NEXT-GENERATION MATH ACCUPLACER TEST REVIEW BOOKLET
Transcript of NEXT-GENERATION MATH ACCUPLACER TEST REVIEW BOOKLET
Next Generation Advanced Algebra and Functions
NEXT-GENERATION MATH ACCUPLACER
TEST REVIEW BOOKLET
2
Property of MSU Denver Tutoring Center
3
Table of Contents
About..................................................................................................................7 Test Taking Tips.................................................................................................9 1 Integers and Rational Numbers.........................................................................11
1.1 Useful Memorization 1.2 Absolute Value 1.3 Order of Operations 1.4 Substitution 1.5 Rules of Fractions 1.6 Mixed Fractions 1.7 Rules of Exponents 1.8 Simplifying Radicals 1.9 Rules of Radicals
1.10 Rationalizing 2 Linear Expressions and Equationsβ¦.................................................................47
2.1 Simplifying Expressions 2.2 Solving Linear Equations 2.3 Solving Inequalities 2.4 Solving Systems of Equations 2.5 Graphing Linear Equations 2.6 Building Lines
3 Logarithms.........................................................................................................79 3.1 Exponential Form vs. Logarithmic Form 3.2 Rules of Logarithms 3.3 Solving Logarithmic Equations 3.4 Common Base Technique 3.5 Exponential Models
4 Functions.........................................................................................................103 4.1 Introduction, Domain, and Range 4.2 Evaluating Functions 4.3 Composition Functions and Inverses 4.4 Transformations of Functions
4
5
5 Quadratic Equations........................................................................................129 5.1 Distribution and Foiling 5.2 Factoring Expressions 5.3 Solving Quadratic Equations
6 Geometry Concepts.........................................................................................145 6.1 Two-Dimensional Shapes 6.2 Three-Dimensional Shapes 6.3 Lines and Angles
7 Trigonometry...................................................................................................165 7.1 Geometric Trigonometry
8 Word Problems................................................................................................175 8.1 Setting up and Equation 8.2 Word Problem Practice
9 Practice Tests...................................................................................................187 9.1 Practice Test 1 9.2 Practice Test 2 9.3 Practice Test 3
Each subsection in this practice book is followed by practice problems and answers. Make sure to check your answers!
This practice book contains a lot of material. The intention of including so much information is to make sure youβre prepared for the test. You can start at the very beginning and work your way up, or you can focus on specific sections you need help with. Make sure you ask a tutor for help if you arenβt sure about something!
Sections with the β symbol at the top of the page are more likely to appear on the
test!
β
6
7
About
Here is some information about the MSU Tutoring Center and the Placement Test. MSU Denver Tutoring Center Mission
The mission of the MSU Denver Tutoring Center is to increase student persistence by offering free student-centered academic support across disciplines. The Tutoring Center is not only for students who are having difficulty with course material, but also for students who are looking to excel. Whether your goal is to catch up, keep up, or do better in your studies, the Tutoring Center is here to help you by offering free one-on-one and group academic support in a variety of subjects. We promote an environment that is welcoming to the diverse student body of MSU Denver by providing professionally trained tutors who are competent in subject material and areas such as diversity, learning styles, and communication. What is the Next-Generation Accuplacer?
The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takersβ ability for selected mathematics content. Questions will focus on a range of topics, including a variety of equations and functions, including linear, quadratic, rational, radical, polynomial, and exponential. Questions will also delve into some geometry and trigonometry concepts. In addition, questions may assess a studentβs math ability via computational or fluency skills, conceptual understanding, or the capacity to apply mathematics presented in a context. All questions are multiple choice in format and appear discretely (stand-alone) across the assessment. The following knowledge and skill categories are assessed: Linear equations Linear applications Factoring Quadratics Functions Radical and rational equations Polynomial equations Exponential and logarithmic equations Geometry concepts Trigonometry
8
9
Test Taking Tips
Take your time on the test. Students tend to get higher scores the longer they spend finding and double checking their answers.
If a calculator is needed for a specific question, a calculator icon will appear in the top right corner. You may not bring your own calculator in to the test.
Read carefully! Pay careful attention to how questions are worded and what they want you to find.
Make use of the scratch paper provided- you are going to make mistakes if you try to solve problems in your head.
The test is multiple choice, so use that to your advantage. If you canβt remember how to approach a problem, maybe youβll find the correct answer by plugging the multiple choice answers into the original question!
If you donβt remember some of the rules for exponents, radicals, or other operations, try constructing a simpler example that you know better and solve that one. The rules may become clear to you once you try the simple example.
Another strategy for remembering the many rules involved with operations is to break things down to their smaller pieces. Under pressure, itβs easier to see whatβs going on with π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯ instead of π₯π₯4.
The test uses an algorithm called branching. This means that it changes its difficulty level each time you answer a question, depending on how well youβve done. If you are getting questions wrong, the test will make itself easier. If you are getting questions correct, the test will make itself harder. Itβs a good sign if the test seems to be getting more challenging. (You earn more points for more difficult questions).
Eat before you take the test. This will help you concentrate better. Get good sleep the night before you take the test. If you donβt sleep well the
night before, postpone it. Donβt be intimidated by the questions or the test in general. Itβs okay to have
anxiety- recognize it and do your best to relax so you can concentrate. Review for this test right before you go to sleep at night.
10
11
Integers and Rational Numbers
12
13
1.1 Useful Memorization
A calculator will not be allowed on the test, so make sure you have some of this memorized or know how to calculate it! Multiplication Table
Γ 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12
2 2 4 6 8 10 12 14 16 18 20 22 24
3 3 6 9 12 15 18 21 24 27 30 33 36
4 4 8 12 16 20 24 28 32 36 40 44 48
5 5 10 15 20 25 30 35 40 45 50 55 60
6 6 12 18 24 30 36 42 48 54 60 66 72
7 7 14 21 28 35 42 49 56 63 70 77 84
8 8 16 24 32 40 48 56 64 72 80 88 96
9 9 18 27 36 45 54 63 72 81 90 99 108
10 10 20 30 40 50 60 70 80 90 100 110 120
11 11 22 33 44 55 66 77 88 99 110 121 132
12 12 24 36 48 60 72 84 96 108 120 132 144
000 000 000 000 000 000 000 000 Perfect Squares
ππππ = ππ
ππππ = ππ
ππππ = ππ
ππππ = ππππ
ππππ = ππππ
ππππ = ππππ
ππππ = ππππ
ππππ = ππππ
ππππ = ππππ
ππππππ = ππππππ
ππππππ = ππππππ
ππππππ = ππππππ
14
15
1.2 Absolute Value
The absolute value | | functions as a grouping and an operation. It groups things together like parentheses, and it also changes negative numbers into positive numbers. One way to think about the absolute value function is a measurement of distance. Consider |3| and |β3|. How far away is 3 from zero on a number line? How far away is β3 from zero on a number line?
Since the number 3 is three units away from zero, |3| = 3. Since the number β3 is three units away from zero, |β3| = 3. Example
An easy way to think about absolute value is that it turns negative numbers positive, and keeps positive numbers positive. Once the absolute value has been evaluated (turned positive), do not continue to write the vertical bars | |.
|β2| = 2 |2| = 2
Practice Problems
Evaluate the following.
1) |β2| 2) |2| 3) |4| 4) |β5| 5) |β12| 6) |370| 7) |β160|
8) |0| 9) |5 β 3| 10) |12 β 1| 11) |3 β 8| 12) |9 β 16| 13) β|6| 14) β|β1|
15) |1 Γ· 3| 16) |β5| + |1| 17) |β3| + |β4| 18) |β9| β |6| 19) |β7 + 4| 20) β|β7 + 4| 21) β|β1 β 13|
-3 -2 -1 0 1 2 3
3 units away from zero 3 units away from zero
16
Answers
1) 2 β2 is two units away from zero.
2) 2 2 is two units away from zero.
3) 4 4 is four units away from zero.
4) 5 β5 is five units away from zero.
5) 12 β12 is twelve units away from zero.
6) 370 370 is 370 units away from zero.
7) 160 β160 is 160 units away from zero.
8) 0 0 is zero units away from itself.
9) 2 |5 β 3| = |2|, and 2 is two units away from zero.
10) 11 |12 β 1| = |11|, and 11 is eleven units away from zero.
11) 5 |3 β 8| = |β5|, and β5 is five units away from zero.
12) 7 |9 β 16| = |β7|, and β7 is seven units away from zero.
13) β6 6 is six units away from zero and the negative is outside of the absolute value, so β|6| = β6.
14) β1 β1 is one unit away from zero and there is a negative outside of the absolute value, so β|β1| = β1.
15) 13
|1 Γ· 3| = οΏ½13οΏ½, and 1
3 is one third of a unit away from zero.
16) 6 |β5| + |1| = 5 + 1 = 6
17) 7 |β3| + |β4| = 3 + 4 = 7
18) 3 |β9| β |6| = 9 β 6 = 3
19) 3 | β 7 + 4| = |β3|, and β3 is three units away from zero.
20) β3 β|β7 + 4| = β|β3|, β3 is three units away from zero, and there is a negative outside of the
absolute value, so β|β3| = β3.
21) β14 β|β1 β 13| = β|β14|, β14 is fourteen units away from zero, and there is a negative outside of
the absolute value, so β|β14| = β14.
17
1.3 Order of Operations
The different operations in mathematical expressions (addition, subtraction, multiplication, division, exponentiation, and grouping) must be evaluated in a specific order. This structure gives us consistent answers when we simplify expressions. Why is it needed? Consider the expression 4 Γ 2 + 1. If multiplication is evaluated before addition, 4 Γ 2 + 1 becomes 8 + 1 which then becomes 9. If addition is evaluated before multiplication, 4 Γ 2 + 1 becomes 4 Γ 3 which then becomes 12. Since 9 β 12, evaluating in any order we want doesnβt give us consistent answers.
The Order of Operations
1) Parentheses (evaluate the inside) ( ) [ ] { } | |
2) Exponents π₯π₯2 βππβππππ ππππππππππππ
3) Multiplication and Division (evaluate from left to right) Γ Γ·
4) Addition and Subtraction (evaluate from left to right) + β
Notice that Multiplication and Division share the third spot and Addition and Subtraction share the fourth spot. In these cases, evaluate from left to right
PEMDAS is the mnemonic used to remember the Order of Operations. Start with Parentheses and end with addition and subtraction. (Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction).
18
Note about exponents
Pay close attention to notation with exponents! The position of a negative sign can change an entire problem.
β22 = β(2)(2) = β4
(β2)2 = (β2)(β2) = 4
Note about absolute value
The absolute value parentheses | | function in two ways. They group an expression just like any other parentheses, but they also turn whatever simplified number they contain into the positive version of that number.
|β3| = 3 and |3| = 3
Note about radicals
Radicals are technically exponents (see section 1.8). So radicals will be evaluated at the same time as exponents:
β11 + 5 + 2
β16 + 2
4 + 2
6
Practice Problems
Evaluate the following. 1) (5 + 3) β 2 2) 5 + (3 β 2) 3) 2 β (8 + 5) β 2
4) 2 β 8 + 5 β 2
5) (2 + 3)2 β |2 β 4|
6) 5 + (3 β 6 + 2)
7) β22 + 32
8) (β2)2 + 32
9) (β2)2β223
10) ((5 β 3)2)2
11) (5β2)(6β3)6Γ·2
12) (4+2)1+23
|2|
13) 0 β (3 + 6 β 8)2 β 14
In the first expression, the negative is sitting outside of the exponent operation. In the second expression, it is included and cancels out.
19
Answers
1) 16 (5 + 3) β 2 = 8 β 2=16
2) 11 5 + (3 β 2) = 5 + 6 = 11
3) 24 2 β (8 + 5) β 2 = 2 β 13 β 2 = 26 β 2 = 24
4) 19 2 β 8 + 5 β 2 = 16 + 5 β 2 = 21 β 2 = 19
5) 23 (2 + 3)2 β |2 β 4| = (5)2 β |β2| = 25 β 2 = 23
6) 25 5 + (3 β 6 + 2) = 5 + (18 + 2) = 5 + (20) = 25
7) 5 β22 + 32 = β4 + 9 = 5
8) 13 (β2)2 + 32 = 4 + 9 = 13
9) 1 (β2)2 β 223
=4 β 2
8=
88
= 1
10) 16 ((5 β 3)2)2 = ((2)2)2 = (4)2 = 16
11) 10 (5 β 2)(6 β 3)6 Γ· 2
=(10)(3)
3=
303
= 10
12) 7 (4 + 2)1 + 23
|2| =(6)1 + 8
2=
6 + 82
=142
= 7
13) β14 0 β (3 + 6 β 8)2 β 14 = 0 β (9 β 8)2 β 14 = 0 β (1)2 β 14 = 0 β 1 β 14 = 0 β 14 = β14
20
21
1.4 Substitution
Simply explained, substitution is when you replace the letters in an expression with numbers. The Process
Replace the variables in the original expression with their assigned values.
Example
Evaluate π₯π₯2π¦π¦ where π₯π₯ = 2 and π¦π¦ = 3.
Starting point:
Replace π₯π₯ with 2 and π¦π¦ with 3
Evaluate the exponent
Multiply
π₯π₯2π¦π¦
(2)2 β (3)
4 β 3
12
Example
Evaluate π₯π₯2π¦π¦ where π₯π₯ = β2 and π¦π¦ = β3.
Starting point:
Replace π₯π₯ with β2 and π¦π¦ with β3
Evaluate the exponent
Multiply
π₯π₯2π¦π¦
(β2)2 β (β3)
4 β β3
β12
Note
It is important to plug in negative numbers carefully! A common mistake in the example above would be writing the second step as β22 β (β3) and therefore ending up with β4 β β3 = 12 as your final answer instead of β12.
22
Practice Problems
Evaluate the following.
1) |π₯π₯| where π₯π₯ = β3 2) |π₯π₯2| where π₯π₯ = β3 3) π₯π₯2π¦π¦ where π₯π₯ = 3 and π¦π¦ = 2 4) π₯π₯2π¦π¦ where π₯π₯ = β4 and π¦π¦ = 2 5) π₯π₯3 where π₯π₯ = 2 6) π₯π₯3 where π₯π₯ = β2 7) βπ₯π₯3 where π₯π₯ = 3 8) βπ₯π₯3 where π₯π₯ = β3 9) (π₯π₯π¦π¦π₯π₯)2 where π₯π₯ = 3,π¦π¦ = β2, and π₯π₯ = 1
2
10) π¦π¦ + βπ₯π₯ where π₯π₯ = 36 and π¦π¦ = β4 11) π₯π₯π¦π¦ β 3
βπ¦π¦ where π₯π₯ = 1
2 and π¦π¦ = 4
12) π₯π₯2
4β π¦π¦2
3 where π₯π₯ = 3 and π¦π¦ = 2
13) 5(π₯π₯+β)β5(π₯π₯)β
where π₯π₯ = 3 and β = 4
14) οΏ½ ππ+ππ|ππβππ|
where ππ = 10 and π π = 6
15) βππ+βππ2β4βππβππ
2βππ where ππ = 4, ππ = 8, and ππ = 3
23
Answers
1) 3 |π₯π₯| = |β3| = 3
2) 9 |π₯π₯2| = |(β3)2| = |9| = 9
3) 18 π₯π₯2π¦π¦ = (3)22 = 9 β 2 = 18
4) 32 π₯π₯2π¦π¦ = (β4)22 = 16 β 2 = 32 5) 8 π₯π₯3 = 23 = 2 β 2 β 2 = 8
6) β8 π₯π₯3 = (β2)3 = (β2)(β2)(β2) = β8
7) β27 βπ₯π₯3 = β(3)3 = β(3)(3)(3) = β27 8) 27 βπ₯π₯3 = β(β3)3 = β(β3)(β3)(β3) = β(β27) = 27
9) 9 (π₯π₯π¦π¦π₯π₯)2 = (3 β (β2) β 12)2 = (β3)2 = 9
10) 2 π¦π¦ + βπ₯π₯ = β4 + β36 = β4 + 6 = 2
11) 12
π₯π₯π¦π¦ β
3
οΏ½π¦π¦= οΏ½
12οΏ½ β 4 β
3β4
=42β
32
=4 β 3
2=
12
12) 1112
π₯π₯2
4βπ¦π¦2
3=
32
4β
22
3=
94β
43
= οΏ½33οΏ½
94β
43οΏ½
44οΏ½ =
2712
β1612
=1112
13) 5 5(π₯π₯ + β) β 5(π₯π₯)β
=5(3 + 4) β 5(3)
4=
5(7) β 5(3)4
=35 β 15
4=
204
= 5
14) 2 οΏ½π π + ππ
|π π β ππ| = οΏ½6 + 10
|6 β 10| = οΏ½16
|β4| = οΏ½164
=β16β4
=42
= 2
15) β12
βππ + βππ2 β 4 β ππ β ππ
2 β ππ=β8 + οΏ½(8)2 β 4 β 4 β 3
2 β 4=β8 + β64 β 48
8=β8 + β16
8=β8 + 4
8
=β48
= β12
24
25
1.5 Rules of Fractions
Fractions can be reduced, added, subtracted, multiplied, and divided.
Multiplication
Rule Example
ππππβππππ
= ππ β ππππ β ππ
23β
12
= 2 β 13 β 2
= 26
= 13
Division
Rule Example
ππππ
Γ·ππππ
= ππππβππππ
= ππ β ππππ β ππ
23
Γ·12
= 23β
21
= 2 β 23 β 1
= 43
Addition and Subtraction
Rule Example
ππππ
+ππππ
= οΏ½πππποΏ½ππππ
+πππποΏ½πππποΏ½ =
ππ β ππ + ππ β ππ ππ β ππ
23
+12
= οΏ½22οΏ½
23
+12οΏ½
33οΏ½ =
2 β 2 + 1 β 3 2 β 3
= 4 + 3
6 =
76
The rule above works for addition or subtraction. Just replace the + with a β to get:
ππππβππππ
= οΏ½πππποΏ½ππππβπππποΏ½πππποΏ½ =
ππ β ππ β ππ β ππ ππ β ππ
26
Practice Problems
1) 12β 37
2) 910β 210
3) β35β β15
9
4) 13
Γ· 19
5) 34
Γ· 12
6) 8 οΏ½15οΏ½
7) 37
+ 23
8) 37β 2
3
9) 54
+ 53
10) 49β 3
2
11) 54
+ 78οΏ½87οΏ½
27
Answers
1) 314
12β
37
=1 β 32 β 7
=3
14
2) 950
9
10β
210
=9 β 2
10 β 10=
18100
=9
50
3) 1 β35ββ15
9=β3 β β15
5 β 9=
4545
= 1
4) 3 13
Γ·19
=13β
91
=1 β 93 β 1
=93
= 3
5) 32 3
4Γ·
12
=34β
21
=3 β 24 β 1
=64
=32
6) 85
8 οΏ½15οΏ½ =
81β
15
=8 β 11 β 5
=85
7) 2321
37
+23
= οΏ½33οΏ½
37
+23οΏ½
77οΏ½ =
3 β 33 β 7
+2 β 73 β 7
=9
21+
1421
=9 + 14
21=
2321
8) β
521
37β
23
= οΏ½33οΏ½
37β
23οΏ½
77οΏ½ =
3 β 33 β 7
β2 β 73 β 7
=9
21β
1421
=9 β 14
21= β
521
9) 3512
54
+53
= οΏ½33οΏ½
54
+53οΏ½
44οΏ½ =
3 β 53 β 4
+5 β 43 β 4
=1512
+2012
=15 + 20
21=
3512
10) β1921
49β
32
= οΏ½22οΏ½
49β
32οΏ½
99οΏ½ =
2 β 42 β 9
β3 β 92 β 9
=8
18β
2718
=8 β 27
21= β
1921
11) 94
54
+78οΏ½
87οΏ½ =
54
+7 β 88 β 7
=54
+5656
= οΏ½1414οΏ½
54
+5656
=7056
+5656
=70 + 56
56=
12656
=94
28
29
1.6 Mixed Fractions
The following are four definitions in regards to fractions. Terminology
A whole number is a fraction where the denominator is 1.
πππππ¦π¦ ππππππππππππ1 β
41 = 4
Terminology
A proper fraction is a fraction where the numerator is smaller than the denominator.
ππππππππππππππππππ π π πππππ π π π ππππππππππππππππππππππππππ π π ππππππππππ β
23
Terminology
An improper fraction is a fraction where the numerator is larger than the denominator.
ππππππππππππππππππ π π ππππππππππππππππππππππππππππππππ π π πππππ π π π ππππ β
72
Terminology
A mixed fraction is a whole number and a proper fraction combined.
π€π€βπππ π ππ ππππππππππππππππππππππππππππππ π π πππππ π π π ππππππππππππππππππππππππππ π π ππππππππππ
β 723
30
Write Improper Fraction as Mixed Fraction
Starting point: 72
Re-write as the sum of fractions: = 22
+ 22
+ 22
+ 12
Simplify to whole numbers: = 1 + 1 + 1 + 12
Combine: = 3 + 12
Write the whole number next to the proper fraction: = 312
Write Mixed Fraction as Improper Fraction
Starting point: 237
Re-write as the sum of fraction and whole number: = 2 + 37
Expand: = 1 + 1 + 37
Write whole numbers as fractions: = 77
+ 77
+ 37
Add across the top (7 + 7 + 3) and keep the bottom the same: = 177
Practice Problems
Write the following improper fractions as mixed fractions.
1) 87
2) 92
3) 32
4) 113
5) 233
6) 175
7) 176
Write the following mixed fractions as improper fractions.
8) 112
9) 413
10) 3 310
11) 612
12) 259
13) 723
14) 147
31
Answers
1) 117 8
7=
77
+17
= 1 +17
= 117
2) 412 9
2=
22
+22
+22
+22
+12
= 1 + 1 + 1 + 1 +12
= 4 +12
= 412
3) 112 3
2=
22
+12
= 1 +12
= 112
4) 323 11
3=
33
+33
+33
+23
= 1 + 1 + 1 +23
= 3 +23
= 323
5) 723 23
3=
33
+33
+33
+33
+33
+33
+33
+23
= 1 + 1 + 1 + 1 + 1 + 1 + 1 +23
= 7 +23
= 723
6) 325 17
5=
55
+55
+55
+25
= 1 + 1 + 1 +25
= 3 +25
= 325
7) 256 17
6=
66
+66
+56
= 1 + 1 +56
= 2 +56
= 256
8) 32
112
= 1 +12
=22
+12
=32
9) 133
413
= 4 +13
=33
+33
+33
+33
+13
=133
10) 3310
3 310
= 3 +3
10=
1010
+1010
+1010
+3
10=
3310
11) 132
612
= 6 +12
=22
+22
+22
+22
+22
+22
+12
=132
12) 239
259
= 2 +59
=99
+99
+59
=239
13) 233
723
= 7 +23
=33
+33
+33
+33
+33
+33
+33
+23
=233
14) 117
147
= 1 +47
=77
+47
=117
32
33
1.7 Rules of Exponents
An exponent is a base raised to a power. A base βππβ raised to the power of βππβ is equal to the multiplication of ππ, ππ times:
a n = a Γ a Γ ... Γ a [ππ times] ππ is the base and ππ is the exponent.
Examples
ππππ = ππ 21 = 2
ππππ = ππ β ππ 22 = 2 β 2
ππππ = ππ β ππ β ππ 23 = 2 β 2 β 2
ππππ = ππ β ππ β ππ β ππ 24 = 2 β 2 β 2 β 2
ππππ = ππ β ππ β ππ β ππ β ππ 25 = 2 β 2 β 2 β 2 β 2 Rules
ππππ β ππππ = ππππ+ππ ππππ β ππππ = (ππ β ππ)ππ ππππ
ππππ= ππππβππ
ππππ
ππππ= οΏ½
πππποΏ½ππ
οΏ½πππποΏ½ππ
=ππππ
ππππ
(ππ β ππ)ππ = ππππ β ππππ (ππππ)ππ = ππππβππ
ππβππ =1ππππ
οΏ½πππποΏ½βππ
= οΏ½πππποΏ½ππ
ππ0 = 1 0ππ = 0, ππππππ ππ > 0
34
Practice Problems
Simplify the following expressions.
1) 22 + 31
2) 32 β 32
3) 32 β 3β2
4) οΏ½32οΏ½3βοΏ½22οΏ½2
31
5) β24 + (β2)4 β 21
6) οΏ½32οΏ½3
7) (32 β 23) β 3β3 + (3 Γ· 33)
8) 25
β22+ 2
23
9) (23)β2(22 β 13)2
10) π₯π₯5
π₯π₯2
11) π₯π₯3π¦π¦2
π₯π₯2π¦π¦5
12) (2π₯π₯2π¦π¦3)0
13) οΏ½3π₯π₯4π¦π¦β2οΏ½β3
(2π₯π₯3π¦π¦2)β2
35
Answers
1) 7 22 + 31 = 4 + 3 = 7
2) 81 32 β 32 = 9 β 9 = 81
3) 1 32 β 3β2 = 32 β 132
= 32
32= 1 ππππ 32 β 3β2 = 32β2 = 30 = 1
4) 23 οΏ½32οΏ½3βοΏ½22οΏ½
2
31= 36β24
31= 729β16
31= 713
31= 23
5) -2 β24 + (β2)4 β 21 = β16 + 16 β 2 = 0 β 2 = β2
6) 278
οΏ½32οΏ½3
= 33
23= 27
8
7) 259
(32 β 23) β 3β3 + (3 Γ· 33) = (9 β 8) β 133
+ οΏ½ 333οΏ½ = 72
27+ 3
27= 75
27= 25
9
8) 334
25
(β2)2+ 2
23= 32
4+ 2
8= οΏ½2
2οΏ½ 324
+ 28
= 648
+ 28
= 64+28
= 668
= 334
9) 14
(23)β2(22 β 13)2 = 2β6(4 β 1)2 = 126β (4)2 = 16
64= 1
4
10) π₯π₯3 π₯π₯5
π₯π₯2= π₯π₯βπ₯π₯βπ₯π₯βπ₯π₯βπ₯π₯
π₯π₯βπ₯π₯= π₯π₯βπ₯π₯βπ₯π₯
1= π₯π₯3 ππππ π₯π₯
5
π₯π₯2= π₯π₯5β2 = π₯π₯3
11) π₯π₯π¦π¦3
π₯π₯3π¦π¦2
π₯π₯2π¦π¦5= π₯π₯βπ₯π₯βπ₯π₯βπ¦π¦βπ¦π¦
π₯π₯βπ₯π₯βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦= π₯π₯
π¦π¦βπ¦π¦βπ¦π¦= π₯π₯
π¦π¦3 ππππ π₯π₯
3π¦π¦2
π₯π₯2π¦π¦5= π₯π₯3β2π¦π¦2β5 = π₯π₯1π¦π¦β3 = π₯π₯ οΏ½ 1
π¦π¦3οΏ½ = π₯π₯
π¦π¦3
12) 1 (2π₯π₯2π¦π¦3)0 = 20π₯π₯0π¦π¦0 = 1 β 1 β 1 = 1 ππππ (2π₯π₯2π¦π¦3)0 = 1
(anything to the power of zero is one) 13) 4π¦π¦10
27π₯π₯6
οΏ½3π₯π₯4π¦π¦β2οΏ½β3
(2π₯π₯3π¦π¦2)β2 = 3β3π₯π₯β12π¦π¦6
2β2π₯π₯β6π¦π¦β4= 22π₯π₯6π¦π¦4π¦π¦6
33π₯π₯12= 22π₯π₯6π¦π¦10
33π₯π₯12= 4π₯π₯6π¦π¦10
27π₯π₯12= 4π¦π¦10
27π₯π₯6
36
37
1.8 Simplifying Radicals
A radical is anything with the root symbol . Radicals are fractional exponents. The notation is very different, but they follow the same basic rules as regular exponents! With practice, you will be able to see the similarities. The following shows how radicals work on a basic level and how to simplify them.
Notation
βππ = βππππ = ππππ ππβ 41 2β = β42 = β2 β 22 = 2
βππππ = ππππ ππβ 81 3β = β83 = β2 β 2 β 23 = 2
βππππ = ππππ ππβ 161 4β = β164 = β2 β 2 β 2 β 24 = 2
βππππ = ππππ ππβ 321 5β = β325 = β2 β 2 β 2 β 2 β 25 = 2
βππππ = ππππ ππβ 641 6β = β646 = β2 β 2 β 2 β 2 β 2 β 26 = 2
Simplification Example
A radical is fully simplified when the radicand (the part under the radical) has no square factors. For instance, β40 = β4 β 5 β 2, and since 4 is a square number, β40 is not simplified. From here, use the rules of radicals to split the root apart and simplify:
β40 = β4 β 10 = β4 β β10 = 2β10
So 2β10 is the most simplified form of β40.
Pattern-Based Approach
Another way to simplify is to factor the radicand completely. For example, β40 =β2 β 2 β 2 β 5. Notice there are three twos. The type of radical we are using is a square root (β2 β 2 β 2 β 5 = β2 β 2 β 2 β 5ππ ). Since that two is there, we find pairs of two and represent the two outside, leaving behind the unpaired numbers under the radical. βππ β ππ β 2 β 5ππ = ππ β β2 β 5ππ . There are more examples on the next page.
38
Pattern-Based Approach Examples
β40 = βππ β ππ β 2 β 5 = ππ β β2 β 5 = 2β10 groups of 2 because of β
β72 = βππ β ππ β ππ β ππ β 2 = ππ β ππ β β2 = 6β2 groups of 2 because of β
β16 = βππ β ππ β ππ β ππ = ππ β ππ = 4 groups of 2 because of β
β403 = βππ β ππ β ππ β 53 = ππ β β53 = 2β53 groups of 3 because of β
β83 = βππ β ππ β ππ3 = ππ = 2 groups of 3 because of β
β2503 = βππ β ππ β ππ β 23 = ππ β β23 = 5β23 groups of 3 because of β
β164 = βππ β ππ β ππ β ππ4 = ππ = 2 groups of 4 because of β
β324 = βππ β ππ β ππ β ππ β 24 = ππ β β24 = 2β24 groups of 4 because of β
If you factor the radicand completely and canβt find any pairs, then the radical is already as simple as it can be. For example, β30 = β2 β 3 β 5. Since there are no pairs, β30 is the most simplified form.
Practice Problems
Simplify the following radicals.
1) β20 2) β16 3) β24 4) β18 5) β180 6) β12 7) β28 8) β50
9) β45 10) β98 11) β48 12) β300 13) β150 14) β80
15) β202
39
Answers
1) 2β5 β20 = β2 β 2 β 5 = 2β5
2) 4 β16 = β4 β 4 = 4
3) 2β6 β24 = β3 β 2 β 2 β 2 = 2β6
4) 3β2 β18 = β3 β 3 β 2 = 3β2
5) 6β5 β180 = β3 β 3 β 2 β 2 β 5 = 6β5
6) 2β3 β12 = β3 β 2 β 2 = 2β3
7) 2β7 β28 = β2 β 2 β 7 = 2β7
8) 5β2 β50 = β2 β 5 β 5 = 5β2
9) 3β5 β45 = β3 β 3 β 5 = 3β5
10) 7β2 β98 = β2 β 7 β 7 = 7β2
11) 4β3 β48 = β4 β 4 β 3 = 4β3
12) 10β3 β300 = β10 β 10 β 3 = 10β3
13) 5β6 β150 = β5 β 5 β 3 β 2 = 5β6
14) 4β5 β80 = β4 β 4 β 5 = 4β5
15) β5 β202
= β2β2β52
= 2β52
= β5
40
41
1.9 Rules of Radicals
Since radicals are fractional exponents, we can use rules of exponents to manipulate them. The following are some of the properties and notation with radicals.
Rules
ππππππ = βππππππ ππ
ππππ = βππππππ
βππππππ = ππ βππππππ = ππππππ = ππππ = ππ
βππππ β οΏ½ππππ = οΏ½ππ β ππππ βππππ β οΏ½ππππ = οΏ½ππππππ
οΏ½ππππ
ππ=βππππ
οΏ½ππππ οΏ½ππππ
ππ=βππππ
οΏ½ππππ
Note On Multiplication And Division
If the radicals are not the same type (number), the multiplication and division rules do not apply.
Rule Examples βππππ β οΏ½ππππ ππππππππππππ ππππππππππππππ βππππ β οΏ½ππππ is already as simple as possible βππππ
οΏ½ππππ ππππππππππππ ππππππππππππππ βππππ
οΏ½ππππ is already as simple as possible
Practice Problems
Evaluate the following radicals.
1) οΏ½β3οΏ½2
2) οΏ½(3)2 3) β2 β β15 4) β6 β β8
5) οΏ½49
6) βοΏ½ 636
7) 1628
8) οΏ½β83 οΏ½2
9) 83β64
10) 2β7 + 5β7 11) ββ1253
42
Answers
1) 3 οΏ½β3οΏ½2
= 322 = 31 = 3
2) 3 οΏ½(3)2 = 322 = 31 = 3
3) β30 β2 β β15 = β30 4) 4β3 β6 β β8 = β6 β 8 = β48 = 4β3
5) 23
οΏ½49
= β4β9
= 23
6) ββ66
βοΏ½ 636
= β β6β36
= ββ66
7) 2 1628 = οΏ½(16)28 = β16 β 168 = β2 β 2 β 2 β 2 β 2 β 2 β 2 β 28 = 2
8) 4 οΏ½β83 οΏ½2
= (2)2 = 4
9) 13
8
3β64= 8
3β8= 1
3
10) 7β7 2β7 + 5β7 = 7β7
11) β5 ββ1253 = β5
43
1.10 Rationalizing
When radicals appear in a fraction, we may have to use algebra to rearrange the fraction so that it is in the proper form. Radicals are allowed to be in the numerator, but not the denominator. If a radical is in the denominator, the process of rearranging the fraction so that the radical is only in the numerator is called rationalization. First some terminology is given, then examples of two situations involving rationalization are outlined. Terminology
A binomial is formed when two terms are added or subtracted from one another. (π₯π₯ + π¦π¦) is a binomial. (2π₯π₯β 7) is also a binomial. The conjugate of a binomial is created when the plus or minus sign in the middle of the binomial is changed to the opposite sign. (π₯π₯ β π¦π¦) is the conjugate of (π₯π₯ + π¦π¦). (2π₯π₯ + 7) is the conjugate of (2π₯π₯ β 7). Example
Consider the following fraction. In order to rationalize a fraction with a radical in the denominator, multiply both the numerator and denominator by that same radical.
3β5
= 3β5
οΏ½β5β5οΏ½ =
3β5
οΏ½β5οΏ½2 =
3β55
In other examples, reduce further if necessary.
Not rationalized Multiply by fraction made up
of the radical
The exponent and radical in the denominator cancel out
Rationalized
44
Example
When a radical in the denominator is part of a binomial, multiply the numerator and denominator by the conjugate.
27 + β5
= 2
7 + β5οΏ½
7 β β57 β β5
οΏ½ = 2οΏ½7 β β5οΏ½
72 β οΏ½β5οΏ½2 =
14 β 2β549 β 5 =
7 β β522
Practice Problems
Rationalize the following fractions.
1) 1β2
2) 1β3
3) 2β5
4) 4β3
5) π₯π₯β2
6) 13+β2
7) 11+β3
8) 52ββ2
9) 34ββ3
10) 2π₯π₯+β5
Not rationalized
Multiply by fraction made up of the conjugate
The exponent and radical in the denominator cancel out
Rationalized
Reduceβ¦
45
Answers
1) β22
1β2
=1β2
οΏ½β2β2οΏ½ =
1 β β2
οΏ½β2οΏ½2 =
β22
2) β33
1β3
=1β3
οΏ½β3β3οΏ½ =
1 β β3
οΏ½β3οΏ½2 =
β33
3) 2β55
2β5
=2β5
οΏ½β5β5οΏ½ =
2 β β5
οΏ½β5οΏ½2 =
2β55
4) 4β33
4β3
=4β3
οΏ½β3β3οΏ½ =
4 β β3
οΏ½β3οΏ½2 =
4β33
5) π₯π₯β22
π₯π₯β2
=π₯π₯β2
οΏ½β2β2οΏ½ =
π₯π₯ β β2
οΏ½β2οΏ½2 =
π₯π₯β22
6) 3 β β27
1
3 + β2=
13 + β2
οΏ½3 β β23 β β2
οΏ½ =3 β β2
32 β οΏ½β2οΏ½2 =
3 β β29 β 2
=3 β β2
7
7) β
1 β β32
1
1 + β3=
11 + β3
οΏ½1 β β31 β β3
οΏ½ =1 β β3
12 β οΏ½β3οΏ½2 =
1 β β31 β 3
=1 β β3β2
= β1 β β3
2
8) 10 + 5β22
5
2 β β2=
52 β β2
οΏ½2 + β22 + β2
οΏ½ =5οΏ½2 + β2οΏ½
22 β οΏ½β2οΏ½2 =
10 + 5β24 β 2
=10 + 5β2
2
9) 12 + 3β313
3
4 β β3=
34 β β3
οΏ½4 + β34 + β3
οΏ½ =3οΏ½4 + β3οΏ½
42 β οΏ½β3οΏ½2 =
12 + 3β316 β 3
=12 + 3β3
13
10) 2π₯π₯ β 2β5π₯π₯2 β 5
2
π₯π₯ + β5=
2π₯π₯ + β5
οΏ½π₯π₯ β β5π₯π₯ β β5
οΏ½ =2οΏ½π₯π₯ β β5οΏ½
π₯π₯2 β οΏ½β5οΏ½2 =
2π₯π₯ β 2β5π₯π₯2 β 5
46
47
Linear Expressions and Equations
48
49
2.1 Simplifying Expressions
Some algebraic expressions can be reduced to a simpler form. Most of the time, the answers to a problem will be in the simplest form. Use the basic rules of algebra to combine alike terms and simplify. Example
Combine like terms when possible.
Starting Point: ππππππ β ππππππ
Group alike terms: ππ β ππ β ππππ β ππππ
Combine terms that can be combined: ππ β ππππ+ππ
Simplify exponent: ππππππ Example
Combine like terms.
Starting Point: ππππ + ππππππ + ππππππ + ππππ
Combine based on number in front: ππππππ + ππππ
50
Example
Cancel terms that are in both the numerator and denominator of a fraction!
Starting Point: ππππππππππ
ππππππππππ
Expand the expression: ππβππβππβππβππβππβππππβππβππβππβππβππβππβππβππ
Cancel out terms that appear on the top and bottom: ππβππβππβππβππβππβππππβππβππβππβππβππβππβππβππ
Write without the cancelled terms: ππππβππβππ
Final answer: ππππππππ
Example
Use the rules of exponents, radicals and fractions to simplify.
Starting Point: ππππβππ
Use rules of exponents: ππ οΏ½ πππππποΏ½
Multiply: ππππππ
51
Example
Factor and use cancellation when you come across polynomials! Refer to Section 5.2 to learn more about factoring.
Starting Point: ππππ+ππππ+ππππππβππ
Factor the numerator and denominator: (ππ+ππ)(ππ+ππ)(ππ+ππ)(ππβππ)
Cancel out terms that appear on the top and bottom: (ππ+ππ)(ππ+ππ)(ππ+ππ)(ππβππ)
Write without the cancelled terms: (ππ+ππ)(ππβππ)
Final answer: ππ+ππππβππ
Practice Problems
Simplify the following algebraic expressions.
1) π₯π₯ + 3π₯π₯ + 4π₯π₯ β 2π₯π₯
2) β7 + 3π₯π₯ + 7 β 5π₯π₯ + 1
3) 3π₯π₯2 β 3π₯π₯4
4) 2π₯π₯ β π₯π₯β2
5) π₯π₯ Γ· π₯π₯3
6) π₯π₯2 β π₯π₯1 + π₯π₯3
7) 4π₯π₯3 β 8π₯π₯3 + π₯π₯3
8) π₯π₯2
+ π₯π₯8
9) π₯π₯2π¦π¦7
2π₯π₯3π¦π¦4
52
Answers
1) 6π₯π₯ π₯π₯ + 3π₯π₯ + 4π₯π₯ β 2π₯π₯ = 4π₯π₯ + 4π₯π₯ β 2π₯π₯ = 8π₯π₯ β 2π₯π₯ = 6π₯π₯
2) 1 β 2π₯π₯ β7 + 3π₯π₯ + 7 β 5π₯π₯ + 1 = β7 + 7 + 1 β 5π₯π₯ + 3π₯π₯ = 1 β 5π₯π₯ + 3π₯π₯ = 1 β 2π₯π₯ 3) 9π₯π₯6 3π₯π₯2 β 3π₯π₯4 = 3 β 3 β π₯π₯2 β π₯π₯4 = 9 β π₯π₯2 β π₯π₯4 = 9 β π₯π₯6 = 9π₯π₯6 4) 2
π₯π₯ 2π₯π₯ β π₯π₯β2 = 2π₯π₯ β 1
π₯π₯2= 2π₯π₯
π₯π₯2= 2
π₯π₯
5) 1π₯π₯2
π₯π₯ Γ· π₯π₯3 = π₯π₯π₯π₯3
= 1π₯π₯2
6) 2π₯π₯3 π₯π₯2 β π₯π₯1 + π₯π₯3 = π₯π₯3 + π₯π₯3 = 2π₯π₯3 7) β3π₯π₯3 4π₯π₯3 β 8π₯π₯3 + π₯π₯3 = β4π₯π₯3 + π₯π₯3 = β3π₯π₯3 8) 5π₯π₯
8
π₯π₯2
+ π₯π₯8
= οΏ½44οΏ½ π₯π₯2
+ π₯π₯8
= 4π₯π₯8
+ π₯π₯8
= 4π₯π₯+π₯π₯8
= 5π₯π₯8
9) π¦π¦3
2π₯π₯
π₯π₯2π¦π¦7
2π₯π₯3π¦π¦4= π₯π₯βπ₯π₯βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦
2βπ₯π₯βπ₯π₯βπ₯π₯βπ¦π¦βπ¦π¦βπ¦π¦βπ¦π¦= π¦π¦βπ¦π¦βπ¦π¦
2βπ₯π₯= π¦π¦3
2π₯π₯
53
2.2 Solving Linear Equations
The overall goal when solving equations is to isolate the variable on one side of the equation and send all of the numbers to the other side. Rules
What you do to one side of the equation you must do to the other!
When you do something, it must be done to the entire side, not just one part.
To undo addition, subtract from both sides.
To undo subtraction, add to both sides.
To undo multiplication, divide on both sides.
To undo division, multiply on both sides.
Example
Consider the equation 3π₯π₯ + 7 = 1.
To solve for π₯π₯, begin by moving single integers to the right-hand side.
3π₯π₯ + 7 = 1 β7 β 7 3π₯π₯ = β6
Now, divide the remaining number away from the variable.
3π₯π₯ = β6 3 3
π₯π₯ = β2
Subtract 7 from each side since subtraction is the opposite of addition.
Divide both sides by 3 since division is the opposite of multiplication.
Final answer! Try plugging π₯π₯ = β2 back into the original equation!
54
Example
Consider the equation 2π₯π₯ β 3 β 2 = 3.
To solve for π₯π₯, begin by combining like terms.
2π₯π₯ β 3 β 2 = 3 2π₯π₯ β 5 = 3
Now proceed by moving single integers to the right-hand side.
2π₯π₯ β 5 = 3 +5 + 5
2π₯π₯ = 8
Now, divide the remaining number away from the variable.
2π₯π₯ = 8 2 2
π₯π₯ = 4
β3β 2 = β5
Divide both sides by 2 since division is the opposite of multiplication.
Final answer! Try plugging π₯π₯ = 4 back into the original equation!
Add 5 to both sides since addition is the opposite of subtraction (the negative 5).
55
Example
Consider the equation β7π₯π₯ + 1 + 2π₯π₯ = 9π₯π₯ β 8 + 1.
To solve for π₯π₯, begin by combining like terms.
β7π₯π₯ + 1 + 2π₯π₯ = 9π₯π₯ β 8 + 1 β5π₯π₯ + 1 = 9π₯π₯ β 7
Now proceed by moving single integers to the right-hand side.
β5π₯π₯ + 1 = 9π₯π₯ β 7 β1 β 1
β5π₯π₯ = 9π₯π₯ β 8
If we want the variable to be isolated on one side, we must now move all variables to the left-hand side.
β5π₯π₯ = 9π₯π₯ β 8 β9π₯π₯ β 9π₯π₯
β14π₯π₯ = β8
Now, divide the remaining number away from the variable.
β14π₯π₯ = β8 β14 β 14
π₯π₯ =47
Make sure like terms are lined up!
Divide both sides by β14 since division is the opposite of multiplication.
Final answer! Try plugging π₯π₯ = 4
7 back into the original equation!
56
Practice Problems
Solve the following equations for the variable.
1) 2π₯π₯ β 3 = 1
2) 4π₯π₯ + 6 = 7
3) 14π₯π₯ β 5 = 9
4) 7π₯π₯ + 8 = 1
5) 2π₯π₯ + 3 = 7π₯π₯ β 1
6) 5π₯π₯ + 2 = 3π₯π₯ β 6
7) π₯π₯ + 5 = π₯π₯ β 6
8) 12π₯π₯ + 3
2= 4
9) 34π₯π₯ + 2 = π₯π₯ β 5
10) 12π₯π₯ + 3
2(π₯π₯ + 1) β 1
4= 5
11) 5π₯π₯ + 2 = 13π₯π₯
12) 5(π₯π₯ + 2) = 3π₯π₯
13) 15π₯π₯ + 1
3= 3
14) 3π₯π₯ + 12
(π₯π₯ + 2) = β2
57
Answers
1) π₯π₯ = 2 2π₯π₯ β 3 = 1 β 2π₯π₯ = 4 β π₯π₯ = 2
2) π₯π₯ =14
4π₯π₯ + 6 = 7 β 4π₯π₯ = 1 β π₯π₯ =14
3) π₯π₯ = 1 14π₯π₯ β 5 = 9 β 14π₯π₯ = 14 β π₯π₯ = 1
4) π₯π₯ = β1 7π₯π₯ + 8 = 1 β 7π₯π₯ = β7 β π₯π₯ = β1
5) π₯π₯ =45
2π₯π₯ + 3 = 7π₯π₯ β 1 β 2π₯π₯ + 4 = 7π₯π₯ β 4 = 5π₯π₯ β45
= π₯π₯
6) π₯π₯ = β4 5π₯π₯ + 2 = 3π₯π₯ β 6 β 5π₯π₯ = 3π₯π₯ β 8 β 2π₯π₯ = β8 β π₯π₯ = β4
7) ππππ π π πππ π ππππππππππ π₯π₯ + 5 = π₯π₯ β 6 β 5 = β6 (ππππππ π π ππππππππ 5 β β6,ππππ π π πππ π ππππππππππ) 8) π₯π₯ = 5 1
2π₯π₯ +
32
= 4 β12π₯π₯ =
52β π₯π₯ = 5
9) π₯π₯ = 28 34π₯π₯ + 2 = π₯π₯ β 5 β 3
4π₯π₯ + 7 = π₯π₯ β 7 = 1
4π₯π₯ β 28 = π₯π₯
10) π₯π₯ =
158
12π₯π₯ + 3
2(π₯π₯ + 1) β 1
4= 5 β 1
2π₯π₯ + 3
2π₯π₯ + 3
2β 1
4= 5 β 2π₯π₯ + 5
4= 5 β 2π₯π₯ = 15
4β 2π₯π₯
= 154β π₯π₯ = 15
8
11) π₯π₯ = β37
5π₯π₯ + 2 = 13π₯π₯ β 14
3π₯π₯ + 2 = 0 β 14
3π₯π₯ = β2 β 14π₯π₯ = β6 β π₯π₯ = β3
7
12) π₯π₯ = β5 5(π₯π₯ + 2) = 3π₯π₯ β 5π₯π₯ + 10 = 3π₯π₯ β 10 = β2π₯π₯ β β5 = π₯π₯
13) π₯π₯ =403
15π₯π₯ + 1
3= 3 β 1
5π₯π₯ = 8
3β π₯π₯ = 40
3
14) π₯π₯ = β67
3π₯π₯ + 12(π₯π₯ + 2) = β2 β 3π₯π₯ + 1
2π₯π₯ + 1 = β2 β 7
2π₯π₯ + 1 = β2 β 7
2π₯π₯ = β3 β π₯π₯ = β6
7
58
59
2.3 Solving Inequalities
Solving inequalities is almost exactly the same as solving equations! The equal sign = has simply been replaced with one of the following: < β€ > β₯. However, there is one extra rule to solving inequalities. If you ever divide or multiply by a negative number in your solving process, you must switch the direction the inequality sign is facing.
Example
Consider the equation 3π₯π₯ + 7 β₯ 1.
To solve for π₯π₯, begin by moving single integers to the right-hand side.
3π₯π₯ + 7 β₯ 1 β7 β 7 3π₯π₯ β₯ β6
Now, divide the remaining number away from the variable.
3π₯π₯ β₯ β6 3 3
π₯π₯ β₯ β2
Subtract 7 from each side since subtraction is the opposite of addition.
You do not have to switch the direction of the β₯ sign when subtracting. Only for multiplying or dividing by a negative number!
The β₯ sign does not switch directions since the 3 is positive.
This is your final answer! It means that any number greater than or equal to β2 will satisfy the original inequality.
Try plugging π₯π₯ = β2 back into the original inequality! Notice that the inequality holds (1 β₯ 1).
Now try plugging in other numbers greater than β2! The inequality will hold!
60
Example
Consider the equation β5π₯π₯ β 8 β₯ 2.
To solve for π₯π₯, begin by moving single integers to the right-hand side.
β5π₯π₯ β 8 β₯ 2 +8 + 8 β5π₯π₯ β₯ 10
Now, divide the remaining number
β5π₯π₯ β₯ 10 β5 β 5
π₯π₯ β€ β2
Example
If your final answer does has > or < involved (instead of β₯ or β€), this excludes the number on the other side of the sign.
For example, π₯π₯ > 5 means the answer is any number greater than 5, but not including ππ.
Add 8 to both sides since addition is the opposite of subtraction.
The β₯ sign MUST change direction now, since we divided by a negative number!
This is your final answer! It means that any number less than or equal to β2 will satisfy the original inequality.
Try plugging π₯π₯ = β2 back into the original inequality! Notice that the inequality holds (2 β₯ 2).
Now try plugging in other numbers less than β2! The inequality will hold!
61
Example
Solve the following inequalities.
1) π₯π₯ β 3 β₯ 12
2) 4π π < β16
3) 5π€π€ + 2 β€ β48
4) 2ππ > 7
5) 2π¦π¦ + 4 β€ π¦π¦ + 8
6) β3ππ β₯ 9
7) β3(π₯π₯ β 6) > 2π₯π₯ β 2
8) β2ππ + 8 β€ 2ππ
62
Answers
1) π₯π₯ β₯ 15 π₯π₯ β 3 β₯ 12 β π₯π₯ β₯ 15
2) π π < β4 4π π < β16 β π π < β4 3) π€π€ β€ β10 5π€π€ + 2 β€ β48 β 5π€π€ β€ β50 β π€π€ β€ β10 4) ππ >
72
2ππ > 7 β ππ >72
5) π¦π¦ β€ 4 2π¦π¦ + 4 β€ π¦π¦ + 8 β 2π¦π¦ β€ π¦π¦ + 4 β π¦π¦ β€ 4
6) ππ β€ β3 β3ππ β₯ 9 β ππ β€ β3 7) π₯π₯ < 4 β3(π₯π₯ β 6) > 2π₯π₯ β 2 β β3π₯π₯ + 18 > 2π₯π₯ β 2 β 18 > 5π₯π₯ β 2 β 20 > 5π₯π₯ β 4 > π₯π₯
8) ππ β₯ 2 β2ππ + 8 β€ 2ππ β 8 β€ 4ππ β 2 β€ ππ
63
β 2.4 Solving Systems of Equations
Sometimes there are two or more equations that are related because their variables are the same. This means that if you find that π₯π₯ = 4 in one equation, the π₯π₯ in the other equation is also guaranteed to be 4. Most often you will have to create two equations from a word problem and then solve them as a system!
Example
The following is a system of equations:
2π₯π₯ + π¦π¦ = 5 ππππππ π₯π₯ β 2π¦π¦ = 15
The solution to this system is π₯π₯ = 5 and π¦π¦ = β5. Try plugging these values into each equation! They work for both!
Example
Consider the system: 2π₯π₯ + π¦π¦ = 5 ππππππ π₯π₯ β 2π¦π¦ = 15
To solve this system, we must isolate one of the variables. In this case, it looks easy to isolate the π₯π₯ in the second equation, so weβll start with that. (You could just as easily choose to isolate the π¦π¦ variable in the first equation- it doesnβt make a difference).
π₯π₯ β 2π¦π¦ = 15
π₯π₯ = 2π¦π¦ + 15
(Continued on next page)
Use algebra here and isolate the variable. For more on solving equations, refer to Section 4.2.
64
Example Continued
Notice, π₯π₯ and 2π¦π¦ + 15 are exactly equal to one another. This means we can replace any π₯π₯ we see with 2π¦π¦ + 15.
We will replace the π₯π₯ in the first equation with 2π¦π¦ + 15.
2π₯π₯ + π¦π¦ = 5
2(2π¦π¦ + 15) + π¦π¦ = 5
4π¦π¦ + 30 + π¦π¦ = 5
5π¦π¦ + 30 = 5
Now that we have substituted that expression in for π₯π₯, our first equation has only one variable to solve for! So now we solve for π¦π¦.
5π¦π¦ + 30 = 5
5π¦π¦ = β25
π¦π¦ = β5
So π¦π¦ = β5! Now we can take that information and plug π¦π¦ = β5 in to either of our beginning equations in order to find π₯π₯. It does not matter which equation we choose because they are both related!
π₯π₯ β 2π¦π¦ = 15
π₯π₯ β 2(β5) = 15
π₯π₯ + 10 = 15
π₯π₯ = 5
So our final answer is π₯π₯ = 5, π¦π¦ = β5.
65
Example
Solve the following systems of equations.
1) 2π₯π₯ + π¦π¦ = 5 ππππππ π₯π₯ β 2π¦π¦ = 15
2) π¦π¦ = 2π₯π₯ + 4 ππππππ π¦π¦ = β3π₯π₯ + 9
3) π¦π¦ = π₯π₯ + 3 ππππππ 42 = 4π₯π₯ + 2π¦π¦
4) 3π₯π₯ + π¦π¦ = 12 ππππππ π¦π¦ = β2π₯π₯ + 10
5) π¦π¦ = β3π₯π₯ + 5 ππππππ 5π₯π₯ β 4π¦π¦ = β3
6) π¦π¦ = β2 ππππππ 4π₯π₯ β 3π¦π¦ = 18
7) β7π₯π₯ β 2π¦π¦ = β13 ππππππ π₯π₯ β 2π¦π¦ = 11
8) 6π₯π₯ β 3π¦π¦ = 5 ππππππ π¦π¦ β 2π₯π₯ = 8
66
Answers
1) π₯π₯ = 5, π¦π¦ = β5
2) π₯π₯ = 1, π¦π¦ = 6
3) π₯π₯ = 6, π¦π¦ = 9
4) π₯π₯ =
25
,π¦π¦ =545
5) π₯π₯ = 1, π¦π¦ = 2
6) π₯π₯ = 3, π¦π¦ = β2
7) π₯π₯ = 3, π¦π¦ = β4
8) ππππ π π πππ π ππππππππππ
67
2.5 Graphing Linear Equations
There are two methods to graph a line: 1) Use the π₯π₯ and π¦π¦ intercepts: If you have the coordinates of the π₯π₯ and π¦π¦ intercepts, plot those two points and connect them with a line.
2) Use slope-intercept form: If you have an equation in slope intercept form (π¦π¦ = πππ₯π₯ + ππ), plot the π¦π¦-intercept and then construct the line using the slope. If the equation is not in slope-intercept form, it can be rearranged to be in slope-intercept form.
3
2
ππ =32
68
Method 1
Consider the equation 6π₯π₯ + 2π¦π¦ = 12.
Find the π₯π₯-intercept:
To find the π₯π₯-intercept, set π¦π¦ equal to zero and solve for π₯π₯.
6π₯π₯ + 2π¦π¦ = 12
6π₯π₯ + 2(0) = 12
6π₯π₯ = 12
π₯π₯ = 2
Find the π¦π¦-intercept:
To find the π¦π¦-intercept, set π₯π₯ equal to zero and solve for π¦π¦.
6π₯π₯ + 2π¦π¦ = 12
6(0) + 2π¦π¦ = 12
2π¦π¦ = 12
π¦π¦ = 6
Now, plot the points (2,0) and (0,6) on the graph and connect them with a line:
This is the π₯π₯-intercept, meaning the point (2,0) is on
the graph.
This is the π¦π¦-intercept, meaning the point (0,6) is on
the graph.
69
Method 2
Consider the equation 6π₯π₯ + 2π¦π¦ = 12.
First, use algebra to rearrange the equation:
6π₯π₯ + 2π¦π¦ = 12
2π¦π¦ = β6π₯π₯ + 12
π¦π¦ = β3π₯π₯ + 6
Then identify your slope and y-intercept:
Compare π¦π¦ = β3π₯π₯ + 6 with π¦π¦ = πππ₯π₯ + ππ.
In this case, ππ = β3 and ππ = 6. In other words, the slope is β3 and the π¦π¦-intercept is at (0,6).
Plot the y-intercept:
Use the slope to draw the rest of the line:
Now the equation is in π¦π¦-intercept form (π¦π¦ = πππ₯π₯ + ππ).
3
1
The slope is β3 = β31
= ππππππππππππππ
The slope is rewritten as a fraction to reveal the rise (numerator) and run (denominator). The rise is the amount the slope changes on the π¦π¦-axis and the
run is the amount the slope changes on the π₯π₯-axis.
70
Graphs with one variable
If a graph only involves one variable, it can be graphed as a horizontal or vertical line. Below are two examples.
Equation Form Resulting Graph π¦π¦ = 5 A horizontal line that crosses
the y-axis at (0,5)
π₯π₯ = 5 A vertical line that crosses the
x-axis at (5,0)
71
Practice Problems
Graph the following equations.
1) π¦π¦ = 4π₯π₯ + 1
2) π¦π¦ = 12π₯π₯ + 3
3) π¦π¦ = β3π₯π₯ + 2
4) π¦π¦ = βπ₯π₯ + 4
5) π¦π¦ = 23π₯π₯ β 2
6) π¦π¦ = π₯π₯
7) 3π¦π¦ β 12 = β6π₯π₯
8) π₯π₯ + 6π¦π¦ = 7
9) π₯π₯ = 4
10) π¦π¦ = β2.5
72
Answers
1)
6)
2)
7)
3)
8)
4)
9)
5)
10)
73
2.6 Building Lines
You may be asked to state the equation of a line based on the points it passes through or its slope. You may also be asked to write the equation of a line parallel or perpendicular to another line. If you donβt know how to graph linear equations, see Section 2.5.
Parallel Lines Lines are parallel if they have the same slope. Below are two examples of parallel lines. The red lines have different π¦π¦-intercepts than the blue lines, but they have matching slopes. In order to construct parallel lines, just find two lines with the same slope.
Example
Find a line parallel to π¦π¦ = 2π₯π₯ + 3.
Any line that has the same slope (ππ = 2) is a correct answer! For instance π¦π¦ = 2π₯π₯ + 4 and π¦π¦ = 2π₯π₯ β 24.
Example
Find a line parallel to π¦π¦ = 57π₯π₯.
Any line that has the same slope (ππ = 57) is a correct answer! For instance
π¦π¦ = 57π₯π₯ + 2 and π¦π¦ = 5
7π₯π₯ β 10.
74
Perpendicular Lines Lines are perpendicular if they intersect at a 90Β° angle. Below is an example of a pair of perpendicular lines. The slopes of perpendicular lines are the negative reciprocals of one another.
Example
Find a line perpendicular to π¦π¦ = 2π₯π₯ + 3.
To find the perpendicular slope, take the negative reciprocal of ππ = 2. The reciprocal of 2 is 1
2. Then make it negative. Therefore, any equation with the
slope ππ = β12 is perpendicular to our original equation. For instance,
π¦π¦ = β12π₯π₯ + 3 and π¦π¦ = β1
2π₯π₯ β 1.
Example
Find a line perpendicular to π¦π¦ = β54π₯π₯ + 6.
To find the perpendicular slope, take the negative reciprocal of ππ = β54.
The reciprocal of β54 is β4
5. Then negate it, causing it to turn positive.
(Continued on the next page)
75
Example Continued
Therefore, any equation with the slope ππ = 45 is perpendicular to our
original equation. For instance, π¦π¦ = 45π₯π₯ + 1
2 and π¦π¦ = 4
5π₯π₯ β 3.
Line Equations When given characteristics of a line, you can build the corresponding equation. When given a π¦π¦-intercept and a slope, use the equation ππ = ππππ + ππ. Plug in your slope for ππ and your π¦π¦-intercept for ππ.
When given a point (π₯π₯1,π¦π¦1) and a slope ππ, use the equation ππ β ππππ = ππ(ππ β ππππ). Then just algebraically rearrange your result into the form π¦π¦ = πππ₯π₯ + ππ.
When given only two points, (π₯π₯1, π¦π¦1) and (π₯π₯2,π¦π¦2), first plug these points into the slope formula ππ = ππππβππππ
ππππβππππ to find your ππ. Then just use π¦π¦ β π¦π¦1 = ππ(π₯π₯ β π₯π₯1) like in
the previous case.
Example
Find the equation of the line which passes through the points (1,2) and (4,9).
First we find the slope using ππ = π¦π¦2βπ¦π¦1π₯π₯2βπ₯π₯1
. Label the points (π₯π₯1,π¦π¦1) and (π₯π₯2,π¦π¦2). In this case π₯π₯1 = 1, π¦π¦1 = 2, π₯π₯2 = 4, and π¦π¦2 = 9. So ππ = 9β2
4β1 = 73.
Now choose either point, and plug it in to π¦π¦ β π¦π¦1 = ππ(π₯π₯ β π₯π₯1). Weβll use the point (1,2) here. π¦π¦ β π¦π¦1 = ππ(π₯π₯ β π₯π₯1)
π¦π¦ β 2 = 73(π₯π₯ β 1)
π¦π¦ β 2 = 73π₯π₯ β 7
3
π¦π¦ = 73π₯π₯ β 1
3
Therefore the equation of the line that passes through the desired points is π¦π¦ = 7
3π₯π₯ β13.
76
Practice Problems
1) Find the equation of a line parallel to π¦π¦ = 3π₯π₯ + 7.
2) Find the equation of a line parallel to π¦π¦ = 12π₯π₯ β 2.
3) Find the equation of a line parallel to π¦π¦ = β32π₯π₯ + 9.
4) Find the equation of a line perpendicular to π¦π¦ = 12π₯π₯ + 1.
5) Find the equation of a line perpendicular to π¦π¦ = 5π₯π₯ + 12.
6) Find the equation of a line perpendicular to π¦π¦ = β27π₯π₯ β 8.
7) Find the equation of the line that has a slope of 2 and intercepts the π¦π¦-axis at
π¦π¦ = 7.
8) Find the equation of the line that has a slope of 45 and intercepts the π¦π¦-axis at
π¦π¦ = β3.
9) Find the equation of the line which passes through the points (4,3) and
(2,2).
10) Find the equation of the line which passes through the points (β4,5) and
(0,0).
77
Answers
1) π¦π¦ = 3π₯π₯ Β± πππππ¦π¦ππβππππππ
2) π¦π¦ =12π₯π₯ Β± πππππ¦π¦ππβππππππ
3) π¦π¦ = β32π₯π₯ Β± πππππ¦π¦ππβππππππ
4) π¦π¦ = β2π₯π₯ Β± πππππ¦π¦ππβππππππ
5) π¦π¦ = β15π₯π₯ Β± πππππ¦π¦ππβππππππ
6) π¦π¦ =72π₯π₯ Β± πππππ¦π¦ππβππππππ
7) π¦π¦ = 2π₯π₯+7
8) π¦π¦ =45π₯π₯ β 3
9) π¦π¦ =12π₯π₯ + 1
10) π¦π¦ = β
54π₯π₯
78
79
Logarithms
80
81
3.1 Exponential Form vs. Logarithmic Form
Logarithms (or logs for short) are the inverse of exponentiation. Consider the following example:
8 = 23 This equation shows that β8 is 2 to the power of 3.β Logs take the left side of the equation and the base of the exponent to identify the power. So in terms of logs, this equation also shows that βlog base 2 of 8 is 3.β This is written as follows:
log2 8 = 3
This relationship is useful for solving equations with a variable in the exponent, such as:
10π₯π₯ = 100 β
log10 100 = π₯π₯ β
2 = π₯π₯
Many logs can be evaluated by hand, but most are evaluated using a calculator. Techniques for evaluating logarithms will be covered in Section 3.2. For this section the focus is changing an equation from one form to the other. The general formula is:
ππππ = ππ β
logππ ππ = ππ Equations of these forms can always be interchanged.
Exponential Form
Logarithmic Form
82
Example
Write the exponential equation 25 = 32 in logarithmic form.
Original equation in exponential form
25 = 32
Identify ππ, ππ, and ππ in the exponential equation
25 = 32
Rearrange into logarithmic form (final answer)
log2 32 = 5
Example
Write the logarithmic equation log9 81 = 2 in exponential form.
Original equation in logarithmic form
log9 81 = 2
Identify ππ, ππ, and ππ in the logarithmic equation
log9 81 = 2
Rearrange into exponential form (final answer)
92 = 81
The next few sections cover more properties of logarithms.
ππ ππ ππ
ππ ππ ππ
83
Practice Problems
Write the following exponential equations as logarithmic equations.
1) 102 = 100 2) 103 = 1000 3) 54 = 625 4) 21 = 2 5) 26 = 64 6) π₯π₯2 = 10 7) π€π€π¦π¦ = π₯π₯
Write the following logarithmic equations as exponential equations.
8) log7 49 = 2
9) log2 8 = 3
10) log5 5 = 1
11) log4 1 = 0
12) log3 27 = 3
13) log10 1000 = π₯π₯
14) logπ₯π₯ π¦π¦ = π₯π₯
84
Answers
1) log10 100 = 2
2) log10 1000 = 3
3) log5 625 = 4
4) log2 2 = 1
5) log2 64 = 6
6) logπ₯π₯ 10 = 2
7) logπ€π€ π₯π₯ = π¦π¦
8) 72 = 49
9) 23 = 8
10) 51 = 5
11) 40 = 1
12) 33 = 27
13) 10π₯π₯ = 1000
14) π₯π₯π§π§ = π¦π¦
85
3.2 Rules of Logarithms
Logarithms are derived from exponents, so all logarithmic rules are based on the rules of exponents. Rules
π₯π₯π₯π₯π₯π₯ππ(ππππ) = π₯π₯π₯π₯π₯π₯ππ(ππ) + π₯π₯π₯π₯π₯π₯ππ(ππ) log2(15) = log2(3 β 5) = log2(3) + log2(5)
π₯π₯π₯π₯π₯π₯ππ οΏ½πππποΏ½
= π₯π₯π₯π₯π₯π₯ππ(ππ) β π₯π₯π₯π₯π₯π₯ππ(ππ) log2(5) = log2 οΏ½102οΏ½ = log2(10) β log2(2)
π₯π₯π₯π₯π₯π₯ππ(ππππ) =π§π§ β π₯π₯π₯π₯π₯π₯ππ(ππ) log2(9) = log2(32) = 2log2(3)
π₯π₯π₯π₯π₯π₯ππ(ππ) =π₯π₯π₯π₯π₯π₯ππππ(ππ)π₯π₯π₯π₯π₯π₯ππππ(ππ) log3(5) =
log10(5)log10(3)
(ππβπππ π πππππππππππ π ππ πππ π πππ π πππππππ π π€π€βππππ πππ π ππππππ ππ πππππ π πππππ π ππππππππ) π₯π₯π₯π₯π₯π₯ππ(ππ) =ππ log6(6) = 1
π₯π₯π₯π₯π₯π₯ππ(ππ) = ππ log4(1) = 0 Notation
π₯π₯π₯π₯π₯π₯ππππ(ππ) = π₯π₯π₯π₯π₯π₯ππ ππππππππππ ππ πππππ π ππ ππππ 10 πππ π π£π£πππππ¦π¦ ππππππππππππ, log10 πππ π ππππππππππ π€π€ππππππππππππ πππ π πππππ π ππ log
π₯π₯π₯π₯π₯π₯ππ(ππ) = π₯π₯π§π§ ππ ππβππ ππππππππππππ ππ (β 2.71828) πππ π πππ π π π ππ ππ π£π£πππππ¦π¦ ππππππππππππ πππππ π ππ, π π ππ
logππ βπππ π πππππ π πππ€π€ππ ππππππππππππππππ πππ π ln
Note On Negatives
For logππ(π₯π₯), π₯π₯ cannot be negative. For example, log10(β4) has no solution. Example
Expand the logarithmic expression log5 π₯π₯π¦π¦2.
Original expression log5 π₯π₯π¦π¦2 Use multiplication rule log5 π₯π₯ + log5 π¦π¦2
Use power rule log5 π₯π₯ + 2log5 π¦π¦ (logarithm is now fully expanded)
Change of base formula
86
Example
Condense the logarithmic expression log2 π₯π₯ β log2 π¦π¦ + log2 π₯π₯.
Original expression log2 π₯π₯ β log2 π¦π¦ + log2 π₯π₯ Rearrange log2 π₯π₯ + log2 π₯π₯ β log2 π¦π¦
Use multiplication rule log2 π₯π₯π₯π₯ β log2 π¦π¦ Use division rule log2
π₯π₯π₯π₯π¦π¦
(logarithm is now fully condensed) Example
Condense the logarithmic expression 3log11 ππ + 2log11 ππ + 12
log11 π π β log11 ππ.
Original expression 3log11 ππ + 2log11 ππ + 12log11 π π β log11 ππ
Use power rule log11 ππ3 + log11 ππ2 + log11 π π 12 β log11 ππ
Rewrite fractional power as root
log11 ππ3 + log11 ππ2 + log11 βπ π β log11 ππ
Use multiplication rule log11 ππ3ππ2βπ π β log11 ππ Use division rule
log11ππ3ππ2βπ π
ππ
(logarithm is now fully condensed) Example
Expand the logarithmic expression log8βππ3 βππβ
.
Original expression log8
βππ3 β ππβ
Use multiplication/division rules log8 βππ3 + log8 ππ β log8 β
Rewrite root as fractional power log8 ππ13 + log8 ππ β log8 β
Use power rule 13log8 ππ + log8 ππ β log8 β
(logarithm is now fully expanded)
87
Example Evaluate the logarithmic expression log2 8. Simplify down to a single number.
Original expression log2 8 Rewrite 8 as 23 log2(23)
Use power rule 3 log2 2
log2 2 = 1 3 β 1 Final answer 3
Example Evaluate the logarithmic expression ln ππ. Simplify down to a single number.
Original expression ln ππ Rewrite ln ππ as logππ ππ logππ ππ
logππ ππ = 1, final answer 1
Example Evaluate the logarithmic expression log10 1. Simplify down to a single number.
Original expression log10 1 Rewrite 1 as 100 log10 100
Use power rule 0 β log10 10
log10 10 = 1 0 β 1 Final answer 0
88
Practice Problems
Expand the logarithmic expressions.
1) log2ππβππππ
2) log1010π₯π₯2
3) ln(π₯π₯3π¦π¦) 4) log 7ππ
ππ
Condense the logarithmic expressions.
5) 5log9 π₯π₯ + log9 π¦π¦ 6) logππ + logππ + 2log ππ β logππ 7) βlog11 π€π€ + 2log11 π₯π₯ β log11 π¦π¦ + 1
2log11 π₯π₯
8) ln π₯π₯ + lnπ¦π¦ β 2ln π₯π₯
Simplify the logarithmic expression down to a single number.
9) 3log4 2 β log4 8 10) log3 3 + log8 8 + log 10 11) ln ππ6 + ln ππ
12) log 100
13) log15(β3)
14) ln 2 + ln 6 β ln 12
Use the change of base formula to rewrite the expression (donβt simplify).
15) log7 5
16) log2 ππ
89
Answers
1) log2 ππ + log2 ππ β log2 π π
2) log10 10 β 2log10 π₯π₯
3) 3logππ π₯π₯ + logππ π¦π¦
4) log 7 + log ππ β log ππ
5) log9 (π₯π₯5π¦π¦)
6) log ππππππ2
ππ
7) log11 π₯π₯2βπ§π§π€π€π¦π¦
8) ln π¦π¦π₯π₯
9) 0
10) 3
11) 7
12) 2
13) ππππ π π πππ π ππππππππππ
14) 0
15) log10(5)log10(7)
16) log10(ππ)log10(2)
90
91
3.3 Solving Logarithmic Equations
Using the rules of logarithms and the ability to rearrange logarithmic and exponential equations, you can solve for variables that might not have been easy to solve for before. Practice Problems
Solve the following logarithmic expressions.
1) logπ₯π₯ 144 = 2 2) logπ₯π₯ 5 = 1
4
3) log9 π₯π₯ = 2 4) log4 π₯π₯ = 3 5) log25 π₯π₯ = β1
2
6) log 110π₯π₯ = β2
7) logπ₯π₯ 49 = 2 8) logπ₯π₯ 5 = 1
3
9) logπ₯π₯ 32 = 52
10) log2 π₯π₯ = 3 11) log10 π₯π₯ = 3 12) log16 π₯π₯ = 1 13) 3 logπ₯π₯ 3 = 1
14) 4 logπ₯π₯ 32 = 5
92
Answers
1) π₯π₯ = 12
2) π₯π₯ = 625
3) π₯π₯ = 81
4) π₯π₯ = 64
5) π₯π₯ = 15
6) π₯π₯ = 100
7) π₯π₯ = 7
8) π₯π₯ = 125
9) π₯π₯ = 4
10) π₯π₯ = 8
11) π₯π₯ = 1000
12) π₯π₯ = 16
13) π₯π₯ = 27
14) π₯π₯ = 16
93
3.4 Common Base Technique
When two exponential expressions with the same base are equal to one another, their exponents can also be set equal to one another. Consider the following example:
2π₯π₯ = 25
Notice how the only possible value for π₯π₯ is 5. Setting the exponents equal to one another, we get π₯π₯ = 5. We are only allowed to set the exponents equal to each other when the bases are the same (in this case, the base on each side of the equation was 2). Now consider another example:
52π₯π₯β3 = 5π₯π₯+5
Since the bases are the same (5), we can set the exponents equal to one another:
2π₯π₯ β 3 = π₯π₯ + 5 β
π₯π₯ = 8
Practice Problems
Solve the equations for π₯π₯:
1) 310 = 3π₯π₯ 2) 116π₯π₯ = 1136 3) 22π₯π₯+9 = 23π₯π₯β9 4) π¦π¦5 = π¦π¦2π₯π₯π¦π¦1
5) 91
9π₯π₯= 9π₯π₯
6) οΏ½ππ
13οΏ½
9
ππβ1= ππ4π₯π₯
ππ12
7) 711π₯π₯β5 = 49π₯π₯
74
94
Answers
1) π₯π₯ = 10
2) π₯π₯ = 6
3) π₯π₯ = 18
4) π₯π₯ = 2
5) π₯π₯ =12
6) π₯π₯ = 4
7) π₯π₯ =19
95
3.5 Exponential Models
Word problems modelling growth or decay using exponents may be on the test. Logarithms can be used to solve these equations Investment Growth
We can use logarithms to determine the growth and decay of certain models. Investment growth is a perfect example when dealing with exponential growth.
Letβs say that $1000 is deposited in an account that pays 10% interest annually, compounded monthly. This means that the interest rate each month is
10%12 ππππππππβπ π
= .83% πππππππππππππ π ππ ππππππ ππππππππβ
At the end of the month, . 83% of the amount deposited in the account is added to the account. Letβs look at some conditions we have specified:
Initial amount: $1000
Time period: 1 month
Monthly growth rate: . 83%
Monthly growth factor: 1 + .83 = 1.83
For the monthly growth factor, 1 is given by the total amount in the account added by the percent of that total (83% = .83).
The amount of money in the account at any given month π₯π₯ can be modeled by the exponential function
π΄π΄(π₯π₯) = 1000(1.83)π₯π₯
In general, if the annual interest rate ππ expressed as a decimal, is compounded ππ amount of times each year then in each period the interest rate is ππ Γ· ππ . In ππ years, there are ππππ time periods with an initial value of ππ deposited into the account. This leads to the formula:
Compound Interest: π΄π΄(ππ) = ππ(1 + ππππ
)ππππ
96
Lets look at our values again:
ππ = $1000
ππ = .83 (decimal percent)
ππ = 1 ππππππππβ
If we substitute these values into the general compound interest formula we get:
π΄π΄(ππ) = 1000(1.83)ππ
This equation is identical to the first one we derived with a replacement of ππ for π₯π₯.
If money is invested and compounded annually, use the following formula:
π΄π΄(ππ) = ππ(1 + ππ)ππ
Example
Majid invests $5000 in a high yield bond that pays 12% annual interest, compounded every 2 months.
1) Find a formula for the amount of money in the account after ππ amount of years. 2) What is the amount of money in the account after 5 years? 3) How long will it take for Majidβs investment to grow to $15,000?
Solution:
1) We use the compound interest formula and plug in our new conditions:
Initial value: ππ = $5000
Interest rate: ππ = 12% = .12 as a decimal
Compound monthly: ππ = 2 months
π΄π΄(ππ) = ππ(1 +ππππ)ππππ
π΄π΄(ππ) = 5000(1 +. 12
2 )2ππ
π΄π΄(ππ) = 5000(1.06)2ππ
97
2) To find the amount in the account after 5 years we replace ππ with 5 in the formula we just derived
π΄π΄(ππ) = 5000(1.06)2(5)
π΄π΄(ππ) = $8954
So in 5 years the balance in the account will sit at $8954.
3) We want to know what time (or value of ππ) the amount of money will reach a value of $15,000. So we need to find the value ππ.
π΄π΄(ππ) = 5000(1.06)2ππ
15,000 = 5000(1.06)2ππ
3 = 1.062ππ
log(3) = log( (1.06)2ππ)
log (3) = (2ππ) log (1.06)
ππ =log (3)
2log (1.06)
ππ = 9.42 π¦π¦πππππππ π
Decay
Sometimes youβll use the formula for exponential decay (using the same variables as before):
π΄π΄(ππ) = ππ(1 β ππ)ππ
A great example of modeling decay is with radioactive material. The amount of radioactive material present at time ππ is given by:
π π (ππ) = π π 0ππππππ ππ < 0
π π 0 πππ π ππβππ πππππππππππππππ π ππππππππππππ ππππ πππππππππππππππ π ππππππ ππ πππππππππππ π πππππππ π ππβππ πππππππππ¦π¦ ππππππππ
98
Example
The half-life of Carbon-12 is 5730 years. How much Carbon-12 is left in an object that has been decaying for 3750 years if the original amount of Carbon-12 is 30,000 atoms?
The half-life of an atom is the time it takes for half of the amount of an atom to decay. Using this value for ππ we can solve for ππ and then use that value along with other initial conditions to find the amount of atoms left after ππ = 3750 years. First, solve for ππ:
π π (ππ) = π π 0ππππππ
1 = 2ππ5730ππ
12 = ππ5730ππ
ln οΏ½12οΏ½
= 5730ππ
ππ =ln οΏ½1
2οΏ½5730
ππ = β.00012
We now use our ππ value to find the amount of atoms left after 3750 years if we initially start with 30,000 atoms.
π π (ππ) = π π 0ππππππ
π π (3750) = 30,000ππβ.00012(3750)
π π = 30,000ππβ.45363
π π = 30,000(. 6353)
π π = 19,059
We conclude that there are 19,059 atoms of Carbon-12 after 3750 years of decay.
If we start with 2 atoms of carbon-12, then we can see that after 5730 years, half of 2 would be our current amount. Thatβs why weβre plugging in 2 and 1. We could use any initial value and come to the same result.
99
Population Growth
We can also see models grow exponentially. Letβs take the example of bacteria. In an idealized environment, the amount of bacteria can be modeled by the equation:
ππ(ππ) = ππππππππ
ππ represents the initial amount of bacteria present and ππ is the growth factor.
Example
Suppose that at the start of an experiment in a lab, 1200 bacteria are introduced into an environment. After some time, a scientist found the amount of bacteria to be 1500. Knowing that the growth rate is 14.46 for this bacteria, what amount of time had passed (in hours)?
Given our initial conditions of
ππ(ππ) = 1500
ππ = 1200
ππ = 14.46
Letβs solve for ππ:
ππ(ππ) = ππππππππ
1500 = 1200ππ14.46ππ
15001200 = ππ14.46ππ
ln οΏ½1512οΏ½
= 14.46ππ
ππ =ln (1.25)
14.46
ππ = .015 βπππππππ π
100
Practice Problems
Set up the following equations. You do not need to solve them.
1) Sam invests $4025 in a high yield bond that pays 5% annual interest, compounded every 2 months.
2) Suppose that at the start of an experiment in a lab, 7700 bacteria are introduced into an environment. After some time, a scientist found the amount of bacteria to be 8000. The growth rate is 42.30 for this bacteria.
3) $10,000 is invested at 7% interest compounded annually. 4) The value of a new $40,000 car decreases by 15% per year.
101
Answers
1) π΄π΄(ππ) = 4025(0.025)2ππ 2) 8000 = 7700ππ42.30ππ
3) π΄π΄(ππ) = 10,000(1.07)ππ
4) π΄π΄(ππ) = 40,000(0.85)ππ
102
103
Functions
104
105
4.1 Introduction, Domain, and Range
A function is a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output. Think of it as taking one number as an initial condition and producing another number as a result. You might feel that a lot of the following information doesnβt seem new. Functions are very similar to equations, but they have slightly different notation and some stricter rules. Functions always have one output for every input. Example
Consider the equation π¦π¦ = π₯π₯ + 1. Here, π₯π₯ is an independent variable. If we choose a number to replace π₯π₯, we get a corresponding π¦π¦ value. So we call the π¦π¦ variable dependent. The value of π¦π¦ will depend on what you choose for your value of π₯π₯. The relationship between input and output can be written as an ordered pair (π₯π₯,π¦π¦):
Ordered pairs π¦π¦ = π₯π₯ + 1 = ? Input Output (0,1) π¦π¦ = 0 + 1 = 1 π₯π₯ = 0 π¦π¦ = 1 (1,2) π¦π¦ = 1 + 1 = 2 π₯π₯ = 1 π¦π¦ = 2 (2,3) π¦π¦ = 2 + 1 = 3 π₯π₯ = 2 π¦π¦ = 3
Domain and Range The domain of a function is the set of values for the independent variable. These input values for π₯π₯ must produce a real number for the output of π¦π¦. Sometimes the domain is specified explicitly and other times you have to find the domain from the given function. The range of a function is the set of π¦π¦ values- the output that is produced by a domain. Ask yourself what the smallest and largest values of π¦π¦ are that will be produced from all the values of π₯π₯. The function on the right has π₯π₯-values between π₯π₯ = 0 and π₯π₯ = 4 and π¦π¦-values between π¦π¦ = 0 and π¦π¦ = 2. So the domain of the function is 0 to 4 and the range is 0 to 2. The notation for domain and range will be explained on the next page.
β
106
Notation for Domain/Range
Set notation is used for domain and range. Parentheses are used to note a set of values in this case. Since the graph to the left uses every single π₯π₯ value between 0 and 4 (including 0 and 4), the notation would be [0,4]. This notation says that every value between 0 and 4 is used, including 0 and 4. If the domain did not include 0 and 4, the notation would be (0,4). Similarly, the range for this graph is [0,2].
Example
What is the domain of π¦π¦ = 2π₯π₯ + 3? (Think of all the values for π₯π₯ that would result in a real number π¦π¦) Thinking through it (or looking at a graph), we can determine that all real numbers substituted in for π₯π₯ will produce a real number for π¦π¦. Therefore the domain of the function π¦π¦ = 2π₯π₯ + 3 is (ββ,β). In this case, all numbers work, so the set of possible values stretches from negative infinity to infinity. Whenever an infinity sign is used, these parentheses ( ) are used instead of [ ]. This is because infinity is a concept, not a number- so it cannot be included in the set of values. All linear equations (equations of the form π¦π¦ = πππ₯π₯ + ππ) have a domain and range of (ββ,β). Example
What is the domain of π¦π¦ = βπ₯π₯ + 15? (Think of all the values for π₯π₯ that would result in a real number π¦π¦) Some values of π₯π₯ would result in an imaginary number (a negative number under the square root, which is not allowed) for π¦π¦, so those π₯π₯-values cannot be included in the domain. For example, π₯π₯ = β16 would produce the π¦π¦-value π¦π¦ = ββ1 which is not a real number, so β16 is not in the domain. The square root of any number less than 0 does not exist. Therefore any value for π₯π₯ that will produce a negative output under the radical will not be part of the domain for this function. (Continued on the next page)
107
Example Continued
So we need to think of the smallest possible π₯π₯-value that produces a zero under the square root. We could solve for this by equating 0 = π₯π₯ + 15 and solving for π₯π₯:
π₯π₯ + 15 = 0
β
π₯π₯ = β15
If we plug β15 in for π₯π₯, the value of π¦π¦ = ββ15 + 15 = β0 = 0. Thus β15 is the smallest value of π₯π₯ that will produce a real number for π¦π¦. We can also see that any positive value for π₯π₯ will yield a real number for π¦π¦; Therefore the domain of π¦π¦ = βπ₯π₯ + 15 is [β15,β). This is reflected on the graph:
Example
Letβs look at the Domain for the two examples: π¦π¦ = 2π₯π₯ + 3 D: (ββ,β) π¦π¦ = βπ₯π₯ + 15 D: [β15,β)
The ranges for each equation are not bounded as the π₯π₯-values increase towards β. So the range for each will approach β. For the first equation, the range will also approach ββ as the π₯π₯-values go toward ββ. But for the second equation, there is no way for π¦π¦ to be a negative number because of the radical, so the range is positive. The resulting ranges are:
π¦π¦ = 2π₯π₯ + 3 R: (ββ,β) π¦π¦ = βπ₯π₯ + 15 R: [0,β)
108
Example
What is the domain of π¦π¦ = 1π₯π₯β2
?
(Think of all the values for π₯π₯ that would result in a real number π¦π¦)
In the case of fractions, the most important rule to remember is that dividing by zero is not allowed. So whatever values of π₯π₯ make the denominator equal to zero must be excluded from the domain. In this case, we can see that π₯π₯ = 2 would cause the denominator to become zero.
π₯π₯ β 2
β
2 β 2
β
0
So the domain is all real numbers except 2.
This domain is notated as (ββ, 2) βͺ (2,β). The ββͺβ symbol in between stands for βunion.β This means we are uniting the set of values (ββ, 2) and the set of values (2,β) into one large set of values. Since the open brackets ( ) are used around the 2, that value is not included. It is interpreted as βall values from negative infinity up to two, and all values after two to infinity,β which was what we concluded from finding the domain.
On the test, you are more likely to see domain written in a form such as βall real numbers except for 2,β which is the same as (ββ, 2) βͺ (2,β).
109
Practice Problems
Find the domain and range for 1-5. Then just find the domain for 6-10.
1) π¦π¦ = 5π₯π₯ + 2
2) π¦π¦ = 12π₯π₯ β 6
3) π¦π¦ = 7π₯π₯
4) π¦π¦ = βπ₯π₯
5) π¦π¦ = βπ₯π₯ β 3
6) π¦π¦ = 1π₯π₯β4
7) π¦π¦ = 1π₯π₯
8) π¦π¦ = π₯π₯+3π₯π₯+7
9) π¦π¦ = π₯π₯2
10) π¦π¦ = |π₯π₯|
110
Answers
1) π·π·: (ββ,β)
π π : (ββ,β)
2) π·π·: (ββ,β)
π π : (ββ,β)
3) π·π·: (ββ,β)
π π : (ββ,β)
4) π·π·: [0,β)
π π : [0,β)
5) π·π·: [3,β)
π π : [0,β)
6) π·π·: (ββ, 4) βͺ (4,β) or all real numbers except for 4
7) π·π·: (ββ, 0) βͺ (0,β) or all real numbers except for 0
8) π·π·: (ββ,β7) βͺ (β7,β) or all real numbers except for β7
9) π·π·: (ββ,β) or all real numbers
10) π·π·: (ββ,β) or all real numbers
111
4.2 Evaluating Functions
Evaluating functions is just like substitution. First, function notation will be introduced. Function Notation
Instead of writing a function as π¦π¦ = π π ππππππππβππππππ, you will see a function expressed as ππ(π₯π₯) = π π ππππππππβππππππ. This is read as βf of x.β Other letters can be used other than ππ in order to distinguish different functions. This notation is meant to indicate that the function output is dependent upon the variable π₯π₯. Here are two examples of function notation:
ππ(π₯π₯) = 2π₯π₯ + 3 ππ(π₯π₯) = βπ₯π₯ + 15
We can also refer to these functions just by their letter name. The function ππ is the top one, ππ is the bottom one. Evaluating Functions
To evaluate a function, substitute the input for the functionβs variable. Consider the function ππ(π₯π₯) = 2π₯π₯ + 3. Below, ππ is evaluated at π₯π₯, 0, 1, 2, and β3.
ππ(π₯π₯) = 2π₯π₯ + 3 ππ(0) = 2(0) + 3 = 3 ππ(1) = 2(1) + 3 = 5 ππ(2) = 2(2) + 3 = 7
ππ(β3) = 2(β3) + 3 = β3 In all of the functions we have seen, the variable has been π₯π₯ and the input has been a number substituted in place of π₯π₯. This is not always the case. A function can be evaluated with any other variable or expression as input. Examples of this are on the next page.
β
112
Evaluating Functions Continued
Below ππ is evaluated at π¦π¦, π₯π₯ + 5, π₯π₯2, and π₯π₯ + β.
ππ(π¦π¦) = 2π¦π¦ + 3 ππ(π₯π₯ + 5) = 2(π₯π₯ + 5) + 3 = 2π₯π₯ + 13
ππ(π₯π₯2) = 2(π₯π₯2) + 3 = 2π₯π₯2 + 3 ππ(π₯π₯ + β) = 2(π₯π₯ + β) + 3 = 2π₯π₯ + 2β + 3
Example
Find ππ(14) for the function ππ(π€π€) = π€π€(π€π€ β 16).
ππ(14) = 14(14 β 16)
ππ(14) = 14(β2)
ππ(14) = β28
Example
Find ππ(5) for the function ππ(π₯π₯) = 3π₯π₯ + 12.
ππ(5) = 3(5) + 12
ππ(5) = 15 + 12
ππ(5) = 27
Example
Find β(2π€π€) for the function β(π₯π₯) = π₯π₯2.
ππ(2π€π€) = (2π€π€)2
ππ(2π€π€) = 4π€π€2
113
Practice Problems
Evaluate the following functions.
1) Find ππ(2) for the function ππ(π₯π₯) = π₯π₯2.
2) Find ππ(β5) for the function ππ(π₯π₯) = π₯π₯2.
3) Find πποΏ½12οΏ½ for the function ππ(π₯π₯) = 2π₯π₯ + 4.
4) Find ππ(β2) for the function ππ(π₯π₯) = 2π₯π₯ + 4.
5) Find β(β5) for the function β(π₯π₯) = βπ₯π₯ + 9.
6) Find ππ(3) for the function ππ(π₯π₯) = π₯π₯+2π₯π₯β2.
7) Find ππ(ππ) for the function ππ(π₯π₯) = 2π₯π₯ + 4.
8) Find ππ(2π₯π₯) for the function ππ(π₯π₯) = 2π₯π₯ + 4.
9) Find ππ(2π₯π₯) for the function ππ(π₯π₯) = π₯π₯2.
10) Find ππ(π₯π₯2) for the function ππ(π₯π₯) = π₯π₯2.
11) Find πποΏ½βπ π οΏ½ for the function ππ(π₯π₯) = π₯π₯2.
12) Find β(π₯π₯ β 3) for the function β(π₯π₯) = βπ₯π₯ + 9.
13) Find ππ(ππ2 + 1) for the function ππ(π₯π₯) = β9 β π₯π₯.
14) Find πποΏ½βπ₯π₯οΏ½ for the function ππ(π₯π₯) = π₯π₯+2π₯π₯β2
.
15) Find ππ(π₯π₯ + 1) for the function ππ(π₯π₯) = π₯π₯+2π₯π₯β2
.
114
Answers
1) ππ(2) = 4 π₯π₯2 = (2)2 = 4
2) ππ(β5) = 25 π₯π₯2 = (β5)2 = 25
3) ππ οΏ½12οΏ½ = 5 2π₯π₯ + 4 = 2 οΏ½1
2οΏ½ + 4 = 1 + 4 = 5
4) ππ(β2) = 0 2π₯π₯ + 4 = 2(β2) + 4 = β4 + 4 = 0
5) β(β5) = 2 βπ₯π₯ + 9 = οΏ½(β5) + 9 = β4 = 2
6) ππ(3) = 5 π₯π₯+2π₯π₯β2
= (3)+2(3)β2
= 51
= 5
7) ππ(ππ) = 2ππ + 4 2π₯π₯ + 4 = 2(ππ) + 4 = 2ππ + 4
8) ππ(2π₯π₯) = 4π₯π₯ + 4 2π₯π₯ + 4 = 2(2π₯π₯) + 4 = 4π₯π₯ + 4
9) ππ(2π₯π₯) = 4π₯π₯2 π₯π₯2 = (2π₯π₯)2 = 4π₯π₯2
10) ππ(π₯π₯2) = π₯π₯4 π₯π₯2 = (π₯π₯2)2 = π₯π₯4
11) πποΏ½βπ π οΏ½ = π π π₯π₯2 = οΏ½βπ π οΏ½2
= π π
12) β(π₯π₯ β 3) = βπ₯π₯ + 6 βπ₯π₯ + 9 = οΏ½(π₯π₯ β 3) + 9 = βπ₯π₯ + 6
13) ππ(ππ2 + 1) = οΏ½8 β ππ2 β9 β π₯π₯ = οΏ½9 β (ππ2 + 1) = οΏ½9 β ππ2 β 1 = οΏ½8 β ππ2
14) πποΏ½βπ₯π₯οΏ½ =
βπ₯π₯ + 2βπ₯π₯ β 2
π₯π₯+2π₯π₯β2
= οΏ½βπ₯π₯οΏ½+2οΏ½βπ₯π₯οΏ½β2
= βπ₯π₯+2βπ₯π₯β2
15) ππ(π₯π₯ + 1) =π₯π₯ + 3π₯π₯ β 1
π₯π₯+2π₯π₯β2
= (π₯π₯+1)+2(π₯π₯+1)β2
= π₯π₯+3π₯π₯β1
115
4.3 Composition of Functions and Inverses
The process of taking one function and using it as the input for another function is known as composition of functions. The notation uses the symbol " β " to show that composition is happening. Composition Notation
The composition of functions ππ and ππ is a function that is denoted by ππ β ππ and defined as
(ππ β ππ)(π₯π₯) = ππ(ππ(π₯π₯))
When the order is reversed, you get (ππ β ππ)(π₯π₯) = ππ(ππ(π₯π₯))
This works exactly like the substitution of an expression into a function from the last section (Section 4.2). Example
Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 2π₯π₯.
(ππ β ππ)(π₯π₯) β
πποΏ½ππ(π₯π₯)οΏ½ β
ππ(2π₯π₯) β
(2π₯π₯)2 β
4π₯π₯2 Visualization of the Example Above
ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 2π₯π₯ πποΏ½ππ(π₯π₯)οΏ½ = (2π₯π₯)2
The function ππ(π₯π₯) is inserted into the function ππ(π₯π₯).
116
If you are asked to evaluate a composite function, first simplify the composite function and then plug in the number. Example
Find (ππ β ππ)(2) when ππ(π₯π₯) = 2π₯π₯ + 3 and ππ(π₯π₯) = 10π₯π₯.
(ππ β ππ)(π₯π₯) β
πποΏ½ππ(π₯π₯)οΏ½ β
ππ(2π₯π₯ + 3) β
10(2π₯π₯ + 3) β
20π₯π₯ + 30 Now that (ππ β ππ)(π₯π₯) has been found, evaluate (ππ β ππ)(2)
(ππ β ππ)(2)
β 20π₯π₯ + 30
β 20(2) + 30
β 40 + 30
β 70
Visualization of the Example Above
ππ(π₯π₯) = 2π₯π₯ + 3 and ππ(π₯π₯) = 10π₯π₯ and π₯π₯ = 2 πποΏ½ππ(2)οΏ½ = 10(2(2) + 3)
In this case, 2 is inserted into the function ππ(π₯π₯), which is then inserted into the function ππ(π₯π₯).
117
Inverse Functions A function that undoes the action of a function ππ is called the inverse of ππ.
Basically, if two functions are inverses of one another, their inputs and outputs are switched. Graphically, this looks like a reflection across the diagonal π¦π¦ = π₯π₯ line.
In the graphs above, the green and purple functions are inverses of one another.
A function ππ is an inverse of the function ππ if the domain of ππ is equal to the range of ππ:
For every π₯π₯ in the domain of ππ and every π¦π¦ in the domain of ππ,
ππ(π¦π¦) = π₯π₯ ππππ ππππππ πππππ π π¦π¦ ππππ ππ(π₯π₯) = π¦π¦
The notation for an inverse function is given by ππβ1, therefore,
ππβ1(π¦π¦) = π₯π₯ ππππ ππππππ πππππ π π¦π¦ ππππ ππ(π₯π₯) = π¦π¦
*The notation ππβ1 does not mean 1ππ
118
Finding an Inverse Function In order to find the inverse of a function, simply switch the π₯π₯ and π¦π¦ values and then solve for π¦π¦. Example
Find the inverse function of ππ(π₯π₯) = 3π₯π₯ + 1.
Original expression ππ(π₯π₯) = 3π₯π₯ + 1 Replace ππ(π₯π₯) with π¦π¦ π¦π¦ = 3π₯π₯ + 1
Interchange the variables π₯π₯ and π¦π¦ π₯π₯ = 3π¦π¦ + 1
Solve for π¦π¦ π¦π¦ = 13π₯π₯ β 1
3
Final answer, replace π¦π¦ with ππβ1(π₯π₯) ππβ1(π₯π₯) = 13π₯π₯ β 1
3
Graphically, ππ(π₯π₯) = 3π₯π₯ + 1 and ππβ1(π₯π₯) = 13π₯π₯ β
13 are shown below.
ππ(π₯π₯) = 3π₯π₯ + 1
ππβ1(π₯π₯) = 13π₯π₯ β 1
3
119
Practice Problems
Evaluate the following.
1) Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 5π₯π₯.
2) Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 5π₯π₯.
3) Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 5π₯π₯.
4) Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = 1π₯π₯ and ππ(π₯π₯) = π₯π₯ + 21.
5) Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ and ππ(π₯π₯) = 2π₯π₯.
6) Find (ππ β ππ)(2) when ππ(π₯π₯) = 2π₯π₯ and ππ(π₯π₯) = 4π₯π₯2.
7) Find (ππ β ππ)(100) when ππ(π₯π₯) = βπ₯π₯ and ππ(π₯π₯) = π₯π₯2.
8) Find the inverse function of ππ(π₯π₯) = βπ₯π₯ + 2
9) Find the inverse function of ππ(π₯π₯) = 3π₯π₯ + 2
10) Find the inverse function of ππ(π₯π₯) = π₯π₯2
120
Answers
1) (ππ β ππ)(π₯π₯) = 25π₯π₯2
2) (ππ β ππ)(π₯π₯) = 5π₯π₯2
3) (ππ β ππ)(π₯π₯) = 25π₯π₯
4) (ππ β ππ)(π₯π₯) = 1π₯π₯+21
5) (ππ β ππ)(π₯π₯) = 4π₯π₯2 + 4π₯π₯
6) (ππ β ππ)(2) = 32
7) (ππ β ππ)(100) = 100
8) ππβ1(π₯π₯) = π₯π₯2 β 2
9) ππβ1(π₯π₯) = 13(π₯π₯ β 2)
10) ππβ1(π₯π₯) = 2π₯π₯
121
4.4 Transformations of Functions
There are ways we can manipulate a function to produce a graph similar to the shape of the original function. We will develop techniques for graphing these functions. These techniques are called transformations. Example
Letβs look at the function ππ(π₯π₯) = π₯π₯2:
We can see that it is centered at the origin (0,0). Now look at the graph of ππ(π₯π₯) = (π₯π₯ β 2)2 + 1:
The function ππ is a transformation of the function ππ.
β
122
We can see that the graph of ππ is identical in shape to that of ππ but has been shifted 2 units to the right and 1 unit up. This is because the values of β2 and 1 in the function ππ determined where the vertex was placed. Therefore:
With the function ππ(π₯π₯) = (π₯π₯ β 2)2 + 1, the value β2 played the role of shifting the function ππ(π₯π₯) = π₯π₯2 to the right 2 units and the value 1 played the role of raising the function ππ(π₯π₯) = π₯π₯2 up 1 unit. Some more examples are below:
Vertical Shifts
ππ(π₯π₯) = ππ(π₯π₯) + ππ, ππππ ππ > 0 π π πππππ π πππ π ππβππ ππππππππβ ππππ ππ πππ¦π¦ ππ πππππππππ π
ππ(π₯π₯) = ππ(π₯π₯) β ππ, ππππ ππ > 0 πΏπΏπππ€π€πππππ π ππβππ ππππππππβ ππππ ππ πππ¦π¦ ππ πππππππππ π
Horizontal Shifts
ππ(π₯π₯) = ππ(π₯π₯ + β), ππππ β > 0 ππβπππππππ π ππβππ ππππππππβ ππππ ππ ππππ ππβππ π π ππππππ β πππππππππ π
ππ(π₯π₯) = ππ(π₯π₯ β β), ππππ β > 0 ππβπππππππ π ππβππ ππππππππβ ππππ ππ ππππ ππβππ ππππππβππ β πππππππππ π
ππ(π₯π₯) = π₯π₯2 + 1 ππ(π₯π₯) = π₯π₯2 β 1
ππ(π₯π₯) = (π₯π₯ + 1)2 ππ(π₯π₯) = (π₯π₯ β 1)2
123
We can also compress and stretch graphs horizontally and vertically. Take a look at the functions:
ππ(π₯π₯) = 2π₯π₯2 ππ(π₯π₯) = (2π₯π₯)2 β(π₯π₯) = 12π₯π₯2 ππ(π₯π₯) = οΏ½12π₯π₯οΏ½
2
The functions ππ and ππ are only different because of the placement of the 2. Similarly, the functions β and ππ are only different because of the placement of the 12. To start off, look at the original parent function, ππ(π₯π₯), without transformations:
On the next page, the different ways this graph can be stretched and compressed will be shown. Compare what you see to this original graph. The differences can be subtle.
ππ(π₯π₯) = π₯π₯2
124
With the function ππ(π₯π₯) = 2π₯π₯2, the value 2 on the outside plays the role of stretching the function π₯π₯2 vertically by a factor of 2. With the function ππ(π₯π₯) = (2π₯π₯)2, the value 2 on the inside plays the role of compressing the function π₯π₯2 horizontally by a factor of 2. With the function β(π₯π₯) = 1
2π₯π₯2, the value 12 on the
outside plays the role of compressing the function π₯π₯2 vertically by a factor of 12. With the function ππ(π₯π₯) = οΏ½12π₯π₯οΏ½
2, the value 12 on the inside plays the role of stretching the function π₯π₯2 horizontally by a factor of 12.
ππ(π₯π₯) = 2π₯π₯2 ππ(π₯π₯) = (2π₯π₯)2
ππ(π₯π₯) = οΏ½12π₯π₯οΏ½
2
Compressing and Stretching Vertically
ππ(π₯π₯) = ππππ(π₯π₯), ππππ ππ > 0
Compressing and Stretching Horizontally
ππ(π₯π₯) = ππ(πππ₯π₯), ππππ ππ > 0
ππ stretches the graph of ππ if ππ > 1
ππ compresses the graph of ππ if 0 < ππ < 1
ππ stretches the graph of ππ if 0 < ππ < 1
ππ compresses the graph of ππ if ππ > 1
β(π₯π₯) = 12π₯π₯2
125
Look again at the function ππ(π₯π₯) = π₯π₯2.
Notice the graph has an upward oriented parabola. Now look at the graph ππ(π₯π₯) = βπ₯π₯2.
We see that this graph still represents a parabola. However, the shape is oriented downward. The difference between the two functions is the negative sign in front of the original function π₯π₯2. This caused a reflection across the π₯π₯-axis. If the negative had been inside the function like (βπ₯π₯)2, the reflection would happen over the π¦π¦-axis. In this case with (βπ₯π₯)2 the π¦π¦-axis reflection looks no different from π₯π₯2.
126
You can also transform a function by taking its absolute value as well. Remember the absolute value takes all negative values of an input and produces a positive output. Look at the graph of ππ(π₯π₯) = π₯π₯ below (left). For every negative value of π₯π₯ we get a negative output for ππ(π₯π₯). However, notice the difference with the absolute value of ππ(π₯π₯) shown below (right). We see that for every negative value of π₯π₯ we get a positive output of ππ(π₯π₯). So basically, taking the absolute value of a function makes all of its output positive.
Reflection about the π₯π₯-axis
ππ(π₯π₯) = βππ(π₯π₯) Multiply ππ(π₯π₯) by β1
Reflection about the π¦π¦-axis
ππ(π₯π₯) = ππ(βπ₯π₯) Replace π₯π₯ with βπ₯π₯
127
Practice Problems
Match the function on the left to its corresponding graph on the right.
1) ππ(π₯π₯) = π₯π₯2
2) ππ(π₯π₯) = π₯π₯2 β 1
3) ππ(π₯π₯) = π₯π₯2 + 4
4) ππ(π₯π₯) = (π₯π₯ β 2)2
5) ππ(π₯π₯) = (π₯π₯ + 2)2
6) ππ(π₯π₯) = 4π₯π₯2
7) ππ(π₯π₯) = 14π₯π₯
2
8) ππ(π₯π₯) = (π₯π₯ β 3)2 β 1
A)
B)
C)
D)
E)
F)
G)
H)
128
Answers
1) πΆπΆ
2) π΄π΄
3) πΉπΉ
4) π·π·
5) π΅π΅
6) π»π»
7) πΊπΊ
8) πΈπΈ
129
Quadratic Equations
130
131
5.1 Distribution and Foiling
When two quantities are multiplied, all of the pieces being multiplied must be taken into account. Distribution and foiling are the same thing, technically- but they are defined separately so that each process is easier to understand. Distribution Example
When multiplying two quantities with no pluses or minuses involved, distribution is simple. Multiply like terms together.
5 β (2π₯π₯) = 5 β 2 β π₯π₯ = 10π₯π₯
3π₯π₯ β (4π₯π₯) = 3 β 4 β π₯π₯ β π₯π₯ = 12π₯π₯2
π¦π¦ β (6π₯π₯) = 6 β π₯π₯ β π¦π¦ = 6π₯π₯π¦π¦
Distribution Example
When multiplying two quantities where one has a plus or minus involved, multiply the outer term to each term connected by the plus or minus.
5 β (2π₯π₯ + 1) = 5 β 2 β π₯π₯ + 5 β 1 = 10π₯π₯ + 5
3π₯π₯ β (4π₯π₯ + 2) = 3 β 4 β π₯π₯ β π₯π₯ + 3 β 2 β π₯π₯ = 12π₯π₯2 + 6π₯π₯
2 β (π₯π₯ β 2) = 2 β π₯π₯ β 2 β 2 = 2π₯π₯ β 4
β2 β (π₯π₯ β 2) = (β2) β π₯π₯ β (β2) β 2 = β2π₯π₯ β β4 = β2π₯π₯ + 4
There are no like terms to combine here, but the multiplication still happens.
The negative is carried through the distribution always.
132
Foiling Example
When multiplying two quantities which both have a plus or minus involved, multiply them in the following way
(π₯π₯ + 1)(π₯π₯ + 2) = π₯π₯2
First
(π₯π₯ + 1)(π₯π₯ + 2) = 2π₯π₯
Outside
(π₯π₯ + 1)(π₯π₯ + 2) = 1π₯π₯
Inside
(π₯π₯ + 1)(π₯π₯ + 2) = 2
Last
First, outside, inside, last is where the name Foiling comes from.
Foiling Example
This is the result of the above method.
(π₯π₯ + 1)(π₯π₯ + 2) = π₯π₯2 + 2π₯π₯ + 1π₯π₯ + 2 = π₯π₯2 + 3π₯π₯ + 2
Thus the distribution results in π₯π₯2 + 3π₯π₯ + 2
133
Practice Problems
Perform the following distributions.
1) 2(3π₯π₯) 2) 10(10π₯π₯) 3) π₯π₯(7π₯π₯) 4) 3π₯π₯(5π₯π₯2) 5) π₯π₯(π₯π₯π¦π¦) 6) 6π¦π¦π₯π₯3(2π₯π₯π¦π¦3) 7) βπ₯π₯(8π₯π₯) 8) β3(4π₯π₯9) 9) (π₯π₯ + 1)(π₯π₯ + 2) 10) (π₯π₯ + 1)(π₯π₯ + 5) 11) (π₯π₯ + 3)(π₯π₯ + 3) 12) (π₯π₯ + 2)(π₯π₯ β 9) 13) (π₯π₯ β 1)(π₯π₯ + 4) 14) (π₯π₯ + 2)(π₯π₯ β 2) 15) (π₯π₯ + 3)(π₯π₯ β 3) 16) (π₯π₯ + 4)(π₯π₯ β 4) 17) (π₯π₯ β 2)(π₯π₯ β 2) 18) (π₯π₯ β 5)(π₯π₯ β 6)
134
Answers
1) 6π₯π₯ 2(3π₯π₯) = 2 β 3 β π₯π₯ = 6 β π₯π₯ = 6π₯π₯
2) 100π₯π₯ 10(10π₯π₯) = 10 β 10 β π₯π₯ = 100 β π₯π₯ = 100π₯π₯ 3) 7π₯π₯2 π₯π₯(7π₯π₯) = 7 β π₯π₯ β π₯π₯ = 7 β π₯π₯2 = 7π₯π₯2 4) 15π₯π₯3 3π₯π₯(5π₯π₯2) = 3 β 5 β π₯π₯ β π₯π₯2 = 15 β π₯π₯3 = 15π₯π₯3 5) π₯π₯2π¦π¦ π₯π₯(π₯π₯π¦π¦) = π₯π₯ β π₯π₯ β π¦π¦ = π₯π₯2 β π¦π¦ = π₯π₯2π¦π¦ 6) 12π₯π₯π¦π¦4π₯π₯3 6π¦π¦π₯π₯3(2π₯π₯π¦π¦3) = 6 β 2 β π₯π₯ β π¦π¦ β π¦π¦3 β π₯π₯3 = 12 β π₯π₯ β π¦π¦4 β π₯π₯3 = 12π₯π₯π¦π¦4π₯π₯3 7) β8π₯π₯2 βπ₯π₯(8π₯π₯) = β8 β π₯π₯ β π₯π₯ = β8 β π₯π₯2 = β8π₯π₯2
8) β12π₯π₯9 β3(4π₯π₯9) = β3 β 4 β π₯π₯9 = β12 β π₯π₯9 = β12π₯π₯9 9) π₯π₯2 + 3π₯π₯ + 2 (π₯π₯ + 1)(π₯π₯ + 2) = π₯π₯ β π₯π₯ + 2 β π₯π₯ + 1 β π₯π₯ + 1 β 2 = π₯π₯2 + 2π₯π₯ + 1π₯π₯ + 2 = π₯π₯2 + 3π₯π₯ + 2 10) π₯π₯2 + 6π₯π₯ + 5 (π₯π₯ + 1)(π₯π₯ + 5) = π₯π₯ β π₯π₯ + 5 β π₯π₯ + 1 β π₯π₯ + 1 β 5 = π₯π₯2 + 5π₯π₯ + 1π₯π₯ + 5 = π₯π₯2 + 6π₯π₯ + 5
11) π₯π₯2 + 6π₯π₯ + 9 (π₯π₯ + 3)(π₯π₯ + 3) = π₯π₯ β π₯π₯ + 3 β π₯π₯ + 3 β π₯π₯ + 3 β 3 = π₯π₯2 + 3π₯π₯ + 3π₯π₯ + 9 = π₯π₯2 + 6π₯π₯ + 9
12) π₯π₯2 β 7π₯π₯ β 18 (π₯π₯ + 2)(π₯π₯ β 9) = π₯π₯ β π₯π₯ β 9 β π₯π₯ + 2 β π₯π₯ β 2 β 9 = π₯π₯2 β 9π₯π₯ + 2π₯π₯ β 18 = π₯π₯2 β 7π₯π₯ β 18
13) π₯π₯2 + 3π₯π₯ β 4 (π₯π₯ β 1)(π₯π₯ + 4) = π₯π₯ β π₯π₯ + 4 β π₯π₯ β 1 β π₯π₯ β 1 β 4 = π₯π₯2 + 4π₯π₯ β 1π₯π₯ β 4 = π₯π₯2 + 3π₯π₯ β 4
14) π₯π₯2 β 4 (π₯π₯ + 2)(π₯π₯ β 2) = π₯π₯ β π₯π₯ β 2 β π₯π₯ + 2 β π₯π₯ β 2 β 2 = π₯π₯2 β 2π₯π₯ + 2π₯π₯ β 4 = π₯π₯2 β 4
15) π₯π₯2 β 9 (π₯π₯ + 3)(π₯π₯ β 3) = π₯π₯ β π₯π₯ β 3 β π₯π₯ + 3 β π₯π₯ β 3 β 3 = π₯π₯2 β 3π₯π₯ + 3π₯π₯ β 9 = π₯π₯2 β 9
16) π₯π₯2 β 16 (π₯π₯ + 4)(π₯π₯ β 4) = π₯π₯ β π₯π₯ β 4 β π₯π₯ + 4 β π₯π₯ β 4 β 4 = π₯π₯2 β 4π₯π₯ + 4π₯π₯ β 16 = π₯π₯2 β 16
17) π₯π₯2 β 4π₯π₯ β 4 (π₯π₯ β 2)(π₯π₯ β 2) = π₯π₯ β π₯π₯ β 2 β π₯π₯ β 2 β π₯π₯ + 2 β 2 = π₯π₯2 β 2π₯π₯ β 2π₯π₯ + 4 = π₯π₯2 β 4π₯π₯ β 4
18) π₯π₯2 β 11π₯π₯ + 30 (π₯π₯ β 5)(π₯π₯ β 6) = π₯π₯ β π₯π₯ β 6 β π₯π₯ β 5 β π₯π₯ + 5 β 6 = π₯π₯2 β 6π₯π₯ β 5π₯π₯ + 30 = π₯π₯2 β 11π₯π₯ + 30
135
5.2 Factoring Expressions
Factoring is the opposite of foiling. Foiling distributes two quantities into one another, factoring breaks a quantity up into its smaller pieces.
Factoring Example
When factoring an expression, take note what each piece of the expression has in common.
5π₯π₯4 + 10π₯π₯3 + 15π₯π₯2
= 5 β π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯ + 5 β 2 β π₯π₯ β π₯π₯ β π₯π₯ + 5 β 3 β π₯π₯ β π₯π₯
= 5 β π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯ + 5 β 2 β π₯π₯ β π₯π₯ β π₯π₯ + 5 β 3 β π₯π₯ β π₯π₯
Now, rewrite the expression using parentheses. On the left side of the parentheses, we will write the numbers and variables each piece has in common. Inside the parentheses we will write the leftovers of each piece.
= 5 β π₯π₯2(π₯π₯ β π₯π₯ + 2 β π₯π₯ + 3)
= 5π₯π₯2(π₯π₯2 + 2π₯π₯ + 3)
The quadratic leftover in the parentheses cannot be simplified further, so we are done.
In conclusion, the factored form of 5π₯π₯4 + 10π₯π₯3 + 15π₯π₯2 is 5π₯π₯2(π₯π₯2 + 2π₯π₯ + 3).
Another factoring example is given on the next page.
Identify what each has in common. In this example, each part of the expression has a 5 and an π₯π₯2 in common.
136
Quadratic Factoring Example
When factoring a quadratic, you can guess and check, or you can use the following process. This process starts by filling out an X diagram that looks like this:
Consider the quadratic equation 2π₯π₯2 β 5π₯π₯ β 3.
Step 1:
Take the first and last coefficients and multiply them together. Place the result in the top part of the X. Then take the middle coefficient and place it in the bottom part of the X.
2π₯π₯2 β 5π₯π₯ β 3
Step 2:
Now, find two numbers that multiply together to make the top number, and add together to make the bottom number. In this case, we need two numbers which multiply to make β6 and add to make β5.
The two numbers that meet those conditions are β6 and 1. These numbers are placed in the sides of the X. It doesnβt matter which side you put them on.
β6 β 1 = β6 ππππππ β 6 + 1 = β5
(Continued on next page)
X
X
2 β β3 = β6
β5 β5
β6
X β6
β5
β6 1
137
Quadratic Factoring Example Continued
Step 3:
Now that the X diagram is filled out, we use that information to rewrite our quadratic equation as follows.
2π₯π₯2 β 5π₯π₯ β 3 Original
2π₯π₯2 β 6π₯π₯ + 1π₯π₯ β 3 New
Notice, the β6 and 1 come from the X diagram.
Step 4:
Our fourth step is to draw a line down the center of the equation and factor each side separately.
2π₯π₯2 β 6π₯π₯ + 1π₯π₯ β 3
Step 5:
Our final step is to write our factors! One of the factors will be the expression inside the parentheses. Notice how they are both the same! So one of our factors is (π₯π₯ β 3). The other factor is made up of the remaining pieces. In this case, the 2π₯π₯ and the positive 1. So our other factor is (2π₯π₯ + 1).
In conclusion, the factored form of 2π₯π₯2 β 5π₯π₯ β 3 is (π₯π₯ β 3)(2π₯π₯ + 1).
β6
β5
β6 1 X
2π₯π₯2 β 6π₯π₯ 1π₯π₯ β 3
2π₯π₯(π₯π₯ β 3) 1(π₯π₯ β 3)
Factor this side. Factor this side.
138
Practice Problems
Factor the following expressions.
1) 3π₯π₯ β 9
2) 5π₯π₯2 + 2π₯π₯
3) 2π₯π₯2 + 4π₯π₯
4) 10π₯π₯3 + 30π₯π₯2
5) π₯π₯2 + 3π₯π₯ + 2
6) π₯π₯2 + 5π₯π₯ + 6
7) π₯π₯2 + 6π₯π₯ + 5
8) π₯π₯2 + 8π₯π₯ + 16
9) π₯π₯2 + 6π₯π₯ + 9
10) π₯π₯2 + 11π₯π₯ + 30
11) π₯π₯2 β π₯π₯ β 2
12) π₯π₯2 β 8π₯π₯ + 12
13) π₯π₯2 β 13π₯π₯ β 30
14) 5π₯π₯2 β 9π₯π₯ β 2
15) 6π₯π₯2 + 13π₯π₯ + 6
16) π₯π₯2 β 4
17) π₯π₯2 β 9
18) π₯π₯2 β 1
19) π₯π₯2 β 16
Hint: When you see quadratics in this form, rewrite them with +0π₯π₯ in the middle.
π₯π₯2 β 4 = π₯π₯2 + 0π₯π₯ β 4
139
Answers
1) 3(π₯π₯ β 3)
2) π₯π₯(5π₯π₯ + 2)
3) 2π₯π₯(π₯π₯ + 2)
4) 10π₯π₯2(π₯π₯ + 3)
5) (π₯π₯ + 1)(π₯π₯ + 2)
6) (π₯π₯ + 2)(π₯π₯ + 3)
7) (π₯π₯ + 1)(π₯π₯ + 5)
8) (π₯π₯ + 4)(π₯π₯ + 4)
9) (π₯π₯ + 3)(π₯π₯ + 3)
10) (π₯π₯ + 5)(π₯π₯ + 6)
11) (π₯π₯ + 1)(π₯π₯ β 2)
12) (π₯π₯ β 2)(π₯π₯ β 6)
13) (π₯π₯ β 10)(π₯π₯ β 3)
14) (5π₯π₯ + 1)(π₯π₯ β 2)
15) (3π₯π₯ + 2)(2π₯π₯ + 3)
16) (π₯π₯ + 2)(π₯π₯ β 2)
17) (π₯π₯ + 3)(π₯π₯ β 3)
18) (π₯π₯ + 1)(π₯π₯ β 1)
19) (π₯π₯ + 4)(π₯π₯ β 4)
140
141
5.3 Solving Quadratic Equations
Solving quadratic equations requires factoring.
Example
First, use algebra to rearrange the quadratic equation so that it is equal to zero.
π₯π₯2 = β3π₯π₯ β 2
π₯π₯2 + 3π₯π₯ + 2 = 0
Next, factor the left side.
π₯π₯2 + 3π₯π₯ + 2 = 0
(π₯π₯ + 2)(π₯π₯ + 1) = 0
Now, set each of those expressions equal to zero and solve.
π₯π₯ + 2 = 0
π₯π₯ = β2
π₯π₯ + 1 = 0
π₯π₯ = β1
The solutions to the quadratic equation are π₯π₯ = β2 and π₯π₯ = β1. Try plugging π₯π₯ = β2 into the original equation. Then try plugging π₯π₯ = β1 into the original equation. Note you will get 0 = 0 each time, meaning both values for π₯π₯ satisfy the equation!
142
Practice Problems
Solve the following equations. (If you have already completed the factoring practice problems in Section 5.2, you can use that work to focus on the solving process here).
1) π₯π₯2 + 5π₯π₯ + 6 = 0
2) π₯π₯2 + 8π₯π₯ + 16 = 0
3) π₯π₯2 β π₯π₯ β 2 = 0
4) π₯π₯2 β 8π₯π₯ = β12
5) π₯π₯2 = 4
6) 5π₯π₯2 = 9π₯π₯ + 2
7) βπ₯π₯2 = 6π₯π₯ + 9
143
Answers
1) π₯π₯ = β2, π₯π₯ = β3 π₯π₯2 + 5π₯π₯ + 6 = 0 β (π₯π₯ + 2)(π₯π₯ + 3) = 0 β π₯π₯ + 2 = 0 ππππππ π₯π₯ + 3 = 0 2) π₯π₯ = β4 π₯π₯2 + 8π₯π₯ + 16 = 0 β (π₯π₯ + 4)(π₯π₯ + 4) = 0 β π₯π₯ + 4 = 0 ππππππ π₯π₯ + 4 = 0
3) π₯π₯ = β1, π₯π₯ = 2 π₯π₯2 β π₯π₯ β 2 = 0 β (π₯π₯ + 1)(π₯π₯ β 2) = 0 β π₯π₯ + 1 = 0 ππππππ π₯π₯ β 2 = 0
4) π₯π₯ = 2, π₯π₯ = 6 π₯π₯2 β 8π₯π₯ + 12 = 0 β (π₯π₯ β 2)(π₯π₯ β 6) = 0 β π₯π₯ β 2 = 0 ππππππ π₯π₯ β 6 = 0 5) π₯π₯ = β2, π₯π₯ = 2 π₯π₯2 β 4 = 0 β (π₯π₯ + 2)(π₯π₯ β 2) = 0 β π₯π₯ + 2 = 0 ππππππ π₯π₯ β 2 = 0
6) π₯π₯ = β15,π₯π₯ = 2 5π₯π₯2 β 9π₯π₯ β 2 = 0 β (5π₯π₯ + 1)(π₯π₯ β 2) = 0 β 5π₯π₯ + 1 = 0 ππππππ π₯π₯ β 2 = 0
7) π₯π₯ = β3 π₯π₯2 + 6π₯π₯ + 9 = 0 β (π₯π₯ + 3)(π₯π₯ + 3) = 0 β π₯π₯ + 3 = 0 ππππππ π₯π₯ + 3 = 0
144
145
Geometry Concepts
146
147
6.1 Two-Dimensional Shapes
Geometry word problems consist of relating two things: the information you are given and your knowledge of formulas. Often you will be asked to recall a formula from memory (such as area or perimeter) and use that formula to solve the problem. The following is a list of common 2D shapes and their formulas.
Terminology
Perimeter: The total distance around the edge of a shape. Finding the perimeter means adding up the lengths of all its sides.
Area: The size of a surface. Hypotenuse: The longest side of a right triangle.
Rectangles
Perimeter: ππ = πΏπΏ + πΏπΏ + ππ + ππ
= 2πΏπΏ + 2ππ
Area: π΄π΄ = πΏπΏ β ππ
Squares
Perimeter: ππ = π₯π₯ + π₯π₯ + π₯π₯ + π₯π₯
= 4π₯π₯ Area: π΄π΄ = π₯π₯ β π₯π₯
= π₯π₯2
β
148
Triangles
Perimeter: ππ = π΄π΄ + π΅π΅ + πΆπΆ
Area: π΄π΄ = 12β π΅π΅πππ π ππ β π»π»ππππππβππ
= 12π΅π΅ β π»π»
Right Triangles
Pythagorean Theorem: ππ2 = ππ2 + ππ2
Circles
Circumference: πΆπΆ = 2πππ π
Area: π΄π΄ = ππ π π 2
Diameter: π·π· = 2π π
149
Practice Problems
Solve the following problems.
1) Find the area of the shaded region.
2) Find the area of the shaded region.
3) Find the area of the shaded region.
4) If the width of this rectangle is 4 times the length, and the area is 36 units squared, what is the width?
150
5) If the hypotenuse of the triangle below is β2, what is the area of the triangle?
6) The area of the figure below is 144 units squared. What is the value of x?
7) If π₯π₯ = 2, find the Area and the Perimeter.
8) Given that the circumference of the circle is 8 ππ , what is the area of the shaded
region?
151
Answers
1) π΄π΄ = 12π₯π₯2
2) π΄π΄ = 4ππ2
3) π΄π΄ = πππ₯π₯2
4) π€π€ = 12
5) π΄π΄ = 12
6) π₯π₯ = 4
7) ππ = 28,π΄π΄ = 24
8) π΄π΄ = 64 β 16ππ
152
153
6.2 Three-Dimensional Shapes
The following is a list of common 3D shapes and their formulas. You are more likely to see cubes and spheres on the test.
Terminology
Surface Area: The total area of the surfaces of a three-dimensional object.
Volume: The amount of space inside of a solid figure.
Rectangular Prisms
Surface Area: πππ΄π΄ = 2(πΏπΏππ + πΏπΏπ»π» + π»π»ππ)
Volume: ππ = πΏπΏ β ππ β π»π»
Cubes
Surface Area: πππ΄π΄ = 6π₯π₯2
Volume: ππ = π₯π₯3
154
Triangular Prisms In prisms there are two triangles (front and back), two surfaces (right and left), and one base surface (bottom).
Surface Area: πππ΄π΄ = 2 β 1
2π΅π΅π»π» + 2πΏπΏππ + π΅π΅πΏπΏ
= π΅π΅π»π» + 2πΏπΏππ + π΅π΅πΏπΏ
Volume: ππ = π΄π΄ππππππ ππππ πππππππππππππ π ππ β π π ππππππππβ
= 12π΅π΅ β π»π» β πΏπΏ
Spheres
Surface Area: πππ΄π΄ = 4πππ π 2
Volume: ππ = 43πππ π 3
Practice Problems Find the surface area and volume of the following shapes. Use the information and diagram provided.
1) The length of the cube is 5 cm.
155
2) The smaller square of a Rubikβs cube has a length of 2 cm.
3) The following is a prism.
4) The soccer ball has a diameter of 9 inches. Find its surface area and volume.
156
Answers
1) πππ΄π΄ = 150 ππππ2,ππ = 125 ππππ3
2) πππ΄π΄ = 216 ππππ2,ππ = 216 ππππ3
3) πππ΄π΄ = 240 ππππ2,ππ = 144 ππππ3
4) πππ΄π΄ = 81ππ ππππ2,ππ = 121.5ππ ππππ3
157
6.3 Lines and Angles
When lines intersect each other, we can find the angles created by the two lines.
Angles and Parallel Lines
When two lines intersect they form two pairs of opposite angles.
In the image above, π΄π΄ and πΆπΆ are opposite angles and π·π· and π΅π΅ are opposite angles. Opposite angles are always congruent, meaning they are equal. If π΄π΄ is 49Β°, then πΆπΆ is 49Β°.
Two angles that are right next to one another are called adjacent angles.
In the image above, the angle π₯π₯π₯π₯π¦π¦ is the sum of angles π΄π΄ and π΅π΅. If π΄π΄ is 30Β° and π΅π΅ is 60Β°, then π₯π₯π₯π₯π¦π¦ is 90Β°.
158
Complementary Angles
Two angles are Complementary when they add up to 90Β° (a right angle ).
Supplementary Angles
Two Angles are Supplementary when they add up to 180Β° (a straight angle).
When there are two parallel lines with a third line intersecting them, eight angles are produced. The crossing line (in blue below) is called a transversal. An example is shown below.
159
The eight angles created by the transversal form four pairs of corresponding angles. Corresponding angles are congruent.
Angles 1 and 5 are corresponding angles. The pairs 3 and 7, 4 and 8, and 2 and 6 are also corresponding. Angles that are in the area between the parallel lines such as 3 and 5 are called interior angles. Angles that are on the outside of the parallel lines like 4 and 7 are called exterior angles. Angles on the opposite sides of the transversal like 4 and 5 are called alternate angles.
Example
Which angles in the diagram to the right measure 65Β°?
Angle 6 is given to be 65Β°.
Angles 6 and 8 are opposite angles and are therefore congruent. So angle 8 is 65Β°.
Angles 6 and 2 are corresponding angles and are therefore congruent. So angle 2 is 65Β°.
Angles 6 and 4 are alternate exterior angles and are therefore congruent. So angle 4 is 65Β°.
160
Distance and Midpoint Formula
On the next pages, two formulas will be given. The distance formula gives the distance between two points and the midpoint formula gives the coordinate point of the center of a line segment.
Distance Formula
Given two points: The distance, ππ, between those points is given by:
(π₯π₯1,π¦π¦1)
(π₯π₯2,π¦π¦2)
Example
Find the distance between the points (2, 5) and (β4, 1).
So here π₯π₯1 = 2, π¦π¦1 = 5, π₯π₯2 = β4, and π¦π¦2 = 1.
ππ = οΏ½(π₯π₯2 β π₯π₯1)2 + (π¦π¦2 β π¦π¦1)2
= οΏ½(β4β 2)2 + (1 β 5)2
= οΏ½(β6)2 + (β4)2
= β36 + 16
= β52
= 2β13
β 7.21
The distance between (2, 5) and (β4, 1) is 2β13 units (or about 7.21 units).
ππ = οΏ½(π₯π₯2 β π₯π₯1)2 + (π¦π¦2 β π¦π¦1)2
161
Midpoint Formula
Given two points: The midpoint, ππ, of those points is given by:
(π₯π₯1,π¦π¦1)
(π₯π₯2,π¦π¦2)
Example
Find the midpoint between the points (4, 1) and (10, 5).
So here π₯π₯1 = 4, π¦π¦1 = 1, π₯π₯2 = 10, and π¦π¦2 = 5.
ππ = οΏ½π₯π₯1 + π₯π₯2
2 ,π¦π¦1 + π¦π¦2
2 οΏ½
= οΏ½4 + 10
2 ,1 + 5
2 οΏ½
= οΏ½142 ,
62οΏ½
= (7, 3)
The point between (4, 1) and (10, 5) is (7, 3).
ππ = οΏ½π₯π₯1 + π₯π₯2
2 ,π¦π¦1 + π¦π¦2
2 οΏ½
This formula is given in the form of a coordinate point. Once all of the numbers are plugged in and it is simplified, you will be left with a point on a graph.
162
Practice Problems Find the distance between the following points.
1) (0,0) and (1,1)
2) (1,1) and (2,9)
3) (β1,β3)and (β9,β9)
4) (6,β1) and (β4,4)
Find the midpoint of the following points.
5) (1,1) and (2,9)
6) (0,3) and (5,2)
Use the diagram to the right.
7) If angle 8 is 40Β°, which other angles are 40Β°?
8) If angle 1 is 100Β°, which other angles are 100Β°?
9) If angle 8 is 40Β°, what is the measure of angle 1?
10) If angle 1 is 100Β°, what is the measure of angle 6?
163
Answers
1) β2
2) β50
3) 10
4) 5β5
5) οΏ½32, 5οΏ½
6) οΏ½52, 52οΏ½
7) π΄π΄πππππ π πππ π 2, 4, 6
8) π΄π΄πππππ π πππ π 3, 5, 7
9) 140Β°
10) 80Β°
164
165
Trigonometry
166
167
7.1 Geometric Trigonometry
Trigonometry is the study of triangles. Here are some important equations that help relate different angle units and triangle sides. Definitions
Both degrees and radians are ways to measure angles. Degrees are a common unit of measurement you have probably seen before. Radians are numerical values that use ππ. Radians are often used in higher level math because much of trigonometry involves relating an angle to a circle.
Conversions
Since degrees and radians both measure the same thing, you can convert from one unit of measurement to another.
Degrees β Radians
πΌπΌ is the starting angle in degrees.
πππππππ π ππ ππππ πππππππππππππ π = πΌπΌ βππ
180
Radians β Degrees
π½π½ is the starting angle in radians.
πππππππ π ππ ππππ πππππππππππππ π = π½π½ β180ππ
Example
Convert 30Β° to radians.
30 βππ
180 =30ππ180 =
ππ6
The angle 30Β° in radians is ππ6.
Example
Convert 2ππ3 to degrees.
2ππ3 β
180ππ =
360ππ3ππ =
3603 = 120
The angle 2ππ3 in degrees is 120Β°.
β
168
Use the following diagram for reference:
Pythagorean Theorem
The Pythagorean Theorem relates the three sides of a right triangle in terms of length. The labelling of π΄π΄ and π΅π΅ doesnβt matter, but πΆπΆ is always the hypotenuse. This only works for right triangles. It is as follows:
π΄π΄2 + π΅π΅2 = πΆπΆ2
Trig Functions
The values of the following trigonometric functions (or just trig functions) can be found as follows. The words on the right might help you remember the fractions!
π π ππππ(ππ) = ππππππππππππππππβπ¦π¦ππππππππππππππππ
= π΄π΄πΆπΆ
πππππ π (ππ) = ππππππππππππππππβπ¦π¦ππππππππππππππππ
= π΅π΅πΆπΆ
ππππππ(ππ) = ππππππππππππππππππππππππππππππππ
= π΄π΄π΅π΅
SOH
CAH
TOA
πΆπΆ π΄π΄
π΅π΅
ππ
169
Example
What is the length of the missing side on the figure to the right? Using the Pythagorean theorem, the missing side can be found:
π΄π΄2 + π΅π΅2 = πΆπΆ2
42 + π₯π₯2 = 52
16 + π₯π₯2 = 25
π₯π₯2 = 25 β 16
π₯π₯2 = 9
π₯π₯ = 3
Example
What is the value of ππππππ(ππ)?
ππππππ(ππ) = ππππππππππππππππππππππππππππππππ
= 2120
Opposite and adjacent are relative to where ππ is. In this case, 21 is directly opposite of ππ and 20 is adjacent to it. The hypotenuse is always the longest side of the triangle, opposite the right angle. This works the same for all of the trig functions. Each of them is simply equal to the ratio (fraction) of the sides. For more explanations and examples, ask a tutor for help!
5 4
π₯π₯
29 ππ
20
21
170
Sometimes youβll have to solve problems for triangles with no right angle.
Use the following diagram for reference:
The Law of Sines and Law of Cosines show relationships between the sides and angles of any triangle (including non-right triangles). Use whichever equations are helpful depending on what information is given!
The Law of Sines
π π ππππ(ππ)π΄π΄
=π π ππππ(ππ)π΅π΅ =
π π ππππ(ππ)πΆπΆ
The Law of Cosines
π΄π΄ = οΏ½π΅π΅2 + πΆπΆ2 β (2π΅π΅πΆπΆ)(πππππ π (ππ))
π΅π΅ = οΏ½π΄π΄2 + πΆπΆ2 β (2π΄π΄πΆπΆ)(πππππ π (ππ))
πΆπΆ = οΏ½π΄π΄2 + π΅π΅2 β (2π΄π΄π΅π΅)(πππππ π (ππ))
πΆπΆ
π΄π΄
π΅π΅ ππ
ππ
ππ
171
Example
Find the measure of side πΆπΆ.
With the given information, using the Law of Sines is the most appropriate. Side π΄π΄ and its opposite angle ππ are there, so we can apply the rule to find side πΆπΆ.
π π ππππ(ππ)π΄π΄
=π π ππππ(ππ)πΆπΆ
β
π π ππππ(50Β°)4
=π π ππππ(20Β°)
πΆπΆ
β
πΆπΆ βπ π ππππ(50Β°)
4= π π ππππ(20Β°)
β
πΆπΆ =4π π ππππ(20Β°)π π ππππ(50Β°)
Then, by using a calculator,
πΆπΆ = 1.9
So the length of side πΆπΆ is 1.9.
The AAF Accuplacer has a calculator built in for you to use if necessary.
π΅π΅
πΆπΆ
4
50Β°
20Β°
172
Example
Find the length of side π΅π΅.
Notice that the Pythagorean Theorem cannot be used here because there is no right angle. With the given information, using the Law of Cosines is the most appropriate. Side π΄π΄, side πΆπΆ, and angle ππ are there, so we can apply the rule to find side π΅π΅.
π΅π΅ = οΏ½π΄π΄2 + πΆπΆ2 β (2π΄π΄πΆπΆ)(πππππ π (ππ))
β
π΅π΅ = οΏ½32 + 22 β (2 β 3 β 2)(πππππ π (100Β°))
β
π΅π΅ = οΏ½9 + 4 β (12)(πππππ π (100Β°))
β
π΅π΅ = οΏ½13β 12πππππ π (100Β°)
Then, by using a calculator,
π΅π΅ = 15.04
So the length of side π΅π΅ is 15.04.
The AAF Accuplacer has a calculator built in for you to use if necessary.
πΆπΆ = 2
π΄π΄ = 3 100Β°
π΅π΅
173
Practice Problems
Convert the following.
1) Convert 90Β° to radians. 2) Convert 180Β° to radians. 3) Convert 225Β° to radians. 4) Convert ππ3 to degrees. 5) Convert 5ππ6 to degrees. 6) Convert 2ππ to degrees.
Use the triangle to the right for reference.
7) If π΅π΅ = 3 and πΆπΆ = 5, what is the length of π΄π΄? 8) If π΄π΄ = 5 and π΅π΅ = 12, what is the length of πΆπΆ? 9) If π΄π΄ = 28, π΅π΅ = 96, and πΆπΆ = 100, what is π π ππππ(ππ)? 10) If π΄π΄ = 28, π΅π΅ = 96, and πΆπΆ = 100, what is πππππ π (ππ)? 11) If π΄π΄ = 28, π΅π΅ = 96, and πΆπΆ = 100, what is ππππππ(ππ)?
Use the triangle to the right for reference. Not to scale. Round to one decimal place.
12) If πΆπΆ = 49, ππ = 17Β°, and ππ = 115Β°, what is π΅π΅? 13) If ππ = 29Β°, ππ = 118Β°, and π΅π΅ = 11, what is π΄π΄? 14) If π΄π΄ = 14, π΅π΅ = 10, and ππ = 44Β°, what is πΆπΆ? 15) If π΄π΄ = 52, π΅π΅ = 16, and ππ = 115Β°, what is πΆπΆ?
174
Answers
1) ππ
2
2) ππ
3) 5ππ4
4) 60Β°
5) 150Β°
6) 360Β°
7) π΄π΄ = 4
8) πΆπΆ = 13
9) π π ππππ(ππ) = 725
10) πππππ π (ππ) = 2425
11) ππππππ(ππ) = 724
12) π΅π΅ = 15.8
13) π΄π΄ = 20.0
14) πΆπΆ = 9.7
15) πΆπΆ = 60.5
175
Word Problems
176
177
8.1 Setting up an equation
An important skill needed on the test is translating sentences into equations. In this section you will practice translating from a phrase into a mathematical expression. Sometimes this can be straightforward and sometimes it can be very challenging. There is no single perfect strategy to translate, so as you go forward think of these questions:
-Can you explain why the answer you chose is correct?
-If you plugged a number into the expression you chose, would that expression manipulate the number in the way you expect it to?
-Are there any multiple choice answers you can eliminate right away? (Ones that donβt make sense or are trying to mislead you).
Addition (+)
When trying to determine if the + operation should be used in an expression, look for key words or phrases like βsumβ, βplusβ, βtotalβ, βandβ, βtogetherβ, βadditionalβ, βadded toβ, βcombined withβ, βmore thanβ, βincreased byβ.
Subtraction (β)
When trying to determine if the β operation should be used in an expression, look for key words or phrases like βminusβ, βreduceβ, βdifferenceβ, βleftβ, βchangeβ, βlessβ, βfewerβ, βdecreased byβ, βless thanβ, βtake awayβ, βsubtract fromβ, βhow much lessβ, βhow much leftβ.
Multiplication (Γ)
When trying to determine if the Γ operation should be used in an expression, look for key words or phrases like βproductβ, βtimesβ, βtwiceβ, βdoubledβ, βtripledβ, βperβ, βsquaredβ, βcubedβ, βofβ, βmultiplied byβ.
Division (Γ·)
When trying to determine if the Γ· operation should be used in an expression, look for key words or phrases like βratioβ, βquotientβ, βaverageβ, βoverβ, βintoβ, βhalf ofβ, βa third ofβ, βsplit evenlyβ, βshared equallyβ, βdivided byβ.
178
Practice Problems
1) Which of the following expressions is twice as much as x?
A) π₯π₯ + 2 B) π₯π₯2 C) π₯π₯ β π₯π₯ D) 2π₯π₯
2) Which of the following expressions is 3 times as much as the sum of y and z?
A) 3 β π¦π¦ + π₯π₯ B) 3 + π¦π¦ + π₯π₯ C) π¦π¦ + π₯π₯ β 3 D) (π¦π¦ + π₯π₯) β 3
3) Which of the following expressions is one less than the sum of x and y?
A) (π₯π₯ + π¦π¦) β 1 B) π₯π₯ β π¦π¦ β 1 C) (π₯π₯ β π¦π¦) + 1 D) β1 + (π₯π₯ β π¦π¦)
4) Which expression represents 10 more than twice the value of x?
A) 2 + π₯π₯ + 10 B) 2π₯π₯ + 10 C) π₯π₯ + 20 D) 10π₯π₯ + 2
5) Which expression represents y minus the sum of x and z?
A) π¦π¦ β (π₯π₯ + π₯π₯) B) π¦π¦ + π₯π₯ + π₯π₯ C) π¦π¦(π₯π₯ β π₯π₯) D) π₯π₯ β π₯π₯ + π¦π¦
6) What is 2 more than the product of x and y?
A) (π₯π₯ + 2) β π¦π¦ B) 2 + π₯π₯ + π¦π¦ C) 2 + (π₯π₯ β π¦π¦) D) (π¦π¦ + 2) β π₯π₯
7) Which expression represents the difference between x and y, multiplied by z?
A) π₯π₯ + π¦π¦ + π₯π₯ B) (π₯π₯ β π¦π¦) β π₯π₯ C) π₯π₯ β (π₯π₯ + π¦π¦) D) π₯π₯ β π₯π₯ β π¦π¦
8) Which expression represents three times the quotient of x and y?
A) 3 β οΏ½π¦π¦
π₯π₯οΏ½
B) 3π₯π₯π¦π¦ C) 3 β π₯π₯2π¦π¦2 D) 3 β οΏ½π₯π₯
π¦π¦οΏ½
179
Answers
1) π·π·
2) π·π·
3) π΄π΄
4) π΅π΅
5) π΄π΄
6) πΆπΆ
7) π΅π΅
8) π·π·
180
181
8.2 Word Problem Practice
The following is word problem practice. A majority of the test will be interpreting a situation and using your math skills to solve for something. When you approach word problems, keep these guidelines in mind:
-What information is a distraction (unneeded)?
-What information is given? Can you list it out or should it be written as an equation?
-What does the word problem want you to solve for in the end? (Keep this in mind as you proceed).
-If you feel like you donβt have the tools to solve the problem given, is there a way to approximate it?
Practice Problems
Solve the following word problems.
1) Alex the farmer wants to fence off a rectangular area of 100 square feet on their farm and wants the side closest to their house to be 25 feet long. How many feet of fencing do they need to purchase to complete the project?
A) 29 feet B) 58 feet
C) 100 feet D) 75 feet
2) Erin makes π₯π₯π¦π¦ dollars selling lemonade, Glenn makes π₯π₯βπ¦π¦
π¦π¦ dollars as a cook,
and James only makes 23 as much as what Glenn makes. Which of the
following equations represents the total combined income of all three people?
A) 2π₯π₯β2π¦π¦3π¦π¦ B) 1 + 2π₯π₯β2π¦π¦
3π¦π¦ C) 1 + 6π₯π₯β3π¦π¦
3π¦π¦ D) 8π₯π₯β5π¦π¦3π¦π¦
β
182
3) If Lee can walk two miles in one hour, how many miles could they walk in four and a half hours?
A) 8 ΒΌ miles B) 8 Β½ miles
C) 9 ΒΌ miles D) 9 miles
4) If the price of an item is represented by ππ, how much is the item worth if it
is discounted by 23%? A) 0.23ππ B) 0.77ππ
C) ππ β 0.23 D) ππ + 0.77ππ
5) Micah drops a ball from a cliff that is 200 meters high. If the distance of
the ball down the cliff is represented by π·π· = 12ππππ
2, how long will it take the ball to travel 122.5 meters down the cliff? (Assume ππ = 9.8ππ π π 2οΏ½ )
A) 5 seconds B) 10 seconds
C) 12 seconds D) 15 seconds
6) If Pat is half the age of Tyler and Tyler is three times the age of Grace,
what is Patβs age in 1995 if Grace was born in 1983?
A) 12 years old B) 16 years old
C) 18 years old D) 36 years old
7) The area of a right triangle is given by the equation π΄π΄ = 1
2ππβ, where ππ represents the base and β represents the height. If a right triangle has an area of 10 and a height of 5, what is the length of its base?
A) 5 B) 2
C) 4 D) 25
183
8) In the beginning of the month, Morgan owes Payton, John, and Susan
$3.50 each. If Morgan receives a paycheck of $250.00 and repays Payton, John, and Susan, how much will she have left afterwards?
A) $239.50 B) $241.50
C) $246.00 D) $243.50
9) Taylor has $90 in their checking account. They make purchases of $22
and $54. Then they overdraft their account with a purchase of $37 resulting in a $35 fee. The next day, Taylor deposits a check for $185. What is the balance of their bank account after the deposit?
A) β$58.00 B) $58.00
C) β$127.00 D) $127.00
10) The temperatures on a certain planet can be as high as 102β during the
day and as low as β35β at night. How many degrees does the temperature drop from day to night?
A) 67β B) 102β
C) 137β D) 35β
11) To calculate the Body Mass Index of an individual, their weight (in
kilograms) is divided by the square of their height (in meters). If Mike weighs 108 kilograms and stands 2 meters tall, what is his BMI?
A) 27 B) 24
C) 22 D) 18
12) Jasmine buys a car at a major dealership. She pays a single payment of
$5,775 for it. If 5% of that payment was taxes and fees, what was the actual price of the car? (Round to the nearest dollar)
A) $6,064 B) $5,500
C) $3,850 D) $5,486
184
13) The perimeter of a rectangle is three times the length. The length is six
inches longer than the width. What are the dimensions of this rectangle?
A) 12 ππππ Γ 6 ππππ B) 15 ππππ Γ 9 ππππ C) 18 ππππ Γ 10 ππππ D) 10 ππππ Γ 4 ππππ
14) What is the area under the curve shown below (above the π₯π₯-axis)?
A) 10 B) 8 C) 16 D) 12
185
Answers
1) π΅π΅
2) π·π·
3) π·π·
4) π΅π΅
5) π΄π΄
6) πΆπΆ
7) πΆπΆ
8) π΄π΄
9) π·π·
10) πΆπΆ
11) π΄π΄
12) π·π·
13) π΄π΄
14) C The trick with this question is to adjust the position of the curve so that the vertex is
over (0,0). Once the graph is in that position, you can approximate the area with a
triangle. After getting this approximation, choose the answer that is larger than the
number you got (since the triangle under-approximates the area).
186
187
Practice Tests
188
189
9.1 Practice Test 1
Answer the following questions.
1) What is 45Β° in radians? A) ππ2 B) ππ4 C) ππ D) 3ππ2
2) Find the Domain and Range of π¦π¦ = βπ₯π₯ A) π·π·: [0,β),π π : (ββ,β) B) π·π·: [0,β), π π : [0,β) C) π·π·: (ββ,β), π π : [0,β) D) π·π·: (ββ,β), π π : (ββ,β)
3) The function ππ is shown on the graph below. For what value of π₯π₯ does ππ reach its maximum value?
A) π₯π₯ = 1 B) π₯π₯ = 5.5 C) π₯π₯ = 2 D) π₯π₯ = 7
190
4) A biologist puts an initial population of 500 bacteria into a growth plate. The population is expected to double every 4 hours. Which of the following equations gives the expected number of bacteria, ππ, after π₯π₯ days? (24 βπππππππ π = 1 πππππ¦π¦)
A) ππ = 500(2)π₯π₯ B) ππ = 500(2)6π₯π₯ C) ππ = 500(6)π₯π₯ D) ππ = 500(6)2π₯π₯
5) Solve for π₯π₯ when log10 100 = π₯π₯.
A) π₯π₯ = 10 B) π₯π₯ = 100 C) π₯π₯ = 2 D) π₯π₯ = 20
6) If π₯π₯ β β2 and π₯π₯ β β32, what is the solution to 5
π₯π₯+2 = π₯π₯2π₯π₯β3?
A) 3 and 5 B) 2 and β3
2 C) β2 and 32 D) β3 and β5
7) Solve the system of equations οΏ½ π₯π₯ + π¦π¦ = 7π₯π₯ + 2π¦π¦ = 11
A) π₯π₯ = 0, π¦π¦ = 1 B) π₯π₯ = 3, π¦π¦ = 2 C) π₯π₯ = 3, π¦π¦ = 4 D) π₯π₯ = 4, π¦π¦ = 4
8) Which of the following expressions is equivalent to οΏ½3π₯π₯4π¦π¦β2οΏ½β3
οΏ½2π₯π₯3π¦π¦2οΏ½β2 ?
A) 27π¦π¦
10
4π₯π₯6
B) 4π₯π₯6
27π¦π¦10
C) 27π₯π₯6
4π¦π¦10
D) 4π¦π¦10
27π₯π₯6
191
9) Use the triangle shown to the right. What is the value of ππ?
A) β7
B) β25
C) 7
D) 12
10) What would ππ(π₯π₯) = π₯π₯2 + 2π₯π₯ look like if it were reflected about the π₯π₯-axis?
A) ππ(π₯π₯) = βπ₯π₯2 β 2π₯π₯
B) ππ(π₯π₯) = π₯π₯2 β 2π₯π₯
C) ππ(π₯π₯) = π₯π₯2 + 2π₯π₯
D) ππ(π₯π₯) = βπ₯π₯2 + 2π₯π₯
3 ππ
4
192
Answers
1) π΅π΅
2) π΅π΅
3) πΆπΆ
4) π΅π΅
5) πΆπΆ
6) π΄π΄
7) πΆπΆ
8) π·π·
9) π΅π΅
10) π΄π΄
193
9.2 Practice Test 2
Answer the following questions.
1) The perimeter of a rectangle is three times the length. The length is six inches longer than the width. What are the dimensions of this rectangle?
A) 12 ππππ Γ 6 ππππ B) 15 ππππ Γ 9 ππππ C) 18 ππππ Γ 10 ππππππ D) 10 ππππ Γ 4 ππππ
2) Find the distance between the points (β2, 2) and (2,β2).
A) β2 B) β4 C) 2β4 D) 4β2
3) What is the value of π π when βπ π = 5?
A) β5 B) β25 C) 52 D) 252
4) Which of the following is not equivalent to π₯π₯+π¦π¦βπ₯π₯+βπ¦π¦
?
A) (π₯π₯+π¦π¦)(βπ₯π₯ββπ¦π¦)(π₯π₯βπ¦π¦)
B) (π₯π₯+π¦π¦)(βπ₯π₯ββπ¦π¦)π₯π₯+π¦π¦+2βπ₯π₯π¦π¦
C) βπ₯π₯(π₯π₯+π¦π¦)ββπ¦π¦(π₯π₯+π¦π¦)(π₯π₯βπ¦π¦)
D) (π₯π₯+π¦π¦)(βπ₯π₯+βπ¦π¦)π₯π₯+π¦π¦+2βπ₯π₯π¦π¦
194
5) What is 3ππ2 in degrees? A) 90Β° B) 180Β° C) 270Β° D) 360Β°
6) Use the triangle shown to the right. What is the value of π¦π¦? A) 45Β° B) 54Β° C) 144Β° D) 36Β°
7) Find ππ(5) for the function ππ(π₯π₯) = 2π₯π₯ + 4. A) 14 B) 15 C) 16 D) 17
8) In the π₯π₯-π¦π¦ plane, a line passes though the points (3,5) and the origin (0,0).
Which of the following is the equation for the line? A) π¦π¦ = 3
5π₯π₯ + 5
B) π¦π¦ = 53π₯π₯
C) π¦π¦ = π₯π₯ + 53
D) π¦π¦ = 53π₯π₯ + 3
9) Determine the point at which the equations 6π₯π₯ + 2π¦π¦ = 2 and π₯π₯ + 1 = π¦π¦
intersect on a graph. A) (0,1) B) (1,0) C) (1,β2) D) (β2,1)
36Β°
π¦π¦
195
10) The function ππ is shown on the graph below.
Which of the following is the graph of |ππ|?
A)
B)
C)
D)
196
Answers
1) π΄π΄
2) π·π·
3) πΆπΆ
4) π΅π΅
5) πΆπΆ
6) π΅π΅
7) π΄π΄
8) π΅π΅
9) π΄π΄
10) π·π·
197
9.3 Practice Test 3
Answer the following questions.
1) What is the equation of the following graph?
A) π¦π¦ = π₯π₯2 β π₯π₯ β 6 B) π¦π¦ = π₯π₯2 + π₯π₯ β 6 C) π¦π¦ = π₯π₯2 β π₯π₯ + 6 D) π¦π¦ = π₯π₯2 + π₯π₯ + 6
2) Use the triangle to the right. What is the value of ππ?
A) β4 B) β2 C) 4 D) 5
3) What is a value of π₯π₯ that satisfies the inequality?
18 < 2π₯π₯2 < 50 A) π₯π₯ = 4 B) π₯π₯ = 5 C) π₯π₯ = 20 D) π₯π₯ = 25
β3 ππ
1
198
4) Set up the equation for when $1000 is invested at 5% interest compounded annually.
A) π΄π΄(ππ) = 1000(0.05)ππ B) π΄π΄(ππ) = 1000(0.05)12ππ C) π΄π΄(ππ) = 1000(1.05)ππ D) π΄π΄(ππ) = 1000(1.05)12ππ
5) The linear equation is in the form πππ₯π₯ + πππ¦π¦ = ππ, where ππ, ππ, and ππ are
constants. If the line is graphed in the π₯π₯-π¦π¦ plane and passes through the origin (0,0), which of the following constants must equal zero?
A) ππ B) ππ C) ππ D) It cannot be determined
6) Which of the following is the graph of a function where π¦π¦ = ππ(π₯π₯)?
A)
B)
C)
D)
199
7) Which of the following best describes the range of π¦π¦ = β2π₯π₯4 + 7? A) π¦π¦ β€ β2 B) π¦π¦ β₯ 7 C) π¦π¦ β€ 7 D) π΄π΄π π π π πππππππ π πππππππππππππ π
8) In the triangle π΄π΄π΅π΅πΆπΆ, angle πΆπΆ is a right angle. If πππππ π (π΄π΄) = 5
8, what is the value of πππππ π (π΅π΅)?
A) 38 B) 58 C) β398 D) β898
9) For which of the following equations is π₯π₯ = 6 the only solution?
A) (6π₯π₯)2 = 0 B) (π₯π₯ β 6)2 = 0 C) (π₯π₯ + 6)2 = 0 D) (π₯π₯ β 6)(π₯π₯ + 6) = 0
10) In the function ππ(π₯π₯) = ππ(π₯π₯ + 2)(π₯π₯ β 3)ππ, ππ and ππ are both integer
constants and ππ is positive. If the end behavior of the graph of π¦π¦ = ππ(π₯π₯) is positive for both very large negative values of π₯π₯ and very large positive values of π₯π₯, what is true about ππ and ππ?
A) ππ is negative, and ππ is even. B) ππ is positive, and ππ is even. C) ππ is negative, and ππ is odd. D) ππ is positive, and ππ is odd.
200
Answers
1) π΄π΄
2) π΄π΄
3) π΄π΄
4) πΆπΆ
5) πΆπΆ
6) πΆπΆ
7) πΆπΆ
8) πΆπΆ
9) π΅π΅
10) π·π·
201