sin150
, .
2
2
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The reflection of the point through the origin is equivalent to a rotation of 210 from the
positive x-axis. The coordinates of the new point are
3 1
( x, y) cos 30, sin 30 cos 210,sin 210
, .
2
2
The reflection of the point over the x-axis is equivalent to a rotation of 330 from the
positive x-axis. The coordinates of the new point are
3 1
( x, y ) cos 30, sin 30 cos 330,sin 330
, .
2
2
If we plot all of the points where 30, 45, and 60 and their reflections, then we get most of
the unit circle. To obtain the rest of the unit circle we have to examine what happens to a point
when 0 .
Notice that the value of x is equal to 1 and that the value of y is zero. Thus, when 0 , the
point ( x, y) 1, 0 . Even though this point does not form a right triangle, any point on a circle
can be found by using cosine and sine. Therefore, cos 0 1 and sin 0 0 .
If we reflect the point over the y-axis, then the new point is ( x, y) 1,0 . This is equivalent
to a rotation of 180 from the positive x-axis. The coordinates of the new point are:
( x, y) cos 0,sin 0 cos180,sin180 1,0 .
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Secondary Mathematics 2
Finally we need to observe what happens when we rotate a point 90 from the positive x-axis.
Notice that the value of y is equal to 1 and that the value of x is zero. Thus, when 90 , the
point ( x, y) 0,1 . Therefore, cos90 0 and sin 90 1 .
If we reflect the point over the x-axis, then the new point is ( x, y) 0, 1 . This is equivalent
to a rotation of 270 from the positive x-axis. The coordinates of the new point are:
( x, y) cos90,sin 90 cos 270,sin 270 0, 1 .
Plotting all of the points, we obtain what is referred to as the unit circle.
The unit circle can be used to find exact values of trigonometric ratios for the angles that relate to
the special right triangle angles.
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Secondary Mathematics 2
Example 1:
Find sin135 .
Answer:
2 2
The point that has been rotated 135 from the positive x-axis has coordinates
,
. The
2
2
2
y-coordinate is the sine value, therefore, sin135
.
2
Example 2:
Find cos 240 .
Answer:
1
3
The point that has been rotated 240 from the positive x-axis has coordinates ,
. The
2
2
1
x-coordinate is the cosine value, therefore, cos 240 .
2
Example 3:
Find all values of , 0 360 , for which sin
1
.
2
Answer:
The points that have been rotated 30 and 150 from the positive x-axis have coordinates
3 1
3 1
, and
, respectively. The y-coordinate is the sine value and both points have a
2
2
2
2
1
y-coordinate of .
2
Example 4:
Find all values of , 0 360 , for which cos 1 .
Answer:
The point that has been rotated 180 from the positive x-axis has coordinates 1, 0 . The xcoordinate is the cosine value which is 1 .
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Secondary Mathematics 2
Defining Tangent Values
Another way to write tan is tan
opposite
sin hypotenuse
tan
adjacent
cos
hypotenuse
opposite
adjacent
hypotenuse hypotenuse
opposite hypotenuse
tan
hypotenuse adjacent
opposite hypotenuse
tan
adjacent hypotenuse
opposite
tan
1
adjacent
tan
sin
. This can be shown algebraically as follows:
cos
opposite
Use the definition sin
and
hypotenuse
adjacent
cos
.
hypotenuse
Rewrite the division problem so that it is easier
to work with.
Dividing by a fraction is the same as
multiplying by its reciprocal.
Use the commutative property of
multiplication to rearrange the terms.
tan
opposite sin
adjacent cos
Example 5:
Find tan 210 .
Answer:
The coordinates of the point that has been rotated 210 from the positive x-axis are
3 1
, .
2
2
sin
tan
cos
sin 210
Use the coordinates of the point to find
tan 210
tan 210 .
cos 210
1
1
3
tan 210 2
sin 210 and sin 210
3
2
2
2
1
3
Rewrite the division problem.
tan 210
2
2
1
2
Dividing by a fraction is the same as
tan 210
multiplying by the reciprocal.
2
3
1
tan 210
Simplify.
3
3
Rationalize the denominator.
tan 210
3
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Secondary Mathematics 2
Tangent Values for the Angles on the Unit Circle
0
30
45
tan
0
1
3
3
180
210
tan
0
3
3
90
120
135
3
undefined
3
1
225
240
270
300
315
1
3
undefined
3
1
60
150
3
3
330
3
3
Previously we defined the six trigonometric functions. Notice that cosecant, secant, and
cotangent are reciprocals of sine, cosine, and tangent, respectively.
The Six Trigonometric Functions
sine sin
opposite
hypotenuse
cosine cos
adjacent
hypotenuse
tangent tan
opposite
adjacent
cosecant csc
secant sec
1
hypotenuse
sin
opposite
1
hypotenuse
cos
adjacent
cotangent cot
1
adjacent
tan opposite
Example 6:
Find sec60 .
Answer:
adjacent
1
and secant is the reciprocal of cosine therefore,
hypotenuse 2
hypotenuse 2
sec 60
2.
adjacent
1
cos 60
Example 7:
Find cot 330 .
Answer:
sin 330
cos 330
cos 330
cot 330
sin 330
tan 330
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3
and cotangent is the reciprocal of tangent therefore,
3
3
3
3 3 3
3.
3
3
3 3
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Secondary Mathematics 2
Example 8:
Find all values of , 0 360 , for which csc
2
.
3
Answer:
Cosecant is the reciprocal of sine, therefore find all the values that satisfy sin
3
. The
2
1 3
and ,
2 2
1 3
angles rotated 60 and 120 from the positive x-axis have coordinates ,
2 2
2
3
respectively. Both have y-coordinates of
. Both will have a cosecant of
.
2
3
Practice Exercises A
Find the value indicated.
1.
sin135
2.
cos 270
3.
tan 300
4.
sin 45
5.
cos 60
6.
tan120
7.
sin180
8.
cos 0
9.
tan 210
10.
sin 240
11.
cos 225
12.
tan 315
13.
csc330
14.
sec30
15.
cot150
16.
csc90
17.
sec180
18.
cot 315
19.
csc 210
20.
sec 225
21.
cot 270
22.
csc 45
23.
sec120
24.
cot 90
28.
sin 1
Find all values of , 0 360 , that make the statement true.
1
2
26.
cos
tan 1
27.
25.
sin
2
2
2
3
csc
31.
tan 0
30.
29.
cos
3
2
33.
cot 3
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34.
csc 1
35.
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sec 1
32. sec 2
36. cot 1
Secondary Mathematics 2
Practice Exercises B
Refer to the diagram. Give the letter that could stand for the function value.
1. cos 180
2. sin 270
3. sin 30
4. cos135
5. sin 0
6. cos330
7. cos90
8. sin 240
9. sin135
10. cos 240
11. sin 330
12. cos 0
For the indicated point, tell if the value for sin or cos is positive, negative, or neither.
13.
cosC
14.
sin G
15.
sin D
16.
cos H
17.
cos E
18.
sin B
19.
cos F
20.
cos B
21.
sin A
22.
cosG
23.
sin C
24.
sin E
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Unit 5 Honors Prove Trigonometric Identities
Trigonometry Proofs
H.5.7 Prove trigonometric identities using definitions, the Pythagorean Theorem, or
other relationships.
H.5.7 Use the relationships to solve problems.
VOCABULARY
A trigonometric identity is a statement of equality that is true for all values of the variable for
which both sides of the equation are defined. The set of values for which the variable is defined
is called the validity of the identity.
sin
is an example of a trigonometric identity. The validity of the
cos
identity would not include values of that would make cos 0 because dividing by zero is
undefined.
The statement tan
Basic Trigonometric Identities
Reciprocal Identities
1
csc
sin
1
sin
csc
Quotient Identities
tan
1
tan
1
tan
cot
1
cos
1
cos
sec
sec
sin
cos
cot
cot
cos
sin
Pythagorean Identities
cos2 sin 2 1
1 tan 2 sec2
cot 2 1 csc2
cos 90 sin
tan 90 cot
cos cos
tan tan
Cofunction Identities
sin 90 cos
Negative Angle Identities
sin sin
Recall the work that you have done with expressions that are quadratic in nature. For example,
( x 3)2 7 x 3 10 is an expression that is quadratic in nature. If u x 3 , then the
expression can be rewritten as u 2 7u 10 . This could then be factored as u 5 u 2 .
Replacing u with x 3 , you get
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x 3 5 x 3 2 .
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Trigonometry expressions can also be
Secondary Mathematics 2
quadratic in nature. For example, the expression cos2 x 5cos x 6 is quadratic in nature. If
u cos x , then it could be rewritten as u 2 5u 6 . This could then be factored as u 2 u 3 .
Substituting cos x back in for u, the factored expression is cos x 2 cos x 3 . You may need
to use this idea when proving trigonometric identities.
Trigonometry Proofs
In a trigonometric proof you manipulate one side of the equation using the known trigonometric
identities until it matches the other side of the equation. Pick the more complicated side to
manipulate.
Example 1:
Prove the identity 1 sec2 x tan 2 x .
Answer:
1 sec2 x tan 2 x
1
sin 2 x
1
cos 2 x cos 2 x
1 sin 2 x
1
cos 2 x
cos 2 x
1
cos 2 x
11
Manipulate the right side of the equation.
Rewrite sec2 x and tan 2 x using the reciprocal
and quotient identities.
The fractions have a common denominator.
Subtract the numerators.
Us the Pythagorean identity cos2 sin 2 1
to replace the numerator with cos2 .
Simplify.
Example 2:
Prove the trigonometric identity cos x cos x sin 2 x cos3 x .
Answer:
cos x cos x sin 2 x cos3 x
cos x 1 sin 2 x cos3 x
cos x cos2 x cos3 x
cos3 x cos3 x
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Manipulate the left side of the equation.
Factor cos x from the two terms cos x and
cos x sin 2 x .
Use the Pythagorean Identity
cos2 x sin 2 x 1 to replace 1 sin 2 x with
cos2 x .
Multiply.
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Secondary Mathematics 2
Example 3:
Prove the identity sin x tan x cos x sec x .
Answer:
sin x tan x cos x sec x
sin x
sin x
cos x sec x
cos x
sin x sin x
cos x sec x
1 cos x
sin 2 x
cos x sec x
cos x
sin 2 x cos x
sec x
cos x
1
2
sin x cos x cos x
sec x
cos x
1 cos x
sin 2 x cos 2 x
sec x
cos x cos x
sin 2 x cos 2 x
sec x
cos x
1
sec x
cos x
sec x sec x
Manipulate the left side of the equation.
Rewrite tan x using a quotient identity.
Rewrite sin x so that it is a fraction.
Multiply the fractions.
Rewrite cos x so that it is a fraction.
The common denominator is cos x 1 cos x .
cos x
Multiply the second fraction by
.
cos x
Simplify.
Add the numerators.
Use the Pythagorean Identity cos2 x sin 2 x 1 .
Use the reciprocal identities to rewrite the
fraction.
Example 4:
Prove the identity tan x cot x sec x csc x .
Answer:
tan x cot x sec x csc x
sin x cos x
sec x csc x
cos x sin x
sin x cos x
sec x csc x
cos x sin x
sin x sin x cos x cos x
sec x csc x
cos x sin x sin x cos x
sin 2 x
cos 2 x
sec x csc x
cos x sin x cos x sin x
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Manipulate the left side of the equation.
Rewrite tan x and cot x using the quotient
identities.
Find a common denominator in order to add
the fractions. The common denominator is
cos x sin x cos x sin x .
Multiply the first fraction by
second fraction by
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sin x
and the
sin x
cos x
.
cos x
Secondary Mathematics 2
Simplify.
Now that the fractions have a common
denominator, add the numerators.
sin x cos x
sec x csc x
cos x sin x
1
sec x csc x
cos x sin x
1
1
sec x csc x
cos x sin x
2
2
Use a Pythagorean Identity, cos2 sin 2 1 ,
to simplify the numerator.
Rewrite the single fraction as a product of two
fractions.
Use the reciprocal identities to rewrite the
fractions.
sec x csc x sec x csc x
Example 5:
Prove the trigonometric identity
cos x 1 sin x
2sec x .
1 sin x
cos x
Answer:
cos x 1 sin x
2sec x
1 sin x
cos x
cos x 1 sin x
2sec x
1 sin x
cos x
cos x cos x 1 sin x 1 sin x
2sec x
1 sin x cos x
cos x 1 sin x
cos 2 x
1 2sin x sin 2 x
2sec x
1 sin x cos x 1 sin x cos x
Manipulate the left side of the equation.
Find a common denominator. The common
denominator will be the product of the two
denominators 1 sin x cos x . Multiply the first
fraction by
1 sin x
.
1 sin x
Simplify.
cos x
and the second fraction by
cos x
1 sin x 1 sin x 1 sin x sin x sin 2 x
1 sin x 1 sin x 2 2sin x sin 2 x
cos 2 x 1 2sin x sin 2 x
2sec x
1 sin x cos x
Add the numerators.
cos 2 x sin 2 x 1 2sin x
2sec x
1 sin x cos x
Rearrange the terms using the properties of
equality.
1 1 2sin x
2sec x
1 sin x cos x
2 2sin x
2sec x
1 sin x cos x
2 1 sin x
2sec x
1 sin x cos x
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Use the Pythagorean Identity cos2 sin 2 1
to simplify the numerator.
Factor a two from both terms in the numerator.
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Secondary Mathematics 2
1 sin x 2 2sec x
1 sin x cos x
1
Simplify.
2
2sec x
cos x
2
Rearrange the terms using the properties of
equality.
Rearrange the terms using the properties of
equality.
1
2sec x
cos x
Rewrite the fraction using a reciprocal identity.
2 sec x 2sec x
Practice Exercises A
Prove the trigonometric identities.
1.
sec x cot x csc x
2.
sin x sec x tan x
3.
tan x cos x sin x
4.
cot x sin x cos x
5.
csc x sin x cot x cos x
6.
tan x csc x cos x 1
7.
cot x sec x sin x 1
8.
9.
sin 2 x 1 cot 2 x 1
10.
11.
cos x cot x
13.
sec x sec x sin 2 x cos x
15.
2sec x
1 sin 2 x
sin x
cos x 1 sin x
1 sin x
cos x
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tan x cot x
sin x
csc x
1 cos 2 x
sin x tan x
cos x
12.
sec2 x csc2 x sec2 x csc2 x
14.
csc x csc x cos2 x sin x
16.
sec x tan x
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2
1 sin x
1 sin x
Secondary Mathematics 2
Solving Trigonometric Equations
Example 6:
Find all values of x if cos x cos x sin x 0 and 0 x 360 .
Answer:
cos x cos x sin x 0
cos x 1 sin x 0
Factor cos x out of both terms.
cos x 0
1 sin x 0
1 sin x
Use the zero product property to set each factor
equal to zero.
x cos1 0
sin 1 1 x
x 90, 270
x 90
To find x, an angle, use the inverse cosine and
sine.
Find all the angles between 0 and 360 that
have a cosine of 0 or a sine of 1.
The angles that satisfy the trigonometric equation are x 90 and x 270 .
Example 7:
Find all values of x if 2cos2 x cos x 1 0 and 0 x 360 .
Answer:
2cos2 x cos x 1 0
2cos x 1 cos x 1 0
Factor.
2 cos x 1 0
2 cos x 1
1
cos x
2
1
x cos 1
2
Use the zero product property to set each factor
equal to zero.
x 60,300
cos x 1 0
cos x 1
x cos1 1
To find x, an angle, use the inverse cosine.
x 180
Find all the angles between 0 and 360 that
have a cosine of 12 or a cosine of 1 .
The angles that satisfy the trigonometric equation are x 60 , x 300 , and x 180 .
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Practice Exercises B
Find all values of x if 0 x 360 .
1.
2 cos x sin x cos x 0
2.
3.
tan x sin 2 x tan x
4.
sin x tan 2 x sin x
5.
tan 2 x 3
6.
2 sin 2 x 1
7.
4 cos 2 x 4 cos x 1 0
8.
2 sin 2 x 3 sin x 1 0
9.
sin 2 x 2 sin x 0
10. 3 sin x 2 cos 2 x
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2 tan x cos x tan x 0
Secondary Mathematics 2
Unit 5 Cluster 6 Honors (F.TF.9)
Prove and Apply Trigonometric Identities
H.5.8
H.5.9
Prove the addition and subtraction formulas for sine, cosine, and tangent and use
them to solve problems.
Justify half-angle and double-angle theorems for trigonometric values.
It is possible to find the exact sine, cosine, and tangent values of angles that do not come from
special right triangles, but you have to use the angles from the unit circle to find them. The
following formulas can be used to find the sine, cosine, and tangent values of angles that are not
on the unit circle.
The Cosine of the Sum of Two Angles
cos A B cos A cos B sin A sin B
The Cosine of the Difference of Two Angles
cos A B cos A cos B sin A sin B
The Sine of the Sum of Two Angles
sin A B sin A cos B cos A sin B
The Sine of the Difference of Two Angles
sin A B sin A cos B cos A sin B
The Tangent of the Sum of Two Angles
tan A tan B
tan A B
1 tan A tan B
The Tangent of the Difference of Two
Angles
tan A tan B
tan A B
1 tan A tan B
Proof of the Cosine Difference Formula: cos A B cos A cos B sin A sin B
Figure 1
Figure 2
Figure 2 shows an angle in standard position. Figure 1 shows the same angle, but it has been
rotated and A B . The chords opposite the angle, , have equal length in both circles.
Therefore, CD is equal to the length of EF . Find the measure of CD and EF .
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Finding CD
CD
cos 1 sin 0
CD
cos 2cos 1 sin
2
2
Use the distance formula to find the distance
between the points cos ,sin and 1, 0 .
2
Expand cos 1 . Remember that
2
2
cos 1
2
cos 1 cos 1 .
Rearrange the terms so that cos2 and sin 2
are next to each other.
Recall that cos2 sin 2 1 (Pythagorean
Identity).
CD cos2 sin 2 2cos 1
CD 1 2cos 1
CD 2 2cos
Simplify.
Finding EF
EF
cos A cos B sin A sin B
2
Use the distance
formula to find the
distance between the
points cos A,sin A
2
and cos B,sin B .
EF
cos
2
A 2cos A cos B cos 2 B sin 2 A 2sin A sin B sin 2 B
EF cos2 A sin 2 A cos2 B sin 2 B 2cos A cos B 2sin A sin B
EF 1 1 2cos A cos B 2sin A sin B
EF 2 2cos A cos B 2sin A sin B
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Expand
2
cos A cos B and
sin A sin B
2
.
Rearrange the terms
so that cos2 A and
sin 2 A and cos2 B
and sin 2 B are next to
each other.
Recall that
cos2 sin 2 1
(Pythagorean
Identity).
Simplify.
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Secondary Mathematics 2
Setting CD EF
2 2cos 2 2cos A cos B 2sin Asin B
Set CD EF .
2 2cos 2 2cos A cos B 2sin A sin B
Square each side to eliminate the square root.
2cos 2cos A cos B 2sin A sin B
Subtract 2 from each side of the equation.
cos cos A cos B sin A sin B
Divide each term on both sides of the equation
by 2 .
cos A B cos A cos B sin A sin B
Recall that A B
The following identities are needed in order to prove the sine of a sum identity.
Negative Angle Identities
sin sin
tan tan
cos cos
Proof of the Cosine of a Sum: cos A B cos A cos B sin A sin B
cos A B cos A B
Rewrite the expression so that it is a difference.
cos A B cos A cos B sin A sin( B)
cos A B cos A cos B sin A sin B
cos A B cos A cos B sin A sin B
Use the cosine of a difference identity to
rewrite the expression.
Use the negative angle identities to eliminate
B .
Use the commutative property of
multiplication to rearrange the terms.
The following identities are needed in order to prove the sine of a sum identity.
Cofunction Identities
sin 90 cos
cos 90 sin
tan 90 cot
Proof of the Sine of a Sum: sin A B sin A cos B cos A sin B
Use the cofunction identity of sine
to rewrite the expression in terms of
cosine.
sin A B cos 90 A B
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sin A B cos 90 A B
Distribute the negative and then
group the first two terms.
sin A B cos 90 A B
sin A B cos 90 A cos B sin 90 A sin B
Use the cosine of a difference
identity to rewrite the expression.
Use the cofunction identities to
rewrite the expression.
sin A B sin A cos B cos A sin B
Proof of the Sine of a Difference: sin A B sin A cos B cos A sin B
sin A B sin A B
Rewrite the expression so that it is a sum.
sin A B sin A cos B cos A sin B
sin A B sin A cos B cos A sin B
sin A B sin A cos B cos A sin B
Use the sine of sum identity to rewrite the
expression.
Use the negative angle identities to eliminate
B .
Use the commutative property of
multiplication to rewrite the expression.
Proof of the Tangent of a Sum Identity: tan A B
tan A B
sin A B
cos A B
sin A cos B cos A sin B
cos A cos B sin A sin B
sin A cos B cos A sin B
cos
A
cos
B
cos A cos B
tan A B
cos A cos B sin A sin B
cos A cos B cos A cos B
sin A cos B cos A sin B
tan A B cos A cos B cos A cos B
cos A cos B sin A sin B
cos A cos B cos A cos B
tan A 1 1 tan B
tan A B
11 tan A tan B
tan A tan B
tan A B
1 tan A tan B
tan A B
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tan A tan B
1 tan A tan B
Use the definition of tangent to rewrite the
expression.
Use the sine and cosine of a sum identities to
rewrite the expressions.
Divide each term in the numerator and the
denominator by cos A cos B .
Rewrite each expression using properties of
equality.
Simplify using the fact that tan
sin
.
cos
Simplify.
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Proof of the Tangent of a Difference Identity: tan A B
tan A B
sin A B
cos A B
sin A cos B cos A sin B
cos A cos B sin A sin B
sin A cos B cos A sin B
cos
A
cos
B
cos A cos B
tan A B
cos A cos B sin A sin B
cos A cos B cos A cos B
sin A cos B cos A sin B
cos
A
cos
B
cos
A
cos B
tan A B
cos A cos B sin A sin B
cos A cos B cos A cos B
tan A 1 1 tan B
tan A B
11 tan A tan B
tan A tan B
tan A B
1 tan A tan B
tan A B
tan A tan B
1 tan A tan B
Use the definition of tangent to rewrite the
expression.
Use the sine and cosine of a difference
identities to rewrite the expressions.
Divide each term in the numerator and the
denominator by cos A cos B .
Rewrite each expression using properties of
equality.
Simplify using the fact that tan
sin
.
cos
Simplify.
Example 1:
Find the exact value of sin165 .
Answer:
sin165 sin 30 135
sin165 sin 30 cos135 cos30 sin135
2 3 2
1
sin165
2 2 2
2
2
6
sin165
4
4
2 6
sin165
4
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Find two special angles that add or subtract to
165 . (There are several possibilities.)
Use the sine of a sum identity to rewrite the
expression.
Substitute known values.
Simplify.
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Secondary Mathematics 2
Example 2:
Find the exact value of tan165 .
Answer:
tan165 tan 210 45
Find two special angles that add or subtract to
165 . (There are several possibilities.)
Use the tangent of a difference identity to
rewrite the expression.
tan 210 tan 45
1 tan 210 tan 45
3
1
3
tan165
3
1
1
3
tan165
Substitute known values.
3 3
3
3
tan165
3
3
3 3
3 3
tan165 3
3 3
3
3 3
tan165
3 3
Simplify.
Practice Exercises A
Use a sum or difference formula to find an exact value.
1.
sin15
2.
cos15
3.
tan 75
4.
sin 75
5.
cos105
6.
tan105
7.
sin195
8.
cos195
9.
tan 255
10.
sin 255
11.
cos 285
12.
tan 285
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Double and Half Angle Formulas
Double-Angle Theorems
cos 2 cos 2 sin 2
sin 2 2sin cos
cos 2 2 cos 2 1
tan 2
cos 2 1 2sin 2
2 tan
1 tan 2
Justification of double angle theorem for sine:
sin 2 sin
Substitute 2 .
sin 2 sin cos cos sin
Use the sine of a sum formula.
sin 2 2sin cos
Simplify.
Justification of double angle theorem for cosine:
cos 2 cos
Substitute 2 .
cos 2 cos cos sin sin
Use the cosine of a sum formula.
sin 2 cos2 sin 2
Simplify.
Justification of double angle theorem for tangent:
tan 2 tan
tan tan
1 tan tan
2 tan tan
tan 2
1 tan 2
tan 2
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Substitute 2 .
Use the tangent of a sum formula.
Simplify.
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Half-Angle Theorems
1 cos
sin
2
2
cos
2
1 cos
2
tan
2
1 cos
1 cos
Justification of half-angle theorem for sine:
Use cos 2 1 2sin 2 . Solve the double angle
cos 2 1 2sin 2
formula for sin .
cos 2 2sin 2 1
2sin 2 1 cos 2
1 cos 2
sin 2
2
1 cos 2
sin
2
1 cos 2
2
sin
2
2
1 cos
sin
2
2
Substitute
2
.
Simplify.
Justification of half-angle theorem for cosine:
Use cos 2 2cos2 1 . Solve the double angle
cos 2 2 cos 2 1
formula for cos .
cos 2 2 cos 2 1
2 cos 2 1 cos 2
1 cos 2
cos 2
2
1 cos 2
cos
2
1 cos 2
2
cos
2
2
1 cos
cos
2
2
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Substitute
2
.
Simplify.
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Secondary Mathematics 2
Justification of half-angle theorem for tangent:
tan
2
sin
cos
2
Use tan
2
1 cos
2
tan
2
1 cos
2
1 cos
2
tan
1 cos
2
2
Use the half-angle theorems for sine and cosine.
1
1 cos
tan 2
1
2
1 cos
2
Simplify using properties of radicals and exponents.
tan
2
sin
. Substitute .
cos
2
1 cos
1 cos
Practice Exercises B
Use the figures to find the exact value of each trigonometric function.
1.
sin 2
2.
cos 2
3.
tan 2
4.
sin 2
5.
cos 2
6.
tan 2
7.
sin 2
8.
cos 2
9.
tan 2
10. sin
13. sin
16. sin
2
2
2
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11. cos
14. cos
17. cos
12. tan
2
15. tan
2
18. tan
2
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2
2
2
Secondary Mathematics 2
Unit 6
Circles With and
Without Coordinates
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Unit 6 Cluster 1 (G.C.1, G.C.2, G.C.3, and Honors G.C.4)
Understand and Apply Theorems about Circles
Cluster 1: Understanding and applying theorems about circles
6.1.1 Prove that all circles are similar.
6.1.2 Understand relationships with inscribed angles, radii, and chords (the relationship
between central, inscribed, circumscribed; the relationship between inscribed
angles on a diameter; the relationship between radius and the tangent).
6.1.3 Construct the inscribed and circumscribed sides of a triangle.
6.1.3 Prove properties of angles for a quadrilateral inscribed in a circle.
6.1.4 (Honors) Construct a tangent line from a point outside a given circle to a circle.
VOCABULARY
A circle is the set of all points equidistant from a given point
which is called the center of the circle.
A radius is any segment with endpoints that are the center of the
circle and a point on the circle. Radii is the plural of radius.
AB is the radius of circle A. The center of the circle is point A.
A segment with endpoints on the circle is called a chord.
DE is a chord of circle A.
A diameter is any chord with endpoints that are on the circle and
that passes through the center of the circle. The diameter is the
longest chord of a circle.
CB is the diameter of circle A.
An angle that intersects a circle in two points and that has its
vertex at the center of the circle is a central angle.
BAF is a central angle of circle A.
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An angle that intersects a circle in two points and that has its
vertex on the circle is an inscribed angle.
DBC is an inscribed angle of circle A.
A polygon that is circumscribed by a circle has all of its vertices
on the circle and the polygon’s interior is completely contained
within the circle.
Circle A is circumscribed about Quadrilateral BCDE.
A planar shape or solid completely enclosed by (fits snugly inside)
another geometric shape or solid is an inscribed figure. Each of
the vertices of the enclosed figure must lie on the “outside” figure.
Quadrilateral BCDE is inscribed in circle A.
A line that intersects a circle in only one point is a tangent line.
The point where the tangent line and the circle intersect is the
point of tangency.
BC is a tangent line to circle A. Point B is the point of tangency.
A line that intersects a circle in two points is a secant line.
BC is a secant line to circle A.
Other helpful sources: http://www.mathgoodies.com/lessons/vol2/geometry.html
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Practice Exercises A
Identify a chord, tangent line, diameter, two radii, the center, and point of tangency and, a central
angle.
1. chord: ______________________________
2. tangent line: _________________________
3. diameter: ____________________________
4. radius: ______________________________
5. point of tangency : ____________________
6. center: ______________________________
7. central angle: ________________________
Identify the term that best describes the given line, segment, or point.
8. AF
9. PF
10. C
11. BD
12. EG
13. PC
14. CE
15. P
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Prove All Circles are Similar
Figures that are similar have corresponding parts that are proportional. To prove that all circles
are similar, you will need to prove that their corresponding parts are proportional.
Proof that all circles are Similar
Given: A D , radius of circle A is x, and radius of circle D is y.
Prove: Circle A is similar to Circle D.
First we need to prove that ABC DEF so we need to establish AA criterion. Since
AB AC , ABC is an isosceles triangle. The base angles of an isosceles triangle are congruent
to one another. Therefore B C . Similarly, DE DF so DEF is also an isosceles
triangle. Therefore E F . The sum of the angles in a triangle is 180 so
A B C 180 and D E F 180 . We know that B C and E F so
A 2B 180 and D 2E 180 . Solving each equation for A and D yields
A 180 2B and D 180 2E . Since A D we know that
180 2B 180 2E and B E . By AA similarity ABC DEF . Because the two
AB x
.
triangles are similar their sides will be proportional. The ratio of proportionality is
DE y
AB is a radius of circle A and DE is a radius of circle D. The ratio of the radii of the circles is
x
. The diameter of circle A is 2x and the diameter of circle D is 2 y . The ratio of the diameter
y
2x x
. The circumference of circle A is 2 x and the
of circle A to the diameter of circle D is
2y y
circumference of circle D is 2 y . The ratio of the circumference of circle A to the
2 x x
. The corresponding parts of circle A are proportional to
circumference of circle D is
2 y y
the corresponding parts of circle D, therefore circle A is similar to circle D.
In addition to having corresponding parts proportional to one another, figures that are similar to one
another are dilations of one another. A dilation is a transformation that produces an image that is the
same shape as the original figure but the image is a different size. The dilation uses a center and a
scale factor to create a proportional figure. The ratio of the corresponding parts is the scale factor of
the dilation.
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Practice Exercises B
1.
Given a circle of a radius of 3 and another circle with a radius of 5, compare the ratios of the
two radii, the two diameters, and the two circumferences.
2.
Given a circle of a radius of 6 and another circle with a radius of 4, compare the ratios of the
two radii, the two diameters, and the two circumferences.
Properties of Central Angles and Inscribed Angles
VOCABULARY
An arc is a portion of a circle's circumference.
An intercepted arc is the arc that lies in the interior of an
angle and has its endpoints on the angle.
A central angle is an angle that intersects a circle in two
points and that has its vertex at the center of the circle. The
measure of the angle is the same as the measure of its
intercepted arc.
An inscribed angle is an angle whose vertex is on a circle and
whose sides contain chords of the circle. The measure of the
angle is half the measure of the intercepted arc.
ACB is a central angle of circle C and its intercepted arc is
AB and mACB mAB . ADB is an inscribed angle of
circle A and its intercepted arc is AB and mADB
1
mAB.
2
If the mACB is less than 180 , then A, B, and all the
points on C that lie in the interior of mACB form a minor
arc. A minor arc is named by two consecutive points. The
measure of a minor arc is the measure of its central angle.
AB is the minor arc of circle C.
Points A, B, and all points on C that do not lie on AB form
a major arc. A major arc is named by three consecutive
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points. The measure of a major arc is 360° minus the measure
of the related minor arc.
ADB is the major arc of circle C.
A semicircle is an arc whose central angle measures 180°. A
semicircle is named by three points.
An inscribed angle that intercepts a semicircle is a right angle.
mABC
1
1
mAC 180 90
2
2
The arc addition postulate states that the measure of an arc
formed by two adjacent arcs is the sum of the measures of the
two arcs.
mML mLN mMN
Congruent arcs are arcs with the same measure either in the
same circle or congruent circles. Congruent central angles or
inscribed angles have congruent arcs and congruent arcs have
congruent central angles or inscribed angles.
ABC CBD AC CD
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Practice Exercises C
FH and JK are diameters. Find the measure of each angle or arc.
1. mFAJ
2. mLAH
3. mKAF
4. mJL
5. mLH
6. mHK
7. mKF
8. mJF
9. mJH
10. mJHF
FH and KJ are diameters, mFHM 40 , mHK 60 , and mJL 50 . Find the measure of
each angle or arc.
11. mJF
12. mLH
13. mJKL
14. mFM
15. mHAK
17. mMK
19. Find the measure of angles 1, 2, and 3.
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16. mKF
18. mJGK
20. Find the measure of angles 1 and 2 if
m1 2 x 13 and m2 x .
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Theorems
If two chords intersect inside a circle, then the measure of
each angle formed is half the sum of the measures of the arcs
intercepted by the angle and its vertical angle.
1
m2 mCD mAB
2
1
m1 mBC mAD
2
If two chords intersect inside a circle, then the product of the
lengths of the segments of one chord is equal to the product
of the lengths of the segments of the other chord.
xy wz
Example 1:
Find the value of x.
Answer:
The two chords forming the angle
that measures x intercepts the
two arcs AB and CD . The
measure of x will be equal to
one-half the sum of the measures
of the intercepted arcs.
1
m AB mCD
2
1
x 106 174
2
1
x 280
2
x 140
x
Example 2:
Find the value of x.
The chords AC and BD intersect
inside the circle, therefore the
product of the lengths of the
segments of each chord are equal
to one another. By setting up this
equation you can solve for x.
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Answer:
9x 3 6
9 x 18
x2
Secondary Mathematics 2
Practice Exercises D
Find the value of x.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
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Theorems
If a line is tangent to a circle, then it is perpendicular to the
radius drawn at the point of tangency.
If l is tangent to
C at B, then l
to CB .
In a plane, if a line is perpendicular to a radius of a circle at
its endpoints on the circle, then the line is tangent to the
circle.
If l
to CB , then l is tangent to
C at B.
If two segments from the same point outside a circle are
tangent to the circle, then they are congruent.
If BD and CD are tangent to circle A at points B and C,
then BD CD .
Example 3: Using Properties of Tangents
AC is tangent to B at point C. Find BC.
Answer:
AC BC AB
2
122 BC 132
2
144 BC 169
2
BC 169 144
2
BC 25
2
2
2
Since AC is tangent to circle B at point C,
AC BC . This makes ABC a right triangle
and the Pythagorean Theorem can be used to
find BC.
BC 5
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Example 4: Finding the Radius of a Circle
BC is tangent to circle A at point B. Solve for r.
Answer:
BC AB AC
2
r 2 152 r 9
2
2
2
r 2 225 r 2 18r 81
225 18r 81
225 81 18r
144 18r
8r
Since BC is tangent to circle A at point B,
AB BC . This makes ABC a right triangle
and the Pythagorean Theorem can be used to
find r.
Example 5: Verify a Line is Tangent to a Circle
Show that AC is tangent to B .
Answer:
9 12 15
81 144 225
225 225
2
2
2
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If you can show that 92 122 152 , then the
lines are perpendicular.
225 = 225 is a true statement, so the lines are
perpendicular.
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Example 6: Using Properties of Tangents
Find x if AB is tangent to circle C at point B
and AD is tangent to circle C at point D.
Answer:
AD AB
2 x 3 11
2x 8
x4
Because AB and AD are tangent to the same
circle and contain the same exterior point, the
measures of each length are equal.
Practice Exercises E
Find the value of x.
1.
3.
6.
2.
4.
5.
Prove that radius AB AC using the Pythagorean
Theorem.
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Circumscribed Angles
Theorem
If two secants, a secant and a tangent, or two tangents intersect in the exterior of a circle, then the
measure of the angle formed is one-half the positive difference of the measures of the intercepted
arcs.
Two Secants
Secant and Tangent
Two Tangents
mA
1
mDE mBC
2
mA
1
mDC mBC
2
mA
1
mCDB mBC
2
Example 7:
Find the value of x.
A is formed by two secants so its measure will be one-
half the positive difference of the two intercepted arcs DE
and BC .
mA
1
mDE mBC
2
1
120 54
2
1
x 66
2
x 33
x
Example 8:
Find the value of x.
A is formed by two tangents so its measure will be onehalf the positive difference of the two intercepted arcs
BDC 360 x and BC x .
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mA
1
mBDC mBC
2
1
360 x x
2
1
30 360 2 x
2
30 180 x
150 x
150 x
30
Practice Exercises F
Find the value of x. Assume lines that appear tangent are tangent.
1.
2.
3.
4.
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5.
6.
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Secondary Mathematics 2
Constructions
See Secondary Math 1 for more help with constructions.
Step 1
Circumscribe a triangle with a circle
Step 1: Create a triangle.
Step 2: Construct the perpendicular bisector of AB by
setting the compass to slightly more than half the
distance between points A and B.
Steps 3 and 4
Step 3: With the compass on point A, draw an arc on both
sides of AB .
Step 4: With the compass on point B, draw an arc on both
sides of AB .
Step 5: With a straightedge, draw the line connecting the
intersection points. This is the perpendicular
bisector of AB .
Step 5
Step 6: Construct the perpendicular bisector of another side
of the triangle.
Step 7: The intersection of the two perpendicular bisectors
will be equidistant from the vertices. Set the
compass to the distance between a vertex and this
intersection point.
Step 8
Step 8: With the compass on the intersection of the
perpendicular bisectors, draw a circle that
circumscribes the triangle.
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Step 2
Inscribe a circle in a triangle
Step 1: Draw a triangle. Label the vertices A, B, and C.
Step 2: Bisect A of the triangle by setting the compass
to a medium length. With the compass
somewhere on point A draw an arc from AB to
AC .
Steps 3 and 4
Step 3: With the compass on AB where the arc intersects
the segment, draw an arc in the interior of A .
Step 4: With the compass on AC where the arc intersects
the segment, draw an arc in the interior of A .
Step 5
Step 5: Connect point A to the intersection of the two
arcs in its interior with a straightedge. This is the
bisector of the angle.
Step 6: Bisect another one of the angles.
Step 6
Step 7: The intersection point of the two bisectors is the
center of the inscribed circle.
Step 8: Construct a perpendicular from the center point to
one side of the triangle.
Step 8
Step 9: Set the compass to be more than the distance from
the center to a side of the triangle AC . With
the compass on the intersection of the two angle
bisectors, draw two arcs that intersect a side of the
triangle AC .
Step 12
Step 10: Set the compass to the distance between the two
arcs that intersect a side of the triangle AC .
With the compass on one of the intersections,
draw an arc below a side of the triangle AC .
Step 11: With the compass on the other intersection, draw
an arc below a side of the triangle AC .
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Step 12: Draw the line connecting the intersection of the
two angle bisectors to the intersection of the two
arcs below a side of the triangle AC .
Step 14
Step 13: Place the compass on the center point; adjust the
length of the compass to touch the intersection of
the perpendicular line and the side of the triangle.
Step 14: Now create the inscribed circle.
Quadrilaterals Inscribed in Circles
Theorem
If a quadrilateral is inscribed in a circle, its opposite angles are supplementary.
Proof of Theorem:
Given: Quadrilateral ABCD is inscribed in circle E
Prove: A and C are supplementary. B and D are
supplementary.
By arc addition and the definitions of arc measure, mBCD mBAD 360 and
mABC mADC 360 . Since the measure of the intercepted arc is twice the measure of the
inscribed angle, 2mA mBCD , 2mC mBAD , 2mB mADC and 2mD mABC . By
substitution 2mA 2mC 360 and 2mB 2mD 360 . Using reverse distribution, the
equations can be rewritten as 2 mA mC 360 and 2 mB mD 360 . Applying
the division property of equality the equations can be rewritten as mA mC 180 and
mB mD 180 . By definition of supplementary, A and C are supplementary and B
and D are supplementary.
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Example 1:
Find x and y.
The opposite angles of a
quadrilateral inscribed in a
circle are supplementary.
Answer:
x 180 mEDC
x 180 80
x 100
y 180 mBCD
y 180 120
y 60
Practice Exercises G
1.
Triangles EFG and EGH are inscribed in A with EF FG .
Find the measure of each numbered angle if m1 12 x 8
and m2 3x 8 .
2.
Quadrilateral LMNO is inscribed in P . If mM 80 and
mN 40 , then find mO and mL .
3.
Find the measure of each numbered angle for the figure if
mJK 120. Diameter JL .
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4.
Find the measure of each numbered angle for the figure if
1
1
mR x and mK x 5.
2
3
5.
Quadrilateral QRST is inscribed in a
circle. If mQ 45 and mR 100, find
mS and mT .
6.
Quadrilateral ABCD is inscribed in a
circle. If mC 28 and mB 110 , find
mA and mD .
(Honors) Construct a tangent line from a point outside a given circle to a circle
Step 1: Start with a circle with center A, and a point B outside
Step 3
of the circle.
Step 2: Draw a line segment with endpoints A and B.
Step 3: Find the midpoint M of AB by constructing the
perpendicular bisector of AB . Depending of the size
of the circle and the location of point B, the midpoint
may be inside or outside of the circle.
Step 4: Place the compass on point M and set the compass
width to the center A of the circle.
Step 5: Without changing the compass width, draw an arc that
intersects the circle in two points. These points (label
them C and D) will be the points of tangency.
Step 6: Draw BC and BD . These lines are tangent to circle A
from a point B outside circle A.
M
Step 5
M
Step 6
M
Other helpful construction resources:
http://www.khanacademy.org/math/geometry/circles-topic/v/right-triangles-inscribed-in-circles-proof;
http://www.mathopenref.com/consttangents.html;
http://www.mathsisfun.com/geometry/construct-triangleinscribe.html;
http://www.mathsisfun.com/geometry/construct-trianglecircum.html;
http://www.benjamin-mills.com/maths/Year11/circle-theorems-proof.pdf
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Unit 6 Cluster 2 (G.C.5)
Circles with Coordinates and Without Coordinates
Cluster 2: Finding arc lengths and areas of sectors of circles
6.2.1 The length of the arc intercepted by the angle is proportional to the radius
6.2.1 The radian measure is the ratio between the intercepted arc and the radius
6.2.1 Derive the formula for the area of the sector
Recall that the measure of an arc is the same as the measure of the
central angle that intercepts it. The measure of an arc is in degrees,
while the arc length is a fraction of the circumference. Thus, the
measure of an arc is not the same as the arc length.
Consider the circles and the arcs shown at the
left. All circles are similar therefore circle 2 can
be dilated so that it is mapped on top of circle 1.
The same dilation maps the slice of the small
circle to the slice of the large circle. Since
corresponding lengths of similar figures are
proportional the following relationship exists.
r1 l1
r2 l2
Solving this proportion for l1 gives the following
equation.
r1l2 l1r2
r1
l2
l1
r2
l2
. We can
r2
find that number by looking at how the central angle compares to the entire circle. Given a
1
30
central angle of 30 it is
or
of the entire circle. The length of the arc that is
12
360
1
intercepted by the central angle of 30 will also be
of the circumference. Therefore the
12
length of the arc depends only on the radius.
This means that the arc length, l1 , is equal to the radius, r1 , times some number,
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To generalize the relationship between the arc length and the radius, set up a proportion showing
that the central angle compared to the whole circle is proportional to the length of the arc
compared to the circumference of the circle.
l
360
2 r
Find the length of an arc by multiplying the central angle ratio by the circumference of the circle
(2πr). In other words solve for l (length).
l
360
2 r
This equation can be simplified because 360, 2 and are all constants.
l
l
2
r
360
180
r
Compare this formula with the formula we obtained earlier r1
where is the measure of the central angle in degrees.
l2
l
l1 . The number 2 is
180
r2
r2
Formula for arc length
If the central angle of a circle with radius r is degrees, then the length, l, of the arc it intercepts
is given by: l
180
r.
Example 1: Finding arc length
Find the arc length if the radius of a circle is 5 centimeters and the central angle is 72 . Write
the answer in terms of and give a decimal approximation to nearest thousandth.
Answer:
l
r
180
72
l
5
180
72
l
5
180
360
l
180
l 2 6.283
Use the formula for arc length. Substitute 5 in
for r and 72 in for .
Simplify.
The arc length is 2 centimeters or approximately 6.283 centimeters.
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Example 2: Finding arc length
Find the arc length if the radius of a circle is 7 inches and the central angle is 120 . Write the
answer in terms of and give a decimal approximation to the nearest thousandth.
Answer:
l
r
180
120
l
7
180
120
l
7
180
840
l
180
14
l 14.661
3
The arc length is
Use the formula for arc length. Substitute 7 in
for r and 120 in for .
Simplify.
14
inches or approximately 14.661inches.
3
Practice Exercises A
1.
Find the arc length if the radius of a circle is 10 yards and the central angle is 44 . Write
the answer in terms of and give a decimal approximation to the nearest thousandth.
2.
Find the arc length if the radius of a circle is 8 meters and the central angle is 99 . Write
the answer in terms of and give a decimal approximation to the nearest thousandth.
3.
Find the arc length if the radius of a circle is 2 feet and the central angle is 332 . Write the
answer in terms of and give a decimal approximation to the nearest thousandth.
4.
Find the arc length if the radius of a circle is 3 kilometers and the central angle is 174 .
Write the answer in terms of and give a decimal approximation to the nearest thousandth.
5.
Find the arc length if the radius of a circle is 9 centimeters and the central angle is 98 .
Write the answer in terms of and give a decimal approximation to the nearest thousandth.
6.
Find the arc length if the radius of a circle is 6 miles the central angle is 125 . Write the
answer in terms of and give a decimal approximation to the nearest thousandth.
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Another way to measure angles is with radians. The radian measure of a central angle is defined
as the ratio of the arc length compared to the radius. If is the radian measure of a central
r
l 180
angle then,
, where is the measure of the central angle in degrees. To
r
r
180
convert any angle in degrees to radian measure multiply the angle in degrees by
.
180
Converting Between Radians and Degrees
radians
To convert degrees to radians, multiply the angle by
.
180
180
To covert radians to degrees, multiply the angle by
.
radians
Example 3: Converting from degrees to radians
Convert an angle of 25 to radian measure. Leave your answer in terms of .
Answer:
radians
25
180
25
radians
180
5
radians
36
Multiply the angle by the conversion factor
radians
.
180
Simplify the fraction.
Example 4: Converting from radians to degrees
Convert an angle of
radians to degrees.
2
Answer:
180
radians
Multiply the angle by the conversion factor
2
radians
180
.
180 radians
radians
2 radians
Simplify the fraction.
90
Practice Exercises B
Find the degree measure of each angle expressed in radians and find the radian measure of each
angle expressed in degrees. (Express radian measures in terms of .)
2
2.
1. 135
3. 45
3
5
5
4.
6.
5. 330
4
2
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Secondary Mathematics 2
The arc length can also be found by using a radian measure for the central angle. When this
happens the formula is l r , where is in radian measure.
Practice Exercises C
Compare your answers found here to those in Practice Exercises A.
11
. Write
45
the answer in terms of and give a decimal approximation to the nearest thousandth.
1.
Find the arc length if the radius of a circle is 10 yards and the central angle is
2.
Find the arc length if the radius of a circle is 8 meters and the central angle is
3.
Find the arc length if the radius of a circle is 2 feet and the central angle is
4.
Find the arc length if the radius of a circle is 3 kilometers and the central angle is
5.
Find the arc length if the radius of a circle is 9 centimeters and the central angle is
6.
Find the arc length if the radius of a circle is 6 miles the central angle is
11
. Write
20
the answer in terms of and give a decimal approximation to the nearest thousandth.
83
. Write the
45
answer in terms of and give a decimal approximation to the nearest thousandth.
29
.
30
Write the answer in terms of and give a decimal approximation to the nearest thousandth.
49
.
90
Write the answer in terms of and give a decimal approximation to the nearest thousandth.
25
. Write the
36
answer in terms of and give a decimal approximation to the nearest thousandth.
Area of a Sector
A region of a circle determined by two radii and the arc intercepted by
the radii is called a sector of the circle (think of a slice of pie). A
sector is a fraction of a circle, so the ratio of the area of the sector to
the area of the entire circle is equal to the measure of the central angle
creating the sector to the measure of the entire circle.
Symbolically this ratio is
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area of a sector measure of central angle
.
area of circle
measure of circle
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Secondary Mathematics 2
Using substitution this becomes,
area of a sector
. Solving for the area of a sector we
2
r
360
r 2 .
360
Another way of looking at area of a sector is shown below.
get: area of a sector
The area of a circle is A r 2 .
The area of one-half a circle is A 12 r 2 .
The area of one-fourth circle is A 14 r 2 .
r2 .
The area of any fraction of a circle is A 360
Area of a sector
If the angle is in degrees then the area of a sector, A, is A
360
r2 .
1 2
r , recall that an angle in radian measure is equal to
(Rewriting the formula for area of a sector you get A
2 180
180
where is in degrees.)
1
If the angle is in radian measure then the area of a sector, A, is A r 2 .
2
Example 5:
Find the area of a sector with radius 5 cm and central angle of 135 . Express your answer in
terms of and approximate it to the nearest thousandths.
Answer:
A
A
Use the formula for the area of a sector with
the angle in degrees.
r2
360
135
5
2
360
3
A 25
8
75
A
8
The area of the sector is
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Substitute known values. 135 and r 5
Simplify.
75
29.452 cm2.
8
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Secondary Mathematics 2
Practice Exercises D
Find the area of the sector given the radius and central angle. Express your answer in terms of
and approximate it to the nearest thousandths.
1.
A radius of 2 feet and a central angle of 180 .
2.
A radius of 5 centimeters and a central angle of 90 .
3.
A radius of 4 inches and a central angle of 60 .
4.
A radius of 10 inches and a central angle of 120 .
5.
A radius of 10 meters and a central angle of 45 .
.
2
5
7. A radius of 2 millimeters and a central angle of .
6
5
8. A radius of 6 feet and a central angle of .
4
3
9. A radius of 3 inches and a central angle of .
2
5
10. A radius of 6 meters and a central angle of .
3
6.
A radius of 7 centimeters and a central angle of
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Unit 6 Cluster 3 (G.GPE.1)
Equation of a Circle
Cluster 3: Translating between descriptions and equations for a conic section
6.3.1 Find the equation of a circle given the center and the radius using the Pythagorean
Theorem; also complete the square to find the center of the circle given the
equation
Deriving the Equation of a Circle Centered at the Origin
A circle is the set of all points equidistant from a given point called the center. In order to derive
the equation of a circle, we need to find the distance from the center to any point x, y on the
circle. This distance is the length of the radius. We can derive the equation of a circle using the
distance formula or the Pythagorean Theorem.
Create a right triangle with the radius as the hypotenuse, the
length x as the horizontal leg, and the length y as the vertical
leg.
Using the Pythagorean Theorem to relate the sides we get:
r 2 x2 y 2 .
Where r is the radius and the circle is centered at the origin.
Example 1:
Identify the center and radius of the circle with equation x 2 y 2 49.
Answer:
Rewrite the equation so that it matches the
standard form of the equation of a circle.
x 2 y 2 49
x2 y 2 72
The square root of 49 is 7, therefore 72 49 .
The center is at the origin and the radius is 7.
Example 2:
Write the equation of a circle centered at the origin with a radius of 2.
Answer:
x 2 y 2 22
Use the standard form of the equation of a
circle.
Substitute known values. r 2
x2 y 2 4
Simplify.
x2 y 2 r 2
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Secondary Mathematics 2
Practice Exercises A
1.
Identify the center and the radius for the circle with the equation x 2 y 2 36.
2.
Write an equation for a circle centered at the origin with a radius of 5.
3.
Write an equation for a circle centered at the origin with a diameter of 14.
4.
Write an equation for a circle centered at the origin that contains the point 3,5 .
Deriving the Equation of a Circle Centered at h, k
Find any point x, y on the circle. Create a right triangle
with the radius as the hypotenuse, the length x h as the
horizontal leg, and the length y k as the vertical leg.
Then using the Pythagorean Theorem to relate the sides we
get:
r 2 x h y k .
2
2
Where r is the radius and the circle is centered at the point
h, k .
Example 3:
Find the center and radius for the circle with equation x 3 y 2 81 .
2
2
Answer:
x 3 y 2
2
2
92
The center is 3, 2 and the radius is 9.
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Rewrite the circle’s equation so that it matches
the standard form of a circle.
The square root of 81 is 9 so 92 81 .
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Secondary Mathematics 2
Example 4:
Write the equation for the circle centered at 1,5 and radius 4.
Answer:
x h y k
2
2
x 1 y 5
2
x 1 y 5
2
2
Start with the standard form of the equation of
a circle.
Substitute the known values. h 1 , k 5 ,
and r 4 .
r2
2
42
16
Simplify.
Example 5:
Write the equation for the circle that has a diameter with endpoints 1,5 and 5, 3 .
Answer:
1 5 5 3
,
2
2
Find the center of the circle by finding the
midpoint of the diameter.
4 2
,
2 2
2,1
r
2 1 1 5
r
2 1 4
2
2
r 3 16
2
r 9 16
2
2
Find the length of the radius by finding the
distance between the center and either of the
endpoints of the diameter.
r 25
r 5
x 2 y 1 52
2
2
x 2 y 1 25
2
2
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Substitute the known values into the standard
form of the equation of a circle centered at
h, k and simplify. h 2 , k 1 , and r 5
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Secondary Mathematics 2
Practice Exercises B
Given the standard form of a circle determine the center and the radius of each circle.
1. ( x 2)2 ( y 3)2 16
2. ( x 1)2 ( y 7)2 25
3. ( x 5)2 ( y 6)2 4
4. ( x 2)2 ( y 9)2 36
5. ( x 10)2 ( y 21)2 196
6. ( x 1)2 ( y 3)2 49
Write the standard form of a circle with the given characteristics.
7.
A circle with radius 10 centered at 8, 6 .
8.
A circle with radius 5 centered at 4,3 .
9.
A circle with diameter endpoints at 9, 2 and 1, 6 .
10. A circle with diameter endpoints at 3, 4 and 5, 2 .
Example 6:
Complete the square to find the center and radius of a circle given by the equation
x2 y 2 6 x 2 y 6 0 .
Answer:
x 6x y 2 y 6
2
x
2
2
6 x ___ y 2 2 y ___ 6
Collect the x terms together, the y terms
together and move the constant to the other
side of the equation.
2
2
2
6 2
2
Group the x and y terms together. Complete
x 6 x y 2 y 6 9 1
the square and simplify.
2
2
x
2
6 x 9 y 2 2 y 1 16
x 3 y 1
2
2
16
Rewrite each trinomial as a binomial squared.
The center is at 3, 1 and the radius is 4.
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Practice Exercises C
Complete the square to find the center and radius of a circle given by the equation.
1. x2 y 2 4 x 6 y 8 0
2. x2 y 2 4 x 10 y 20 0
3. x2 y 2 6 x 2 y 15 0
4. x2 y 2 6 x 4 y 9 0
Challenge Problem 2 x2 2 y 2 6 x 8 y 12 0
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Unit 6 Cluster 4 (G.GPE.4)
Proving Geometrical Theorems Algebraically
Cluster 4: Using coordinates to prove theorems algebraically
6.4.2 Use coordinates to prove geometric theorems algebraically (include simple proofs
involving circles)
A trapezoid is a quadrilateral with only one set of parallel
sides.
An isosceles trapezoid is a trapezoid with congruent legs and
congruent base angles. The diagonals of an isosceles
trapezoid are congruent.
A parallelogram is a quadrilateral with opposite sides
parallel and congruent. The diagonals of a parallelogram
bisect each other.
A rectangle is a special parallelogram with four right angles.
The diagonals of a rectangle are congruent.
A rhombus is a special parallelogram with four congruent
sides. The diagonals bisect each other and are perpendicular
to one another.
A square is a special rectangle and rhombus with four
congruent sides. The diagonals are congruent, bisect each
other, and are perpendicular to each other.
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Example 1:
Prove FGHJ with vertices F 2,1 , G 1, 4 , H 5, 4 , and J 6,1 is a parallelogram.
Plot the points on a coordinate plane.
GF
2 1 1 4
HJ
5 6 4 1
FJ
6 2 1 1
GH
2
2
2
2
2
2
5 1 4 4
2
12 3 1 9 10
2
1
2
32 1 9 10
Find the distance of GF , HJ , FJ ,
and GH .
42 02 16 4
2
42 02 16 4
1 4 3
3
2 1 1
4 1 3
mHJ
3
5 6 1
1 1 0
mFJ
0
62 4
44 0
mGH
0
5 1 4
Find the slopes of GF , HJ , FJ , and
mGF
GH .
Since opposite sides are congruent and parallel the quadrilateral FGHJ is a parallelogram.
Example 2:
Prove that the point (5,2) is on the circle with center (3,2) and radius 2.
Create the circle on a coordinate plane.
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Secondary Mathematics 2
r
5 3 2 2
2
Use the distance formula to show that the
radius is 2. Use the center as x1 , y1 and 5, 2
2
r 22 02
as x2 , y2 .
r 4
r2
The radius of the circle is 2 therefore the point must lie on the circle.
Practice Exercises A
1.
Prove that quadrilateral EFGH is an
isosceles trapezoid given the following
vertices: E 3, 2 , F 2, 2 , G 3, 2 ,
2.
B 3, 6 , C 6,8 and D 5,5 .
and H 4, 2 .
3.
Prove that ABCD is a parallelogram given
vertices: A 3, 2 , B 0, 4 , C 1,8 and
4.
D 2, 6 using distance and slope.
5.
Prove that ABCD is a parallelogram given
vertices: A 2, 4 , B 6, 4 , C 5, 0 , and
Prove that the point
6.
Prove that ABCD is a rhombus given
vertices: A 1, 2 , B 0, 6 , C 4, 7 , and
D 3,3 .
3, 3 lies on the
8.
Given a circle with center 2,3
determine whether or not the points
4, 1 and 3,5 are on the same circle.
11. Prove that the line containing the points
5,3 and 3,3 is tangent to the circle
10. Given a circle with center at the origin
determine whether or not the points
1, 3 and 0, 2 lie on the same circle.
12. Prove that the line containing the points
0,8 and 4,11 is tangent to the circle
with equation x 2 y 3 100 .
with equation x 1 y 1 16 .
2
Prove that the point 2, 5 lies on the
circle with radius 2 and center 2, 3 .
circle with radius 2 and center 2, 0 .
9.
Prove that PQRS is not a rectangle given
vertices: P 0, 2 , Q 2,5 , R 5,5 , and
S 4, 2 .
D 1, 0 .
7.
Prove that quadrilateral ABCD is a
parallelogram given vertices: A 2,3 ,
2
2
2
Other helpful resources:
http://www.regentsprep.org/regents/math/geometry/GCG4/CoordinatepRACTICE.htm
http://staff.tamhigh.org/erlin/math_content/Geometry/PolyQuad/CoordinateGeometryNotes.pdf
http://mtprojectmath.com/Resources/Coordinate%20Geometry%20Proofs.pdf
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Unit 6 Cluster 5 (G.GMD.1 and G.GMD.3, Honors G.GMD.2)
Formulas and Volume
Cluster 5: Explaining and using volume formulas
6.5.1 Informal arguments for circumference of circle, volume of a cylinder, pyramid,
and cone (use dissection arguments, Cavalieri’s principle) (use the relationship of
scale factor k, where k is for a single length, k2 is area, k3 is volume)
H.6.1 Give an informal argument using Cavalieri’s principle for the formulas for the
volume of a sphere and other solid figures.
6.5.2 Use volume formulas for cylinders, pyramids, cones, and spheres to solve
problems
Formula for Circumference
The circumference of a circle is 2 r . The perimeter of a regular polygon inscribed in a circle
gives an estimate of the circumference of a circle. By increasing the number of sides, n, of the
regular polygon we increase the accuracy of the approximation.
n3
n4
n5
n6
n 8
If we had a regular polygon with an infinite number of sides, then we would have the exact value
of the circumference of a circle. To prove this numerically, we need a formula for the perimeter
of a regular polygon with n sides. The perimeter of a regular polygon is found by multiplying
the number of sides by the side length. Symbolically that is P ns . If we can find the value for
any side length, s, then we can find the perimeter for any regular polygon with n sides.
Given a regular polygon with side length s and radius r we can divide
the polygon into n congruent isosceles triangles.
Each triangle has a hypotenuse of length r and a base of length s. If we
construct a perpendicular bisector from the central angle to the base,
we can use trigonometric ratios to find the length of s.
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Since the perpendicular bisector, a, divides the side of length, s, in
half s 2 x . Thus, we need to find the value x. The opposite side is x
x
and the hypotenuse is r therefore sin . By multiplying each side
r
of the equation by r we can isolate x and we get x r sin . Since
s 2 x , s 2r sin . Substituting this value into our perimeter
equation yields P ns n 2r sin .
The central angle is found by dividing 360 by the number of sides, n. The angle is half of the
360
central angle. Thus,
.
2n
We can use this equation to show that as n increases the perimeter approaches 2 r 6.283r .
The following table demonstrates this relationship.
n
3
360
2n
360
60
2 3
Perimeter Expression P n 2r sin
Approximation
P 3 2r sin 60
5.196r
4
360
45
2 4
P 4 2r sin 45
5.656r
5
360
36
2 5
P 5 2r sin 36
5.878r
6
360
30
2 6
P 6 2r sin 30
6.000r
10
360
18
2 10
P 10 2r sin 18
6.180r
360
1.8
2 100
P 100 2r sin 1.8
6.282r
360
0.18
2 1000
P 1000 2r sin 0.18
6.283r
100
1000
A regular polygon with 1000 sides is accurate to four decimal places for the approximation of
2 r . With an infinite number of sides, the regular polygon would essentially be a circle and the
perimeter would equal 2 r . Therefore, the circumference of a circle is 2 r .
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Area of a Circle
A similar process can be used to show that the area of a circle is r . The area of a regular
polygon inscribed in a circle gives an estimate of the area of a circle. By increasing the number
of sides, n, of the regular polygon we increase the accuracy of the approximation of the area of a
circle.
2
n3
n4
n5
n6
n 8
Because a regular polygon with n sides of side length s can be divided
into n isosceles triangles, the area of the regular polygon can be found
by multiplying the area of one isosceles triangle by the number of
triangles formed. Symbolically, A area of triangle n or
1
A bh n .
2
The base is length s and we know that s 2r sin . We need to find
the height, which is the altitude of the triangle and is sometimes called
the apothem. This can be done by using trigonometric ratios.
The altitude is adjacent to the angle and the radius is the hypotenuse.
a
So, cos . By multiplying each side by r we can isolate a and we
r
get a r cos .
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Secondary Mathematics 2
1
1
By substituting these values into our area formula we get A bh n 2r sin r cos n
2
2
2
simplifying it we get A r n sin cos . We can use this equation to show that as n increases the
2
2
area approaches r 3.142r . The following table demonstrates this relationship.
n
3
360
2n
360
60
2 3
2
Area Expression A r n sin cos
Approximation
A r 2 3 sin 60 cos 60
1.299r 2
4
360
45
2 4
A r 2 4 sin 45 cos 45
2.000r 2
5
360
36
2 5
A r 2 5 sin 36 cos 36
2.378r 2
6
360
30
2 6
A r 2 6 sin 30 cos 30
2.598r 2
10
360
18
2 10
A r 2 10 sin 18 cos 18
2.939r 2
360
1.8
2 100
A r 2 100 sin 1.8 cos 1.8
3.140r 2
360
0.18
2 1000
A r 2 1000 sin 0.18 cos 0.18
3.142r 2
100
1000
A regular polygon with 1000 sides is accurate to four decimal places for the approximation of
r 2 . With an infinite number of sides the regular polygon would essentially be a circle and the
2
2
area would equal to r . Therefore, the area of a circle is r .
A simpler way to look at this is by slicing the circle into infinitely many slices and arranging
those slices so that they form a parallelogram. The smaller the slice, the more linear the
intercepted arc becomes and the more it looks like a parallelogram.
n6
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n 8
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n 10
Secondary Mathematics 2
1
2 r r . The
2
height is the radius. The area of a parallelogram can be found by multiplying the base times the
height. Symbolically, A bh r r r 2 .
The base of the parallelogram is half of the circumference. Therefore, b
Volume
VOCABULARY
The volume of a three dimensional figure is the space that it occupies.
A prism is a three-dimensional figure with two congruent and parallel
faces that are called bases.
A cylinder is a three-dimensional figure with parallel bases that are
circles.
A cone is a three-dimensional figure that has a circle base and a vertex
that is not in the same plane as the base. The height of the cone is the
perpendicular distance between the vertex and the base.
A pyramid is a three-dimensional figure with a polygon as its base
and triangles as its lateral faces. The triangles meet at a common
vertex.
A sphere is the set of all points in space that are the same distance
from the center point.
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The three figures above have the same volume. Each figure has the same number of levels and
each level has the same volume. This illustrates Cavalieri’s Principle.
Cavalieri’s Principle
If two space figures have the same height and the same cross-sectional area at every level, then
they have the same volume.
The volume of any prism or cylinder can be found by multiplying the area of the base times the
height, V Bh .
Volume of a Cylinder
Extending the idea of Cavalieri’s principle, the volume of a cylinder is the area of its base, a
2
circle, times the perpendicular height. Symbolically, V Bh r h .
Formula for the Volume of a Cylinder
The volume of a cylinder with radius r and height h is V r h .
2
Example 1: Volume of a cylinder
The radius of a circular container is 4 inches and the height is 10 inches. Find the volume of the
container.
Answer:
V r 2h
2
V 4 10
V 16 10
V 160 502.655
The volume is approximately 502.655 in3.
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Use the formula for the volume of a cylinder.
Substitute in known values. r 4 and h 10
Simplify.
Volume is measured in cubic units.
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Secondary Mathematics 2
Volume of a Pyramid
The volume of a cube can help us find the volume of a pyramid. Since a cube has 6 faces that
2
3
are all squares, the volume of a cube is V Bh b b b , where b is the length of the side.
Inside the cube, place 6 pyramids that have a face of the cube as a
base and share a vertex at the center of the cube. The six pyramids
are equally sized square pyramids.
Using the formula for the volume of a cube, we can derive the volume of one of the pyramids.
V b3
V Bh
V b2 b
Begin with the volume of a cube. Rewrite it so
that it is in the form V Bh .
1
V b2 b
6
There are 6 equally sized pyramids within the
cube. The volume of one pyramid will be equal
to one-sixth the volume of the cube.
1
V b2 b
6
1
V b 2 2h
6
2
V b2h
6
1
V b2h
3
Rewrite the formula using the height of the
pyramids. Use b 2h because the height of
the cube is equivalent to the height of two
pyramids.
Simplify.
1
1
In general, the volume of a pyramid is V b2 h Bh . In other words, the volume of a
3
3
pyramid is the area of its base, regardless of whether or not it is a square, times the height.
Formula for the Volume of a Pyramid
1
For any pyramid with area of base, B, and height h, the volume is V Bh .
3
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Secondary Mathematics 2
Example 2: Volume of a Pyramid
Find the volume of the pyramid with a rectangular base that is 4 ft by 5 ft
and height of 6 ft.
Answer:
1
V Bh
3
1
V 20 6
3
1
V 120
3
V 40
The volume of the pyramid is 40 ft3.
Use the formula for the volume of a pyramid.
The base is a rectangle so the area of the base
is A lw 4 5 20 . The height is 6.
Simplify.
Volume is measured in cubic units.
Volume of a Cone
The relationship between the volume of a cone and the volume of a cylinder is similar to the
relationship between the volume of a pyramid and the volume of a prism. We already know that
the volume of a pyramid is one-third the volume of a prism. We want to show that the volume of
a cone is also one-third the volume of a cylinder. To do this, we are going to look at a pyramid
with a base that has n sides. If we increase the number of sides until there are an infinite number
of sides, the pyramid will have a circular base and be a cone.
6 sided base
8 sided base
Inifinite sided base
When we increase the number of sides of the base of pyramid, we also change the shape of the
prism that holds the pyramid. The prism that contains the pyramid with an infinite number of
sides is a cylinder.
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Secondary Mathematics 2
6 sided base
8 sided base
Infinite sided base
Since a cone is essentially a pyramid with an infinite number of sides and a cylinder is a prism
with a base that has an infinite number of sides, the volume of a cone is one-third the volume of
a cylinder.
Formula for the Volume of a Cone
1
For any cone with radius, r, and height h, the volume is V r 2 h .
3
Example 3: Volume of a cone
Find the volume of the cone with radius 8 cm and height 12 cm.
Answer:
1
V r 2h
3
1
2
V 8 12
3
1
V 64 12
3
1
V 768
3
V 256 804.248
The volume of the cone is approximately
804.248 cm3.
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Use the formula for the volume of a cone.
Substitute in known values. r 8 and h 12
Simplify.
Volume is measured in cubic units.
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Secondary Mathematics 2
Volume of a Sphere
In order to derive the formula for the volume of a sphere, we need to review Cavalieri’s
Principle. That is, that if two figures have the same height and same cross-sectional area at every
1
level, then they have the same volume. We will also use the volume of a cone V r 2 h and the
3
2
volume of a cylinder V r h .
Consider the cylinder and hemisphere given below, both with a radius of r and height r. From the
cylinder, a cone is cut out that shares the same base as the cylinder and also has a radius of r and
a height of r. We are going to prove that this “cone-less cylinder”, or the part of the cylinder that
remains after the cone has been removed, has the same volume as the hemisphere. Once we have
proven that they have the same volume, we can merely double the formula for the “cone-less
cylinder” to obtain the volume for a sphere.
Cylinder
Hemisphere
We start calculating the volume of the “cone-less cylinder” by subtracting the volume of the
cone from the volume of the cylinder. Since the height is r, the volume of this cone would be
1
1
1
2
2
3
V r 2 h r 2 r r 3 and the volume of the cylinder would be V r h r r r .
3
3
3
1
2
Upon subtracting their volumes we get V r 3 r 3 r 3 . So the volume of the “cone-less
3
3
2
cylinder” is V r 3
3
In order to use Cavalieri’s Principle, we are going to create circular cross sections of the figures
by slicing them at height h and comparing their areas. The cross section of the cylinder and the
hemisphere are shown below.
Hemisphere
Cylinder with cone inside
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Secondary Mathematics 2
Cross sectional slice of cylinder with cone
Cross sectional slice of hemisphere
If we can prove that these cross sections have the same area, then by Cavalieri’s Principle the
volume of the hemisphere and the volume of the “cone-less” cylinder are the same. Start by
finding the area of the cross section of the hemisphere, or the disk shaped cross section. We need
to find its radius, z. Since the cross section is h units above the base of the hemisphere, we can
use the Pythagorean Theorem and the radius of the hemisphere’s base to calculate the cross
sectional radius z to be z r 2 h2 . Using this radius we can now calculate the area of the
cross section. A z 2
r 2 h2
2
(r 2 h 2 ) . To find the area of the cross-section of the
“cone-less cylinder”, or the washer shaped cross section, we need to subtract the small circle
from the large circle. Since the radius of the cone at any height h is proportional to the height
with a 1:1 ratio, the large circle (cross-section of the cylinder) has a radius of r while the small
2
circle (cross-section of the cone) has a radius of h. Thus the area of the large circle is r and the
2
area of the small circle is h . Their difference is r 2 h2 (r 2 h2 ) . Therefore, the area of
the cross section from the hemisphere is the same as the area of the cross section from the “coneless” cylinder. Cavalieri’s Principle states that if we sum up all of our equal slices or cross
sectional areas then we obtain equivalent volumes for the hemisphere and “cone-less cylinder.”
2
Recall that the volume of the cone-less cylinder is V r 3 . We can then infer that the volume
3
2 3
of the hemisphere is V r and the that volume of the sphere would be twice the volume of
3
2
4
the hemisphere or V 2 r 3 r 3 .
3
3
Formula for the Volume of a Sphere
4
For any sphere with radius r the volume is V r 3 .
3
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Secondary Mathematics 2
Example 4: Volume of a sphere
Find the volume of a sphere with radius of 2 ft.
Answer:
4
V r3
3
4
3
V 2
3
4
V 8
3
32
V 33.5210
3
The volume of the sphere is approximately
33.5210 ft3.
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Use the formula for the volume of a sphere.
Substitute in known values. r 2
Simplify.
Volume is measured in cubic units.
Page 379
Secondary Mathematics 2
Practice Exercises A
Find the volume of the following.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13. Find the volume of a cylindrical pool if
the diameter is 20 feet and the depth is 6
feet.
14. A fish tank is in the shape of a cylinder.
The tank is 22 feet deep and 40 feet in
diameter. Find the volume of the tank.
15. Popcorn is served in a conical container
that has a radius of 3 inches and a height
of 6 inches. What is the volume of the
small container?
16. A tank is in the shape of a pyramid. The
base of the tank is a rectangle with length
3 feet and width 7 feet. It is 9 feet high.
Find the volume of the tank.
17. Find the volume of a pyramid if the base
is a square with side 6 feet and the height
is 4 feet.
18. How much ice cream is needed to fill a
conical shaped sugar cone that is 4 inches
deep and 6 inches in diameter?
19. A soccer ball has a radius of 4.3 inches.
What is the volume of soccer ball?
20. A softball has a diameter of 9.6 cm. What
is its volume?
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Secondary Mathematics 2
Relating Scale Factor, Length, Area, and Volume
The drawings below are dilations of one another. The length and area of each figure are
compared.
Original Drawing
Dilated Drawing
Length Scale Factor
Dilated Length
k
Original Length
Area Scale Factor
Dilated Area
k2
Original Area
6
2
3
18
4 22
4.5
6 cm
3 cm
3 cm
1
A 3 3 4.5 cm2
2
6 cm
1
A 6 6 18 cm2
2
By comparing the side of the dilated triangle to the corresponding side of the original triangle, we
see that the ratio is 2. Thus, the scale fator k 2 . If we compare the area of the dilated triangle to
the area of the original triangle, the ratio is 4. If we write this in terms of k we get 4 22 or k 2 .
Original Drawing
4 in
Dilated Drawing
Length Scale Factor
Dilated Length
k
Original Length
Area Scale Factor
Dilated Area
k2
Original Area
2 1
6 3
4
1 1
36 9 3
2in
2
A 22 4 in 2
A 62 36 in 2
By comparing the radius of the dilated circle to the radius of the original figure, we see that the
1
1
ratio is . Thus, the scale factor k . If we compare the area of the dilated circle to the area of
3
3
2
1
1 1
the original circle, the ratio is . If we write this in terms of k we get or k 2 .
9
9 3
2
Therefore, if a figure is dilated by a scale factor of k, its area is k times the area of the original
figure.
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Secondary Mathematics 2
The three dimensional figures below are dilations of one another, the length and the volume scale
factors of each figure are compared.
Original Drawing
Dilated Drawing
Length Scale Factor
Dilated Length
k
Original Length
Volume Scale Factor
Dilated Volume
k3
Original Volume
2.5
2
5
60
8 23
7.5
V 2.5 3 1 7.5 ft 3
V 5 6 2 60 ft 3
By comparing one side of the dilated prism to the corresponding side of the original prism, we
find that the ratio is 2. Thus, the scale factor k 2 . If we compare the volume of the dilated
prism to the volume of the original prism, then the ratio is 8. If we write this in terms of k, we
8 23 get or k 3 .
Original Drawing
V 82 12 768 m3
Dilated Drawing
Length Scale Factor
Dilated Length
k
Original Length
Volume Scale Factor
Dilated Volume
k3
Original Volume
3 1
12 4
12
1 1
768 64 4
3
V 22 3 12 m3
By comparing the radius of the dilated cylinder to the radius of the original cylinder, we find that
1
1
the ratio is . Thus, the scale factor k . If we compare the volume of the dilated cylinder to
4
4
1
the volume of the original cylinder, then the ratio is
. If we write this in terms of k we get
64
3
1 1
or k 3 .
64 4
3
Therefore, if a figure is dilated by a scale factor of k, its volume is k times the volume of the
original figure.
Jordan School District
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Secondary Mathematics 2
Example 5:
2
The volume of a cylinder is V r h . If the original cylinder has radius 4 inches and height 10
inches how will the volume compare to a cylinder with the same radius but double the height?
Answer:
V r 2h
V r 2h
2
V 4 10
Find the volume of each cylinder.
For the original cylinder r 4 and h 10 .
V 16 10
V 160 502.655 in
3
V r 2h
2
V 4 20
For the cylinder with double the height r 4
and h 20 .
V 16 20
V 320 1005.310 in
3
volume of original
160 1
Compare the volumes.
volume of double height 320 2
The volume of the original cylinder is half the volume of the cylinder with double the height.
The dilation was applied to only one length of the cylinder so it doubled the volume. If the
dilation had been applied to the radius and the height, then the volume of the cylinder would
have been one-eighth of the cylinder that was double the radius and double the height.
k
Example 6:
If the original cylinder has radius 4 inches and height 10 inches how will the volume compare to
a cylinder with the same height but double the radius?
Answer:
V r 2h
V r 2h
2
V 4 10
Find the volume of each cylinder.
For the original cylinder r 4 and h 10 .
V 16 10
V 160 502.655 in
3
V r 2h
2
V 8 10
For the cylinder with double the radius r 8
and h 10 .
V 64 10
V 640 2010.619 in
3
volume of original
160 1
Compare the volumes.
volume of double height 640 4
The volume of the original cylinder is one-fourth the volume of the cylinder with double the
radius.
k
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Secondary Mathematics 2
Practice Exercises B
Use the trapezoids to answer questions 1–3.
1.
What is the ratio of the perimeter of the larger trapezoid to the perimeter of the smaller
trapezoid?
2.
What is the ratio of the area of the larger trapezoid to the area of the smaller trapezoid?
1
A b1 b2 h
2
3.
What is the scale factor k?
Use the prisms to answer questions 4–6.
4.
What is the ratio of the surface area of the smaller prism to the surface area of the larger
prism? S 2lw 2lh 2wh
5.
What is the ratio of the volume of the smaller prism to the volume of the larger prism?
6.
What is the scale factor k?
7.
You have a circular garden with an area of 32 square feet. If you increase the radius by a
scale factor of 5, what is the area of the new garden?
8.
Jan made an enlargement of an old photograph. If the ratio of the dimensions of the
photograph to the enlargement is 1:3, what will be the ratio of the area of the original
photograph to the area of the enlargement?
9.
You and your friend are both bringing cylindrical thermoses of water on a camping trip.
Your thermos is twice as big as his in all dimensions. How much more water will your
thermos carry than your friend’s? If your friend’s thermos has a diameter of 10 cm and a
height of 18 cm, what are the dimensions of your thermos?
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Secondary Mathematics 2
10. A cylinder has radius 3 cm and height 5 cm. How does the volume of the cylinder change if
both the radius and the height are doubled?
11. A pyramid has a square base with length 4 ft and a height of 7 ft. How does the volume of
the pyramid change if the base stays the same and the height is doubled?
12. A pyramid has a square base with length 4 ft and a height of 7 ft. How does the volume of
the pyramid change if the height stays the same and the side length of the base is doubled?
13. A movie theater sells a small cone of popcorn for $2. A medium cone of popcorn is sold for
$4 and comes in a similar container but it is twice as tall as the small container. Which
popcorn size gives you more for your money? Explain your answer.
14. Another movie theater sells a small cone of popcorn for $2. A medium cone of popcorn is
sold for $4 and comes in a similar container as the small but the radius is twice the length of
the small container. Which popcorn size gives you more for your money? Explain your
answer.
15. How much does the volume of a sphere increase if the radius is doubled? Tripled?
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Secondary Mathematics 2
Selected
Answers
Jordan School District
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Secondary Mathematics 2
Secondary Mathematics 2 Selected Answers
Unit 1 Cluster 1 (N.RN.1 and N.RN.2)
Practice Exercises A
1. 53/4
1
5. 2/3
y
9. 1
3. yes, 72
5. no
7. When adding two rational numbers the
result is rational. When adding a
rational number and an irrational
number the result is irrational.
1/3
3. k
1
7.
6
11. y1/3
Practice Exercises B
1.
3
5.
9.
5
84
3.
k3
7.
9
Unit 1 Cluster 4 (A.APR.1)
a5
3
64
2
4
2
2x
Practice Exercises C
1. 111/3
5. x 7/6
9. z 2/5
Practice Exercises D
1. 3 p 5
5. 5 x 3 y
3. 10
8/3
7. w5/7
3. 2 xy 3 3
7. 24 y
y
Unit 1 Cluster 2 (N.RN.3)
3.
3
7. 3 5 2 6
You Decide
Polynomials are closed under addition,
subtraction and multiplication. All of the
answers to Practice Exercises B were
polynomials.
9. 12 2 4 3
Practice Exercises B
1. 56x 2
3. 8 x 3 10
7. 2
5. 2 3 20
9. 10 13 5 20
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11. 2 y
11/28
Practice Exercises A
1. not a polynomial; it has a variable raised
to a negative exponent
3. polynomial; degree 5, leading
coefficient 2
5. not a polynomial; there is a cube root of
a variable
Practice Exercises B
1. 4 x2 2 x 4 ; polynomial
3. 3x3 x2 9 x 3 ; polynomial
5. 2 x3 2 x2 6 x ; polynomial
7. 12u 2 3u ; polynomial
9. x2 4 x 21 ; polynomial
11. 8x2 14 x 3 ; polynomial
13. 4 x2 28x 49 ; polynomial
15. 25x6 10 x3 1 ; polynomial
17. x3 2 x2 5x 12 ; polynomial
19. x4 2 x3 x2 2 x 3 ; polynomial
9. 14m 3 12
Practice Exercises A
1. 2 7
5. 9 3 2
You Decide
1. yes, 145
28
2y
11
Page 387
Secondary Mathematics 2
Unit 2 Cluster 1 and 2 (F.IF.4, 5, 7 and
9)
Practice Exercises D
1. a. D: , ; R: ,
b. 2.5, 0 , 0,5
c. no symmetry
d. inc. ,
Practice Exercises A
1. Domain: , ; Range: ,
3. Domain: 3, ; Range: 2,
5. Domain: x k where k is an integer and
0k
Range: y 0.1k 10 k is an integer and
0k
e. positive 2.5, ; negative , 2.5
f. no relative extrema
g. lim f ( x) ; lim f ( x)
3.
e. positive 14, ; negative 2,14
5. x-int: 3,0 , 6,0 ; y-int: 0, 18
f. rel. min. 2, 4
7. maximum 1,3
g. lim f ( x)
9. maximum 2, 2
x
5.
e. positive 124, ; negative
, 124
e. negative , 6
b. decreasing , 4
c. never constant
d. ,
e. never negative
a. D: , ; R: ,
b. 124, 0 , 0, 4
c. no symmetry
d. inc. ,
Practice Exercises C
1. a. increasing ,
b. never decreasing
c. never constant
d. positive 6,
3. a. increasing 3,
b. never decreasing
c. never constant
d. positive 3,
e. never negative
5. a. increasing 4,
f. no relative extrema
g. lim f ( x) ; lim f ( x)
x
x
You Decide
Group A should win because the rocket
reached a maximum height of 487 feet and
was in the air for 11 seconds. Group B’s
rocket reached a maximum height of 450
feet and was in the air for 10.5 seconds.
Group C’s rocket reached a maximum height
of 394 feet and was in the air for 10 seconds.
Unit 2 F.IF.6 and F.LE.3
Practice Exercises A
1. 5
61
5. 10
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c. no symmetry
d. inc. 2,
x-int: 6,0 , 5,0 ; y-int: 0, 30
11. minimum 3,12
a. D: 2, ; R: 4,
b. 14, 0 ; 0,5.3
Practice Exercises B
1. x-int: 5, 0 ; y-int: 0, 2
3.
x
x
Page 388
3. 6
7. 32,000
Secondary Mathematics 2
Practice Exercises B
Table 1: -4, -2, 0, 2, 4, 6, 8, 10
Table 3: 14 , 12 , 1, 2, 4, 8, 16, 32
Practice Exercises C
1. f(x)
3. h(x)
Practice Exercises E
1. 2 x 5 x 5
3. 3x 4 x 3
5. 5x 8 y 5 x 8 y
7. 4 y y 1
5. f(x)
9. 2 x 3 2 x 3
11. 3x 2 3x 2
Practice Exercises D
1.
Unit 2 Cluster 2 F.IF.8, A.SSE.1,
A.SSE.3
Practice Exercises A
1. 0,0 7,0
3. 8 x 2 x 2 4 x 2
Practice Exercises B
1. x 7 x 3
3. x 1 x 2
3. x 2 3x 4
5. 2 x 1 x 1
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11. 65 ,0 2,0
5. 2 x 2 6 x 9
Practice Exercises C
1. 5, 3
5. 9 3x 2 4 x 2
Practice Exercises D
1. 5, 45 ; min
5. 4, 6 ; min
3. 2, 29 ; min
9. x 8 y x 9 y
Practice Exercises E
1. 1, 11 ; min
5. 6 x 11 6 x 11
7. no factors
9. 4 x 3 3x 2
5. 2,3
3. 0,5
7. x 4 y x y
3. 3x 2 3x 2
Practice Exercises D
1. x 6 2 x 1
5. 6, 0 1, 0
3. x 22 x 121
5. x 9 x 8
Practice Exercises C
1. 7 x 5 7 x 5
9. 0,0 3,0
2
Factoring
Practice Exercises A
1. 4 x 2 12 x 16
3. 13,0 4,0
Practice Exercises B
1. x 2 10 x 25
You Decide
1. h(x) and r(x)
7. 6,0 2,0
3. 3, 15 ; min
You Decide
The zeros are 0, 0 and 10, 0 . The first zero
means that the rocket starts its launch from the
ground at time zero. The second zero represents
when the rocket comes back down to the ground 10
seconds later. The vertex is 5, 400 . This means
that the rocket reaches its maximum height of 400
feet after 5 seconds.
Page 389
Secondary Mathematics 2
3. V: 0, 6 0,6 ; F: 0, 3 0,3
Practice Exercises F
1. 2 seconds, 144 feet
3. 7 seconds
5. 4 seconds
Unit 6 Cluster 3 G.GPE.2
Practice Exercises A
1. y 201 x 2
5. y 16 x 32 2.5
3. y 181 x 2 1.5
Practice Exercises B
1. x 161 y 2
5. V: 0, 3 0,3 ; F: 0, 5 0, 5
5. x 18 y 42 4
3. x 121 y 2
Practice Exercises C
1. v: 5,1 , f: 6,1 , d: x 4
3. v: 5,1 , f: 6,1 , d: x 4
5. v: 2, 3 , f: 5, 3 , d: x 1
You Decide
(2, 1) is a point on the parabola because the
distance to the point from the focus is equal
to the distance from the point to the
directrix.
x2 y 2
1
11.
16 36
x2 y 2
1
7.
64 39
x2 y 2
1
9.
7 16
Practice Exercises B
1. C: 2,1 V: 5,1 1,1 ;
F: 2 5,1 2 5,1
Unit 6 Cluster 3 G.GPE.3 (HONORS)
Practice Exercises A
1. V: 4,0 4,0 ; F: 3,0 3,0
3.
C: 3, 1 V: 3,3 3, 5 ;
F: 3, 1 7 3, 1 7
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Secondary Mathematics 2
C: 1, 3 V: 1, 0 1, 6 ;
5.
F: 1, 3 5 1, 3 5
5. C: 0, 0 V: 0, 4 0, 4
F: 0, 2 5 0, 2 5
A: y 2 x, y 2 x
7.
9.
x 3
2
9
x 5
2
x 1
16
2
1
5
9
11.
y 4
y 2
2
1
8
2
y 2
2
9
11.
x2 y 2
1
16 25
1
Practice Exercises C
1. C: 0, 0 V: 2,0 2,0
y 2 x2
1
1 3
y 2 x2
1
9.
25 24
7.
F: 2 5, 0 2 5, 0
Practice Exercises D
1. C: 6,5 , V: 6, 4 6,0
F: 6,5 41 6,5 41
A: y
5
4
x 6 5,
y 54 x 6 5
A: y 2 x, y 2 x
3.
3. C: 0, 0 V: 1,0 1,0
F: 10, 0
10, 0
C: 6,5 , V: 6, 4 6,0
F: 6,5 41 6,5 41
A: y 54 x 6 5, y 54 x 6 5
A: y 3x, y 3x
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Page 391
Secondary Mathematics 2
5.
C: 4, 6 , V: 4, 4 4,16
F: 4, 6 2 30 4, 6 2 30
A: y
2105
x 4 6,
y
10
2 5
x 4 6
Practice Exercises E
1. 35
3. -31
5. 182
7. 54
9. -1
Practice Exercises G
1. P x 0.75x 2 50 x 19,900 ;
$750,000,000,000
3. C x 0.5x 2 34 x 1213 ; $2813
2
7.
9.
11.
Unit 2 Cluster 4 F.BF.3 and F.BF.4
y 5 x 1 1
4
12
x 2
2
25
y 4
4
2
y 4
2
1
11
2
x 3
Practice Exercises A
1. shifted up 6 units, reflected over the line
y6
3. shifted 4 units to the right and a vertical
stretch by a factor of 3
5. axis of symmetry x = 1, vertex (1, -3)
2
1
16
Unit 2 Cluster 3 F.BF.1
Practice Exercises A
1. quadratic
function
3. exponential
function
5. exponential
function
D: , ; R: , 3
7. axis of symmetry x = -6, vertex (-6, -4)
Practice Exercises B
1. f ( x) 2.5x 2 24.7 x 3
3. f ( x) 119.60 1.096
x
Practice Exercises C
x
1. exponential, f ( x) 49.34 0.85
3. quadratic, f ( x) 0.015x2 0.24 x 4.73
Practice Exercises D
1. 12 x2 5x 3
11.
D: , ; R: 4,
9. axis of symmetry x = 3, vertex (3, 0)
1
4x 3
3. 12 x2 8 x
5. 12 x2 2 x 4
7. 36 x3 21x2 20 x 12
9. 144 x4 24 x3 143x2 12 x 36
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D: , ; R: 0,
Page 392
Secondary Mathematics 2
Practice Exercises B
1. Shifted 5 units up and reflected over the
line y = 5.
3. Shifted 5 units down, 2 units to the left
and stretched by a factor of 3.
5. vertex: 2, 4 , D: , R: 4,
7. vertex: 0, 5 , D: , R: 5,
Practice Exercises F
1
2
1. f 1 x x
3
3
1
5
3. f 1 x x
6
6
3
5. f 1 x x 6
2
x5
7. f 1 x
3
1
9. f x x 9 7
Unit 3 Cluster 1 A.SSE.2
Practice Exercises A
1. yes
3. no
5. yes
Practice Exercises B
1. 12 x 7 y 2 12 x 7 y 2
3. 10 x 4 11y 3 10 x 4 11y 3
5. 2 x1/2 1 x1/2 1
9. vertex: 4, 0 , D: , R: 0,
7. 3x5 y 3x5 y
9. x1/3 2 3x1/3 4
Practice Exercises C
2
2
1. u 2 x ; 2 x 3 2 x 2
2
2
3. u 2 x 5 ; 2 x 5 5 5 2 x 5 4
Practice Exercises C
1. odd
3. even
Practice Exercises D
1. neither
3. even
5. neither
1/4
5. even
Unit 1 Cluster 3 N.CN.1 and N.CN.2
6, 10 , 9,3 , 4, 1 , 1, 7 , 8,6
Jordan School District
1/4
2
2
9. u x 2 1 ; 2 x 2 1 3 2 x 2 1 3
Practice Exercises E
1. 1,1 , 2, 2 , 3,3 , 4, 4 , 5,5
3.
5. u x ; x 5 x 2
7. u x1/2 ; 2 x1/2 5 2 x1/2 5
1/4
Practice Exercises A
1. 5i, 5i
3. 12i, 12i
Page 393
5. 4i 13, 4i 13
Secondary Mathematics 2
Practice Exercises B
1. 13 7i
3. 3 2i
5. 40 10i
Practice Exercises E
5i 7
1. x
4
3. x 2 i 2
3 i 19
5. x
2
7. 15 9i
9. 9 17i
11. 90 22i
Unit 1 Cluster 3 HONORS N.CN.3
Practice Exercises A
1. 6 6i
Practice Exercises B
1. 53 i
3 i
Practice Exercises C
1. 34 14 i
3. 3 i
3. 2 3i
Unit 3 Cluster 5 HONORS N.CN.8 and
N.CN.9
5. 2 2i
Practice Exercises A
1. x 3i ; x 3i x 3i
3.
5.
5
4
125 i
Practice Exercises A
1. 5, 5
3. 3, 3
5. 2, 2
7. 3 6,3 6
9. 1 15, 1 15
Practice Exercises B
1. 2 3, 2 3
3. 1, 15
5. 2, 4
x 1 i ; x 1 i x 1 i
7.
x 2 i ; x 2 i x 2 i
x i, 2i ;
9.
x i x i x 2i x 2i
x 1, i ; x 1 x 1 x i x i
5.
Unit 3 Cluster 3 and 4 A.REI.4 and
N.CN.7
Unit 3 Cluster 3 A.CED.1 and A.CED.4
Practice Exercises A
1. width 12 in, length 15 in
3. height 30 in, base 50 in
5. 2 in
7. 14, 34
Practice Exercises B
1. 10.102 seconds
3. 227.5 meters
Practice Exercises C
1. 2 real
3. 2 real
5. 1 real
Practice Exercises D
5 109
1. x
6
14
3. x 1,
3
5. x 3 6
1 i 3
7. x
2
1 i 11
9. x
3
Jordan School District
9
7. x i
2
9. x 1 7i
11. x 2i
5.
7.
22.0625 feet
1.118 seconds
Practice Exercises C
1. , 10 8,
3.
5.
7.
9.
11.
Page 394
, 4 2,
0, 2
,5
12 ,3
40, 200
Secondary Mathematics 2
5.
Practice Exercises D
1. b c 2 a 2
3. s
A
6
3 9 8N
2
1 1 8N
7. n
2
5. k
Unit 3 Cluster 3 Honors
Practice Exercises A
1. 4, 1 3,
7. , 2 6,
3. , 5 0,5
9. , 3 2,
5. 1,
11. , 3 2,
7.
You Decide
7.236 hours
9.
Unit 3 Cluster 3 A.CED.2
Practice Exercises A
1. f x x 2 3
7. f x x 2 1
3. f x x 3 1
9. f x x 1 2
5. f x x 1 x 7
11. f x x x 4
2
2
2
2
Practice Exercises B
1.
Unit 3 Cluster 6 A.REI.7
Practice Exercises A
1. 1, 2 and 6, 3
3.
5.
3.
7.
9.
11.
1, 4 and 114 , 54
2,1
0, 0 and 1,1
8,5 and 5, 8
2, 2
Unit 3 Cluster 6 Honors A.REI.8 and
A.REI.9
Practice Exercises A
3 3 1 6
1. 0 6 9 3
8 5 5 1
Jordan School District
Page 395
Secondary Mathematics 2
3 3 1 6
3. 8 1 14 2
0 6 9 3
5. R2 : 2R1 3R2
Practice Exercises C
1. 6,3,5
Practice Exercises D
1. x 1, y 2, z 3 , consistent
3. inconsistent
5. x 3, y 2, z 1, w 1 , consistent
1
5. 14
14
1
12
1
2
3
7
4
34
3. 24.79%
You Decide
1.93%; the Utah population can’t grow
indefinitely at this rate. Students should
include ideas such as housing, jobs, access to
water etc. to justify their conclusion.
Unit 4 Cluster 1 S.CP.1
Practice Exercises A
1.
7. (1, 2, -1)
9. (-1, -2, 1)
Unit 2 Cluster 2b F.IF.8b, A.SSE.1b,
A.SSE.3c
Practice Exercises A
1. $13,140.67
3. Analeigh should choose the
compounded monthly because after 3
years it is $14,795.11 while the
continuous is $14,737.67.
Jordan School District
Practice Exercises C
1. 7.36 %
Practice Exercises E
1. a. decrease 6.57%
t
b. B t 1400 0.934
c. 707 or 708 birds
d. 38.65 years
t
3. a. P t 124,009,000 1.031
b. 497,511,091 people
c. 15.66 years
3. 1, 2,3
Practice Exercises F
1. (1, 2, -1)
3. (2, -1, 5)
5. (2, 3, -5)
3. $7,126.24
Practice Exercises D
1. a. 574,000,000 people
b. 0.026 people/year
c. 2,461,754,033 people
d. 40 years, 2014
Practice Exercises B
1 0 3
1.
; 3, 5
0
1
5
Practice Exercises E
4 5
15
1. 12 3 4
4 1 1
0 1 2
3. 14 12 14
14 12 54
Practice Exercises B
1. 1.23%
A. {A, B, C, D, E, F, G, H, I, J, K, L, M, N, O,
P, Q, R, S, T, U, V, W, X, Y, Z}
B. Answers will vary {F, I, R, S, T}
C. Answers will vary {A, L, S, T}
D. Answers will vary {A, F, I, L, R, S, T}
E. Answers will vary {S, T}
3.
A. {hearts: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A;
diamonds: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K,
A}
B. {diamonds: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K,
A}
C. {spades: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A;
clovers: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A}
D. the empty set
Page 396
Secondary Mathematics 2
Unit 4 Cluster 1–2 S.CP.2-7
Practice Exercises A
25 1
1. A.
50 2
7
B.
50
10 1
C.
50 5
41
D.
50
7
3. A.
10
8 4
B.
10 5
5 1
C.
10 2
Practice Exercises B
54 27
1.
100 50
38 19
3.
100 50
Practice Exercises C
100 1
1.
200 2
42
21
3.
200 100
9
5.
200
12 6
50 25
34 17
F.
50 25
15 3
G.
50 10
3
H.
50
7
D.
10
8 4
E.
10 5
E.
5.
154 22
403 29
11. no, one is
0.5625 and the
other is 0.45
Practice Exercises F
1.Yes,
P grades male
P male
P grades 0.52
3. Yes, P cold|vitamin C 0.122 while
P cold|no vitamin C 0.221
Unit 4 Clusters 2-3 Honors S.CP.8-9,
and S.MD.6-7
27
5.
100
Practice Exercises A
1. A.
85 17
200 40
0
9.
200
7.
Practice Exercises D
1.
Yes, because
3.
A. answers will vary
B. answers will vary
C. answers will vary
D. none of the activities and gender are
independent
P A P B 0.7 0.3 P A B 0.21
Jordan School District
Practice Exercises E
64
1.
403
146 73
3.
394 197
7. 0.5625
9. 0.4375
1
12
1
C.
12
0
D.
12
B.
Practice Exercises B
8 7 10 9
7
1.
18 17 16 15 102
33 32 16
3.
55 54 45
Page 397
Secondary Mathematics 2
5. A. 0.15
C. 0.45
Practice Exercises C
1. 336
3. 1
5. 4,989,600
Practice Exercises D
1. 462
3. 43,680
B. 0.2
D. 0.2
3.
7. 362,880
9. 210
5.
5. 3060
7. 15,504
You Decide
A. 44,878,080
B. 40,320
C. 5040
7.
Practice Exercises E
C
2
1. 8 3
55
22 C3
C C
20
3. 14 3 10 2
703
39 C5
C C
275
5. 12 3 6 2
714
18 C5
9. false
11. false
Practice Exercises F
268
1.
0.825
325
3. No, a student is more likely to score above
the minimum requirement if he or she takes
the prep class.
Unit 5 Cluster 1 G.SRT.1, G.SRT.2, and
G.SRT.3
Practice Exercises A
1.
Practice Exercises B
1. ABC DEF
5. QRU STU
9. GKJ IKH
3. KNO
7. DGH
11. No
KLM
DEF
Practice Exercises C
1.
JKL NML ; x 7 , ML 6 ,
LK 10
3.
JKM
NOP ; x 2.8 , KM 1.8 ,
OP 4.8
Unit 5 Cluster 2 G.CO.9
Practice Exercises A
1. Vertical Angles: 1 and 4 ,
2 and 3 , 5 and 8 , 6 and 7
Corresponding Angles: 1 and 5 ,
2 and 6 , 3 and 7 , 4 and 8
Alternate Interior Angles: 3 and 6 ,
4 and 5
Jordan School District
Page 398
Secondary Mathematics 2
3.
Vertical Angles: 2 and 8 ,
3 and 9 , 4 and 6 , 5 and 7 ,
10 and 16 , 11 and 17 ,
12 and 14 , 13 and 15
Corresponding Angles: 2 and 10 ,
3 and 11 , 4 and 12 ,
5 and 13 , 6 and 14 ,
7 and 15 , 8 and 16 ,
9 and 17
Alternate Interior Angles: 8 and 10 ,
6 and 12 , 9 and 11,
7 and 13
5.
7.
We are given that l m . We know that
2 3 because vertical angles are congruent.
We know that 3 7 because corresponding
angles are congruent. We can conclude that
2 7 because of the transitive property of
equality.
We are given that l m . We know that
m2 m4 180 because they form a
straight angle. We know that m4 m8
because corresponding angles are congruent.
We can conclude that m2 m8 180
because of the substitution property of equality.
Practice Exercises B
1. x 1
3.
Statements
PA PB
1.
PM AB
mPMA 90
2.
mPMB 90
3. mPMA mPMB
4. mPAM mPBM
5. PAM PBM
6. AM BM
7. PM is the
perpendicular bisector
of AB
Jordan School District
Unit 5 Cluster 2 G.CO.10
Practice Exercises A
1. 106
3.
m1 128 , m2 52 , m3 68 ,
m4 60 , m5 116
5.
7.
9.
11.
13.
6
39, 39
9
6
a. D 2.5,0 , E 6.5,0 , F 4, 4
8
4
b. the medians intersect at the point ,
3 3
equations: y 2 x 4 , y
y
8
52
,
x
23
23
8
20
x
31
31
Unit 5 Cluster 2 G.CO.11
Practice Exercises A
1. a. Given b. Definition of consecutive
angles c. L and M are consecutive
angles d. J and K are supplementary
e. J L f. supplementary angles of the
same angle are congruent.
55
18.3 b. y 12 z 4 c. x 22
3
5. a. x 4, QS 4, RT 4
5
29
29
b. x , QS , RT
3
3
3
c. x 3, QS 7, RT 7
3. a. b
Reasons
1. Given
2. Definition of
perpendicular
3. substitution property
of equality
4. base angles of an
isosceles triangle are
congruent
5. AAS congruence
6. CPCTC
7. definition of
perpendicular bisector
Unit 5 Cluster 3 G.SRT.4, G.SRT.5
Practice Exercises A
1. x
128
, y 16.5
3
3. x 8
5. Yes, the scale factor for corresponding
parts of HNJ to HMK is 1:3.
Page 399
Secondary Mathematics 2
Practice Exercises B
1. 3 2
3. 60
Unit 5 Cluster 5 Honors N.CN.3N.CN.6
5. 1345 36.7
Practice Exercises A
1. 2,3
Unit 5 Cluster 4 G.GPE.6
Practice Exercises A
1. 3, 2
5. 1, 1
3. 1, 2
5. 2.6, 1.2
3. 2, 2
Unit 5 Cluster 5 G.SRT.6, G.SRT.7,
G.SRT.8
#3
Practice Exercises A
5
4
4
3
5
4
3
5
3
5
sec , cot
4
3
11
7
7
3. sin
, cos
, tan ,
11
170
170
1. sin , cos , tan , csc ,
170
170
11
, sec
, cot
11
7
7
9
2
22
9
5. sin , cos
, tan
,
13
13
2 22
csc
csc
13
2 22
13
, sec
, cot
9
9
2 22
7. 0.156
9. 1.111
11. 3.078
Practice Exercises B
1. 30
3. 80.538
5. 48.590
7. 53.130
9. 32.471
11. 43.813
Practice Exercises C
1. 60
3. 67
Practice Exercises D
1. 649.721 feet
3. 16.960
5. 437.66 feet
Jordan School District
#2
#1
#5
#4
#6
7. 13
9.
5 2.236
Practice Exercises B
1. 13 cos112.620 i sin112.620
3. 2 cos90 i sin 90
5. 0, 3
9.
3 2 3 2
,
7.
2
2
a. 10, 24
b. 26 cos67.380 i sin 67.380
c. 10, 24
d. they are the same
Practice Exercises C
1. 9i
3. 9 6i
5. 15 11i
7. 8,001.037 feet
9. 19.151 feet
Page 400
Secondary Mathematics 2
7. 1 12i
Unit 5 Cluster 6 F.TF.8
Practice Exercises A
3
5
1. sin , cos
4
5
15
15
, tan
4
1
12
5
5. cos , tan
12
13
8
5
7. sin
, cos
89
89
1
9. cos , tan 3
2
3. sin
9. 4 15i
Unit 5 Honors Unit Circle
Practice Exercises A
11. 1 3i
13. 13 11i
Practice Exercises D
1. 16 16i
5. 8 8 3i
3. 119 120i
2
2
1
5.
2
3
9.
3
1.
13. -2
17. -1
21. 0
25. 225, 315
29. 150, 210
33. 30, 210
Practice Exercises B
1. i
5. b
9. h
13. neither
17. negative
21. neither
3. 3
7. 0
2
2
15. 3
11.
19. -2
23. -2
27. 135, 315
31. 150, 210
35. 180
3. d
7. e
11. q
15. positive
19. negative
23. positive
Practice Exercises E
5
5
5
3. i , 586 24.207
2
2
9
5. 8i , 29 5.385
2
1. 1 i , 17 4.123
2
Jordan School District
Page 401
Secondary Mathematics 2
Unit 5 Honors Prove Trigonometric
Identities
cos x cot x
13.
sec x sec x sin 2 x cos x
Practice Exercises A
sec x cot x csc x
1.
1 cos x
csc x
cos x sin x
1
csc x
sin x
csc x csc x
3.
tan x cos x sin x
sec x 1 sin 2 x cos x
tan x cos x sin x
sin x
cos x sin x
cos x
sin x sin x
5.
7.
9.
sec x cos 2 x cos x
1
cos2 x cos x
cos x
cos x cos x
csc x sin x cot x cos x
1
sin 2 x
cot x cos x
sin x sin x
1 sin 2 x
cot x cos x
sin x
cos 2 x
cot x cos x
sin x
cos x
cos x cot x cos x
sin x
cot x cos x cot x cos x
cot x sec x sin x 1
cos x 1
sin x 1
sin x cos x
cos x 1
sin x 1
cos x sin x
11
sin 2 x 1 cot 2 x 1
sin 2 x csc 2 x 1
1
sin 2 x 2 1
sin x
11
Jordan School District
1 sin 2 x
sin x
cos 2 x
cos x cot x
sin x
cos x
cos x cot x cos x
sin x
sin x tan x cos x cot x
11.
15.
cos x cos x 1 sin x 1 sin x
1 sin x cos x
cos x 1 sin x
cos 2 x
1 2sin x sin 2 x
2sec x
1 sin x cos x 1 sin x cos x
2sec x
2sec x
cos 2 x 1 2sin x sin 2 x
1 sin x cos x
2sec x
1 1 2sin x
1 sin x cos x
2sec x
2 2sin x
1 sin x cos x
2sec x
2 1 sin x
1 sin x cos x
2
cos x
2sec x 2sec x
2sec x
Practice Exercises B
1. 30, 90, 150, 270
3. 0, 90, 180, 270, 360
5. 60, 120, 240, 300
7. 60, 300
9. 0, 180, 360
Page 402
Secondary Mathematics 2
Practice Exercises D
1. 40
Unit 5 Cluster 6 Honors F.TF.9
Practice Exercises A
6 2
1.
4
3 3
3.
3 3
2 6
5.
4
7.
9.
2 6
4
3 3
5. 41
9. 7.5
3 3
6 2
11.
4
Practice Exercises E
1. 10
3. 3
5. 3
Practice Exercises B
24
25
119
5.
169
240
9.
161
1.
13.
17.
1
26
26
26
5
5 34
34
34
Practice Exercises F
1. 20
5. 5
24
7
240
7.
289
2
3.
15.
1
5
Unit 6 Cluster 2 G.C.5
Unit 6 Cluster 1 G.C.1-4
Practice Exercises A
1. BD or CG
5. G
9. radius
13. radius
3. CG
7. CAE or GAE
11. tangent line
15. center
Practice Exercises B
3 6 3 6 3
1. ,
,
5 10 5 10 5
Practice Exercises C
1. 65
3. 115
5. 65
7. 115
9. 115
11. 60
13. 25
15. 30
17. 40
19. m1 27.5 , m2 27.5 , m3 30
Jordan School District
3. 130
Practice Exercises G
1. m1 64 , m2 26 , m3 45 ,
m4 45
3. m1 30 , m2 30 , m3 60 ,
m4 30 , m5 60 , m6 60 ,
m7 30 , m8 60
5. mS 135 , mT 80
2 5
5
5
11.
3. 120
16
7.
3
11. 98.5
Practice Exercises A
22
1.
7.679 yds
9
166
3.
11.589 ft
45
49
5.
15.394 cm
10
Practice Exercises B
3
1.
4
11
5.
6
1
3.
4
Practice Exercises C
22
7.679 yds
1.
9
49
5.
15.394 cm
10
Page 403
3.
166
11.589 ft
45
Secondary Mathematics 2
Practice Exercises D
1. 2 6.283 ft
2
25
39.270 m2
2
27
9.
21.206 in 2
4
5.
8
3.
8.378 in 2
3
5
7.
5.236 mm2
3
9. d1
2 4 3 1
2
2
20 which
doesn’t equal
d2
2 3 3 5
2
2
29
11. there is only one point of intersection at
(1, 9)
Unit 6 Cluster 5 G.GMD.1-3
Unit 6 Cluster 3 G.GPE.1
Practice Exercises A
1. 12 37.699 ft 3
Practice Exercises A
1. center (0, 0), radius 6
3. x 2 y 2 49
5. 75 in 3
9.
Practice Exercises B
1. center (2, 3) radius 4
3. center (-5, 6) radius 2
5. center (10, -21) radius 14
2
2
7. x 8 y 6 100
9. x 4 y 4 29
2
2
Practice Exercises C
1. center (2, 3) radius 5
3. center (-3, 1) radius 5
Challenge: center 32 , 2 radius
1
2
Unit 6 Cluster 4 G.GPE.4
Practice Exercises A
1. mEF mGH 0 , d EF dGH 4.123
2
3. mAB mCD , mBC mAD 4 ,
3
d AB dCD 3.606 , d BC d AD 4.123
5.
7.
400 1256.637 cm3
3. 588 1847.256 m3
7. 96 301.593 cm3
11. 4000
4188.790 cm3
3
13. 60 188.496 ft 3
15. 18 56.55 in3
17. 48 ft 3
19. 333.038 in3
Practice Exercises B
30 5
5
1.
3. k
18 3
3
81 27
5.
7. 800 square feet
192 64
9. 8 times as much, 20 cm diameter and 36
cm height
11. 2 times as much volume
13. They are the same. The larger popcorn
has twice as much popcorn as the
smaller popcorn for twice the price.
15. 8 times as much volume, 27 times as
much volume
mAB mCD 0 , mBC mAD 4 ,
d AB dCD 4 , d BC d AD 4.123
x 2
3 2
2
2
y2 4
3
2
4
12 3 4
44
Jordan School District
Page 404
Secondary Mathematics 2
