Milestones

Georgia

Assessment System

Study/Resource Guide

Study/Resource Guide for Students and Parents
Analytic Geometry

The Study/Resource Guides are intended to serve as a resource for parents and students.
They contain practice questions for the course. The standards identified in the Study/Resource Guides address a sampling of the state-mandated content standards.
For the purposes of day-to-day classroom instruction, teachers should consult the wide array of resources that can be found at www.georgiastandards.org.

Table of Contents
THE GEORGIA MILESTONES ASSESSMENT SYSTEM	3
GEORGIA MILESTONES END-OF-COURSE (EOC) ASSESSMENTS	4
HOW TO USE THIS GUIDE	5
OVERVIEW OF THE ANALYTIC GEOMETRY EOC ASSESSMENT	6
ITEM TYPES	6
DEPTH OF KNOWLEDGE DESCRIPTORS	7
DEPTH OF KNOWLEDGE EXAMPLE ITEMS	10
DESCRIPTION OF TEST FORMAT  AND ORGANIZATION	16
PREPARING FOR THE ANALYTIC GEOMETRY EOC ASSESSMENT	17
STUDY SKILLS	17
ORGANIZATION—OR TAKING CONTROL OF  YOUR WORLD	17
ACTIVE PARTICIPATION	17
TEST-TAKING STRATEGIES	17
PREPARING FOR THE ANALYTIC GEOMETRY EOC ASSESSMENT	18
CONTENT OF THE ANALYTIC GEOMETRY EOC ASSESSMENT	19
SNAPSHOT OF THE COURSE	20
UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS	21
UNIT 2: RIGHT TRIANGLE TRIGONOMETRY	64
UNIT 3: CIRCLES AND VOLUME	72
UNIT 4: EXTENDING THE  NUMBER SYSTEM	97
UNIT 5: QUADRATIC FUNCTIONS	109
UNIT 6: GEOMETRIC AND  ALGEBRAIC CONNECTIONS	166
UNIT 7: APPLICATIONS OF PROBABILITY	177
ANALYTIC GEOMETRY ADDITIONAL PRACTICE ITEMS	189
ADDITIONAL PRACTICE ITEMS ANSWER KEY	210
ADDITIONAL PRACTICE ITEMS SCORING RUBRICS AND EXEMPLAR RESPONSES	214

THE GEORGIA MILESTONES ASSESSMENT SYSTEM

Dear Student,
The Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents is intended as a resource for parents and students .
This guide contains information about the core content ideas and skills that     are covered in the course .  There  are  practice  sample  questions  for  every unit . The questions are fully explained and  describe  why  each  answer  is either correct or incorrect . The explanations also help illustrate how each question connects to the Georgia state standards .
In addition, the guide includes activities that you can try to help you better understand the concepts taught in the course . The standards and additional instructional resources can be found on the Georgia Department of Education website, www .georgiastandards .org .
Get ready—open this guide—and get started!

GEORGIA MILESTONES END-OF-COURSE (EOC) ASSESSMENTS
The EOC assessments serve as the final exam in certain courses . The courses are:
English Language Arts
•	Ninth Grade Literature and Composition
•	American Literature and Composition

Mathematics
•	Algebra I
•	Geometry
•	Analytic Geometry
•	Coordinate Algebra

Science
•	Physical Science
•	Biology

Social Studies
•	United States History
•	Economics/Business/Free Enterprise

All End-of-Course tests accomplish the following:
•	Ensure that students are learning
•	Count as part of the course grade
•	Provide data to teachers, schools, and school districts
•	Identify instructional needs and help plan how to meet those needs
•	Provide data for use in Georgia’s accountability measures and reports

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HOW TO USE THIS GUIDE

Let’s get started!
First, preview the entire guide . Learn what is discussed and where to find helpful information . Even though the focus of this guide is Analytic Geometry, you need to keep in mind your overall good reading habits .    ß Start reading with a pencil or a highlighter in your hand and sticky notes nearby .
ß Mark the important ideas, the things you might want to come back to, or the explanations you have
questions about . On that last point, your teacher is your best resource .
ß You will find some key ideas and important tips to help you prepare for the test .
ß You can learn about the different types of items on the test .
ß When you come to the additional practice items, don’t just read them, do them . Think about strategies you can use for finding the right answer . Then read the analysis of the item to check your work . The
reasoning behind the correct answer is explained for you . It will help you see any faulty reasoning in     the ones you may have missed .
ß  For constructed-response questions, you will be directed to a rubric or scoring guide so you can see what is expected . The rubrics provide guidance on how students earn score points, including criteria for
how to earn partial credit for these questions . Always do your best on these questions . Even if you do    not know all of the information, you can get partial credit for your responses .
ß  Use the activities in this guide to get hands-on understanding of the concepts presented in each unit . ß With the Depth of Knowledge (DOK) information, you can gauge just how complex the item is . You will see that some items ask you to recall information and others ask you to infer or go beyond simple
recall . The assessment will require all levels of thinking .
ß Plan your studying and schedule your time .
ß Proper preparation will help you do your best!

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 5 of 230

OVERVIEW OF THE ANALYTIC GEOMETRY EOC ASSESSMENT

ITEM TYPES
The Analytic Geometry EOC assessment consists of selected-response, technology-enhanced, constructed-response, and extended constructed-response items .
A selected-response item, sometimes called a multiple-choice item, is a question, problem, or statement   that is followed by four answer choices . These questions are worth one point .
A technology-enhanced (TE) item has a question, problem, or statement .  You  may  be  asked  to  select more than one right answer . This type of question is called a multi-select TE item . Or, the question may         have two parts, where you will need to provide one answer in each part . This is called a multi-part TE item . These types of questions are worth two points . Partial credit may be awarded if you select some but not all         of the correct answers or if you get one part of the question correct but not the other part .
A constructed-response item asks a question and you provide a response that you construct on your own . These questions are worth two points . Partial credit may be awarded if part of the response is correct .
An extended constructed-response item is a specific type of constructed-response item that requires a longer, more detailed response . These items are worth four points . Partial credit may be awarded .

Strategies for Answering Constructed-Response Items

Read the question or prompt carefully .

Think about what the question is asking you to do .

Add details, examples, or reasons that help support and explain your response . Reread your response and be sure you have answered all parts of the question .       Be sure that the evidence you have provided supports your answer .

Your response will be scored based on the accuracy of your response and how well you have supported your answer with details and other evidence .

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DEPTH OF KNOWLEDGE DESCRIPTORS
Items found on the Georgia Milestones assessments, including the Analytic Geometry EOC assessment,    are developed with a particular emphasis on the kinds of thinking required to answer questions . In current educational terms, this is referred to as Depth of Knowledge (DOK) . DOK is measured on a scale of 1 to 4 and refers to the level of cognitive demand (different kinds of thinking) required to complete a task, or in this case, an assessment item . The following table shows the expectations of the four DOK levels in detail .
The DOK table lists the skills addressed in each level as well as common question cues . These question cues not only demonstrate how well you understand each skill but also relate to the expectations that are part of the state standards .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 7 of 230

Level 1—Recall of Information
Level 1 generally requires that you identify, list, or define . This level usually asks you to recall facts, terms, concepts, and trends and may ask you to identify specific information contained in documents, maps, charts, tables, graphs, or illustrations . Items that require you to  “describe”  and/or  “explain” could be classified as Level 1 or Level 2 . A Level 1 item requires that you just recall, recite, or reproduce information .	
Skills Demonstrated	Question Cues	
•	Make observations
•	Recall information
•	Recognize formulas, properties, patterns, processes
•	Know vocabulary, definitions
•	Know basic concepts
•	Perform one-step processes
•	Translate from one representation to another
•	Identify relationships	•	Find
•	List
•	Define
•	Identify; label; name
•	Choose; select
•	Compute; estimate
•	Express
•	Read from data displays
•	Order	

Level 2—Basic Reasoning
Level 2 includes the engagement (use) of some mental processing beyond recalling or reproducing a response . A Level 2 “describe” and/or “explain” item would require that you go beyond a description or explanation of recalled information to describe and/or explain a result or “how” or “why .”	
Skills Demonstrated	Question Cues	
•	Apply learned information to abstract and real-life situations
•	Use methods, concepts, and theories in abstract and real-life situations
•	Perform multi-step processes
•	Solve problems using required skills or knowledge (requires more than habitual response)
•	Make a decision about how to proceed
•	Identify and organize components of a whole
•	Extend patterns
•	Identify/describe cause and effect
•	Recognize unstated assumptions; make inferences
•	Interpret facts
•	Compare or contrast simple concepts/ideas	•	Apply
•	Calculate; solve
•	Complete
•	Describe
•	Explain how; demonstrate
•	Construct data displays
•	Construct; draw
•	Analyze
•	Extend
•	Connect
•	Classify
•	Arrange
•	Compare; contrast	

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Level 3—Complex Reasoning
Level 3 requires reasoning, using evidence, and thinking on a higher and more abstract level than Level    1 and Level 2 . You will go beyond explaining or describing “how and why” to justifying the “how and  why” through application and evidence . Level 3 items often involve making connections across time and place to explain a concept or a “big idea .”	
Skills Demonstrated	Question Cues	
•	Solve an open-ended problem with more than one correct answer
•	Create a pattern
•	Relate knowledge from several sources
•	Draw conclusions
•	Make predictions
•	Translate knowledge into new contexts
•	Compare and discriminate between ideas
•	Assess value of methods, concepts, theories, processes, and formulas
•	Make choices based on a reasoned argument
•	Verify the value of evidence, information, numbers, and data	•	Plan; prepare
•	Predict
•	Create; design
•	Generalize
•	Justify; explain why; support; convince
•	Assess
•	Rank; grade
•	Test; judge
•	Recommend
•	Select
•	Conclude	

Level 4—Extended Reasoning
Level 4 requires the complex reasoning of Level 3 with the addition of planning, investigating, applying significant conceptual understanding, and/or developing that will most likely require an extended period of time . You may be required to connect and relate ideas and concepts within the content area or among content areas in order to be at this highest level . The Level 4 items would be a show of evidence, through a task, a product, or an extended response, that the cognitive demands have been met .	
Skills Demonstrated	Question Cues	
•	Analyze and synthesize information from multiple sources
•	Apply mathematical models to illuminate a problem or situation
•	Design a mathematical model to inform and solve a practical or abstract situation
•	Combine and synthesize ideas into new  concepts	•	Design
•	Connect
•	Synthesize
•	Apply concepts
•	Critique
•	Analyze
•	Create
•	Prove	

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DEPTH OF KNOWLEDGE EXAMPLE ITEMS
Example items that represent the applicable DOK levels across various Analytic Geometry content domains are provided on the following pages .
All example and sample items contained in this guide are the property of the Georgia Department of Education .
Example Item 1
Selected-Response
DOK Level 1: This is a DOK Level 1 item because it requires the student to demonstrate an understanding of dilations and determining the scale factor .
Analytic Geometry Content Domain: Congruence and Similarity
Standard: MGSE9-12 .G .SRT .1 . Verify experimentally the properties of dilations given by a center and a scale factor .
b . The dilation of a line segment is longer or shorter according to the ratio given by the scale factor .
The smaller triangle is transformed to create the larger triangle . Which of these is the scale factor of the dilation centered at the point (0, 0)?
y

x

A .  4

B .  2

C .  1
D .  1
2

Correct Answer: B

Explanation of Correct Answer: The correct answer is choice (B) 2 . Since the length of each segment has doubled, the scale factor is 2 . Choice (A) is incorrect because a scale factor of 4 would make the horizontal side of the image a length of 8 . Choice (C) is incorrect because a scale factor of 1 does not change the size of the pre-image . Choice (D) is incorrect because it represents the scale factor when the pre-image and image are reversed .
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Example Item 2
Constructed-Response
DOK Level 2: This is a DOK Level 2 item because it requires the student to apply a formula .
Analytic Geometry Content Domain: Geometry
Standard: MGSE9-12 .G .GMD .3 . Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems .

A sand castle mold is in the shape of a cylinder with a diameter of 6 inches and a height of 8 inches .

To the nearest cubic inch, how much sand will fit in the sand castle mold? Explain how you determined your answer . Write your answer in the space provided .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 11 of 230

Scoring Rubric

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of finding the volume of a cylinder .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of finding the volume of a cylinder .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of finding the volume of a cylinder .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Exemplar Response

Points Awarded	Sample Response	

2	The mold will hold 226 cubic inches of sand .
AND
To find the radius of the circular base, I divided the diameter by 2 . To find the area of the  circular base, I squared the radius and multiplied by pi . To find the volume of the cylinder, I multiplied the area of the base by the height . Or other valid explanation .	

1	The mold will hold 226 cubic inches of sand with no explanation or an incorrect explanation .
OR
an explanation that contains a computation error but contains the correct process	
0	Response is irrelevant, inappropriate, or not provided.	

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Example Item 3
Extended Constructed-Response
DOK Level 3: This is a DOK Level 3 item because it requires complex reasoning .
Analytic Geometry Content Domain: Equations and Measurement
Standard: MGSE9-12 .G .GPE .4 . Use coordinates to prove simple geometric theorems algebraically .

ABCD is a parallelogram . Prove that the diagonals of ABCD bisect each other and justify each step . Write your answer in the space provided .

y

	

											
	A						B				
											
				E							

D						C					

10
9
8
7
6
5
4
3
2
1
0	1 2 3 4

5 6 7

x
8 9 10

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 13 of 230

Scoring Rubric

Points	Rationale	

4	The response achieves the following:
•	The response demonstrates a complete understanding of using coordinates to prove simple geometric theorems algebraically .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

3	The response achieves the following:
•	The response demonstrates a nearly complete understanding of using coordinates to prove simple geometric theorems algebraically .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

2	The response achieves the following:
•	The response demonstrates a partial understanding of using coordinates to prove simple geometric theorems algebraically .
•	The response is only partially correct .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

1	The response achieves the following:
•	The response demonstrates a minimal understanding of using coordinates to prove simple geometric theorems algebraically .
•	The response is only minimally correct .
•	The response shows the incomplete or inaccurate application of a relevant strategy .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of geometric proofs .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

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Exemplar Response

Points Awarded	Sample Response	

4	Line segments AB and DC are parallel, making angle ABE congruent to angle CDE and angle
BAE congruent to angle DCE because they are alternate interior angles .
AND
Sides AB and DC are congruent because opposite sides of parallelograms  are  congruent . That means that triangle ABE and triangle CDE are congruent by angle-side-angle .
AND
Line segment AE is equal to CE and BE is equal to DE because Corresponding Parts of Congruent Triangles are Congruent (CPCTC) .
AND
By definition, AC and BD bisect each other . Or other valid explanation .	
3	The student correctly answers three of the four parts .	
2	The student correctly answers two of the four parts .	
1	The student correctly answers one of the four parts .	
0	Response is irrelevant, inappropriate, or not provided .	
Note: If a student makes an error in one part that is carried through to subsequent parts, then the student is not penalized again for the same error.

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 15 of 230

DESCRIPTION OF TEST FORMAT AND ORGANIZATION
The Georgia Milestones Analytic Geometry EOC assessment consists of a total of 73 items . You will be asked to respond to selected-response (multiple-choice), technology-enhanced, constructed-response, and extended constructed-response items .
The test will be given in two sections .
•	You may have up to 85 minutes per section to complete Sections 1 and 2 .
•	The total estimated testing time for the Analytic Geometry EOC assessment ranges from approximately 120 to 170 minutes . Total testing time describes the amount of time you have to complete the assessment . It does not take into account the time required for the test examiner to complete pre- administration and post-administration activities (such as reading the standardized directions to students) .
•	Sections 1 and 2 may be administered on the same day or across two consecutive days, based on the district’s testing protocols for the EOC measures (in keeping with state guidance) .
•	During the Analytic Geometry EOC assessment, a formula sheet will be available for you to use . Another feature of the Analytic Geometry assessment is that you may use a graphing calculator in calculator-approved sections .

Effect on Course Grade
It is important that you take this course and the EOC assessment very seriously .
•	For students in Grade 10 or above beginning with the 2011–2012 school year, the final grade in each course is calculated by weighing the course grade 85% and the EOC score 15% .
•	For students in Grade 9 beginning with the 2011–2012 school year and later, the final grade in each course is calculated by weighing the course grade 80% and the EOC score 20% .
•	A student must have a final grade of at least 70% to pass the course and to earn credit toward graduation .

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PREPARING FOR THE ANALYTIC GEOMETRY EOC ASSESSMENT

STUDY SKILLS
As you prepare for this test, ask yourself the following questions:
✽	How would you describe yourself as a student?
✽	What are your study skills strengths and/or weaknesses?
✽	How do you typically prepare for a classroom test?
✽	What study methods do you find particularly helpful?
✽	What is an ideal study situation or environment for you?
✽	How would you describe your actual study environment?
✽	How can you change the way you study to make your study time more productive?

ORGANIZATION—OR TAKING CONTROL OF YOUR WORLD

Establish a study area that has minimal distractions . Gather your materials in advance .

Develop and implement your study plan .

ACTIVE PARTICIPATION
The most important element in your preparation is you . You and your actions are the key ingredient . Your active studying helps you stay alert and be more productive . In short, you need to interact with the course content . Here’s how you do it .

Carefully read the information and then DO something with it . Mark the important material with a highlighter, circle it with a pen, write notes on it, or summarize the information in your own words .

Ask questions . As you study, questions often come into your mind . Write them down and actively seek    the answers .

Create sample test questions and answer them .

Find a friend who is also planning to take the test and quiz each other .

TEST-TAKING STRATEGIES
Part of preparing for a test is having a set of strategies you can draw from . Include these strategies in your plan:
✽	Read and understand the directions completely . If you are not sure, ask a teacher .
✽	Read each question and all of the answer choices carefully .
✽	If you use scratch paper, make sure you copy your work to your test accurately .
✽	Underline important parts of each task . Make sure that your answer goes on the answer sheet .
✽	Be aware of time . If a question is taking too much time, come back to it later .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 17 of 230

✽	Answer all questions . Check your answers for accuracy . For constructed-response questions, do as much as you can . Remember, partially correct responses will earn a partial score .
✽	Stay calm and do the best you can .

PREPARING FOR THE ANALYTIC GEOMETRY EOC ASSESSMENT
Read this guide to help prepare for the Analytic Geometry EOC assessment .
The section of the guide titled “Content of the Analytic Geometry EOC Assessment” provides a snapshot of the Analytic Geometry course . In addition to reading this guide, do the following to prepare to take the assessment:
•	Read your resources and other materials .
•	Think about what you learned, ask yourself questions, and answer them .
•	Read and become familiar with the way questions are asked on the assessment .
•	Look at the sample answers for the constructed-response items to familiarize yourself with the elements of the exemplary responses . The rubrics will explain what is expected of you, point by point .
•	Answer some practice Analytic Geometry questions .
•	There are additional items to practice your skills available online . Ask your teacher about online practice sites that are available for your use .

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CONTENT OF THE ANALYTIC GEOMETRY EOC ASSESSMENT

Up to this point in the guide, you have been learning how to prepare for taking the EOC assessment . Now you will learn about the topics and standards that are assessed in the Analytic Geometry EOC assessment and will see some sample items .

The first part of this section focuses on what will be tested . It also includes sample items that will let you apply what you have learned in your classes and from this guide .

The second part contains additional items to practice your skills .

The third part contains a table that shows the standard assessed for each item, the DOK level, the correct answer (key), and a rationale/explanation of the right and wrong answers .

You can use the additional practice items to familiarize yourself with the item formats found on the assessment .

All example and sample items contained in this guide are the property of the Georgia Department of Education .

The Analytic Geometry EOC assessment will assess the Analytic  Geometry  standards  documented  at www .georgiastandards .org .
The content of the assessment is organized into five groupings, or domains, of standards for the purpose of providing feedback on student performance .

A content domain is a reporting category that broadly describes and defines the content of the course,       as measured by the EOC assessment .

On the actual test the standards for Analytic Geometry are grouped into five domains that follow your classwork: Congruence and Similarity; Circles; Equations and Measurement; Expressions, Equations, and Functions (including Number); and Statistics and Probability .

Each domain was created by organizing standards that share similar content characteristics .

The content standards describe the level of understanding each student is expected to achieve . They include the knowledge, concepts, and skills assessed on the EOC assessment, and they are used to plan instruction throughout the course .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 19 of 230

SNAPSHOT OF THE COURSE
This section of the guide is organized into seven units that review the material taught within the five domains of the Analytic Geometry course . The material is presented by concept rather than by category or standard . In each unit you will find sample items similar to what you will see on the EOC assessment . The next section of the guide contains additional items to practice your skills followed by a table that shows    for each item the standard assessed, the DOK level, the correct answer (key), and a rationale/explanation about the key and distractors .
All example and sample items contained in this guide are the property of the Georgia Department of Education .
The more you understand about the concepts in each unit, the greater your chances of getting a good score on the EOC assessment .

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UNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS
This unit introduces the concepts of similarity and congruence . The definition  of  similarity  is  explored through dilation transformations . The concept of scale factor with respect to dilations allows figures to be enlarged or reduced . Rigid motions lead to the definition of congruence . Once congruence is established, various congruence criteria (e .g ., ASA, SSS, SAS) can be explored . Once similarity is established,
various similarity criteria (e .g ., AA) can be explored . These criteria, along with other postulates and definitions, provide a framework for solving various geometric proofs . In this unit, various geometric figures are constructed . These topics allow students a deeper understanding of formal reasoning, which will be beneficial throughout the remainder of Geometry . Students are asked to prove theorems about parallelograms . Theorems include opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and, conversely, rectangles are parallelograms with congruent diagonals . The method for proving is not specified, so it could be done by using knowledge of congruency and establishing a formalized proof, it could be proven by constructions, or it could be proved algebraically by using the coordinate plane .

1 .1 Understand Similarity in Terms of Similarity Transformations
MGSE9-12 .G .SRT .1 Verify experimentally the properties of dilations given by a center and a scale factor . a . The dilation of a line not passing through the center of the dilation results in a parallel line and
leaves a line passing through the center unchanged .
b . The dilation of a line segment is longer or shorter according to the ratio given by the scale factor .

MGSE9-12 .G .SRT .2 Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain, using similarity transformations, the meaning of similarity for triangles         as the equality of all  corresponding  pairs  of  angles  and  the  proportionality  of  all  corresponding  pairs  of sides .
MGSE9-12 .G .SRT .3 Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 21 of 230

KEY IDEAS
A dilation is a transformation that changes the size of a figure, but not the shape, based on a ratio given by a scale factor with respect to a fixed point called the center . When the scale factor is greater than 1, the figure is made larger . When the scale factor is between 0 and 1, the figure is made smaller . When the scale factor is 1, the figure does not change . When the center of dilation is the origin, you can multiply each coordinate of the original figure, or pre-image, by the scale factor to find the coordinates of the dilated figure, or image .
Example: The diagram below shows △ABC dilated about the origin with a scale factor of 2 to create
△A′B′C′.

6
4
2

–6

–4

C'

–2

C

6

4

2

0

2

–

4

–

6

–

A

B

A'

B'

y

x

When the center of dilation is not the origin, you can use a rule that is derived from shifting the center of dilation, multiplying the shifted coordinates by the scale factor, and then shifting the center of dilation back to its original location . For a point (x, y) and a center of dilation (xc, yc), the rule for finding the coordinates of the dilated point with a scale factor of k is (xc + k(x – xc), k(y – yc) + yc) .
y

	

)	

(xc	+	k(x	–	xc)	, k	(y	– y	c)	+ yc		

					(x,	y)					
											
	(xc	, y	c)								

10
9
8
7
6
5
4
3

2
1
0	1 2 3 4

5 6 7

x
8 9 10

Page  22 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

When a figure is transformed under a dilation, the corresponding angles of the pre-image and the image have equal measures .
For △ABC and △A′B′C′ below, /A ≅ /A′, /B ≅ /B′, and /C ≅ /C′ .
When a figure is transformed under a dilation, the corresponding sides of the pre-image and the image are proportional .

A′B′

For △ABC and △A′B′C′ on the coordinate grid below, AB

= BC
B′C′

= AC .
A′C′

So when a figure is under a dilation transformation, the pre-image and the image are similar .       For △ABC and △A′B′C′ below, △ABC ∿ △A′B′C′ .

6
4
2

–6

–4

C'

–2

C

6

4

2

0

2

–

4

–

6

–

A

B

A'

B'

y

x

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 23 of 230

When a figure is dilated, a segment of the pre-image that does not pass through the center of dilation is parallel to its image . In the figure below, A}C ǁ A}′C′ since neither segment passes through the center of dilation . The same is true about A}B and A}′B′ as well as B}C and B}′C′  .

8
6
4
2

–8

–6

–4

C'

–2

C

8

6

4

2

0

2

–

4

–

6

–

8

–

A

B

A'

B'

y

x

center of dilation (0,0) scale factor = 2

When the segment of a figure does pass through the center of dilation, the segment of the pre-image and image are on the same line . In the figure below, the center of dilation is on A}C , so A}C and A}′C′ are on the
same line .

8
6
4
2

–8

–6

–4

C'

–2

C

8

6

4

2

0

2

–

4

–

6

–

8

–

A

B

A'

B'

y

x

center of dilation (–1,0) scale factor = 2

Page  24 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Draw a triangle with vertices at A(0, 1), B(–3, 3), and C(1, 3) . Dilate the triangle using a scale factor of               1 .5 and a center of (0, 0) . Sketch and name the dilated triangle A′B′C′ .
Solution:
Plot points A(0, 1), B(–3, 3), and C(1, 3) . Draw A}B, A}C , and B}C  .

5
4
3
2
1

2 –1 0

–1
–2
–3
–4

1 2 3 4

–5

5

3 –

4 –

5 –

–

A

C

B

y

x

The center of dilation is the origin, so to find the coordinates of the image, multiply the coordinates of the pre-image by the scale factor, 1 .5 .
Point A′: (1 .5 ∙ 0, 1 .5 ∙ 1) = (0, 1 .5)
Point B′: (1 .5 ∙ (–3), 1 .5 ∙ 3) = (–4 .5, 4 .5)
Point C′: (1 .5 ∙ 1, 1 .5 ∙ 3) = (1 .5, 4 .5)
Plot points A′(0, 1 .5), B′(–4 .5, 4 .5), and C′(1 .5, 4 .5) . Draw A}′B′ , A}′C′ , and B}′C′  .

5
4
3
2
1

2 –1 0

–1
–2
–3
–4

1 2 3 4

–5

5

3 –

4 –

5 –

–

A

A'

C

B

B'

'

C

y

x

Note: Since no part of the pre-image passes through the center of dilation, B}C ǁ B}′C′ , A}B ǁ A}′B′ , and
A}C ǁ A}′C′  .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 25 of 230

◆	Line segment CD is 5 inches long . If line segment CD is dilated to form line segment C′D′ with a scale factor of 0 .6, what is the length of line segment C′D′ ?
Solution:
The ratio of the length of the image and the pre-image is equal to the scale factor .
C′D′ = 0 .6
CD

Substitute 5 for CD .

Solve for C′D′ .

The length of line segment C′D′ is 3 inches .
◆	Figure A′B′C′D′ is a dilation of figure ABCD .

C′D′ = 0 .6 5

C′D′= 0 .6 ∙ 5
C′D′= 3

6
4
2

–6

–4

C

–2

D

6

4

2

0

2

–

4

–

6

–

C'

D'

B'

A'

B

A

y

x

a . Determine the center of dilation .
b . Determine the scale factor of the dilation .
c . What is the relationship between the sides of the pre-image and the corresponding sides of the image?

Page  26 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Solution:
a .    To  find the center of dilation, draw lines connecting each corresponding vertex from the pre-image     to the image . The lines meet at the center of dilation .

6
4
2

–6

–4

C

–2

D

6

4

2

0

2

–

4

–

6

–

C'

D'

B'

A'

B

A

y

x

The center of dilation is (4, 2) .

b . Find the ratios of the lengths of the corresponding sides .
A′B′ =  6  =  1
AB	12	2
B′C′ = 3 = 1
BC	6	2
C′D′ =  6  =  1
CD	12	2
A′D′ = 3 = 1
AD	6	2
1	1
The ratio for each pair of corresponding sides is 2, so the scale factor is 2 .

c . Each side of the image is parallel to the corresponding side of its pre-image and is 1 the length .
2
Note: Lines connecting corresponding points pass through the center of dilation .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 27 of 230

SAMPLE ITEMS
1 . Figure A′B′C′D′F′
at (–4, –1) .

1
is a dilation of figure ABCDF by a scale factor of 2 . The dilation is centered

5
4
3
2
1

2 –1 0

–1
–2
–3
–4

1 2 3 4

–5

F

F'

A

A'

5

3 –

4 –

5 –

–

B'

D'

C'

D

B

C

y

x

Which statement is true?

A . AB A′B′
B . AB A′B′
C . AB A′B′
D . AB A′B′

= B′C′
BC
= BC
B′C′
= BC
D′F′
= D′F′
BC

2 . Which transformation results in a figure that is similar to the original figure but has a greater area?
A .   a dilation of triangle QRS by a scale factor of 0 .25   B .   a dilation of triangle QRS by a scale factor of 0 .5     C . a dilation of triangle QRS by a scale factor of 1
D .   a dilation of triangle QRS by a scale factor of 2

Page  28 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

3 .   In the coordinate plane, segment P}Q is the result of a dilation of segment X}Y by a scale factor 1
of 2 .

8
7
6
5
4
3
2
1

1 2 3 4 5 6 7

–8

–7

–6

–5

–4

–3

–2

8

1

1– 0

2 –

3 –

4 –

5 –

6 –

7 –

8 –

–

P

X

Q

Y

y

x

Which point is the center of dilation?

A .  (–4, 0)
B .   (0, –4)
C .   (0, 4)
D .  (4, 0)

Note: Draw lines connecting corresponding points to determine the point of intersection (center of dilation) .

Answers to Unit 1 .1 Sample Items  1 .   B	2 .   D	3 . A

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 29 of 230

1 .2 Prove Theorems Involving Similarity

MGSE9-12 .G .SRT .4 Prove theorems about triangles . Theorems include: a line parallel to one side of a triangle divides the other two proportionally, (and its converse); the Pythagorean Theorem using triangle similarity .
MGSE9-12 .G .SRT .5 Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures .

KEY IDEAS
When proving that two triangles are similar, it is sufficient to show that two pairs of corresponding angles        of the triangles are congruent . This is called Angle-Angle (AA) Similarity .
Example: The triangles below are similar by AA Similarity because each triangle has a 60° angle and a 90° angle . The similarity statement is written as △ABC ∿ △DEF, and the order in which the vertices are written indicates which angles/sides correspond to each other .
A	
60°

60°

F	E

B	C	D

When a triangle is dilated, the pre-image and the image are similar triangles . There are three cases of triangles being dilated:
•	The image is congruent to the pre-image (scale factor of 1) .
•	The image is smaller than the pre-image (scale factor between 0 and 1) .
•	The image is larger than the pre-image (scale factor greater than 1) .

When two triangles are similar, all corresponding pairs of angles are congruent . When two triangles are similar, all corresponding pairs of sides are proportional . When two triangles are congruent, the triangles are also similar .
A two-column proof is a series of statements and reasons often displayed in a chart that works from given information to the statement that needs to be proven . Reasons can be given information, can be based on definitions, or can be based on postulates or theorems .
A paragraph proof also uses a series of statements and reasons that work from given information to the statement that needs to be proven, but the information is presented as running text in paragraph form .

Page  30 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	In the triangle shown, A}C ǁ D@##E$ .

   D	E	

B

A	C
Prove that D@##E$ divides A}B and C}B proportionally . Solution:
Step	Statement	Reason	
1	A}C ǁ D@##E$	Given	
2	/BDE ≅ /BAC	If two parallel lines are cut by a transversal, then corresponding angles are congruent .	
3	/DBE ≅ /ABC	Reflexive Property of Congruence because they are the same angle	
4	△DBE ∿ △ABC	Angle-Angle (AA) Similarity	

5	BA = BC BD	BE	
Corresponding sides of similar triangles are proportional .	
6	BD + DA = BA BE + EC = BC	Segment Addition Postulate	

7	BD + DA = BE + EC BD	BE	
Substitution	

8	BD + DA = BE + EC BD	BD	BE	BE	
Rewrite each fraction as a sum of two fractions .	

9	1 + DA = 1 + EC BD	BE	
Simplify .	

10	DA = EC BD	BE	
Subtraction Property of Equality	
11	@D##E$ divides A}B and
C}B proportionally .	Definition of proportionality	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 31 of 230

◆	Gale is trying to prove the Pythagorean Theorem using similar triangles . Part of her proof is shown below .
B

A	D	C

Step	Statement	Reason	
1	/ABC ≅ /BDC	All right angles are congruent .	
2	/ACB ≅ /BCD	Reflexive Property of Congruence	
3	△ABC ∿ △BDC	Angle-Angle (AA) Similarity	
4	BC = AC DC	BC	Corresponding sides of similar triangles are proportional .	
5	BC2 = AC ∙ DC	In a proportion, the product of the means equals the product of the extremes .	
6	/ABC ≅ /ADB	All right angles are congruent .	
7	/BAC ≅ /DAB	Reflexive Property of Congruence	
8	△ABC ∿ △ADB	Angle-Angle (AA) Similarity	
9	AB = AC AD	AB	Corresponding sides of similar triangles are proportional .	
10	AB2 = AC ∙ AD	In a proportion, the product of the means equals the product of the extremes .	

What should Gale do to finish her proof? Solution:
Step	Statement	Reason	
11	AB2 + BC2 = AC ∙ AD + AC ∙ DC	Addition Property of Equality	
12	AB2 + BC2 = AC(AD + DC)	Distributive Property	
13	AC = AD + DC	Segment Addition Postulate	
14	AB2 + BC2 = AC ∙ AC	Substitution	
15	AB2 + BC2 = AC2	Definition of exponent	
AB2 + BC2 = AC2 is a statement of the Pythagorean Theorem, so Gale’s proof is complete .

Page  32 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS

3

1 . In the triangles shown, △ABC is dilated by a factor of 2 to form △XYZ .

A
X

B	C	Y	Z

Given that m/A = 50° and m/B = 100°, what is m/Z ?

A .  15°
B .  25°
C .  30°
D .  50°

2 .   In the triangle shown, @G##H$ ǁ D}F  .

6
 	G	H	

3

4

E

D	F
What is the length of G}E ? A . 2 .0
B .  4 .5
C .  7 .5
D . 8 .0

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 33 of 230

3 . Use this triangle to answer the question .

H

J

I

G	K

This is a proof of the statement “If a line is parallel to one side of a triangle and intersects the other two sides at distinct points, then it separates these sides into segments of proportional lengths .”

Step	Statement	Reason	
1	G}K is parallel to H}J  .	Given	
2	/IGK ≅ /IHJ
/IKG ≅ /IJH	
?	
3	△GIK ∿ △HIJ	AA Similarity	
4	IG = IK IH	IJ	Corresponding sides of similar triangles are proportional .	
5	HG + IH = JK + IJ IH	IJ	Segment Addition Postulate	
6	HG = JK IH	IJ	Subtraction Property of Equality	
Which reason justifies Step 2?

A . Alternate interior angles are congruent .  B .  Alternate exterior angles are congruent .  C . Corresponding angles are congruent .
D . Vertical angles are congruent .

Answers to Unit 1 .2 Sample Items   1 .   C	2 .   B	3 . C

Page  34 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

1 .3 Understand Congruence in Terms of Rigid Motions

MGSE9-12 .G .CO .6 Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms        of rigid motions to decide if they are congruent .
MGSE9-12 .G .CO .7 Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent .
MGSE9-12 .G .CO .8 Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the definition of congruence in terms of rigid motions . (Extend to include HL and AAS .)

KEY IDEAS
A rigid motion is a transformation of points in space consisting of a sequence of one or more translations, reflections, and/or rotations (in any order) . This transformation leaves the size and shape of the original figure unchanged .
Two plane or solid figures are congruent if one can be obtained from the other by rigid motion (a sequence of rotations, reflections, and translations) . Congruent figures have the same corresponding side lengths and the same corresponding angle measures as each other .
Two triangles are congruent if and only if their corresponding sides and corresponding angles are congruent . This is sometimes referred to as CPCTC, which means Corresponding Parts of Congruent Triangles are Congruent .
When given two congruent triangles, you can use a series of translations, reflections, and rotations to show        the triangles are congruent .
You  can use ASA (Angle-Side-Angle) to show two triangles are congruent . If two angles and the included    side of a triangle are congruent to two angles and the included side of another triangle, then the triangles       are congruent .
△ABC ≅ △DEF by ASA .

D

C	F
E

A
B

You can use SSS (Side-Side-Side) to show two triangles are congruent . If three sides of a triangle are congruent to three sides of another triangle, then the triangles are congruent .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 35 of 230

△GIH ≅ △JLK by SSS .
G

L
I	J

K

H

You can use SAS (Side-Angle-Side) to show two triangles are congruent . If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent .
△MPN ≅ △QSR by SAS .
M	Q
P	S

N	R

You  can use AAS (Angle-Angle-Side) to show two triangles are congruent . If two angles and a non-included side of a triangle are congruent to two angles and the corresponding non-included side of another triangle,    then the triangles are congruent .
△VTU ≅ △YWX by AAS .
Y

V

X
T

U
W

Page  36 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Important Tips

If two sides and a non-included angle of one triangle are congruent to two sides and a non-included angle of a second triangle, the triangles are not necessarily congruent . Therefore, there is no way to show triangle congruency by Side-Side-Angle (SSA) .

If two triangles have all three angles congruent to each other, the triangles  are  similar,  but  not necessarily congruent . Thus, you can show similarity by Angle-Angle-Angle (AAA), but you cannot show congruence by AAA .

REVIEW EXAMPLES
◆	Is △ABC congruent to △MNP ? Explain .

10
8
6
4
2

2	4	6

–10

–8

A

C

–6

P

M

–4

0

8 1

2

2– 0

4 –

6 –

8 –

10 –

–

B

N

y

x

(scale unit = 2)

Solution:
A}C corresponds to M}P  . Both segments are 6 units long . B}C corresponds to N}P. Both segments are
9 units long . Angle C (the included angle of A}C and B}C ) corresponds to angle P (the included angle of
M}P and N}P ) . Both angles measure 90° . Because two sides and an included angle are congruent, the
triangles are congruent by SAS .
Or, △ABC is a reflection of △MNP over the y-axis . This means that all of the corresponding sides  and corresponding angles are congruent, so the triangles are congruent . (Reflections preserve angle measurement and lengths; therefore, corresponding angles and sides are congruent .)

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 37 of 230

◆	Rectangle WXYZ has coordinates W(1, 2), X(3, 2), Y(3, –3), and Z(1, –3) .
a . Graph the image of rectangle WXYZ after a rotation of 90° clockwise about the origin . Label the  image W′X′Y′Z′ .
b .  Translate rectangle W′X′Y′Z′ 2 units left and 3 units up . Label the image W″X″Y″Z″ . c . Is rectangle WXYZ congruent to rectangle W″X″Y″Z″ ? Explain .
Solution:
a . For a 90° clockwise rotation about the origin, use the rule (x, y) → (y, –x) .
W(1, 2) → W′(2, –1)	y

5
4
3
2
1

1	2	3	4

–5

Y

X'

Z

–4

Y'

–3

W'

–2

Z'

5

1

1– 0

2 –

3 –

4 –

5 –

–

X

W

X(3, 2) → X′(2, –3)
Y(3, –3) → Y′(–3, –3)
Z(1, –3) → Z′(–3, –1)

x

b . To translate rectangle W′X′Y′Z′ 2 units left and 3 units up, use the rule (x, y) → (x – 2, y + 3) .
W′(2, –1) → W″(0, 2)	y

W''

5
4
3

X''

2
1

1– 0

1
–2
–3
–4
–5

1	2	3	4

Y

X'

Z

Y'

W'

Z'

5

2 –

3 –

4 –

5 –

–

Y''

X

W

Z''

X′(2, –3) → X″(0, 0)
Y′(–3, –3) → Y″(–5, 0)
Z′(–3, –1) → Z″(–5,  2)

x

c . Rectangle W″X″Y″Z″ is the result of a rotation and a translation of rectangle WXYZ . These are both rigid transformations, so the shape and the size of the original figure are unchanged . All of the corresponding parts of WXYZ and W″X″Y″Z″ are congruent, so WXYZ and W″X″Y″Z″ are congruent .

Page  38 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 . Parallelogram FGHJ was translated 3 units down to form parallelogram F′G′H′J′ . Parallelogram
F′G′H′J′ was then rotated 90° counterclockwise about point G′ to obtain parallelogram F″G″H″J″.

5
4
3
2
1

H''
1	2	3	4

J''

F''

–5

–4

–3

–2

5

1

1– 0

2 –

3 –

4 –

5 –

–

G''

H

J

G

F

y

x

Which statement is true about parallelogram FGHJ and parallelogram F″G″H″J″?
A .   The figures are both similar and congruent .    B .  The figures are neither similar nor congruent .  C . The figures are similar but not congruent .
D .   The figures are congruent but not  similar .

2 . Consider the triangles shown .

C

B

D
A

F	E

Which can be used to prove the triangles are congruent?

A .  SSS
B .  ASA
C .  SAS
D .  AAS

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 39 of 230

3 .   In this diagram, D}E ≅ J}I and /D ≅ /J .
E	I

G

D	H	F	J

Which additional information is sufficient to prove that △DEF is congruent to △JIH? A .   }ED ≅ }IH
B .   }DH ≅ }JF
C .   }HG ≅ }GI
D .  }HF ≅ }JF

Answers to Unit 1 .3 Sample Items   1 .   A	2 .   D	3 . B

Page  40 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

1 .4 Prove Geometric Theorems

MGSE9-12 .G .CO .9 Prove theorems about lines and angles . Theorems include: vertical angles are congruent; when a transversal crosses parallel lines,  alternate  interior  angles  are  congruent  and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints .
MGSE9-12 .G .CO .10 Prove theorems about triangles . Theorems include: measures of interior angles of a triangle sum to 180 degrees; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point .
MGSE9-12 .G .CO .11 Prove theorems about parallelograms . Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals .

KEY IDEAS
A two-column proof is a series of statements and reasons often displayed in a chart that works from given information to the statement that needs to be proven . Reasons can be given information, can be based on definitions, or can be based on postulates or theorems .
A paragraph proof also uses a series of statements and reasons that work from given information to the statement that needs to be proven, but the information is presented as running text in paragraph form .
It is important to plan a geometric proof logically . Think through what needs to be proven and decide how         to get to that statement from the given information . Often a diagram or a flow chart will help organize your thoughts .
An auxiliary line is a line drawn in a diagram that makes other figures, such as congruent triangles or angles formed by a transversal . Many times, an auxiliary line is needed to help complete a proof .
Once a theorem in geometry has been proven, that theorem can be used as a reason in future proofs .   Some important key ideas about lines and angles include the following:
•	Vertical Angle Theorem: Vertical angles are congruent .

•	Alternate Interior Angles Theorem: If two parallel lines are cut by a transversal, then alternate interior angles formed by the transversal are congruent .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 41 of 230

•	Corresponding Angles Postulate: If two parallel lines are cut by a transversal, then corresponding angles formed by the transversal are congruent .

•	Points on a perpendicular bisector of a line segment are equidistant from both of the segment’s endpoints .

Some important key ideas about triangles include the following:
•	Triangle Angle-Sum Theorem: The sum of the measures of the angles of a triangle is 180° .
•	Isosceles Triangle Theorem: If two sides of a triangle are congruent, then the angles opposite those sides are also congruent .

•	Triangle Midsegment Theorem: If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side and half its length .

D 	E

A

B	C

2

mDE = 1 mBC

•	Points of Concurrency: The point where three or more lines intersect . There are 4 points of concurrency: incenter, centroid, orthocenter, and circumcenter .

Page  42 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Some important key ideas about parallelograms include the following:
•	Opposite sides are congruent and opposite angles are congruent .

•	The diagonals of a parallelogram bisect each other .

•	If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram .
•	A rectangle is a parallelogram with congruent diagonals .

REVIEW EXAMPLES
◆	In this diagram, line m intersects line n .

1

2
4

3

m

n

Write a two-column proof to show that the vertical angles /1 and /3 are congruent . Solution:
Construct a proof using intersecting lines .

Step	Statement	Reason	
1	Line m intersects line n.	Given	
2	/1 and /2 form a linear pair .
/2 and /3 form a linear pair .	Definition of a linear pair	
3	m/1 + m/2 = 180°
m/2 + m/3 = 180°	Angles that form a linear pair have measures that sum to 180° .	
4	m/1 + m/2 = m/2 + m/3	Substitution	
5	m/1 = m/3	Subtraction Property of Equality	
6	/1 ≅ /3	Definition of congruent angles	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 43 of 230

◆	In this diagram, @X##Y$ is parallel to A}C , and point B lies on X@##Y$ .
   X	B	Y 

A	C

Write a paragraph to prove that the sum of the angles in a triangle is 180° .

Solution:
A}C and X@##Y$ are parallel, so A}B is a transversal . The alternate interior angles formed by the transversal are congruent . So, m/A = m/ABX . Similarly, B}C is a transversal, so m/C = m/CBY . The sum of the
angle measures that make a straight line is 180° .
So, m/ABX + m/ABC + m/CBY = 180° . Now, substitute m/A for m/ABX and m/C for m/CBY to get m/A + m/ABC + m/C = 180° .

◆	In this diagram, ABCD is a parallelogram and B}D is a diagonal .
A	B

D	C
Write a two-column proof to show that A}B and C}D are congruent . Solution:
Construct a proof using properties of the parallelogram and its diagonal .

Step	Statement	Reason	
1	ABCD is a parallelogram .	Given	
2	B}D is a diagonal .	Given	

3	A}B is parallel to D}C  .
A}D is parallel to B}C  .	
Definition of a parallelogram	
4	/ABD ≅ /CDB
/DBC ≅ /BDA	Alternate interior angles are congruent .	
5	B}D ≅ B}D	Reflexive Property of Congruence	
6	△ADB ≅ △CBD	ASA	
7	A}B ≅ C}D	CPCTC	
Note: Corresponding parts of congruent triangles are congruent .

Page  44 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 .   In this diagram, C}D is the perpendicular bisector of A}B . The two-column proof shows that A}C is congruent to B}C  .

C

A	D	B

Step	Statement	Reason	
1	C}D is the perpendicular bisector of A}B .	Given	
2	}AD ≅ }BD	Definition of a bisector	
3	}CD ≅ }CD	Reflexive Property of Congruence	
4	/ADC and /BDC are right angles .	Definition of perpendicular lines	
5	/ADC ≅ /BDC	All right angles are congruent .	
6	△ADC ≅ △BDC	?	
7	}AC ≅ }BC	CPCTC	

Which of the following would justify Step 6?

A .  AAS
B .  ASA
C .  SAS
D .  SSS

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 45 of 230

2 .   In this diagram, STU is an isosceles triangle where }ST is congruent to }UT . The two-column proof shows that /S is congruent to /U .
T

S	U

Step	Statement	Reason	
1	S}T ≅ U}T	Given	
2	Construct T}V , the angle bisector for /T, where V is on S}U .	Every angle has a bisector .	
3	/STV ≅ /UTV	Definition of an angle bisector	
4	T}V ≅ T}V	Reflexive Property of Congruence	
5	△STV ≅ △UTV	SAS	
6	/S ≅ /U	?	

Which reason is missing in the proof?

A . CPCTC
B . Reflexive Property of Congruence
C . Definition of right angles
D . Angle Congruence Postulate

Answers to Unit 1 .4 Sample Items  1 .   C	2 . A

Page  46 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

1 .5 Make Geometric Constructions

MGSE9-12 .G .CO .12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc .) . Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing  perpendicular  lines, including the perpendicular bisector of a line segment; and constructing a line parallel  to  a  given  line through a point not on the line .
MGSE9-12 .G .CO .13 Construct an equilateral triangle, a square, and a regular hexagon, each inscribed in a circle .

KEY IDEAS
To copy a segment, follow the steps given:
Given: A}B	A	B
Construct: P}Q congruent to A}B
Procedure:
1 . Use a straightedge to draw a line, l .
2 .   Choose a point on line l and label it point P .    3 . Place the point of a compass on point A .
4 .   Adjust the compass width to the length of A}B .
5 . Without changing the compass, place the compass point on point P and draw an arc intersecting line l . Label the point of intersection as point Q .
6 .   }PQ ≅ }AB

A 	B

  P	Q	l

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 47 of 230

C

To copy an angle, follow the steps given:
Given: /ABC	B	A
Construct: /QRY congruent to /ABC
Procedure:
1 . Draw a point, R, that will be the vertex of the new angle .
2 .   From point R, use a straightedge to draw #R##Y$, which will become one side of the new angle . 3 . Place the point of a compass on vertex B and draw an arc through point A .
4 .   Without changing the compass, place the compass point on point R, draw an arc intersecting #R##Y$,
and label the point of intersection point S .
5 .   Place the compass point on point A and adjust its width to where the arc intersects #B##C$ .
6 . Without changing the compass width, place the compass point on point S and draw another arc  across the first arc . Label the point where both arcs intersect as point Q .

C

A

Q

S

R

Y

7 . Use a straightedge to draw #R##Q$ . 8 .   /QRY ≅ /ABC

B

Page  48 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

A

To bisect an angle, follow the steps given:
Given: /ABC	B	C
Construct: #B##Y$, the bisector of /ABC
Procedure:
1 . Place the point of a compass on vertex B .
2 . Open the compass and draw an arc that crosses both sides of the angle .
3 .   Set the compass width to more than half the distance from point B to where the arc crosses #B##A$ .
Place the compass point where the arc crosses #B##A$ and draw an arc in the angle’s interior .
4 .   Without changing the compass width, place the compass point where the arc crosses #B##C$ and draw an arc so that it crosses the previous arc . Label the intersection point Y .

A

Y

C

5 . Using a straightedge, draw a ray from vertex B through point Y . 6 .   #B##Y$ is the bisector of /ABC, and /ABY ≅ /YBC.

B

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 49 of 230

To construct a perpendicular bisector of a line segment, follow the steps given:
Given: A}B	A	B
Construct: The perpendicular bisector of A}B
Procedure:
1 .   Adjust a compass to a width greater than half the length of A}B .
2 .   Place the compass point on point A and draw an arc passing above A}B and an arc passing below A}B .
3 . Without changing the compass width, place the compass point on point B and draw an arc passing above A}B and an arc passing below A}B .
4 . Use a straightedge to draw a line through the points of intersection of these arcs . 5 .   The segment is the perpendicular bisector of A}B .

A	B

Note: To bisect A}B, follow the same steps listed above to construct the perpendicular bisector . The point where the perpendicular bisector intersects A}B is the midpoint of A}B .

Page  50 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

To construct a line perpendicular to a given line through a point not on the line, follow the steps given:

P

Given: Line l and point P that is not on line l	l
Construct: The line perpendicular to line l through point P
Procedure:

1 . Place the point of a compass on point P .
2 . Open the compass to a distance that is wide enough to draw two arcs across line l, one on each side of point P . Label these points Q and R .
3 . From points Q and R, draw arcs on the opposite side of line l from point P so that the arcs intersect . Label the intersection point S .
4 .   Using a straightedge, draw @P#S#$ .
5 .   P}S is perpendicular to line l .

P

Q

R

S

l

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 51 of 230

To construct a line parallel to a given line through a point not on the line, follow the steps given:

P

Given: Line l and point P that is not on line l	l
Construct: The line parallel to line l through point P

Procedure:
1 . Draw a transversal line through point P crossing line l at a point . Label the point of intersection Q .

P

Q

l

2 .   Open a compass to a width about half the distance from point P to point Q . Place the compass    point on point Q and draw an arc that intersects both lines . Label the intersection of the arc and
P}Q as point M and the intersection of the arc and line l as point N .

M

P

Q

N

l

Page  52 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

3 . Without changing the compass width, place the compass point on point P and draw an arc that crosses P}Q above point P . Note that this arc must have the same orientation as the arc drawn from
point M to point N . Label the point of intersection R .

R

M

P

Q

N

l

4 . Set the compass width to the distance from point M to point N .
5 . Place the compass point on point R and draw an arc that crosses the upper arc . Label the point of intersection S .

R

M

P

S

Q

N

l

R

M

P

S

Q

N

6 . Using a straightedge, draw a line through points P and S . 7 .   @P#S#$ ǁ l

l

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 53 of 230

O

Given: Circle O
Construct: Equilateral △ABC inscribed in circle O
Procedure:
1 .    Mark a point anywhere on the circle and label it point P .    2 . Open a compass to the radius of circle O .
3 .    Place the compass point on point P and draw an arc that intersects the circle at two points . Label     the points A and B .
4 .   Using a straightedge, draw A}B .
5 .   Open the compass to the length of A}B .
6 .  Place the compass point on point A . Draw an arc from point A that intersects the circle . Label this point C .
7 .   Using a straightedge, draw A}C and B}C  .
8 . Equilateral △ABC is inscribed in circle O .

A

P

O

C

B

Page  54 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

O

Given: Circle O
Construct: Square ABCD inscribed in circle O
Procedure:

A

O

1 . Mark a point anywhere on the circle and label it point A .

A

O

C

2 . Using a straightedge, draw a diameter from point A . Label the other endpoint of the diameter as point C . This is diameter A}C  .

B

A

O

C

D

3 .   Construct a perpendicular bisector of A}C through the center of circle O . Label the points where it intersects the circle as point B and point D .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 55 of 230

B

A

O

C

D

4 .   Using a straightedge, draw A}B, B}C , C}D , and A}D  .

5 . Square ABCD is inscribed in circle O .

O

To construct a regular hexagon inscribed in a circle, follow the steps given:

Given: Circle O
Construct: Regular hexagon ABCDEF inscribed in circle O
Procedure:
1 .   Mark a point anywhere on the circle and label it point A .    2 . Open a compass to the radius of circle O .
3 . Place the compass point on point A and draw an arc across the circle . Label this point B .
4 .    Without changing the width of the compass, place the compass point on point B and draw another    arc across the circle . Label this point C .
5 .   Repeat this process from point C to a point D, from point D to a point E, and from point E to a  point F .
6 .   Use a straightedge to draw A}B, B}C , C}D , D}E , E}F , and A}F  . 7 . Regular hexagon ABCDEF is inscribed in circle O .
B
A
C
O	F

D
E

Page  56 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Allan drew angle BCD .

B

C	D

a .    Copy angle BCD . List the steps you used to copy the angle . Label the copied angle RTS .         b . Without measuring the angles, how can you show they are congruent to one another?
Solution:
a .    Draw point T . Draw #T#S#$ .

B

C	D	T	S

Place the point of a compass on point C . Draw an arc . Label the intersection points X and Y . Keep the compass width the same, and place the point of the compass on point T . Draw an arc and label the intersection point V .
B

V

S

X

C	Y	D	T

Place the point of the compass on point Y and adjust the width to point X . Then place the point of the compass on point V and draw an arc that intersects the first arc . Label the intersection point U .

U

V

S

B
X

C	Y	D	T

Draw #T#U#$ and point R on #T#U#$ . Angle BCD has now been copied to form angle RTS .

R
U V

S

B
X

C	Y	D	T

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 57 of 230

△	△

b .    Connect points X and Y and points U and V to form	XCY and	UTV. C}Y and T}V , X}Y and U}V , and C}X and T}U are congruent because they were drawn with the same compass width . So
△XCY ≅ △UTV by SSS, and /C ≅ /T because congruent parts of congruent triangles are
congruent .

R

U

V

S

B
X

C	Y	D	T

◆	Construct a line segment perpendicular to M}N from a point not on M}N  . Explain the steps you used to make your construction .
M	N

Solution:
Draw a point P that is not on M}N  . Place the point of a compass on point P . Draw an arc that intersects M}N at two points . Label the intersection points Q and R . Without changing the width of the compass, place the compass point on point Q and draw an arc under M}N  . Place the compass point on point R and draw another arc under M}N  . Label the intersection point S . Draw P}S . Segment PS is perpendicular to and bisects M}N  .

M Q

R	N

S

P

Page  58 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	Construct equilateral △HIJ inscribed in circle K . Explain the steps you used to make your construction .
Solution:
(This is an alternate method from the method shown in the Key Ideas .) Use a compass to draw
circle K . Draw segment F}G through the center of circle K . Label the points where F}G intersects circle K
as points I and P . Using the compass setting you used when drawing the circle, place the compass on   point P and draw an arc passing through point K . Label the points where the arc intersects circle K as
points H and J . Draw H}J , I}J , and H}I  . Triangle HIJ is an equilateral triangle inscribed in circle K .
F I

K

H	J
P

G

SAMPLE ITEMS
1 . Consider the construction of the angle bisector shown .

X

A

Y

C

B

Z

Which could have been the first step in creating this construction?

A .   Place the compass point on point A and draw an arc inside /Y .
B .   Place the compass point on point B and draw an arc inside /Y .
C .   Place the compass point on vertex Y and draw an arc that intersects Y}X and Y}Z  .
D . Place the compass point on vertex Y and draw an arc that intersects point C.

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 59 of 230

2 . Consider the beginning of the construction of a square inscribed in circle Q .
Step 1:	Label point R on circle Q .
Step 2:	Draw a diameter through R and Q .
Step 3:	Label the point where the diameter intersects the circle as point T .

R

Q

T

What is the next step in this construction?
A .   Draw radius S}Q  .
B . Label point S on circle Q.
C .   Construct a line segment parallel to R}T  .
D .  Construct the perpendicular bisector of R}T  .

Answers to Unit 1 .5 Sample Items   1 .   C	2 . D

Page  60 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

1 .6 Use Coordinates to Prove Simple Geometric Theorems Algebraically

}

MGSE9-12 .G .GPE .4 Use coordinates to prove simple geometric theorems algebraically . For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, Ï3) lies on the circle centered at the origin and containing the point (0, 2) .
(Focus on quadrilaterals, right triangles, and circles .)

KEY IDEAS
To prove properties about special parallelograms on a coordinate plane, you can use the midpoint, distance, and slope formulas:
•	The midpoint formula is (x1  + x2, y1  + y2) . This formula is used to find the coordinates of the midpoint
—	2	2
of AB , given A(x1, y1) and B(x2, y2) .
}}	}
•	The distance formula is d = Ï(x2  – x1)2  + (y2  – y1)2  . This formula is used to find the length of AB , given
A(x1, y1) and B(x2, y2) .
•	
x2 – x1

The slope formula is m = (y2  – y1) . This formula is used to find the slope of a line or line segment, given any two points on the line or line segment A(x1, y1) and B(x2, y2) .
You can use properties of quadrilaterals to help prove theorems, such as the following:
•	To prove a quadrilateral is a parallelogram, show that the opposite sides are parallel using the slope formula .
•	To prove a quadrilateral is a rectangle, show that the opposite sides are parallel and the consecutive sides are perpendicular using the slope formula .
•	To prove a quadrilateral is a rhombus, show that all four sides are congruent using the distance formula .
•	To prove a quadrilateral is a square, show that all four sides are congruent and consecutive sides are perpendicular using the distance and slope formulas .

You can also use diagonals of a quadrilateral to help prove theorems, such as the following:
•	To prove a quadrilateral is a parallelogram, show that its diagonals bisect each other using the midpoint formula .
•	To prove a quadrilateral is a rectangle, show that its diagonals bisect each other and are congruent using the midpoint and distance formulas .
•	To prove a quadrilateral is a rhombus, show that its diagonals bisect each other and are perpendicular using the midpoint and slope formulas .
•	To prove a quadrilateral is a square, show that its diagonals bisect each other, are congruent, and are perpendicular using the midpoint, distance, and slope formulas .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 61 of 230

Important Tips

When using the formulas for midpoint, distance, and slope, the order of the points does not matter . Either point can be (x1, y1), but be careful to always subtract in the same order .

Parallel lines have the same slope . Perpendicular lines have slopes that are the negative reciprocal of     each other .

REVIEW EXAMPLE
◆	Quadrilateral ABCD has vertices A(–1, 3), B(3, 5), C(4, 3), and D(0, 1) . Is ABCD a rectangle? Explain      how you know .
Solution:
First determine whether the figure  is  a  parallelogram .  If  the  figure  is  a  parallelogram,  then  the diagonals bisect each other . If the diagonals bisect each other, then the midpoints of the diagonals are        the same point . Use the midpoint formula to determine the midpoints for each diagonal .

2

2

2

2

2

2

Midpoint A}C : (x1  + x2, y1  + y2) = (–1 + 4, 3 + 3) = (3, 6) = (1 .5, 3)

2

2

2

2

2

2

Midpoint B}D : (x1  + x2, y1  + y2) = (3 + 0, 5 + 1) = (3, 6) = (1 .5, 3)
The diagonals have the same midpoint; therefore, the diagonals bisect each other and the figure is a parallelogram .
A parallelogram with congruent diagonals is a rectangle . Determine whether the diagonals are congruent .
Use the distance formula to find the length of the diagonals .
}}	}	}	}
AC = Ï(4 – (–1))2 + (3 – 3)2 = Ï(5)2 + (0)2 = Ï25 + 0 = Ï25 = 5
}}	}}	}	}
BD = Ï(0 – 3)2 + (1 – 5)2 = Ï(–3)2 + (–4)2 = Ï9 + 16 = Ï25 = 5
The diagonals are congruent because they have the same length .
The figure is a parallelogram with congruent diagonals, so the figure is a rectangle .

Page  62 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 . Which information is sufficient to show that a parallelogram is a rectangle?
A . The diagonals bisect each other .
B . The diagonals are congruent .
C . The diagonals are congruent and perpendicular .
D . The diagonals bisect each other and are perpendicular .

2 . Look at quadrilateral ABCD .

A	B

D	C

Which information is needed to show that quadrilateral ABCD is a parallelogram?

A . Use the distance formula to show that diagonals AC and BD have the same length .
B . Use the slope formula to show that segments AB and CD are perpendicular and segments AD and
BC are perpendicular .
C . Use the slope formula to show that segments AB and CD have the same slope and segments AD
and BC have the same slope .
D . Use the distance formula to show that segments AB and AD have the same length and segments
CD and BC have the same length .

Answers to Unit 1 .6 Sample Items   1 .   B	2 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 63 of 230

UNIT 2: RIGHT TRIANGLE TRIGONOMETRY
This unit investigates the properties of right triangles . The trigonometric ratios sine, cosine, and tangent along with the Pythagorean Theorem are used to solve right triangles in applied problems . The relationship between the sine and cosine of complementary angles is identified .

2 .1 Right Triangle Relationships

MGSE9-12 .G .SRT .6 Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles .
MGSE9-12 .G .SRT .7 Explain and use the relationship between the sine and cosine of complementary angles .
MGSE9-12 .G .SRT .8 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems .

KEY IDEAS
The trigonometric ratios sine, cosine, and tangent are defined as ratios of the lengths of the sides in a   right triangle with a given acute angle measure . These terms are usually seen abbreviated as sin, cos, and tan .

hypotenuse	opposite θ
adjacent

•	
θ

sin	= length of opposite side
length of hypotenuse
•	
θ

cos	= length of adjacent side
length of hypotenuse
•	
θ

tan	=  length of opposite side
length of adjacent side

C

opposite

B

•	sin A =

•	cos A =

•	tan A =

adjacent	A

hypotenuse

length of opposite side length of hypotenuse
length of adjacent side length of hypotenuse
length of opposite side length of adjacent side

The two acute angles of any right triangle are complementary . As a result, if angles P and Q are complementary, sin P = cos Q and sin Q = cos P .
When solving problems with right triangles, you can use both trigonometric ratios and the Pythagorean Theorem (a2 + b2 = c2) . There may be more than one way to solve the problem, so analyze the given information to help decide which method is the most efficient .

Important Tip

The tangent of angle A is also equivalent to sin A .
cos A
Page  64 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Triangles ABC and DEF are similar .

10

B

5

6	E
3

A	8	C	D	4	F

a .    Find the ratio of the side opposite angle B to the hypotenuse in △ABC .       b . What angle in △DEF corresponds to angle B?
c .    Find the ratio of the side opposite angle E to the hypotenuse in △DEF .       d . How does the ratio in part (a) compare to the ratio in part (c)?
e . Which trigonometric ratio does this represent? Solution:

a .

A}C is opposite angle B . B}C is the hypotenuse . The ratio of the side opposite angle B to the

hypotenuse in △ABC is  8	4
10 = 5 .
b . Angle E in △DEF corresponds to angle B in △ABC .

c .

D}F is opposite angle E . E}F is the hypotenuse . The ratio of the side opposite angle E to the

hypotenuse in △	4
DEF is 5 . d . The ratios are the same .

 .

e . This represents sin B and sin E, because both are the ratio opposite 
hypotenuse

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 65 of 230

◆	Ricardo is standing 75 feet away from the base of a building . The angle of elevation from the ground where Ricardo is standing to the top of the building is 32° .

x

32°
75 ft
Note: figure not drawn to scale.

What is x, the height of the building, to the nearest tenth of a foot?

3	4

sin 32° ≈ 0 .5299 cos 32° ≈ 0 .8480 tan 32° ≈ 0 .6249

Solution:
When finding the length of the side opposite the 32° angle and given the length of the side adjacent to         the 32° angle, use the tangent ratio . Substitute x for the opposite side, 75 for the adjacent side, and           32° for the angle measure . Then solve .
tan 32° =   x 
75
75tan 32° = x
75 ∙ 0 .6249 ≈ x
46 .9  ≈ x
The building is about 46 .9 feet tall .

Page  66 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	An airplane is at an altitude of 5,900 feet .  The airplane descends at an angle of 3°, called the angle  of depression .

3°

5,900 ft.	x

3°
Note: figure not drawn to scale.

About how far will the airplane travel in the air until it reaches the ground?

3	4

sin 3° ≈ 0 .0523 cos 3° ≈ 0 .9986 tan 3° ≈ 0 .0524

Solution:
Use sin 3° to find the distance the airplane will travel until it reaches the ground, x . Substitute x for the hypotenuse, 5,900 for the opposite side, and 3° for the angle measure . Then solve .

x

sin 3° = 5,900
x sin 3° = 5,900
x = 5,900
sin 3°

≈

x	 5,900 0 .0523
x ≈ 112,811
The airplane will travel about 113,000 feet until it reaches the ground .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 67 of 230

◆	Triangle ABC is a right triangle .
A

5

B

13

12	C

What is the best approximation for m/C ?

3	4

sin 67 .4° ≈ 0 .923 cos 22 .6° ≈ 0 .923 tan 42 .7° ≈ 0 .923

Solution:
Find the trigonometric ratios for angle C .

≈

sin C =  5	0 .385 13

≈

cos C = 12	0 .923 13

≈

tan C =  5	0 .417 12
Using the table, cos 22 .6° ≈ 0 .923, so m/C ≈ 22 .6°, or using trigonometric inverses, sin−1 5 = 22 .6, cos−1 12 = 22 .6, or tan−1 5 = 22 .6 .
13	13	12

Page  68 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS

 .

1 .  In right triangle ABC, angle A and angle B are complementary angles . The value of cos A is 5 
13
What is the value of sin B ?
A . 5 
13
B .  12
13
C .  13
12
D .  13
5

2 . Triangle ABC is given below .

15 m

A

9 m

B	12 m	C

What is the value of cos A?
A .  3
5
B .  3
4
C .  4
5
D .  5
3

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 69 of 230

3 .   In right triangle HJK, /J is a right angle and tan /H = 1 . Which statement about triangle HJK must be true?
A .   sin /H = 1
2
B .   sin /H = 1

C .   sin /H = cos /H
D .  sin /H =  	1	
cos /H

4 . A 12-foot ladder is leaning against a building at a 75° angle to the ground .

12 ft.

x

75°

x

Which equation can be used to find how high the ladder reaches up the side of the building? A . sin 75° = 12

x

B .   tan 75° = 12
C . cos 75° = x 
12
D . sin 75° = x 
12

Page  70 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

5 . A hot-air balloon is 1,200 feet above the ground . The angle of depression from the basket of the hot-air balloon to the base of a monument is 54° .

54°

1,200 ft.

d

Note: figure not drawn to scale.

Which equation can be used to find the distance, d, in feet, from the basket of the hot air balloon to the base of the monument?
A .   sin 54° =     d	
1200
B .   sin 54° =  1200
d
C .   cos 54° =     d	
1200
D .   cos 54° =  1200
d

Answers to Unit 2 .1 Sample Items
1 .   A	2 .   A	3 .   C	4 .   D	5 . B

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 71 of 230

UNIT 3: CIRCLES AND VOLUME
This unit investigates the properties of circles and addresses finding the volume of solids . Properties of circles are used to solve problems involving arcs, angles, sectors, chords, tangents, and secants . Volume formulas are derived and used to calculate the volumes of cylinders, pyramids, cones, and spheres .

3 .1 Understand and Apply Theorems about Circles

MGSE9-12 .G .C .1 Understand that all circles are similar .
MGSE9-12 .G .C .2 Identify and describe relationships among inscribed angles, radii, chords, tangents, and secants . Include  the  relationship  between  central,  inscribed,  and  circumscribed  angles;  inscribed  angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent  where  the  radius intersects the circle .
MGSE9-12 .G .C .3 Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle .
MGSE9-12 .G .C .4 Construct a tangent line from a point outside a given circle to the circle .

KEY IDEAS
A circle is the set of points in a plane equidistant from a given point, which is the center of the circle . All circles are similar .

A radius is a line segment from the center of a circle to any point on the circle . The word radius is also used to describe the length, r, of the segment . A}B is a radius of circle A .

r

A

B

A chord is a line segment whose endpoints are on a circle . B}C is a chord of circle A .

C

A

B

Page  72 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

B

d

A

C

A diameter is a chord that passes through the center of a circle . The word diameter is also used to describe the length, d, of the segment . B}C is a diameter of circle A .

C

B

A

A secant line is a line that is in the plane of a circle and intersects the circle at exactly two points . Every chord lies on a secant line . B@#C#$ is a secant line of circle A .

A tangent line is a line that is in the plane of a circle and intersects the circle at only one point, the point of tangency . D@##F$ is tangent to circle A at the point of tangency, point D.

A

D

F

A

D

F

If a line is tangent to a circle, the line is perpendicular to the radius drawn to the point of tangency . @D##F$ is tangent to circle A at point D, so A}D ⊥ D}F  .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 73 of 230

Tangent segments drawn from the same point are congruent . In circle A, }CG ≅ }BG .

C

A

G

B

Circumference is the distance around a circle . The formula for the circumference, C, of a circle is C = πd, where d is the diameter of the circle . The formula is also written as C = 2πr, where r is the length of the   radius of the circle . π is the ratio of circumference to diameter of any circle .

X	X	X

An arc is a part of the circumference of a circle . A minor arc has a measure less than 180° . Minor arcs are written using two points on a circle . A semicircle is an arc that measures exactly 180° . Semicircles  are written using three points on a circle . This is done to show which half of the circle is being described . A major arc has a measure greater than 180° . Major arcs are written with three points to distinguish them from the corresponding minor arc . In circle A, CB is a minor arc, CBD is a semicircle, and CDB is a major arc .

C

A

D

B

A

P

B

X

A central angle is an angle whose vertex is at the center of a circle and whose sides are radii of the circle . The measure of a central angle of a circle is equal to the measure of the intercepted arc . /APB is a central angle for circle P, and AB is the intercepted arc .
m/APB = mXAB

Page  74 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

X

An inscribed angle is an angle whose vertex is on a circle and whose sides are chords of the circle . The measure of an angle inscribed in a circle is half the measure of the intercepted arc . For circle D, /ABC is an inscribed angle, and AC is the intercepted arc .

A

D

C

B

m/ABC = 1 (   X)	1

/ADC)

2 mAC = 2 (m
m/ADC = mXAC = 2(m/ABC)
A circumscribed angle is an angle formed by two rays that are each tangent to a circle . These rays are perpendicular to radii of the circle . In circle O, the measure of the circumscribed angle is equal to 180° minus the measure of the central angle that forms the intercepted arc . The measure of the circumscribed angle can also be found by using the measures of two intercepted arcs .

A

B

O

C

m/ABC = 180° – m/AOC

X

When an inscribed angle intercepts a semicircle, the inscribed angle has a measure of 90° . For circle O,
/RPQ intercepts semicircle  RSQ as shown .

P

O

S

R

Q

m/RPQ = 1(  X)	1
2  mRSQ  = 2(180°) = 90°

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 75 of 230

The measure of an angle formed by a tangent and a chord with its vertex on the circle is half the measure of the intercepted arc . A}B is a chord for the circle, and #B##C$ is tangent to the circle at point B . So /ABC is
formed by a tangent and a chord .

A

B	C
m/ABC = 1(mXAB)
2
When two chords intersect inside a circle, two pairs of vertical angles are formed . The measure of any one of the angles is half the sum of the measures of the arcs intercepted by the pair of vertical angles .
E	C
B

A	D
F
m/ABE = 1(mXAE + mXCD)
2
m/ABD = 1(mXAFD + mXEC)
2
When two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord .
E	C
B

A	D
F

AB ∙ BC = EB ∙ BD

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Angles outside a circle can be formed by the intersection of two tangents (circumscribed angle),
two secants, or a secant and a tangent . For all three situations, the measure of the angle is half the difference of the measure of the larger intercepted arc and the measure of the smaller intercepted arc .

F

D

A

D

C

A

E
D

B		C	B B
A
m/ABD = 1(mXAFD – mXAD)	m/ACE = 1(mXAE – mXBD)	m/ABD = 1(mXAD – mXAC)
2	2	2
When two secant segments intersect outside a circle, part of each secant segment is a segment formed outside the circle . The product of the length of one secant segment and the length of the segment formed outside the circle is equal to the product of the length of the other secant segment and the length of the segment formed outside the circle .
E
D

C
B
A

EC ∙ DC = AC ∙ BC

When a secant segment and a tangent segment intersect outside a circle, the product of the length of     the secant segment and the length of the segment formed outside the circle is equal to the square of the length of the tangent segment .

D

C

A

B

DB ∙ CB = AB2
An inscribed polygon is a polygon whose vertices all lie on a circle . This diagram shows a triangle, a quadrilateral, and a pentagon each inscribed in a circle .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 77 of 230

In a quadrilateral inscribed in a circle, the opposite angles are supplementary .
B	C

A
D

m/ABC + m/ADC = 180°
m/BCD + m/BAD = 180°

When a triangle is inscribed in a circle, the center of the circle is the circumcenter of the triangle . The circumcenter is equidistant from the vertices of the triangle . Triangle ABC is inscribed in circle Q, and point Q is the circumcenter of the triangle .
A	B
Q

C

AQ = BQ = CQ

An inscribed circle is a circle enclosed in a polygon, where every side of the polygon is tangent to the circle . Specifically, when a circle is inscribed in a triangle, the center of the circle is the incenter of the triangle . The incenter is equidistant from the sides of the triangle . Circle Q is inscribed in triangle ABC, and point Q is the incenter of the triangle . Notice also that the sides of the triangle form circumscribed angles with the circle .

Q

A

C	B

Page  78 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	/PNQ is inscribed in circle O, and mXPQ = 70° .

Q

O

N

P

a .    What is the measure of /POQ ?
b .    What is the relationship between /POQ and /PNQ ? c .    What is the measure of /PNQ ?
Solution:

X

a . The measure of a central angle is equal to the measure of the intercepted arc .
m/POQ = mPQ = 70° .
b .    /POQ is a central angle that intercepts XPQ . /PNQ is an inscribed angle that intercepts XPQ . The

2

measure of the central angle is equal to the measure of the intercepted arc . The measure of the inscribed angle is equal to one-half the measure of the intercepted arc . So m/POQ = mXPQ, m/PNQ = 1(mXPQ), and m/POQ = 2(m/PNQ) .
c .    From part (b), m/POQ = 2(m/PNQ) Substitute:	70° = 2(m/PNQ)
Divide:	35° = m/PNQ

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 79 of 230

◆	In circle P below, A}B is a diameter .
A
P

C	B

If m/APC = 100°, find the following:
a .    m/BPC
b .    m/BAC
c .    mXBC
d .    mXAC
Solution:
a .    /APC and /BPC are supplementary, so m/BPC = 180° – m/APC = 180° – 100° = 80° . b .    /BAC is an angle in △APC. The sum of the measures of the angles of a triangle is 180° .
For △APC: m/APC + m/BAC + m/ACP = 180°
You are given that m/APC = 100° . Substitute: 100° + m/BAC + m/ACP = 180°
Subtract 100° from both sides: m/BAC + m/ACP = 80°
Because two sides of △APC are radii of the circle, △APC is an isosceles triangle . This means that the two base angles are congruent, so m/BAC = m/ACP .
Substitute: m/BAC for m/ACP: m/BAC + m/BAC = 80° Add: 2(m/BAC) = 80°
Divide: m/BAC = 40°

X

You could also use the answer from part (a) to solve for m/BAC . Part (a) shows m/BPC = 80° . Because the central angle measure is equal to the measure of the intercepted arc,
m/BPC = mBC = 80° .

Because an inscribed angle is equal to one-half the measure of the intercepted arc,
m/BAC = 1(mXBC) .
2
Substitute: m/ 	1
BAC = 2(80°)
Therefore, m/BAC = 40° .

Page  80 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

c .    /BAC is an inscribed angle intercepting XBC . The intercepted arc is twice the measure of the inscribed angle .
mXBC = 2(m/BAC)
From part (b), m/BAC = 40° . Substitute: mXBC = 2 ∙ 40°
mXBC = 80°

X	X

You could also use the answer from part (a) to solve . Part (a) shows m/BPC = 80° . Because /BPC
is a central angle that intercepts  BC, m/BPC = mBC = 80° .
d .    /APC is a central angle intercepting XAC . The measure of the intercepted arc is equal to the measure of the central angle .
mXAC = m/APC
You are given m/APC = 100° . Substitute: mXAC = 100°
◆	In circle P below, D}G is a tangent, AF = 8, EF = 6, BF = 4, and EG = 8 .

P

F

E

A

B

G
D	C
Find CF and DG .
Solution:
First, find CF . Use the fact that C}F is part of a pair of intersecting chords .
AF ∙ CF = EF ∙ BF
8 ∙ CF = 6 ∙ 4
8 ∙ CF = 24
CF = 3
Next, find DG . Use the fact that D}G is tangent to the circle .
EG ∙ BG = DG2
8 ∙ (8 + 6 + 4) = DG2
8 ∙ 18 = DG2
144 = DG2
±12 = DG
12 = DG (since length cannot be negative)

CF = 3 and DG = 12

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 81 of 230

◆	In this circle, A}B is tangent to the circle at point B, A}C is tangent to the circle at point C, and point D
lies on the circle . What is m/BAC ?

A

B

C

O

48°

D

Solution:
Method 1
First, find the measure of angle BOC . Angle BDC is an inscribed angle, and angle BOC is a central  angle .
m/BOC = 2(m/BDC)
= 2 ∙ 48°
= 96°
Angle BAC is a circumscribed angle . Use the measure of angle BOC to find the measure of angle BAC . m/BAC = 180° – m/BOC
= 180° – 96°
= 84°

Page  82 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Method 2
Angle BDC is an inscribed angle . First, find the measures of XBC and XBDC .
m/BDC = 1(mXBC)
2
48° = 1(mXBC)
2
2 ∙ 48° = mXBC
96° = mXBC
mXBDC = 360° – mXBC
= 360° – 96°
= 264°
Angle BAC is a circumscribed angle . Use the measures of XBC and XBDC to find the measure of angle BAC .
m/BAC = 1 (mXBDC – mXBC)
2
1
= 2 (264° – 96°)
1
= 2 (168°)
= 84°

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 83 of 230

SAMPLE ITEMS
1 . Circle P is dilated to form circle P′ . Which statement is ALWAYS true?
A . The radius of circle P is equal to the radius of circle P′ .
B . The length of any chord in circle P is greater than the length of any chord in circle P′ .
C . The diameter of circle P is greater than the diameter of circle P′ .
D . The ratio of the diameter to the circumference is the same for both circles .

2 .   In the circle shown, }BC is a diameter and mXAB = 120° .
A

120°	C

B

What is the measure of /ABC?

A .  15°
B .  30°
C .  60°
D . 120°

Answers to Unit 3 .1 Sample Items   1 .   D	2 . B

Page  84 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

3 .2 Find Arc Lengths and Areas of Sectors of Circles

MGSE9-12 .G .C .5 Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality; derive the formula for the area of a sector .

KEY IDEAS
Circumference is the distance around a circle . The formula for the circumference, C, of a circle is C = 2πr,  where r is the length of the radius of the circle .
Area is a measure of the amount of space a circle covers . The formula for the area, A, of a circle is A = πr2,  where r is the length of the radius of the circle .

θ

Arc length is a portion of the circumference of a circle . To find the length of an arc, divide the number of degrees in the central angle of the arc by 360 and then multiply that amount by the circumference of the circle . The formula for the arc length, s, is s =  2πθ , where	is the degree measure of the central angle and
360
r is the radius of the circle .

A

θ
r

s

Important Tip

Do not confuse arc length with the arc measure in degrees . Arc length depends on the size of the circle because it is part of the circumference of the circle . The measure of the arc is independent of (does not depend on) the size of the circle .
The following shows one way to remember the formula for arc length:
arc length = fraction of the circle × circumference = s = 2πrθ .
360

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 85 of 230

A sector of a circle is the region bounded by two radii of a circle and the resulting arc between them . To

θ

find the area of a sector, divide the number of degrees in the central angle of the arc by 360 and then    multiply that amount by the area of the circle . The formula for the area of a sector is πr2θ, where	is the
360
degree measure of the central angle and r is the radius of the circle .

A

θ
r

Important Tip

The following shows one way to remember the formula for the area of a sector: area of a sector = fraction of the circle × area = θ (πr2) .
360

Page  86 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	
50
9
10 ft.
100°	
B

20
12 ft.	3
100°	
C

Circles A, B, and C have a central angle measuring 100° . The length of each radius and the length of each intercepted arc are shown .

9

35	ft. 7 ft. 100° 
A

ft.

ft.

a . What is the ratio of the radius of circle B to the radius of circle A?
b . What is the ratio of the length of the intercepted arc of circle B to the length of the intercepted arc of circle A?
c . Compare the ratios in parts (a) and (b) .
d . What is the ratio of the radius of circle C to the radius of circle B?
e . What is the ratio of the length of the intercepted arc of circle C to the length of the intercepted arc of circle B?
f .	Compare the ratios in parts (d) and (e) .
g . Based on your observations of circles A, B, and C, what conjecture can you make about the length  of the arc intercepted by a central angle and the radius?
h . What is the ratio of arc length to radius for each circle?
Solution:
a . Divide the radius of circle B by the radius of circle A: circle B = 10
circle A	7
b . Divide the length of the intercepted arc of circle B by the length of the intercepted arc of circle A:
50π
  9    = 50π ∙    9   = 10

c . The ratios are the same .

35π 9

9	35π	7

d .    Divide the radius of circle C by the radius of circle B:  circle C  =  12  = 6
circle B	10	5
e . Divide the length of the intercepted arc of circle C by the length of the intercepted arc of circle B:
20 π
 3	= 20π ∙    9   = 6

f .	The ratios are the same .

50 π
9

3	50π	5

g . When circles, such as circles A, B, and C, have the same central angle measure, the ratio of the lengths of the intercepted arcs is the same as the ratio of the radii .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 87 of 230

35 π
h .   Circle A:  9	= 35 π = 5 π
7	63	9
50 π
Circle B:  9	= 50 π = 5 π
10	90	9

20 π
Circle C:  3	= 20 π = 5 π
12	36	9

◆	Circle A is shown .

8 m

A

θ

x°

If x = 50, what is the area of the shaded sector of circle A? Solution:
To find the area of the sector, divide the measure of the central angle of the arc in degrees by 360 and then multiply that amount by the area of the circle . The arc measure, x, is equal to the measure of the central angle, θ. The formula for the area of a circle is A = πr2 .

Asector

= πr2θ
360
50π(8)2

Area of sector of a circle with radius r and central angle θ in degrees

Asector =

360

Substitute 50 for θ and 8 for r .

Asector Asector
Asector

= 5π(64)
36
= 320π
36
= 80π
9

Rewrite the fraction and the power . Multiply .
Rewrite .

The area of the sector is 80 π square meters .
9

Page  88 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 . Circle E is shown .

12 yd.

145°

E

D

C

What is the length of CD ?
A . 29 π yd .
72
B .   29 π yd .
6
C .   29 π yd .
3
D .   29 π yd .
2

2 . Circle Y is shown .

Y   75°	

9 cm

X

Z

What is the area of the shaded part of the circle?
A . 57 π cm2
4
B .   135 π cm2
8
C .   405 π cm2
8
D .  513 π cm2
8

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 89 of 230

3 . The spokes of a bicycle wheel form 10 congruent central angles . The diameter of the circle formed by the outer edge of the wheel is 18 inches .

What is the length, to the nearest tenth inch, of the outer edge of the wheel between two consecutive spokes?

A .  1 .8 inches B . 5 .7  inches C . 11 .3 inches D . 25 .4 inches

Answers to Unit 3 .2 Sample Items   1 .   C	2 .   D	3 . B

Page  90 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

3 .3 Explain Volume Formulas and Use Them to Solve Problems

MGSE9-12 .G .GMD .1 Give informal arguments for geometric formulas .
a . Give informal arguments for the formulas of the circumference of a circle and area of a circle using dissection arguments and informal limit arguments .
b . Give informal arguments for the formula of the volume of a cylinder, pyramid, and cone using Cavalieri’s principle .
MGSE9-12 .G .GMD .2 Give an informal argument using Cavalieri’s principle for the formulas for the volume of a sphere and other solid figures .
MGSE9-12 .G .GMD .3 Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems .

KEY IDEAS
The volume of a figure is a measure of how much space it takes up . Volume is a measure of capacity .

r

The formula for the volume of a cylinder is V = πr2h, where r is the radius and h is the height . The volume formula can also be given as V = Bh, where B is the area of the base . In a cylinder, the base is a circle and           the area of a circle is given by A = πr2 . Therefore, V = Bh = πr2h .

h

B

When a cylinder and a cone have congruent bases and equal heights, the volume of exactly three cones will fit into the cylinder . So, for a cone and cylinder that have the same radius r and height h, the volume of the cone is one-third of the volume of the cylinder .
The formula for the volume of a cone is V = 1 πr2h, where r is the radius and h is the height .

r

3

h	h
 	
B	B

r

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 91 of 230

1
The formula for the volume of a pyramid is V = 3 Bh, where B is the area of the base and h is the height .

h

B

The formula for the volume of a sphere is V = 4 πr3, where r is the radius .
3

r

Cavalieri’s principle states that if two solids are between parallel planes and all cross sections at equal distances from their bases have equal areas, the solids have equal volumes .  For example, this cone     and this pyramid have the same height and the cross sections have the same area, so they have equal volumes .

h

h

Page  92 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	What is the volume of the cone shown below?

17 cm

16 cm

Solution:
The diameter of the cone is 16 cm . So the radius is 16 cm ÷ 2 = 8 cm . Use the Pythagorean Theorem,
a2 + b2 = c2, to find the height of the cone . Substitute 8 for b and 17 for c and solve for a:
a2 + 82 = 172
a2  + 64 = 289
a2  = 225
a = 15

The formula for the volume of a cone is V = 1 πr2h . Substitute 8 for r and 15 for h:
3
V = 1 πr2h = 1 π(8)2(15)
3	3
The volume is 320π cm3 .
◆	A sphere has a radius of 3 feet . What is the volume of the sphere?
Solution:
The formula for the volume of a sphere is V = 4 πr3 . Substitute 3 for r and solve .
3

V = 4 πr3
3
V = 4 π(3)3
3
V = 4 π(27) 3
V = 36π ft3

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 93 of 230

◆	A cylinder has a radius of 10 cm and a height of 9 cm .  A cone has a radius of 10 cm and a height of     9 cm . Show that the volume of the cylinder is three times the volume of the cone .
Solution:
The formula for the volume of a cylinder is V = πr2h . Substitute 10 for r and 9 for h:
V = πr2h
= π(10)2(9)
= π(100)(9)
= 900π cm3
The formula for the volume of a cone is V = 1πr2h. Substitute 10 for r and 9 for h:
3

V = 1 πr2h
3
= 1 π(10)2(9)
3
= 1 π(100)(9)
3
= 300π cm3 Divide: 900π ÷ 300π = 3

Page  94 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	
10 m

10 m

Cylinder A and Cylinder B are shown below . What is the volume of each cylinder?

12 m	12 m

Cylinder A	Cylinder B

Solution:
To find the volume of Cylinder A, use the formula for the volume of a cylinder, which is V = πr2h . Divide the diameter by 2 to find the radius: 10 ÷ 2 = 5 . Substitute 5 for r and 12 for h:

V	= πr h

2
Cylinder A
= π(5)2(12)
= π(25)(12)
= 300π m3
≈ 942 m3

10 m

10 m

10 m

10 m

Notice that Cylinder B has the same height and the same radius as Cylinder A . The only difference is that Cylinder B is slanted . For both cylinders, the cross section at every plane parallel to the bases is a circle with the same area . By Cavalieri’s principle, the cylinders have the same volume; therefore, the volume of Cylinder B is 300π m3, or about 942 m3 .

12 m	12 m

Cylinder A	Cylinder B

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 95 of 230

SAMPLE ITEMS
1 . Jason constructed two cylinders using solid metal washers . The cylinders have the same height, but one of the cylinders is slanted as shown .

Which statement is true about Jason’s cylinders?

A . The cylinders have different volumes because they have different radii .
B . The cylinders have different volumes because they have different surface areas .
C . The cylinders have the same volume because the washers are solid .
D .  The cylinders have the same volume because they have the same cross-sectional area at every  plane parallel to the bases .

2 .   What is the volume of a cylinder with a radius of 3 in . and a height of 9 in .?
2
A . 81 π in .3 2
B . 27 π in .3 4
C . 27 π in .3 8
D . 9 π in .3 4

Answers to Unit 3 .3 Sample Items  1 .   D	2 . A

Page  96 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

UNIT 4: EXTENDING THE NUMBER SYSTEM
This unit investigates properties of square roots  and  rewriting  expressions  involving  radicals .  Sum  and product of rational and irrational numbers are explored . Closure properties are explored in terms of number systems as well as polynomials .

4 .1 Use Properties of Rational and Irrational Numbers

MGSE9-12 .N .RN .2 Rewrite expressions involving radicals .
MGSE9-12 .N .RN .3 Explain why the sum or product of rational numbers is rational; why the sum of a rational number and an irrational number is irrational; and why the product of a nonzero rational number   and an irrational number is irrational .

KEY IDEAS
The nth root of a number is the number that must be used as a factor n times to equal a given value . It can be notated with radicals and indices or with rational exponents . When a root does not have an index, the index is assumed to be 2 .

Examples:

indexÏ}radicand

—	}
Ï49 = Ï7 ∙ 7 = 7
—	}	}
Ï18 = Ï9 ∙ 2 = 3Ï2
—	}	}
Ïx5  = Ïx4 ∙ x = x2 Ïx

q

A rational number is a real number that can be represented as a ratio p such that p and q are both integers and q ≠ 0 . All rational numbers can be expressed as a decimal that stops or repeats .
3	}
Examples: –0 .5, 0, 7, 2, 0 .26

}

q

An irrational number is a real number that cannot be expressed as a ratio p such that p and q are both integers and q ≠ 0 . Irrational numbers cannot be represented by decimals that stop or repeat .
Examples: Ï}3, π, Ï5
2
The sum, product, or difference of two rational numbers is always a rational number . The quotient of two
rational numbers is always rational when the divisor is not zero . The sum of an irrational number and a rational number is always irrational . The product of a nonzero rational number and an irrational number is always irrational . The sum or product of rational numbers is rational .
Example: The sum is irrational since it cannot be written as a fraction and the sum of a rational number and an irrational number is irrational .
Let a be an irrational number, and let b be a rational number . Suppose that the sum of a and b  is a rational number, c . If you can show that this is not true, it is the same as proving the original statement .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 97 of 230

p

n

q	n

Let b = q, where p and q are integers and q ≠ 0 . Let c = Substitute p and m for b and c . Then subtract to find a .
a + b = c

m, where m and n are integers and n ≠ 0 .

p	m a + q = n
a =  m – p
n	q
a = mq – pn
nq	nq

a =

mq – pn nq

nq

The set of integers is closed under multiplication and subtraction, so mq – pn

is an integer divided

by an integer . This means that a is rational . However, a was assumed to be irrational, so this is a contradiction . This means that c must be irrational . So, the sum of an irrational number and a rational number is irrational .
Example: Is the sum of 0 .75 and –2 .25 a rational or an irrational number?
The sum is a rational number . The sum is –1 .50, which can be rewritten as the fraction –150 .
100
Example: Is the sum of 1 and Ï}2 a rational or an irrational number?
2

The sum is an irrational number . The square root of 2 is a decimal that does not terminate or repeat .
Therefore, the actual sum can be written only as 1 + Ï}2 .
2

}

Example: Is the product of –0 .5 and Ï3 a rational or an irrational number? Explain your reasoning .

}

The product is an irrational number . The square root of 3 is a decimal that does not terminate or repeat . Therefore, the product can be written only as –0 .5Ï3 .
To rewrite square root expressions, you can use properties of square roots where a and b are real numbers with a > 0 and b > 0 .

—	} }
•	Product Property: Ïab = ÏaÏb
•	

Ï

}a
•	Quotient  Property:   b

Ï}a

}

=
Ïb

Examples:

—	}	}	}	}

Ï32 = Ï16 ∙ 2 = Ï16 ∙ Ï2 = 4Ï2

—	}	}	}
3Ï700 = 3Ï7 ∙ 100 = 3 ∙ 10Ï7 = 30Ï7
—	}

}

25

Ï25

5

Ï 9  =  Ï9  = 3
Page  98 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Two  radical expressions that have the same index and the same radicand are called like radicals . To  add          or subtract like radicals, you can use the Distributive Property .
Example:
—	}
Ï8 + Ï2
—	}
Ï4 ∙ 2 + Ï2	Factor out the perfect square .
—	}	}
Ï4 ∙ Ï2 + Ï2	Use the product property of square roots .
—	}

}

2Ï2 + Ï2	Compute the square root .

}

(2 + 1)Ï2	Use the Distributive Property . 3Ï2	Add .

REVIEW EXAMPLES
◆	Rewrite Ï}2(Ï}12  – Ï}3) .
Solution:
Ï}2(Ï}12  – Ï}3)	Original expression
—	}	}	}
Ï2 ∙ Ï12  – Ï2 ∙ Ï3	Distributive Property
—	}	}	}
Ï2 ∙ Ï4 ∙ 3  – Ï2 ∙ Ï3	Factor out the perfect square .
—	}	}	}	}
Ï2 ∙ Ï4 ∙ Ï3 – Ï2  ∙ Ï3	Product Property
—	}	}	}
2 ∙ Ï2 ∙ Ï3 – Ï2  ∙ Ï3	Evaluate the square root .
—	}
2Ï6 – Ï6	Product Property

}

(2 – 1)Ï6	Distributive Property

}

Ï6	Subtract .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 99 of 230

◆	18Ï

}

Write		in an equivalent form where no radical has a perfect square factor 25
and there is no radical in the denominator .
Solution:

Ï

}

18
25

}

}

Ï18 Ï25

}

Ï18 5

}

Ï9 ∙ 2
5

Original expression Quotient Property
Evaluate the square root .

Factor out the perfect square .

—	}

Ï9 ∙ Ï2
5

}

3Ï2
5

Product Property

Evaluate the square root .

◆	
}

Write Ï(4p2)3   in an equivalent form without a square root . Assume that p is nonnegative .
Solution:

}

Ï(4p2)3	Original expression

}

Ï64p6	Power of Powers Rule

— }
Ï64Ïp6	Product Property

8p3	Evaluate  the  square  roots .

◆	Look at the expression below .

(5 + Ï}2) + 2Ï}2

Is the value of the expression rational or irrational? Explain . Solution:
Use the associative property to add like terms .

}

(5 + Ï}2) + 2Ï}2 = 5 + (Ï}2  + 2Ï}2)
= 5 + 3Ï2

—	}

}

}

Since 3Ï2 is the product of a rational number, 3, and an irrational number, Ï2, and the product of an irrational number and a rational number is always irrational, 3Ï2 is irrational . Since 5 is a rational

}

number and 3Ï2 is an irrational number and the sum of a rational number and an irrational number is always irrational, the value of the expression 5 + 3Ï2 is irrational .
Page  100 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	Explain why the product of π ∙ 5 is irrational .
Solution:
The product of π ∙ 5 is 5π . Assume 5π is a rational number . The quotient of two rational numbers is rational . Therefore, 5π would be a rational number, and 5π = π . This would mean that π is a rational
5	5
number . This is a contradiction because π is an irrational number . So 5π is irrational .

◆	Is the value of the expression Ï}8(5Ï}8  + Ï}2) rational or irrational? Explain how you found your answer .
Solution:
Ï}8(5Ï}8  + Ï}2) = 5Ï}8  ∙ Ï}8  + Ï}2  ∙ Ï}8

—	}
= 5Ï64 + Ï16
= 5 ∙ 8 + 4
= 40 + 4
= 44

The value of the expression is 44, which is rational .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 101 of 230

SAMPLE ITEMS
—	}
1 . Which expression is equivalent to Ï32 – Ï8?

}

}

A .  2Ï2

}

B .  6Ï2

}

C .  2Ï6
D . 2Ï10

27Ï

}
2 . Which expression is equivalent to	?
16

}

A . 4Ï3 3

}

B . 2Ï3 3

}

C . 3Ï3 4

}

D . 4Ï3 9

3 . Which expression has a value that is a rational number?

}

A . Ï10 + 16
B .   2(Ï}5  + Ï}7)
—	}

}

C . Ï9 + Ï4
D . Ï3 + 0

Page  102 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

4 .   Which statement is true about the value of (Ï}8 + 4) ∙ 4?
A . It is rational because the product of two rational numbers is rational .
B . It is rational because the product of a rational number and an irrational number is rational .
C . It is irrational because the product of two irrational numbers is irrational .
D . It is irrational because the product of an irrational number and a rational number is irrational .

5 . Let a be a nonzero rational number and b be an irrational number . Which of these MUST be a rational number?
A . b + 0 B . a + a C . a + b D . b + b

Answers to Unit 4 .1 Sample Items
1 .   A	2 .   C	3 .   C	4 .   D	5 . B

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 103 of 230

4 .2 Perform Arithmetic Operations on Polynomials

MGSE9-12 .A .APR .1 Add, subtract, and multiply polynomials; understand that polynomials form a system analogous to the integers in that they are closed under these operations .

KEY IDEAS
A polynomial is an expression made from one or more terms that involve constants, variables, and exponents . Examples: 3x, x3 + 5x2 + 4, a2b – 2ab + b2
To  add and subtract polynomials, combine like terms . In a polynomial, like terms have the same variables      and are raised to the same powers .
Examples:
7x + 6 + 5x – 3 = 7x + 5x + 6 – 3	13a + 1 – (5a – 4) = 13a + 1 – 5a + 4
= 12x + 3	= 13a – 5a + 1 + 4
= 8a + 5

To multiply polynomials, use the Distributive Property . Multiply every term in the first polynomial by every term in the second polynomial . To completely simplify, add like terms after multiplying .
Example:
(x + 5)(x – 3) = (x)(x) + (–3)(x) + (5)(x) + (5)(–3)
= x2  – 3x + 5x – 15
= x2  + 2x – 15

The multiplication can also be represented with tiles and area models .

x2			
5x			
						
–3x			–15			

x	+	5
x
– 3

x	+	5

x2	
5x	

–3x	
–15	

x
– 3

Polynomials are closed under addition, subtraction, and multiplication, similar to the set of integers . This means that the sum, difference, or product of two polynomials is always a polynomial .

Page  104 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	The dimensions of a rectangle are shown .

3x + 8

5x + 2

What is the perimeter, in units, of the rectangle? Solution:
Substitute 5x + 2 for l and 3x + 8 for w in the formula for the perimeter of a rectangle:
P = l + l + w + w P = 2l + 2w
P = 2(5x + 2) + 2(3x + 8)
P = 10x + 4 + 6x + 16
P = 10x + 6x + 4 + 16
P = 16x + 20 units

◆	Rewrite the expression (x3 + 2x2 – x) – (–x3 + 2x2 + 6) .
Solution:
Distribute the negative and then combine like terms:
(x3 + 2x2 – x) – (–x3 + 2x2 + 6) = x3 + 2x2 – x + x3 – 2x2 – 6
= x3  + x3  + 2x2  – 2x2  – x – 6
= 2x3  – x – 6

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 105 of 230

◆	The dimensions of a patio, in feet, are shown below .

2x – 3

4x + 1

What is the area of the patio, in square feet? Solution:
Substitute 4x + 1 for b and 2x – 3 for h in the formula for the area of a rectangle:
A = bh
A = (4x + 1)(2x – 3)
A = 8x2  – 12x + 2x – 3
A = 8x2  – 10x – 3 square feet

Page  106 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 . What is the product of 7x – 4 and 8x + 5 ?
A .   15x + 1
B .   30x + 2
C .   56x2  +  3x – 20
D .   56x2  –  3x + 20

2 . A model of a house is shown .

6x – 4

12x + 3

14x + 13

What is the perimeter, in units, of the model?
A . 32x + 12 B . 46x +  25 C . 50x + 11 D . 64x + 24

3 . Which expression has the same value as the expression (8x2 + 2x – 6) – (5x2 – 3x + 2)?
A .  3x2  – x – 4 B . 3x2  + 5x – 8 C . 13x2 – x – 8
D . 13x2 – 5x – 4

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 107 of 230

4 . Kelly makes two different-sized ceramic tiles in the shape of a right isosceles triangle . This diagram shows the leg lengths of the smaller tile .

3 inches

3 inches

Kelly makes the larger tile by increasing the length of each leg of the smaller tile by x inches . Which expression represents the length, in inches, of the hypotenuse of the larger tile?

A . 18 + x
B . (x + 3)2

}

}

C . (x + 3)Ï2
D . 3Ï2 + x

Answers to  Unit  4 .2  Sample  Items 1 .   C	2 .   C	3 .   B	4 . C

Page  108 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

UNIT 5: QUADRATIC FUNCTIONS
This unit investigates quadratic functions . Students study the structure of expressions and write expressions in equivalent forms . They solve quadratic equations by inspection, by completing the square, by factoring, and by using the quadratic formula . Students also graph quadratic functions and analyze characteristics of those functions, including end behavior . They write functions for various situations and build functions from other functions, using operations as needed . Given two-variable data, students fit a function to the data and use it to make predictions .
5 .1 Interpret the Structure of Expressions

MGSE9-12 .A .SSE .1 Interpret expressions that represent a quantity in terms of its context .
MGSE9-12 .A .SSE .1a Interpret parts of an expression, such as terms, factors, and coefficients, in context .
MGSE9-12 .A .SSE .1b Given situations which utilize formulas or expressions with multiple terms and/or factors, interpret the meaning (in context) of individual terms or factors .

MGSE9-12 .A .SSE .2 Use the structure of an expression to rewrite it in different equivalent forms . For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2) (x2 + y2) .

KEY IDEAS
An algebraic expression contains variables, numbers, and operation symbols . For example, two expressions are 5x2 – 10x + 15 and 30x2 + 6x .
A term in an algebraic expression can be a constant, a variable, or a constant multiplied by a variable or variables . Every term is separated by a plus sign . For example, in the expression 5x2 – 3x + 8, the terms are 5x2, –3x, and 8 .
A coefficient is the constant number that is multiplied by a variable in a term . For example, in the expression 5x2 – 3x + 8, the coefficient of the x2 term is 5 and the coefficient of the x term is –3 .
A common factor is a variable or number that terms can be divided by without a remainder . Factors are numbers multiplied together to get another number . Example: For the terms 30x2 and 6x, the common factors are 1, 2, 3, 6, and x .
A common factor of an expression is a number or term that the entire expression can be divided by without a remainder . Example: For the expression 30x2 + 6x, a common factor of the expression is 6x because 30x2 + 6x = 6x (5x + 1) . Notice that any of the common factors discussed in the previous example would be common factors of the expression .
If parts of an expression are independent of each other, the expression can be interpreted in different
ways . Example: In the expression 1 h(b + b ), the factors h and (b + b ) are independent of each other . It
2	1	2	1	2
can be interpreted as the product of h and a term that does not depend on h .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 109 of 230

There are many methods for factoring quadratic expressions to help find equivalent forms . We will look at just two possible ways . Remember that the standard form of a quadratic function is ax2 + bx + c .
Suppose we have the expression 2x2 + 4x – 30 . Each term has a common factor of 2 . We can factor out that 2 to have a quadratic function that is easier to factor .
2x2 + 4x – 30 = 2(x2 + 2x – 15)
Now, looking at the quadratic (x2 + 2x – 15), there is no coefficient on the x2 . We consider the factors of the constant term, c = –15, that would add up to the b coefficient, 2 .  The factors of –15 are 1 and –15,     or –1 and 15, or 3 and –5, or –3 and 5 . Of these factors, only one pair adds to 2: –3 and 5 . Therefore, the quadratic function can be factored to (x2 + 2x – 15) = (x + 5)(x – 3) . So going back to what we began with: 2x2 + 4x – 30 = 2(x + 5)(x – 3) .
The second method we will explore is called the ac method . We will use the expression 8x2 + 10x – 3 to explain how this method works . The terms have no common factor and there is a coefficient on the x2 term . Using the standard form of a quadratic function, ax2 + bx + c, we take the coefficients a and c and multiply them .
8 ∙ (–3) = –24
Now we consider the factors of –24 that would add to 10 (the b coefficient), which are 12 and –2 . Now we rewrite our bx term using these factors as shown . Then we can use grouping and common factors to factor further .
8x2 + 10x – 3 = 8x2 + (12x – 2x) – 3
= (8x2  + 12x) + (–2x – 3)
= 4x(2x + 3) + (–1)(2x + 3)
= (4x – 1) (2x + 3)

The structure of some expressions can be used to help rewrite them . For example, some fourth-degree expressions are in quadratic form .
Example: x4 + 5x2 + 4 = (x2 + 4)(x2 + 1)
Example: x4 – y4 = (x2 – y2)(x2 + y2) = (x – y)(x + y)(x2 + y2)

Page  110 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Consider the expression 3n2 + n + 2 . a . What is the coefficient of n?
b . What terms are being added in the expression?
Solution:
a . 1
b . 3n2, n, and 2
◆	Factor the expression 16a2 – 81 .
Solution:
The expression 16a2 – 81 is quadratic in form because it is the difference of two squares (16a2 = (4a)2 and 81 = 92) and both terms of the binomial are perfect squares . The difference of squares can be factored as
x2 – y2 = (x + y)(x – y)
16a2  – 81	Original expression
(4a +  9)(4a – 9)	Factor the binomial (difference of two squares) .
◆	Factor the expression 12x2 + 14x – 6 .
Solution:
12x2  +  14x – 6	Original expression
2(6x2   +  7x – 3)	Factor the trinomial (common factor) . 2(3x –  1)(2x + 3)	Factor .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 111 of 230

SAMPLE ITEMS
1 . In which expression is the coefficient of the n term –1?
A . 3n2 + 4n – 1  B . –n2  + 5n + 4  C . –2n2  – n + 5 D . 4n2 + n – 5

2 . Which expression is equivalent to 121x2 – 64y2 ?
A . (11x – 16y)(11x + 16y)  B . (11x – 16y)(11x – 16y)  C . (11x + 8y)(11x + 8y)
D .   (11x +  8y)(11x – 8y)

3 . The expression s2 is used to calculate the area of a square, where s is the side length of the square . What does the expression (8x)2 represent?
A .   the area of a square with a side length of 8     B .   the area of a square with a side length of 16   C .  the area of a square with a side length of 4x   D . the area of a square with a side length of 8x

Answers to Unit 5 .1 Sample Items   1 .   C	2 .   D	3 . D

Page  112 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

5 .2 Write Expressions in Equivalent Forms to Solve Problems

MGSE9-12 .A .SSE .3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression .
MGSE9-12 .A .SSE .3a Factor any quadratic expression to reveal the zeros of the function defined by the expression .
MGSE9-12 .A .SSE .3b Complete the square in a quadratic expression to reveal the maximum and minimum value of the function defined by the expression .

KEY IDEAS
The zeros, roots, or x-intercepts of a function are the values of the variable that make the function equal          to zero . When the function is written in factored form, the Zero Product Property can be used to find the    zeros of the function . The Zero Product Property states that if the product of two factors is zero, then one        or both of the factors must be zero . So, the zeros of the function are the values that make either factor       equal to zero .
Example:
x2  – 7x + 12  = 0	Original equation (x – 3)(x – 4) = 0	Factor .
Set each factor equal to zero and solve .
x – 3 = 0	x – 4 = 0
x = 3	x = 4
The zeros of the function y = x2 – 7x + 12 are x = 3 and x = 4 .
To complete the square of a quadratic function means to write a function as the square of a sum . The standard form for a quadratic expression is ax2 + bx + c, where a ¹ 0 . When a = 1, completing the square

of the function x2  + bx = d gives (x +  b)2  = d + (b)2

¹ 1, factor

2
the value of a from the expression .

Example:

2 . To complete the square when the value a

To complete the square, take half of the coefficient of the x-term, square it, and add it to both sides of the equation .
x2  + bx = d	Original expression

x2  + bx + (b)2  = d + (b)2	b	b

2	2

The coefficient of x is b . Half of b is	. Add the  square of	to both sides of
2	2	the equation .
(x +  b)2  = d + (b)2	The expression on the left side of the equation is a perfect square trinomial .
2	2	Factor to write it as a binomial squared .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 113 of 230

This figure shows how a model can represent completing the square of the expression x2 + bx, where

x2	
bx
2	
bx
2		

b is positive .

This model represents the expression x2 + bx . To complete the square, create a model that is a square .	

x		x

x2			b

bx		

Split the rectangle for bx into two rectangles that represent b x .
2	

x		x

x2		
bx
2	b	
bx
2	

Rearrange the two rectangles that represent b x .
2		

x

b
2		
x		b
2		
The missing piece of the square measures b by b . Add a square
2	2
with these dimensions to complete the model of the square . The
large square has a side length of x  +  b, so this model represents
2
(x +  b)2	(b)2
2    minus   2    .

x2  + bx = (x +  b)2  – (b)2
2	2   .		

x

b
2		
x	

	
b
2		

x2	
bx
2	
bx
2	b 2
2	

Page  114 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Examples:
Complete the square:
x2 + 3x + 7
(x2  + 3x + (3)2) + 7 – (3)2

Complete the square:
x2 + 3x + 7 = 0
x2  + 3x + (3)2  = –7 + (3)2

2	2

2

4

(x + 3)2  + 19

2	2

2

4

(x + 3)2  = – 19

Every quadratic function has a minimum or a maximum . This  minimum  or  maximum  is  located  at  the vertex (h, k) . The vertex (h, k) also identifies the axis of symmetry and the minimum or maximum value of    the function . The axis of symmetry is x = h .
Example: The quadratic equation f(x) = x2 – 4x – 5 is shown in this graph . The minimum of the function occurs at the vertex (2, –9) . The zeros or x-intercepts of the function are (–1, 0) and (5, 0) . The axis of symmetry is x = 2 .

10

8

6

4

2

–10

2	4	6	8

10

–

–8

–6

–4

–2

0

1

0

2

–

4

–

6

–

8

–

y

x

The vertex form of a quadratic function is f(x) = a(x – h)2 + k where (h, k) is the vertex . One way to convert an equation from standard form to vertex form is to complete the square .

(	)

The vertex of a quadratic function can also be found by using the standard form and determining the
value  –b  . The vertex is   –b , f (–b )  .

2a
Important Tips

2a	2a

When you complete the square, make sure you are only changing the form of the expression and not changing the value .

When completing the square in an expression, add and subtract half of the coefficient of the x-term squared .

When completing the square in an equation, add half of the coefficient of the x-term squared to both sides of the equation .
Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 115 of 230

REVIEW EXAMPLES
◆	Write f(x) = 2x2 + 12x + 1 in vertex form .
Solution Method 1:
The function is in standard form, where a = 2, b = 12, and c = 1 .
2x2  + 12x + 1	Original expression
2(x2  +  6x) + 1	Factor out 2 from the quadratic and linear terms .
2(x2  +  6x +  (3)2  – (3)2) + 1	Add and subtract the square of half of the coefficient of the
linear term .
2(x2  + 6x + (3)2) – 2(9) + 1	Remove the subtracted term from the parentheses .
Remember to multiply by a . 2(x2   +  6x +  (3)2) – 17	Combine the constant terms .
2(x +  3)2 – 17	Write the perfect square trinomial as a binomial squared . The vertex of the function is (–3, –17) .
Solution Method 2:

(	)

2a

2a

2a

The vertex of a quadratic function can also be found by writing the polynomial in standard form and determining the value of  –b  . The vertex is   –b , f (–b )  .
For f(x) = 2x2 + 12x + 1, a = 2, b = 12, and c = 1 .
–b = –12 = –12 = –3
2a	2(2)	4
f (–3) = 2(–3)2 + 12(–3) + 1
= 2(–3)2 – 36 + 1
= 18 – 36 + 1
= –17

The vertex of the function is (–3, –17) .
◆	The function h(t) = –t 2 + 8t + 2 represents the height, in feet, of a stream of water being squirted out of a fountain after t seconds . What is the maximum height of the water?
Solution:
The function is in standard form, where a = –1, b = 8, and c = 2 .

The x-coordinate of the vertex is –b =	–8

= 4 .

2a	2(–1)
The y-coordinate of the vertex is h(4) = –(4)2 + 8(4) + 2 = 18 .
The vertex of the function is (4, 18) . So, the maximum height of the water occurs at 4 seconds and is 18 feet .

Page  116 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	What are the zeros of the function represented by the quadratic expression x2 + 6x – 27?
Solution:
Factor the expression: x2 + 6x – 27 = (x + 9)(x – 3) .
x   +  9

x2	9x	
–3x	–27	

x
– 3
Set each factor equal to 0 and solve for x . x +  9 = 0	x – 3 = 0
x = –9	x = 3

The zeros are x = –9 and x = 3 . This means that f(–9) = 0 and f(3) = 0 .
◆	What are the zeros of the function represented by the quadratic expression 2x2 – 5x – 3?
Solution:
Factor the expression: 2x2 – 5x – 3 = (2x + 1)(x – 3) .  Set each factor equal to 0 and solve for x .
2x + 1 = 0	x – 3 = 0

x = – 1
2
1

x = 3

The zeros are x = – 2 and x = 3 .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 117 of 230

SAMPLE ITEMS
1 . What are the zeros of the function represented by the quadratic expression 2x2 + x – 3?

A .

3
x = – 2 and x = 1

2 .	B .

C .

D .

Wh	x = – 3  and x = 1
x = –1 and x =  2
3
x = –1 and x = – 3
2

at is the vertex of the graph of f ( x) = x2 + 10x – 9?	
	A .
B .
C .
D .	(5, 66)
(5, –9)
(–5, –9)
(–5, –34)	

2

3 . Which of these is the result of completing the square for the expression x2 + 8x – 30?
A .   (x + 4)2  – 30
B .   (x + 4)2  – 46
C .   (x + 8)2  – 30
D .  (x + 8)2  – 94

Page  118 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

4 . The expression –x2 + 70x – 600 represents a company’s profit for selling x items . For which number(s) of items sold is the company’s profit equal to $0?
A .	0 items
B . 35 items
C . 10  items  and  60  items
D .   20  items  and  30 items

Answers to  Unit  5 .2  Sample  Items 1 .   A	2 .   D	3 .   B	4 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 119 of 230

5 .3 Create Equations That Describe Numbers or Relationships

MGSE9-12 .A .CED .1 Create equations and inequalities in one variable and use them to solve problems . Include equations arising from linear, quadratic, and exponential functions (integer inputs only) .
MGSE9-12 .A .CED .2 Create linear, quadratic, and exponential equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes  with  labels  and  scales . (The phrase “in two or more variables” refers to formulas like the compound interest formula, in which

n

A = P(1 +  r )nt   has multiple variables .)
MGSE9-12 .A .CED .4 Rearrange formulas to highlight a quantity of interest using the same reasoning as in solving equations . Examples: Rearrange Ohm’s law V = IR to highlight resistance R; Rearrange area of a circle formula A = πr 2 to highlight the radius r.

KEY IDEAS
Quadratic equations can be written to model real-world situations .
Here are some examples of real-world situations that can be modeled by quadratic functions:
•	Finding the area of a shape: Given that the length of a rectangle is 5 units more than the width, the   area of the rectangle in square units can be represented by A = x(x + 5), where x is the width and x + 5 is the length .
•	Finding the product of consecutive integers: Given a number, n, the next consecutive number is n + 1 and the next consecutive even (or odd) number is n + 2 . The product, P, of two consecutive numbers is P = n(n + 1) .
•	Finding the height of a projectile that is dropped: When heights are given in metric units, the equation
•	
used is h(t) = –4 .9t2 + v

t + h , where v is the initial velocity, in meters per second, and h

is the

o	o	o	o
initial height, in meters . The coefficient –4 .9 represents half the force of gravity . When heights are

given in customary units, the equation used is h(t) = –16t2 + v

t + h , where v

is the initial velocity,

o	o	o
in feet per second, and ho is the initial height, in feet . The coefficient –16 represents half the force of gravity . For example, the height, in feet, of a ball thrown with an initial velocity of 60 feet per second and an initial height of 4 feet can be represented by h(t) = –16t2 + 60t + 4, where t is seconds .

You can use the properties of equality to solve for a variable in an equation . Use inverse operations on both sides of the equation until you have isolated the variable .
Example: What is the value of r when S = 0 for the equation S = 2πr 2 + 2πrh for r ? First, factor the expression 2πr2 + 2πrh .

Next, set each factor equal to 0 .

2πr(r + h)

2πr = 0,	r + h = 0
r = 0,	r = –h

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To graph a quadratic equation, find the vertex of the graph and the zeros of the equation . The axis of symmetry goes through the vertex and divides the graph into two sides that are mirror images of each other . To draw the graph, you can plot points on one side of the parabola and use symmetry to find the corresponding points on the other side of the parabola .
Example: Graph the quadratic equation y = x2  + 5x +  6 .
First, we can find the zeros by solving for x when y = 0 . This is where the graph crosses the x-axis .
0 = x2 + 5x + 6
0 = (x + 2)(x + 3)
x + 2 = 0, x + 3 = 0
x = –2,	x = –3; this gives us the points (–2, 0) and (–3, 0) .

Next, we can find the axis of symmetry by finding the vertex . The axis of symmetry is the

equation x = –

 b 
2a . To find the vertex, we first find the axis of symmetry .

x = – 5 2(1)

= – 5
2

Now we can find the value of the y-coordinate of the vertex .

2

2

y = (– 5)2  + 5(– 5 ) + 6

4

2

= 25 + (– 25) + 6
= 25 + – 50 + 24
4	4	4
= – 25 + 24
4	4
= – 1
4

So, the vertex is located at

– 2, – 4 .

( 5	1)
Next, we can find two more points to continue the curve . We  can use the y-intercept to find the first of   the two points .
y = (0)2 + 5(0) + 6 = 6 . The y-intercept is at (0, 6) .

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This point is 2 .5 more than the axis of symmetry, so the last point will be 2 .5 less than the axis of symmetry . The point 2 .5 less than the axis of symmetry with a y-value of 6 is (–5, 6) .

6		
x	= – 5
2		
4		
			
2		
					
–	4	–	2	0		

y

x

The axis of symmetry is the midpoint for each corresponding pair of x-coordinates with the same
y-value . If (x1, y) is a point on the graph of a parabola and x = h is the axis of symmetry, then (x2, y) is

also a point on the graph, and x2

can be found using this equation: x1 + x2 = h . In the example shown,
2

we can use the zeros (–3, 0) and (–2, 0) to find the axis of symmetry .
–3 + –2	5

REVIEW EXAMPLES

2	= – 2 = – 2 .5, so x = –2 .5

◆	The product of two consecutive positive integers is 132 .       a . Write an equation to model the situation .
b . What are the two consecutive integers?
Solution:
a . Let n represent the lesser of the two integers . Then n + 1 represents the greater of the two integers . So, the equation is n(n + 1) = 132 .
b . Solve the equation for n .
n(n + 1) = 132	Original equation
n2  +  n = 132	Distributive Property
n2  + n – 132  = 0	Subtraction Property of Equality (n + 12)(n – 11)  = 0	Factor .

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Set each factor equal to 0 and solve for n .
n + 12 = 0	n – 11 = 0
n = –12	n = 11

Because the two consecutive integers are both positive, n = –12 cannot be the solution .  So, n = 11  is the solution, which means that the two consecutive integers are 11 and 12 .
◆	The formula for the volume of a cylinder is V = πr2h.
a . Solve the formula for r .
b . If the volume of a cylinder is 200π cubic inches and the height of the cylinder is 8 inches, what is  the radius of the cylinder?
Solution:
a . Solve the formula for r .
V = πr2h	Original formula
 V = r2	Division Property of Equality
πh

 VÏ

}
±	= r	Take the square root of both sides .
πh

 VÏ

}
= r	Choose the positive value because the radius cannot be negative .
πh

b . Substitute 200π for V and 8 for h and evaluate .

—	}	}
r = Ï V  = Ï200π  = Ï200  = Ï}25  = 5
πh	π(8)	8
The radius of the cylinder is 5 inches .
◆	Graph the function represented by the equation y = 3x2 – 6x – 9 .
Solution:
Find the zeros of the equation .
0 = 3x2  – 6x – 9	Set the equation equal to 0 . 0 = 3(x2  – 2x – 3)	Factor out 3 .
0 = 3(x – 3)(x + 1)	Factor .
0 = (x – 3)(x + 1)	Division Property of Equality Set each factor equal to 0 and solve for x .
x – 3 = 0	x + 1 = 0
x = 3	x = –1

The zeros are at x = –1 and x = 3 .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 123 of 230

Find the vertex of the graph .

–b = –(–6) = 6 = 1

2a	2(3)	6

Substitute 1 for x in the original equation to find the y-value of the vertex:
3(1)2 – 6(1) – 9 = 3 – 6 – 9
= –12

Graph the two x-intercepts (3, 0) and (–1, 0) and the vertex (1, –12) .
Another descriptive point is the y-intercept . You can find the y-intercept by substituting 0 for x .
y = 3x2  – 6x – 9
y = 3(0)2  – 6(0) – 9
y = –9

You can find more points for your graph by substituting x-values into the function . Find the y-value when
x = –2 .
y = 3x2  – 6x – 9
y = 3(–2)2  – 6(–2) – 9
y = 3(4) + 12 – 9
y = 15

Page  124 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Graph the points (0, –9) and (–2, 15) . Then use the concept of  symmetry  to  draw  the  rest  of  the function . The axis of symmetry is x = 1 . So, the mirror image of (0, –9) is (2, –9) and the mirror image          of (–2, 15) is (4, 15) .

16
14
12
10
8
6
4
2

1 2 3 4

16

–

14

–

12

–

10

–

–8

–6

–4

5

2

1– 0

2 –

3 –

4 –

5 –

–

y

x

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 125 of 230

◆	A garden measuring 8 feet by 12 feet will have a walkway around it . The walkway has a uniform width, and the area covered by the garden and the walkway is 192 square feet . What is the width of the walkway?

8 feet

12 feet

x

The length of the garden and walkway is x + 12 + x = 12 + 2x . The width of the garden and walkway is
x + 8 + x = 8 + 2x . The area covered by the garden and the walkway is shown .
192 = (12 + 2x)(8 + 2x)
192 – 192 = 4x2  + 40x + 96 –  192
0 = 4x2  + 40x – 96
0 = 4(x2  + 10x – 24)
0 = 4(x + 12)(x – 2)
This means x could be –12 or 2 . The walkway cannot be a negative length so the width of the walkway  must be 2 feet .

Page  126 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEM
1 . The formula for the area of a circle is A = πr2 . Which equation shows the formula in terms of r ?

π

A .  r = 2A

B .	ÏAr =

}

}

π

C . r =

A

Ïπ
D .   r = A 
2π

Answer to Unit 5 .3 Sample Item  1 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 127 of 230

5 .4 Solve Equations and Inequalities in One Variable

MGSE9-12 .A .REI .4 Solve quadratic equations in one variable .
MGSE9-12 .A .REI .4a Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)2  = q that has the same solutions . Derive the quadratic formula     from ax2 + bx + c = 0 .
MGSE9-12 .A .REI .4b Solve quadratic equations by inspection (e .g ., for x2 = 49), taking square roots, factoring, completing the square, and the quadratic formula, as appropriate to the initial form of the equation (limit to real number solutions) .

KEY IDEAS
When quadratic equations do not have  a  linear  term,  you  can  solve  the  equation  by  taking  the  square root of each side of the equation . Remember, every square root has a positive value and a negative value . Earlier in the guide, we eliminated the negative answers when they represented length or distance .
Example:
3x2 – 147 = 0
3x2  = 147	Addition Property of Equality
x2  = 49	Multiplicative Inverse Property
x = ±7	Take the square root of both sides .

Check your answers:
3(7)2 – 147 = 3(49) – 147
= 147 – 147
= 0

3(–7)2 – 147 = 3(49) – 147
= 147 – 147
= 0

You  can factor some quadratic equations to find the solutions . To  do this, rewrite the equation in standard  form set equal to zero (ax2 + bx + c = 0) . Factor the expression, set each factor to 0 (by the Zero Product Property), and then solve for x in each resulting equation . This will provide two rational values for x .
Example:
x2 − x = 12
x2  − x − 12  = 0	Addition Property of Equality (x − 4)(x + 3)  = 0	Factor .
Set each factor equal to 0 and solve .
x − 4 = 0	x + 3 = 0
x = 4	x = −3
Page  128 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Check your answers:
42 − 4 = 16 − 4	(−3)2 − (−3) = 9 + 3
= 12	= 12
You can complete the square to solve a quadratic equation . First, write the equation that represents the function in standard form, ax2 + bx + c = 0 . Subtract the constant from both sides of the equation:

ax2

+ bx = –c . Divide both sides of the equation by a: x2

b
+ a x =

–c . Add the square of half the coefficient

a

of the x-term to both sides: x

+ a x +

2a

=

a +

2a

 . Write the perfect square trinomial as a binomial

2	b	( b )2	–c	( b )2

2a

4a2

squared: (x +  b )2  = b2  – 4ac .   Take the square root of both sides of the equation and solve for x . This
method works best when a is 1 and b is even .

Example:
x2 – 6x – 7 = 0
x2  – 6x = 7	Addition Property of Equality

2

2

x2  – 6x + (–6)2  = 7 + (–6)2	Addition Property of Equality
x2 – 6x + (–3)2 = 7 + (–3)2
(x – 3)2  = 7 + 9	Distribution Property
(x – 3)2 = 16
x – 3 = ± 4	Take the square root of both sides .
x = 3 ± 4	Addition Property of Equality
x = 3 + 4 = 7; x = 3 – 4 = –1	Solve for x for both operations .

All quadratic equations can be solved using the quadratic formula . The quadratic formula
}

is x =

–b ± Ï b2 – 4ac 2a

, where ax2

+ bx + c = 0 . The quadratic formula will yield real solutions . We can solve

the previous equation using the quadratic formula .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 129 of 230

Example: 5x2 – 6x – 8 = 0, where a = 5, b = –6, and c = –8 .

Ï

}}
–(–6) ± (–6)2 – 4(5)(–8)

}}

x =	2(5)
x = 6 ± Ï36 – 4(–40)

}}

10
x = 6 ± Ï36 – (–160)

}

10
x = 6 ± Ï36 + (160)
10
}
x = 6 ± Ï196
10
x =  6 ± 14
10
x = 6 + 14 = 20 = 2; x = 6 – 14 = –8 = –4
10	10	10	10	5

Important Tip

While there may be several methods that can be used to solve a quadratic equation, some methods may be easier than others for certain equations .

REVIEW EXAMPLES
◆	Solve the equation x2 – 10x + 25 = 0 by factoring .
Solution:
Factor: x2 – 10x + 25 = (x – 5)(x – 5) .
Both factors are the same, so solve the equation:
x – 5 = 0
x = 5
◆	Solve the equation x2 – 100 = 0 by using square roots .
Solution:
Solve the equation using square roots .
x2  = 100	Add 100 to both sides of the equation .

}

x = ±Ï100	Take the square root of both sides of the equation .
x = ±10	Evaluate .

Page  130 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	Solve the equation 4x2 – 7x + 3 = 0 using the quadratic formula .

}

Solution:

Solve the equation using the quadratic formula x =

–b ± Ïb2 – 4ac 2a

, where ax2

+ bx + c = 0 .

Given the equation in standard form, the following values will be used in the formula:
a = 4, b = –7, and c = 3

Ï

}}
–(–7) ± (–7)2 – 4(4)(3)

x =	2(4)

Substitute each value into the quadratic formula .

}

x = 7 ± Ï1
8
x = 7 + 1 = 1 and x = 7 – 1 = 6 = 3

Simplify the expression .

Evaluate .

8	8	8	4

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 131 of 230

SAMPLE ITEMS
1 . What are the solutions to the equation 2x2 – 2x – 12 = 0 ?
A . x = –4, x = 3 B . x = –3, x = 4 C . x = –2, x = 3 D . x = –6, x = 2

2 . What are the solutions to the equation 6x2 – x – 40 = 0 ?

A .	8	5
x = – 3, x = – 2
B .	8	5
x = – 3, x = 2
C .	5	8

D .

x = 2, x = 3
5	8

x = – 2, x = 3

3 . What are the solutions to the equation x2 – 5x = 14 ?
A . x = –7, x = –2
B . x = –14, x = –1
C . x = –2, x = 7
D . x = –1, x = 14

Page  132 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

4 . An object is thrown into the air with an initial velocity of 5 m/s from a height of 9 m . The equation
h(t) = –4 .9t2  + 5t + 9 models the height of the object in meters after t seconds .

About how many seconds does it take for the object to hit the ground? Round your answer to the nearest tenth of a second .
A .  0 .9
B .  1 .5
C .  2 .0
D . 9 .0

Answers to  Unit  5 .4  Sample  Items 1 .   C	2 .   D	3 .   C	4 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 133 of 230

5 .5 Interpret Functions That Arise in Applications in Terms of the Context

MGSE9-12 .F .IF .4 Using tables, graphs, and verbal descriptions, interpret the key characteristics of a function which models the relationship between two quantities . Sketch a graph showing key features including: intercepts; interval where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior .
MGSE9-12 .F .IF .5 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes . For example, if the function h(n) gives the number of person-hours it takes
to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function.
MGSE9-12 .F .IF .6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval . Estimate the rate of change from a graph .

KEY IDEAS
An x-intercept, root, or zero of a function is the x-coordinate of a point where the function crosses the  x-axis . A function may have multiple x-intercepts . To find the x-intercepts of a quadratic function, set the function equal to 0 and solve for x . This can be done by factoring, completing the square, or using the quadratic formula .
The y-intercept of a function is the y-coordinate of the point where the function crosses the  y-axis .  A function may have zero y-intercepts or one y-intercept . To  find the y-intercept of a quadratic function, find   the value of the function when x equals 0 .
A function is increasing over an interval when the values of y increase as the values of x increase over that interval . The interval is represented in terms of x .
A function is decreasing over an interval when the values of y decrease as the values of x increase over     that interval . The interval is represented in terms of x .
Every quadratic function has a minimum or maximum, which is located at the vertex . When the function       is written in standard form, the x-coordinate of the vertex is –b . To find the y-coordinate of the vertex,

substitute the value of –b
2a

2a
into the function and evaluate . When the value of a is positive, the graph opens

up, and the vertex is the minimum point . When the value of a is negative, the graph opens down, and the vertex is the maximum point .
The end behavior of a function describes how the values of the function change as the x-values approach negative infinity and positive infinity .
The domain of a function is the set of values for which it is possible to evaluate the function . The domain of a quadratic function is typically all real numbers, although in real-world applications it may only make sense to look at the domain values on a particular interval . For example, time must be a nonnegative number .

Page  134 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

The average rate of change of a function over a specified interval is the change in the y-value divided by

x – x

the change in the x-value for two distinct points on a graph . To calculate the average rate of change of a function over the interval from a to b, evaluate the expression m = y2 – y1 .
2	1

end behavior (increasing)	y

end behavior
(increasing)	y

x-intercept

6
5
4
3
2
1

–6 –5 –4 –3 –2 –1– 0

x-intercept
1	2	3	4	5	6

y-intercep–t

1
–2
–3

4

–5
–6

minimum

y-intercept

6
5
4

maximum

x-intercept

3
2
1

–6 –5 –4 –3 –2 –1– 0

1
–2
–3
–4
–5
–6

x-intercept
1	2	3	4	5	6

x	x

end behavior (decreasing)

end behavior (decreasing)

Parabolas have this shape when a > 0.	Parabolas have this shape when a < 0.

REVIEW EXAMPLES
◆	A ball is thrown into the air from a height of 4 feet at time t = 0 . The function that models this situation is h(t) = –16t 2 + 63t + 4, where t is measured in seconds and h is the height in feet .
a .    What is the height of the ball after 2 seconds?    b . When will the ball reach a height of 50 feet?
c .   What is the maximum height of the ball?    d . When will the ball hit the ground?
e . What domain makes sense for the function? Solution:
a . To find the height of the ball after 2 seconds, substitute 2 for t in the function .
h(2) = –16(2)2 + 63(2) + 4 = –16(4) + 126 + 4 = –64 + 126 + 4 = 66
The height of the ball after 2 seconds is 66 feet .
b . To find when the ball will reach a height of 50 feet, find the value of t that makes h(t) = 50 .
50 = –16t 2 + 63t + 4
0 = –16t 2  + 63t – 46

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 135 of 230

}

Use the quadratic formula with a = –16, b = 63, and c = –46 .
–b ± Ïb2 – 4ac
t =	2a
}}
t = –63 ± Ï(63)2 – 4(–16)(–46)

}}

2(–16)
t = –63 ± Ï3969 – 2944

}

–32
t = –63 ± Ï1025
–32
t ≈ 0 .97 or t ≈ 2 .97 . So, the ball is at a height of 50 feet after approximately 0 .97 second and 2 .97 seconds .
c . To find the maximum height, find the vertex of h(t) .

The x-coordinate of the vertex is equal to –b : –63 

≈ 1 .97 . To find the y-coordinate, find

h(1 .97):

2a 2(–16)

h(1 .97) = –16(1 .97)2 + 63(1 .97) + 4 ≈ 66
The maximum height of the ball is about 66 feet .
d . To find when the ball will hit the ground, find the value of t that makes h(t) = 0 (because 0 represents 0 feet from the ground) .
0 = –16t 2 + 63t + 4
Using the quadratic formula (or by factoring), t = –0 .0625 or t = 4 .
Time cannot be negative, so t = –0 .0625 is not a solution . The ball will hit the ground after 4 seconds .
e . Time  must always be nonnegative and can be expressed by any fraction or decimal . The ball is  thrown at 0 seconds and reaches the ground after 4 seconds . So, the domain 0 ≤ t ≤ 4 makes sense for function h(t) .
◆	This table shows a company’s profit, p, in thousands of dollars, over time, t, in months .

Time, t
(months)	Profit, p
(thousands of dollars)	
3	18	
7	66	
10	123	
15	258	
24	627	
a . Describe the average rate of change in terms of the given context .
b .    What is the average rate of change of the profit between 3 and 7 months?  c . What is the average rate of change of the profit between 3 and 24 months?

Page  136 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Solution:
a . The average rate of change represents the rate at which the company earns a profit .
b . Use the expression for average rate of change . Let x1 = 3, x2 = 7, y 1 = 18, and y 2 = 66 .

m = y2 – y1

= 66 – 18 = 48 = 12

x2 – x1

7 – 3	4

The average rate of change between 3 and 7 months is 12 thousand dollars ($12,000) per month .     c . Use the expression for average rate of change . Let x1 = 3, x2 = 24, y1 = 18, and y2 = 627 .
m = y2 – y1 = 627 – 18 = 609 = 29

x2 – x1

24 – 3	21

The average rate of change between 3 and 24 months is 29 thousand dollars ($29,000) per month .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 137 of 230

SAMPLE ITEMS
1 . A flying disk is thrown into the air from a height of 25 feet at time t = 0 . The function that models this situation is h(t) = –16t2 + 75t + 25, where t is measured in seconds and h is the height in feet . What values of t best describe the time when the disk is flying in the air?
A . 0 < t < 5
B . 0 < t < 25
C . all real numbers
D . all positive integers

2 . Use this table to answer the question .

x	f ( x)	
–2	15	
–1	9	
0	5	
1	3	
2	3	
What is the average rate of change of x over the interval –2 ≤ x ≤ 0? A . –10
B .	–5
C .	5
D .	10

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3 .   What is the end behavior of the graph of f(x) = –0 .25x2  – 2x + 1?
A .   As x increases, f(x) increases .  As x decreases, f(x) decreases .   B .   As x increases, f(x) decreases .  As x decreases, f(x) decreases .   C .   As x increases, f(x) increases .  As x decreases, f(x) increases .    D . As x increases, f(x) decreases . As x decreases, f(x) increases .

Answers to Unit 5 .5 Sample Items   1 .   A	2 .   B	3 . B

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 139 of 230

5 .6 Analyze Functions Using Different Representations

MGSE9-12 .F .IF .7 Graph functions expressed algebraically and show key features of the graph both by hand and by using technology .
MGSE9-12 .F .IF .7a Graph linear and quadratic functions and show intercepts, maxima, and minima (as determined by the function or by context) .
MGSE9-12 .F .IF .8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function .
MGSE9-12 .F .IF .8a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context . For example, compare and contrast quadratic functions in standard, vertex, and intercept forms.
MGSE9-12 .F .IF .9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions) . For example, given a graph of one function        and an algebraic expression for another, say which has the larger maximum .

KEY IDEAS
Functions can be represented algebraically, graphically, numerically (in tables), or verbally (by description) . Examples:
Algebraically: f(x) = x2 + 2x
Verbally (by description): a function that represents the sum of the square of a number and twice the number
Numerically (in a table):

x	f(x)	
–1	–1	
0	0	
1	3	
2	8	

Page  140 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Graphically:

6
5
4
3
2
1

+ 2

1	2	3	4	5

–4

–3

–2

6

1

1– 0

2 –

3 –

4 –

5 –

6 –

–

x

x2

x) =

f(

y

x

You can compare key features of two functions represented in different ways . For example, if you are given an equation of a quadratic function and a graph of another quadratic function, you can calculate the vertex of the first function and compare it to the vertex of the graphed function .

REVIEW EXAMPLES
◆	Graph the function f(x) = x2 – 5x – 24 .
Solution:
Use the algebraic representation of the function to find the key features of the graph of the function .     Find the zeros of the function .
0 = x2  – 5x – 24	Set the function equal to 0 . 0 = (x – 8)(x + 3) Factor .

Set each factor equal to 0 and solve for x .
x – 8 = 0	x + 3 = 0
x = 8	x = –3

The zeros are at x = –3 and x = 8 . Find the vertex of the function .

x = –b = –(–5) = 5 = 2 .5

2a	2(1)	2

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 141 of 230

Substitute 2 .5 for x in the original function to find f(2 .5):
f(x) = x2 – 5x – 24
f(2 .5) = (2 .5)2 – 5(2 .5) – 24 = 6 .25 – 12 .5 – 24 = –30 .25

The vertex is (2 .5, –30 .25) .
Find the y-intercept by finding f(0) .
f(x) = x2 – 5x – 24
f(0) = (0)2 – 5(0) – 24 = –24

The y-intercept is (0, –24) . Use symmetry to find another point . The line of symmetry is x = 2 .5 .
0 + x = 2 .5
2
0 + x = 5

x = 5

So, point (5, –24) is also on the graph .
Plot the points (–3, 0), (8, 0), (2 .5, –30 .25), (0, –24), and (5, –24) . Draw a smooth curve through the   points .
We can also use the value of a in the function to determine if the graph opens up or down . In f(x) = x2 – 5x – 24, a = 1 . Since a > 0, the graph opens up .

16
12
8
4

1 2 3 4 5 6 7 8

32

–

28

–

24

–

20

–

16

–

12

–

–8

9

4

1– 0

2 –

3 –

4 –

–

f(x)

x

Page  142 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	This graph shows a function f(x) .

8
6
4
2

–6
–8

18

–

16

–

14

–

12

–

10

–

–4

–2

6

4

2

0

2

–

4

–

6

–

y

x

Compare the graph of f(x) to the graph of the function given by the equation g(x) = 4x2 + 6x – 18 . Which function has the lesser minimum value? How do you know?
Solution:
The minimum value of a quadratic function that opens up is the y-value of the vertex .   The vertex of the graph of f(x) appears to be (2, –18) . So, the minimum value is –18 .   Find the vertex of the function g(x) = 4x2 + 6x – 18 .
To find the vertex of g(x), use (–b, g (–b )) with a = 4 and b = 6 .
2a	2a
x = –b = –(6) = –6 = –0 .75
2a	2(4)	8
Substitute –0 .75 for x in the original function g(x) to find g(–0 .75):
g(x) = 4x2  + 6x – 18
g(–0 .75) = 4(–0 .75)2 + 6(–0 .75) – 18 2 .25 – 4 .5 – 18
= –20 .25
The minimum value of g(x) is –20 .25 .
–20 .25 < –18, so the function g(x) has the lesser minimum value .
Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 143 of 230

SAMPLE ITEMS
1 . Use this graph to answer the question .

16

12

8

4

16

–

12

–

–8

–4

8

6

4

2

0

2

–

4

–

6

–

8

–

f(x)

x

Which function is shown in the graph? A . f(x) = x2 – 3x – 10
B . f(x) = x2 + 3x – 10  C .   f(x) = x2  + x – 12 D . f(x) = x2 – 5x – 8

Page  144 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

2 . The function f(t) = –16t2 + 64t + 5 models the height of a ball that was hit into the air, where t is measured in seconds and h is the height in feet .

This table represents the height, g(t), of a second ball that was thrown into the air .

Time, t
(seconds)	Height, g(t) (feet)	
0	4	
1	36	
2	36	
3	4	
Which statement BEST compares the length of time each ball is in the air?
A .   The ball represented by f(t) is in the air for about 5 seconds, and the ball represented by g(t) is in    the air for about 3 seconds .
B .   The ball represented by f(t) is in the air for about 3 seconds, and the ball represented by g(t) is in     the air for about 5 seconds .
C .   The ball represented by f(t) is in the air for about 3 seconds, and the ball represented by g(t) is in     the air for about 4 seconds .
D .   The ball represented by f(t) is in the air for about 4 seconds, and the ball represented by g(t) is in    the air for about 3 seconds .

Answers to Unit 5 .6 Sample Items  1 .   A	2 . D

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 145 of 230

5 .7 Build a Function That Models a Relationship between Two Quantities

MGSE9-12 .F .BF .1 Write a function that describes a relationship between two quantities .
MGSE9-12 .F .BF .1a Determine an explicit expression and the recursive process (steps for calculation) from context . For example, if Jimmy starts out with $15 and earns $2 a day, the explicit expression “2x + 15” can be described recursively (either in writing or verbally) as “to find out how much money Jimmy will have tomorrow, you add $2 to his total today.” Jn = Jn – 1 + 2, J0 = 15
KEY IDEAS
An explicit expression contains variables, numbers, and operation symbols and does not use an equal sign to relate the expression to another quantity .
A recursive process can show that a quadratic function has second differences that are equal to one another .
Example: Consider the function f(x) = x2  + 4x – 1 . This table of values shows five values of the function .
x	f(x)	
–2	–5	
–1	–4	
0	–1	
1	4	
2	11	
The first and second differences are shown . The first differences are the differences between the consequence terms . The second differences are the differences between the consequence terms of the      first differences .
First differences	Second differences

11

2

4

1

–1

0

–4

–1

–5

–2

f(x)

x

–4 – (–5) = 1
–1 – (–4) = 3

4 – (–1) = 5
11 – 4 = 7

3 – 1 = 2
5 – 3 = 2
7 – 5 = 2

A recursive function is one in which each function value is based on a previous value (or previous values)         of the function .
When building a model function, functions can be added, subtracted, or multiplied together . The result will    still be a function . This includes linear, quadratic, exponential, and constant functions .

Page  146 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Annie is framing a photo with a length of 6 inches and a width of 4 inches . The distance from the edge of the photo to the edge of the frame is x inches . The combined area of the photo and frame is 63 square inches .

	6 in.

4 in.			
		x	

a .    Write a quadratic function to find the distance from the edge of the photo to the edge of the frame .           b . How wide are the photo and frame together?
Solution:
a . The length of the photo and frame is x + 6 + x = 6 + 2x . The width of the photo and frame is
x + 4 + x = 4 + 2x . The area of the frame is (6 + 2x)(4 + 2x) = 4x2 + 20x + 24 . Set this expression equal to the area: 63 = 4x2 + 20x + 24 .
b . Solve the equation for x .
63 = 4x2  + 20x + 24
0 = 4x2  + 20x – 39
x = –6 .5 or x = 1 .5

Length cannot be negative, so the distance from the edge of the photo to the edge of the frame is    1 .5 inches . Therefore, the width of the photo and frame together is 4 + 2x = 4 + 2(1 .5) = 7 inches .
◆	A scuba diving company currently charges $100 per dive . On average, there are 30 customers per day .        The company performed a study and learned that for every $20 price increase, the average number of customers per day would be reduced by 2 .
a . The total revenue from the dives is the price per dive multiplied by the number of customers . What is the revenue after 4 price increases?
b .    Write a quadratic equation to represent x price increases .    c . What price would give the greatest revenue?

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 147 of 230

Solution:
a . Make a table to show the revenue after 4 price increases .

Number of Price Increases	Price per Dive ($)	Number of Customers per Day	Revenue per Day ($)	
0	100	30	(100)(30) = 3,000	
1	100 + 20(1) = 120	30 – 2(1) = 28	(120)(28) = 3,360	
2	100 + 20(2) = 140	30 – 2(2) = 26	(140)(26) = 3,640	
3	100 + 20(3) = 160	30 – 2(3) = 24	(160)(24) = 3,840	
4	100 + 20(4) = 180	30 – 2(4) = 22	(180)(22) = 3,960	
The revenue after 4 price increases is ($180)(22) = $3,960 .
b . The table shows a pattern . The price per dive for x price increases is 100 + 20x . The number of customers for x price increases is 30 – 2x . So, the equation y = (100 + 20x)(30 – 2x) =
–40x2 + 400x + 3,000 represents the revenue for x price increases .
c .     To  find the price that gives the greatest revenue, first find the number of price increases that gives    the greatest value . This occurs at the vertex .
Use –b with a = –40 and b = 400 .
2a
–b = –400 = –400 = 5
2a	2(–40)	–80
The maximum revenue occurs after 5 price increases . 100 + 20(5) = 200
The price of $200 per dive gives the greatest revenue .

Page  148 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	Consider the sequence 2, 6, 12, 20, 30, . . . .
a	What explicit expression can be used to find the next term in the sequence?
b	What is the tenth term of the sequence?
Solution:
a . The difference between terms is not constant, so the operation involves multiplication Make a
table to try to determine the relationship between the number of the term and the value of the term .

Term number	Term value	Relationship	
1	2	1 ∙ 2	
2	6	2 ∙ 3	
3	12	3 ∙ 4	
4	20	4 ∙ 5	
5	30	5 ∙ 6	
Notice the pattern: The value of each term is the product of the term number and one more than the term number . So, the expression is n(n + 1) or n2 + n .
b . The tenth term is n2 + n = (10)2 + (10) = 110 .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 149 of 230

SAMPLE ITEMS
1 . What explicit expression can be used to find the next term in this sequence?

2, 8, 18, 32, 50,  .  . .

A . 2n
B . 2n + 6
C . 2n2
D . 2n2 + 1

2 .   The function s(t) = vt + h – 0 .5at2  represents the height of an object, s, in feet, above the ground in relation to the time, t, in seconds, since the object was thrown into the air with an initial velocity of v feet per second at an initial height of h feet and where a is the acceleration due to gravity (32 feet per second squared) .

A baseball player hits a baseball 4 feet above the ground with an initial velocity of 80 feet per second . About how long will it take the baseball to hit the ground?
A .   2 seconds
B .   3 seconds
C .   4 seconds
D .   5 seconds

3 . A café’s annual income depends on x, the number of customers . The function I(x) = 4x2 – 20x describes the café’s total annual income . The function C(x) = 2x2 + 5 describes the total amount the café spends in a year . The café’s annual profit, P(x), is the difference between the annual income and the amount spent in a year .
Which function describes P(x)? A . P(x) = 2x2 – 20x – 5
B . P(x) = 4x3 – 20x2
C . P(x) = 6x2 – 20x + 5
D . P(x) = 8x4 – 40x3 – 20x2 – 100x

Answers to Unit 5 .7 Sample Items  1 .   C	2 .   D	3 . A

Page  150 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

5 .8 Build New Functions from Existing Functions

MGSE9-12 .F .BF .3 Identify the effect on the graph of replacing f (x) by f(x) + k, k f(x), f(kx), and f(x + k)  for specific values of k (both positive and negative); find the value of k given the graphs . Experiment with
cases and illustrate an explanation of the effects on the graph using technology . Include recognizing even and odd functions from their graphs and algebraic expressions for them .

KEY IDEAS
A parent function is the basic function from which all the other functions in a function family are modeled . For the quadratic function family, the parent function is f(x) = x2 .
For a parent function f(x) and a real number k,
•	the function f(x) + k will move the graph of f(x) up by k units .
•	the function f(x) – k will move the graph of f(x) down by k units .

10

8

6

4

2

x2

–6

–4

–2

8

6

4

2

0

2

–

4

–

6

–

8

–

– 3

=

(x)

h

x2

=

(x)

f

3

2 +

= x

x)

g(

y

x

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 151 of 230

•	the function f(x + k) will move the graph of f(x) left by k units .
•	the function f(x – k) will move the graph of f(x) right by k units .

10

8

6

4

2

x)

–6

–4

= x2

f(

–2

8

6

4

2

0

2

–

4

–

6

–

8

–

x –

= (

x)

h(

)2

+ 4

(x

) =

y

g(x

4)2
x

For a parent function f(x) and a real number k,
•	the function kf(x) will vertically stretch the graph of f(x) by a factor of k units for |k| > 1 .
•	the function kf(x) will vertically shrink the graph of f(x) by a factor of k units for |k| < 1 .
•	the function kf(x) will reflect the graph of f(x) over the x-axis for negative values of k .

10

8

6

4

2

1 x2

2

2

2x

= –

x)

p(

–6

–4

–2

8

6

4

2

0

2

–

4

–

6

–

8

–

=

h(x)

x2

=

x)

f(

x2

2

x) =

g(

y

x

Page  152 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

•	
>

the function f(kx) will horizontally shrink the graph of f(x) by a factor of 1 units for |k|	1 .
k
•	
<

the function f(kx) will horizontally stretch the graph of f(x) by a factor of 1 units for |k|	1 .
k
•	the function f(kx) will reflect the graph of f(x) over the y-axis for negative values of k .

10

8

6

4

2

1 x

2

–6

–4

–2

8

6

4

2

0

2

–

4

–

6

–

8

–

) =

h(x

2

x2

) =

f(x

2x)

= (

x)

g(

y

2

x

You can apply more than one of these changes at a time to a parent function .
Example: f(x) = 5(x + 3)2 – 1 will translate f(x) = x2 left 3 units and down 1 unit and stretch the function vertically by a factor of 5 .

10

8

6

4

2

4	–

–6

–4

–2

–1

3)2

+

5(x

=

(x)

g

8

6

4

2

0

2

–

6

–

8

–

x2

) =

f(x

y

x

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 153 of 230

•	If a graph is symmetric to the y-axis, then it is an even function . That is, if f(–x) = f(x), then the function is even .
•	If a graph is symmetric to the origin, then it is an odd function . That is, if f(–x) = –f(x), then the function is odd .

6
5

–x) = f(x) 4

3
2
1

1	2	3	4	5

–6

–5

–4

–3

–2

6

1

1– 0

2 –

3 –

4 –

5 –

6 –

–

f(x)

–x)

f(

f(

6
5
4
f(–x)	3
2
1

f(–x) = –f(x)

1	2	3	4	5

–6

–5

–4

x)

–f(

–3

–2

6

1

1– 0

2 –

3 –

4 –

5 –

6 –

–

y	y

x	x

Symmetric about the y-axis Even Function

Symmetric about the Origin Odd Function

Important Tip

Remember that when you change f(x) to f(x + k), move the graph to the left when k is positive and      to the right when k is negative . This may seem different from what you would expect, so be sure to understand why this occurs in order to apply the shift or translation correctly .

REVIEW EXAMPLES
◆	Compare the graphs of the following functions to f(x) .
◆	
a . 1
2

f(x)

b . f(x) – 5
c . f(x – 2) + 1 Solution:
a . The graph of 1	1
2 f(x) is a vertical shrink of f(x) by a factor of 2 .
b . The graph of f(x) – 5 is a shift or vertical translation of the graph of f(x) down 5 units .
c . The graph of f(x – 2) + 1 is a shift or vertical translation of the graph of f(x) right 2 units and up 1 unit .
Page  154 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	Is f(x) = 2x3 + 6x even, odd, or neither? Explain how you know .
Solution:
Substitute –x for x and evaluate:
f(–x) = 2(–x)3  + 6(–x)
= 2(–x)3  – 6x
= –(2x3  + 6x)

f(–x) is the opposite of f(x), so the function is odd . Substitute –3 for x and evaluate:
f(–3) = 2(–3)3 + 6(–3)	f(3) = 2(3)3 + 6(3)
= 2(–3)3 – 18	= 2(27) + 18
= –(2(27) + 18)	= 72
= –(72)

f(–3) is the opposite of f(3), so the function is odd .
◆	
2

How does the graph of f(x) compare to the graph of f (1 x)?
Solution:
The graph of f (1 x) is a horizontal stretch of f(x) by a factor of 2 . The graphs of f(x) and g(x) = f (1 x) are
2	2
shown .
For example, at y = 4, the width of f(x) is 4 and the width of g(x) is 8 . So, the graph of g(x) is wider than
f(x) by a factor of 2 .

f(x) = x2

y
8

6

4

2

g(x) =

x

–4

–2

6

4

2

0

2

–

4

–

6

–

1 x 2
2

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 155 of 230

SAMPLE ITEMS
1 . Which statement BEST describes the graph of f(x + 6)?
A .   The graph of f(x) is shifted up 6 units .    B .   The graph of f(x) is shifted left 6 units .   C .   The graph of f(x) is shifted right 6 units .
D .   The graph of f(x) is shifted down 6  units .

2 . Which of these is an even function?
A . f(x) = 5x2 – x B . f(x) = 3x3 + x C . f(x) = 6x2 – 8
D . f(x) = 4x3 + 2x2

3 . Which statement BEST describes how the graph of g(x) = –3x2 compares to the graph of f(x) = x2 ?
A . The graph of g(x) is a vertical stretch of f(x) by a factor of 3 .
B . The graph of g(x) is a reflection of f(x) across the x-axis .
C . The graph of g(x) is a vertical shrink of f(x) by a factor of 1 and a reflection across the x-axis .
3
D . The graph of g(x) is a vertical stretch of f(x) by a factor of 3 and a reflection across the x-axis .

Answers to Unit 5 .8 Sample Items   1 .   B	2 .   C	3 . D

Page  156 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

5 .9 Construct and Compare Linear, Quadratic, and Exponential Models to Solve Problems

MGSE9-12 .F .LE .3 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function .

KEY IDEAS
Exponential functions have a fixed number as the base and a variable number as the exponent .
The value of an exponential function with a base greater than 1 will eventually exceed the value of a quadratic function . Similarly, the value of a quadratic function will eventually exceed the value of a linear function .

Exponential	
x	y = 2x	
1	2	
2	4	
3	8	
4	16	
5	32	
6	64	

Quadratic	
x	y = x2 + 2	
1	3	
2	6	
3	11	
4	18	
5	27	
6	38	

Linear	
x	y = x + 2	
1	3	
2	4	
3	5	
4	6	
5	7	
6	8	

Example:

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 157 of 230

REVIEW EXAMPLES
◆	This table shows that the value of f(x) = 5x2 + 4 is greater than the value of g(x) = 2x over the interval [0, 8] .

x	f(x)	g(x)	
0	5(0)2 + 4 = 4	20 = 1	
2	5(2)2 + 4 = 24	22 = 4	
4	5(4)2 + 4 = 84	24 = 16	
6	5(6)2 + 4 = 184	26 = 64	
8	5(8)2 + 4 = 324	28 = 256	
As x increases, will the value of f(x) always be greater than the value of g(x)? Explain how you know . Solution:
For some value of x, the value of an exponential function will eventually exceed the value of a quadratic function . To demonstrate this, find the values of f(x) and g(x) for another value of x, such as x = 10 .
f(x) = 5(10)2 + 4 = 504
g(x) = 210 = 1,024

In fact, this means that for some value of x between 8 and 10, the value of g(x) becomes greater than       the value of f(x) and remains greater for all subsequent values of x .

Page  158 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

◆	How does the growth rate of the function f(x) = 2x + 3 compare with g(x) = 0 .5x2 – 3?
Use a graph to explain your answer . Solution:
Graph f(x) and g(x) over the interval x ≥ 0 .

1 2 3 4 5 6 7 8

g(x)

x)

f(

y

24
22
20
18
16
14
12
10
8
6
4
2
0	x
–2
–4

The graph of f(x) increases at a constant rate because it is linear .

The graph of g(x) increases at an increasing rate because it is quadratic .

The graphs can be shown to intersect at (6, 15), and the value of g(x) is greater than the value of f(x) for x > 6 .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 159 of 230

SAMPLE ITEMS

1 . A table of values is shown for f(x) and g(x) .

x	f(x)	
0	0	
1	1	
2	4	
3	9	
4	16	
5	25	

x	g(x)	
0	-–2	
1	–1	
2	1	
3	5	
4	13	
5	29	

Which statement compares the graphs of f(x) and g(x) over the interval [0, 5]? A . The graph of f(x) always exceeds the graph of g(x) over the interval [0, 5] .
B .   The graph of g(x) always exceeds the graph of f(x) over the interval [0, 5] .
C . The graph of g(x) exceeds the graph of f(x) over the interval [0, 4], the graphs intersect at a point between 4 and 5, and then the graph of f(x) exceeds the graph of g(x) .
D . The graph of f(x) exceeds the graph of g(x) over the interval [0, 4], the graphs intersect at a point between 4 and 5, and then the graph of g(x) exceeds the graph of f(x) .

2 . Which statement is true about the graphs of exponential functions?
A . The graphs of exponential functions never exceed the graphs of linear and quadratic functions .
B . The graphs of exponential functions always exceed the graphs of linear and quadratic functions .
C . The graphs of exponential functions eventually exceed the graphs of linear and quadratic functions .
D . The graphs of exponential functions eventually exceed the graphs of linear functions but not quadratic functions .

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3 . Which statement BEST describes the comparison of the function values for f(x) and g(x)?

x	f(x)	g(x)	
0	0	–10	
1	2	–9	
2	4	–6	
3	6	–1	
4	8	6	

A .   The values of f(x) will always exceed the values of g(x) .
B .   The values of g(x) will always exceed the values of f(x) .
C . The values of f(x) exceed the values of g(x) over the interval [0, 5] .
D . The values of g(x) begin to exceed the values of f(x) within the interval [4, 5] .

Answers to Unit 5 .9 Sample Items   1 .   D	2 .   C	3 . D

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5 .10 Summarize, Represent, and Interpret Data on Two Categorical and Quantitative Variables

MGSE9-12 .S .ID .6 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related .
MGSE9-12 .S .ID .6a Decide which type of function is most appropriate by observing graphed data, charted data, or by analysis of context to generate a viable (rough) function of best fit . Use this function to solve problems in context . Emphasize quadratic models .

KEY IDEAS
A quadratic regression equation is a curve of best fit for data given in a scatter plot . The curve most likely will not go through all the data points but should come close to most of them .
Example:

										

										

										

										

y
10

Population (thousands)

9
8
7
6
5
4
3
2
1
x
0	Year

A quadratic regression equation can be used to make predictions about data . To do this, evaluate the equation for a given input value .

Page  162 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	Amery recorded the distance and height of a basketball when shooting a free throw .

Distance (feet), x	Height (feet), f(x)	
0	4	
2	8 .4	
6	12 .1	
9	14 .2	
12	13 .2	
13	10 .5	
15	9 .8	
The height of the basketball at a distance of x feet can be approximated by the quadratic function f(x) = –0 .118x2 + 2 .112x + 4 .215 . Using this function, what is the approximate maximum height  of the basketball?
Solution:
Find the vertex of the function .
–b = –(2 .112) ≈ 8 .949
2a	2(–0 .118)
Substitute 8 .949 for x in the original function:
f(8 .949) = –0 .118(8 .949)2 + 2 .112(8 .949) + 4 .215 ≈ 13 .665 ≈ 13 .7
The maximum height of the basketball predicted by the function is about 13 .7 feet .

◆	This table shows the population of a city every 10 years since 1970 .

Years since 1970, x	Population (thousands), y	
0	489	
10	801	
20	1,202	
30	1,998	
40	2,959	
a . Make a scatter plot showing the data .
b . Which type of function better models the relationship between 1970 and 2010: quadratic or linear?

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 163 of 230

Solution:
a . Plot the points on a coordinate grid .
y

3,200

2,800

Population (thousands)

2,400

2,000

1,600

1,200

800

400

0	10

x
20	30	40

Years since 1970

b . A quadratic model represents the population better than a linear model .

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SAMPLE ITEMS

1 . This scatter plot shows the height, in feet, of a ball launched in the air from an initial height of 3 feet and the time, in seconds, the ball traveled .
y

110
100
90

Height (feet)

80
70
60
50
40
30
20
10
0

x
1 2 3 4 5 6
Time (seconds)

Based on an estimated quadratic regression curve, which height is the BEST estimate for the maximum height of the ball?

A .	75  feet
B .	85  feet
C .   100 feet
D .   120 feet

2 . The quadratic function f(x) = –45x2 + 350x + 1,590 models the population of a city, where x is the number of years after 2005 and f(x) is the population of the city in thousands of people . What is the estimated population of the city in 2015?

A .	45,000	
B .	77,000	
C .	590,000	
D .	670,000	

Answers to Unit 5 .10 Sample Items   1 .   C	2 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 165 of 230

UNIT 6: GEOMETRIC AND ALGEBRAIC CONNECTIONS
This unit investigates coordinate geometry . Students look at equations for circles and parabolas and       use given information to derive equations for representations of these figures on a coordinate plane .
Students also use coordinates to prove simple geometric theorems using the properties of distance, slope, and midpoints . Students will verify whether a figure is a special quadrilateral by showing that sides of a figure are parallel or perpendicular . Students will model situations by using geometric shapes and their properties to describe objects . Students will also be able to use area and volume to apply concepts of density such as determining a number of units per volume .

6 .1 Translate between the Geometric Description and the Equation for a Conic Section

MGSE9-12 .G .GPE .1 Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation .

KEY IDEAS
A circle is the set of points in a plane equidistant from a given point, or center, of the circle .
The standard form of the equation of a circle is (x – h)2 + (y – k)2 = r2, where (h, k) is the center of the circle and r is the radius of the circle .
y

r

(x, y)

(h, k)

x

The equation of a circle can be derived from the Pythagorean Theorem; a2 + b2 = c2 .
Example: Given a circle with a center at (h, k) and a point (x, y) on the  circle,  draw  a  horizontal  line segment from (h, k) to (x, k) . Label this line segment a . Draw a vertical line segment from (x, y) to (x, k) . Label this line segment b . Label the radius c . A right triangle is formed .
The length of line segment a is given by (x – h) . The length of line segment b is given by (y – k) .

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y

y

(x, y)

c

b

k 	(h, k)

a

(x, k)

h	x

x

Using the Pythagorean Theorem, substitute (x – h) for a, (y – k) for b, and r for c in the equation .
a2  +  b2 = c2	Use the Pythagorean Theorem . (x – h)2  + (y – k)2 = r2	Substitution
The equation for a circle with a center at (h, k) and a radius r is (x – h)2 + (y – k)2 = r2 .

REVIEW EXAMPLES
◆	What is the equation of the circle with a center at (4, 5) and a radius of 2?

7
6
5
4
3
2
1

2

1	2	3	4	5	6

–2

7

1

1– 0

2 –

–

, 5)

(4

)

x, y

(

y

x

Solution:
Use the standard form for the equation for a circle: (x – h)2 + (y – k)2 = r2 . Substitute the values into the equation, with h = 4, k = 5, and r = 2 .
(x – h)2  + ( y – k)2  = r2	Equation for a circle
(x – (4))2  + ( y – (5))2  = (2)2	Substitute the values in the equation of a circle . (x – 4)2  + ( y – 5)2  = 4	Evaluate .
The equation of a circle with a center at (4, 5) and a radius of 2 is (x – 4)2  + (y – 5)2  = 4,   or x2 + y2 – 8x – 10y + 37 = 0 when expanded .
Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 167 of 230

◆	What is the center and radius of the circle given by 8x2 + 8y2 – 16x – 32y + 24 = 0?
Solution:
Write the equation in standard form to identify the center and radius of the circle . First, write the equation so the x-terms are next to each other and the y-terms are next to each other, both on the left side of the equation, and the constant term is on the right side of the equation .
8x2  + 8y2  – 16x – 32y + 24 = 0	Original equation
8x2  + 8y2  – 16x – 32y = –24	Subtract 24 from both sides . 8x2  – 16x + 8y2  – 32y = –24	Commutative Property
x2  – 2x + y2  – 4y = –3	Divide both sides by 8 . (x2  – 2x) + (y2  – 4y) = –3	Associative Property

(	)

Next, to write the equation in standard form, complete the square for the x-terms and the y-terms . Using ax2 + bx + c = 0, find b 2 for the x- and y-terms .
2a

2a

2(1)

x-term: ( b )2  = ( –2  )2  = (–1)2  = 1

2a

2(1)

y-term: ( b )2  = ( –4  )2  = (–2)2  = 4
(x2  – 2x + 1) + (y2  – 4y + 4) = –3 + 1  + 4	Add 1 and 4 to each side of the equation .
(x – 1)2  + (y – 2)2  = 2	Write the trinomials as squares of binomials .

}

This equation for the circle is written in standard form, where h = 1, k = 2, and r2 = 2 . The center of the circle is (1, 2), and the radius is Ï2 .

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SAMPLE ITEMS
1 . Which is an equation for the circle with a center at (–2, 3) and a radius of 3?
A .   x2  + y2  + 4x – 6y + 22 = 0  B .  2x2  + 2y2  + 3x – 3y + 4 = 0  C . x2 + y2 + 4x – 6y + 4 = 0
D . 3x2 + 3y2 + 4x – 6y + 4 = 0

2 . What is the center of the circle given by the equation x2 + y2 – 10x – 11 = 0?
A .   (5, 0)
B .   (0, 5)
C .   (–5, 0)
D .  (0, –5)

Answers to Unit 6 .1 Sample Items   1 .   C	2 . A

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6 .2 Use Coordinates to Prove Simple Geometric Theorems Algebraically

}

MGSE9-12 .G .GPE .4 Use coordinates to prove simple geometric theorems algebraically . For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, Ï3) lies on the circle centered at the origin and containing the point (0, 2) .
(Focus on quadrilaterals, right triangles, and circles .)

KEY IDEAS
Given the equation of a circle, you can verify whether a point lies on the circle by substituting the coordinates of the point into the equation . If the resulting equation is true, then the point lies on the figure . If the resulting equation is not true, then the point does not lie on the figure .
Given the center and radius of a circle, you can verify whether a point lies on the circle by determining whether the distance between the given point and the center is equal to the radius .

REVIEW EXAMPLES
◆	Circle C has a center of (–2, 3) and a radius of 4 . Does point (–4, 6) lie on circle C?
Solution:
The distance from any point on the circle to the center of the circle is equal to the radius . Use the distance formula to find the distance from (–4, 6) to the center (–2, 3) . Then see if it is equal to the radius, 4 .

Ï

}}
(–4 – (–2))2  + (6  – 3)2	Substitute the coordinates of the points in the distance formula .

}

Ï(–2)2  + (3)2	Evaluate within parentheses .

}

Ï4 + 9	Evaluate the exponents .

}

Ï13	Add .

}

The distance from (–4, 6) to (–2, 3) is not equal to the radius, so (–4, 6) does not lie on the circle . (In fact, since Ï13 < 4, the distance is less than the radius, so the point lies inside the circle .)

◆	Circle C has a diameter of 10 and a center at (2, 2) .  Point A is located at (–1, 6) and point B is located  at (5, –2) . Is segment AB a diameter of the circle?
Solution:
Both points A and B must be a distance of 5 units from the center and the length of segment AB must be 10 units . We can use the distance formula to determine this .

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Determine if points A and B are a distance of 5 units from (2, 2) .

Ï

}}
(–1 – 2)2  + (6  – 2)2	Substitute the coordinates of the points
into the distance formula .

Ï

}

}

(–3)2  + (4)2	Evaluate within parentheses .

}

Ï9 + 16	Evaluate the exponents .
Ï25	Add .
5	Find the square root .

Ï

}}
(5 – 2)2  + (–2  – 2)2	Substitute the coordinates of the points
into the distance formula .

Ï

}

}

(3)2  + (–4)2	Evaluate within parentheses .

}

Ï9 + 16	Evaluate the exponents .
Ï25	Add .
5	Find the square root .

Now we will determine if AB = 10 to determine if it is a diameter of the circle .

Ï

}}
(–1 – 5)2  + (6  – (–2))2	Substitute the coordinates of the points in
the distance formula .

Ï

}

}

(–6)2  + (8)2	Evaluate within parentheses .

}

Ï36 + 64	Evaluate the exponents .
Ï100	Add .
10	Find the square root .

The distance from (–1, 6) to (5, –2) is 10 units and both points are 5 units away from the center of the circle . This means both points lie on the circle and the measure of segment AB equals twice the radius . This makes segment AB a diameter of the circle .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 171 of 230

SAMPLE ITEMS

1 . Which point is on a circle with a center of (3, –9) and a radius of 5?
A .  (–6, 5)
B .   (–1, 6)
C . (1, 6)
D . (6, –5)

2 . Which two points can form the diameter of a circle with a center at the origin and a radius of 10?
A . (6, –8) and (–6, 8)
B .   (–1, 0) and (9,  0)
C .   (0, 5) and (0,  –5)
D .   (0, 0) and (10,  0)

Answers to Unit 6 .2 Sample Items   1 .   D	2 . A

Page  172 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

6 .3 Apply Geometric Concepts in Modeling Situations

MGSE9-12 .G .MG .1 Use geometric shapes, their measures, and their properties to describe objects (e .g ., modeling a tree trunk or a human torso as a cylinder) .
MGSE9-12 .G .MG .2 Apply concepts of density based on area and volume in modeling situations (e .g ., persons per square mile, BTUs per cubic foot) .
MGSE9-12 .G .MG .3 Apply geometric methods to solve design problems (e .g ., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios) .

KEY IDEAS
Modeling can be applied to describe real-life objects with geometric shapes .
Density is the mass of an object divided by its volume .
Population density can be determined by calculating the quotient of the number of people in an area and the area itself .
Apply constraints to maximize or minimize the cost of a cardboard box used to package a product that represents a geometric figure . Apply volume relationships of cylinders, pyramids, cones, and spheres .

REVIEW EXAMPLE
◆	A city has a population of 6,688 people . The area of the city is approximately 7 .267 square miles .
How many people per square mile live in the city? Solution:
Find the quotient of 6,688 and 7 .267 to find the number of people per square mile in the city .

≈

People per square mile = 6,688	920 people per square mile
7 .267

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 173 of 230

◆	This is a hand drawing of a mountain .

Explain which geometric shape could be used to estimate the total amount of earth the mountain is made of .
Solution:
The most accurate shape that could be used to model the mountain is a cone because, to determine the total amount of earth the mountain is made from, a three-dimensional shape is needed, which is why a triangle is not as accurate as a cone .

◆	A construction company is preparing 10 acres of land for a new housing community . The land contains   large rocks that need to be removed . A machine removes 10 rocks from 360 square feet of land .

1 acre = 43,560 square feet

About how many rocks will need to be removed from the 10 acres of land? Solution:
If there are 10 rocks in 360 square feet, then we can predict that there will be about 10 rocks every 360 square feet of land .
We will need to determine how many 360 square feet are in 10 acres . 10(43,560) = 435,600, so 435,600 square feet are in 10 acres .
435,600 = 1,210, so 1,210 parcels of 360 square feet are on the 10 acres .
360
(1,210)(10) = 12,100, so there should be about 12,100 rocks on the 10 acres of land .

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◆	A company needs to package this bell in a rectangular box .

8 inches

6 inches

What are the smallest dimensions (length, width, and height) the rectangular box can have so that the lid of the box can also close?
Solution:
Since the diameter of the base of the bell is 6 inches, the width and length of the box cannot be smaller than 6 inches . Since the height of the bell is 8 inches, then the height of the box cannot be smaller than 8 inches .
This gives us a rectangular box with these dimensions: Length = 6 inches
Width = 6 inches Height = 8 inches

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 175 of 230

SAMPLE ITEMS
1 . Joe counts 250 peach trees on 25% of the land he owns . He determines that there are 10 trees for every 1,000 square feet of land . About how many acres of land does Joe own?

1 acre = 43,560 square feet

A . 2 .3
B . 10
C .   43 .56
D .   2,500

2 . A square pyramid is packaged inside a box .

5 inches

5 inches
5 inches

The space inside the box around the pyramid is then filled with protective foam . About how many cubic inches of foam is needed to fill the space around the pyramid?

A . 8
B .  41
C .  83
D . 125

Answers to Unit 6 .3 Sample Items   1 .   A	2 . C

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UNIT 7: APPLICATIONS OF PROBABILITY
This unit investigates the concept of probability . Students look at sample spaces and identify unions, intersections, and complements . They identify ways to tell whether events are independent . The concept      of conditional probability is related to independence, and students use the concepts to solve real-world problems, including those that are presented in two-way frequency tables . Students find probabilities of compound events using the rules of probability .

7 .1 Understand Independence and Conditional Probability and Use Them to Interpret Data

MGSE9-12 .S .CP .1 Describe categories of events as subsets of a sample space using unions, intersections, or complements of other events (or, and, not) .
MGSE9-12 .S .CP .2 Understand that if two events A and B are independent, the probability of A and B occurring together is the product of their probabilities, and that if the probability of two events A and B occurring together is the product of their probabilities, the two events are independent .
MGSE9-12 .S .CP .3 Understand the conditional probability of A given B as P(A and B)/P(B) . Interpret independence of A and B in terms of conditional probability; that is, the conditional probability of A given           B is the same as the probability of A, and the conditional probability of B given A is the  same  as  the probability of B .
MGSE9-12 .S .CP .4 Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified .  Use the two-way table as a sample space to decide if events        are independent and to approximate conditional probabilities . For example, use collected data from a random sample of students in your school on their favorite subject among math, science, and English.
Estimate the probability that a randomly selected student from your school will favor science given that the student is in tenth grade. Do the same for other subjects and compare the results.
MGSE9-12 .S .CP .5 Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations . For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer.

KEY IDEAS
In probability, a sample space is the set of all possible outcomes . Any subset from the sample space is an
event .
If the outcome of one event does not change the possible outcomes of the other event, the events are independent . If the outcome of one event does change the possible outcomes of the other event, the events are dependent .
The intersection of two or more events is all of the outcomes shared by both events . The intersection is denoted with the word “and” or with the ⋂ symbol . For example, the intersection of A and B is shown as       A ⋂ B.
The union of two or more events is all of the  outcomes  for  either  event .  The  union  is  denoted  with  the word “or” or with the ⋃ symbol . For example, the union of A and B is shown as A ⋃ B . The probability of the union of two events that have no outcomes in common is the sum of each individual probability .

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The complement of an event is the set of outcomes in the same sample space that are not included              in the outcomes of the event . The complement is denoted with the word “not” or with the ′ symbol . For example, the complement of A is shown as A′ .  The set of outcomes and its complement make up the   entire sample space .
Conditional probabilities are found when one event has already occurred and a second event is being analyzed . Conditional probability is denoted P(A | B) and is read as “the probability of A given B .”

P(A | B) = P(A and B)
P(B)
Two  events—A and B—are independent if the probability of the intersection is the same as the product         of each individual probability . That is, P(A and B) = P(A) ∙ P(B) . This is called the Multiplication Rule for Independent Events .
If two events are independent, then P(A | B) = P(A) and P(B | A) = P(B).
Two-way frequency tables summarize data in two categories . These tables can be used to show whether   the two events are independent and to approximate conditional probabilities .
Example: A random survey was taken to gather information about grade level and car ownership status of students at a school . This table shows the results of the survey .
Car Ownership by Grade

	Owns a Car	Does Not Own a Car	Total	
Junior	6	10	16	
Senior	12	8	20	
Total	18	18	36	
Estimate the probability that a randomly selected student will be a junior given that the student owns           a car .
Let P(J) be the probability that the student is a junior . Let P(C) be the probability that the student owns a         car .

P(J | C) = P(J and C) =

6

36 = 6 = 1

P(C)

18	18	3

36
1

The probability that a randomly selected student will be a junior given that the student owns a car is 3 .

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REVIEW EXAMPLES
◆	This chart shows the names of students in Mr . Leary’s class sorted by bicycle and skateboard ownership .
Bicycle and Skateboard Ownership

Owns a Bicycle	Owns a Skateboard	Owns a Bicycle AND a Skateboard	Does NOT Own a Bicycle OR a Skateboard	
Ryan	Brett	Joe	Amy	
Sarah	Juan	Mike	Gabe	
Mariko	Tobi	Linda	Abi	
Nina		Rose		
Dion				
Let set A be the names of students who own bicycles, and let set B be the names of students who own skateboards .
a .  Find A and B. What does the set represent?   b . Find A or B. What does the set represent?
c . Find (A or B)′ . What does the set represent?
Solution:
a .    The intersection is the set of elements that are common to both set A and set B, so A and            B is {Joe, Mike, Linda, Rose} . This set represents the students who own both a bicycle and a skateboard .
b .    The union is the set of elements that are in set A or set B or in both set A and set B . You  need          to list the names in the intersection only one time, so A or B is {Ryan, Sarah, Mariko, Nina, Dion,  Brett, Juan, Tobi, Joe, Mike, Linda, Rose} . This set represents the students who own a bicycle, a skateboard, or both .
c . The complement of A or B is the set of names that are not in A or B . So the complement of A or B
is {Amy, Gabe, Abi} . This set represents the students who own neither a bicycle nor a skateboard .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 179 of 230

◆	In a certain town, the probability that a person plays sports is 65% . The probability that a person is between the ages of 12 and 18 is 40% . The probability that a person plays sports and is between the ages of 12 and 18 is 25% . Are the events independent? How do you know?
Solution:
Let P(S) be the probability that a person plays sports .
Let P(A) be the probability that a person is between the ages of 12 and 18 . If the two events are independent, then P(S and A) = P(S) ∙ P(A) .
Because P(S and A) is given as 25%, find P(S) ∙ P(A) and then compare .
P(S) ∙ P(A) = 0 .65 ∙ 0 .4
= 0 .26
Because 0 .26 ¹ 0 .25, the events are not independent .

◆	A random survey was conducted to gather information about employment status and age . This table shows the data that were collected .

Employment Survey Results

	Age (years)		
	Younger Than 18	18 or Older	Total	
Status	Has Job	20	587	607	
	Does Not Have Job	245	92	337	
	Total	265	679	944	
a . What is the probability that a randomly selected person surveyed has a job given that the person is younger than 18 years old?
b .    What is the probability that a randomly selected person surveyed has a job given that the person is       18 years old or older?
c . Are having a job (A) and being 18 years old or older (B) independent events? Explain .
•	P(A) = has a job
•	P(A′) = does not have a job
•	P(B) = 18 years old or older
•	P(B′) = younger than 18 years old

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Solution:
a . Find the total number of people surveyed younger than 18 years old: 265 . Divide the number of people who have a job and are younger than 18 years old, 20, by the number of people younger

than 18 years old, 265: 20
265

» 0 .08 . The probability that a person surveyed has a job given that

the person is younger than 18 years old is about 0 .08 .

20
P(A | B′) = P(A and B′) = 944 = 20 = 0 .08

P(B′)

265

944

265

b .   Find the total number of people surveyed 18 years old or older: 679 . Divide the number of people   who have a job and are 18 years  old  or  older,  587,  by  the  number  of  people  18  years  old  or older, 679: 587 » 0 .86 . The probability that a person surveyed has a job given that the person is
679
18 years old or older, is about 0 .86 .
587

P(A | B) = P(A and B) = 944 = 587 = 0 .86

P(B)

679

944

679

c . The events are independent if P(A | B) = P(A) and P(B | A) = P(B) .
From part (b), P(A | B) ≈ 0 .86 .

≈

P(A) = 607	0 .64 944
P(A ǀ B) ≠ P(A), so the events are not independent .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 181 of 230

SAMPLE ITEMS
1 . In a particular state, the first character on a license plate is always a letter . The last character is always a digit from 0 to 9 .

E23 PC8

If V represents the set of all license plates beginning with a vowel and O represents the set of all license plates that end with an odd number, which license plate belongs to the set V and O′?

A .

MG4 3F5

B .

AR8 8X9

C .

P7M Z56

D .

2 . For which set of probabilities would events A and B be independent?
A .   P(A) = 0 .25; P(B) = 0 .25; P(A and B) = 0 .5
B .   P(A) = 0 .08; P(B) = 0 .4; P(A and B) = 0 .12
C .   P(A) = 0 .16; P(B) = 0 .24; P(A and B) = 0 .32
D .   P(A) = 0 .3; P(B) = 0 .15; P(A and B) = 0 .045

Page  182 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

3 . Assume that the following events are independent:
•	The probability that a high school senior will go to college is 0 .72 .
•	The probability that a high school senior will go to college and live on campus is 0 .46 .
What is the probability that a high school senior will live on campus given that the high school senior will go to college?

A . 0 .26 B . 0 .33 C . 0 .57 D . 0 .64

4 . A random survey was conducted about gender and hair color . This table records the data .

Hair Color

	Brown	Blond	Red	Total	
Male	548	876	82	1,506	
Female	612	716	66	1,394	
Total	1,160	1,592	148	2,900	

What is the probability that a randomly selected person has blond hair given that the person selected is male?

A . 0 .51 B . 0 .55 C . 0 .58 D . 0 .63

Answers to  Unit  7 .1  Sample  Items 1 .   A	2 .   D	3 .   D	4 . C

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 183 of 230

7 .2 Use the Rules of Probability to Compute Probabilities of Compound Events in a Uniform Probability Model

MGSE9-12 .S .CP .6 Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in context .
MGSE9-12 .S .CP .7 Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answers in context .

KEY IDEAS
Two events are mutually exclusive if the events cannot occur at the same time .
When two events A and B are mutually exclusive, the probability that event A or event B will occur is the sum of the probabilities of each event: P(A or B) = P(A) + P(B) .
When two events A and B are not mutually exclusive, the probability that event A or B will occur is the sum of the probability of each event minus the intersection of the two events . That is,
P(A or B) = P(A) + P(B) – P(A and B) . This is called the Addition Rule .
You can find the conditional probability, P(A | B), by finding the fraction of B’s outcomes that also belong to A .
Example: Event A is choosing a heart card from a standard deck of cards . Event B is choosing a face card from a standard deck of cards .
P(A |  B) is the probability that a card is a heart given that the card is a face card .  You  can look at B’s  outcomes and determine what fraction belongs to A; there are 12 face cards, 3 of which are also hearts:
P(A |  B) =   3   = 1
12	4

	Heart	Not a Heart	Total	
Face Card	3	9	12	
Not a Face Card	10	30	40	
Total	13	39	52	

Page  184 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

REVIEW EXAMPLES
◆	In Mr . Mabry’s class, there are 12 boys and 16 girls . On Monday, 4 boys and 5 girls were wearing white shirts .
a .    If a student is chosen at random from Mr . Mabry’s class, what is the probability of choosing a boy        or a student wearing a white shirt?
b .    If a student is chosen at random from Mr . Mabry’s class, what is the probability of choosing a girl       or a student not wearing a white shirt?
Solution:
a .    Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), where A is the set of boys and B is the      set of students wearing a white shirt .
(A and B) is the set of boys wearing a white shirt . There are 12 + 16 = 28 students in Mr . Mabry’s class .

So P(A) =	, P(B) =	=

12	4 + 5	 9	 4 
,  and  P(A and B) =	.
28	28	28	28
P(a boy or a student wearing a white shirt) = 12 +  9  –  4  = 17
28	28	28	28
b .    Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), where A is the set of girls and B is the      set of students not wearing a white shirt .
(A and B) is the set of girls not wearing a white shirt . There are 12 + 16 = 28 students in Mr . Mabry’s class .
So P(A) = 16, P(B) = 8 + 11 = 19, and P(A and B) = 11 .
28	28	28	28
P(a girl or a student not wearing a white shirt) = 16 + 19 – 11 = 24 = 6
28	28	28	28	7

◆	Terry has a number cube with sides labeled 1 through 6 . He rolls the number cube twice .
a .    What is the probability that the sum of the two rolls is a prime number given that at least one of     the rolls is a 3?
b .    What is the probability that the sum of the two rolls is a prime number or at least one of the rolls is          a 3?

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 185 of 230

Solution:
a .    This is an example of a mutually exclusive event . Make a list of the combinations where at least    one of the rolls is a 3 . There are 11 such pairs .

1, 3	2, 3	3, 3	4, 3	5, 3	6, 3	
3, 1	3, 2	3, 4	3, 5	3, 6		
Then identify the pairs that have a prime sum .

2, 3	3, 2	3, 4	4, 3	

 .

Of the 11 pairs of outcomes, there are 4 pairs whose sum is prime . Therefore, the probability that     the sum is prime of those that show a 3 on at least one roll is 4 
11
b .  This is an example of events that are NOT mutually exclusive . There are 36 possible outcomes   when rolling a number cube twice .
List the combinations where at least one of the rolls is a 3 .

1, 3	2, 3	3, 3	4, 3	5, 3	6, 3	
3, 1	3, 2	3, 4	3, 5	3, 6		
P(at least one roll is a 3) = 11
36
List the combinations that have a prime sum .

1, 1	1, 2	1, 4	1, 6	
2, 1	2, 3	2, 5		
3, 2	3, 4			
4, 1	4, 3			
5, 2	5, 6			
6, 1	6, 5			
P(a prime sum) =  15
36
Identify the combinations that are in both lists .

2, 3	3, 2	3, 4	4, 3	
The combinations in both lists represent the intersection . The probability of the intersection is the number of outcomes in the intersection divided by the total possible outcomes .
P(at least one roll is a 3 and a prime sum) = 4 
36
If two events share outcomes, then outcomes in the intersection are counted twice when the
probabilities of the events are added . So you must subtract the probability of the intersection from the    sum of the probabilities .
P(at least one roll is a 3 or  a prime sum) = 11 + 15 –   4   = 22 = 11
36	36	36	36	18

Page  186 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

SAMPLE ITEMS
1 . Mrs . Klein surveyed 240 men and 285 women about their vehicles . Of those surveyed, 155 men and 70 women said they own a red vehicle . If a person is chosen at random from those surveyed, what is the probability of choosing a woman or a person who does NOT own a red vehicle?
A .   14
57
B .    71 
105
C .   74 
105
D .	88
105

2 . Bianca spins two spinners that have four equal sections numbered 1 through 4 . If she spins a 4 on at least one spin, what is the probability that the sum of her two spins is an odd number?
A .	1
4
B .   7 
16
C .	4
7
D .  11
16

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 187 of 230

3 . Each letter of the alphabet is written on separate cards in red ink . The cards are placed in a container . Each letter of the alphabet is also written on separate cards in black ink . The cards are placed in the same container . What is the probability that a card randomly selected from the container has a letter written in black ink or the letter is A or Z?
A .	1
2
B .   7 
13
C .  15
26
D .   8 
13

Answers to Unit 7 .2 Sample Items  1 .   C	2 .   C	3 . B

Page  188 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

ANALYTIC GEOMETRY ADDITIONAL PRACTICE ITEMS

This section has two parts . The first part is a set of 24 sample items for Analytic Geometry . The second part contains a table that shows for each item the standard assessed, the DOK level, the correct answer (key), and a rationale/explanation about the key and distractors . The sample items can be utilized as a mini-test to familiarize students with the item formats found on the assessment .
All example and sample items contained in this guide are the property of the Georgia Department of Education .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 189 of 230

Milestones

Georgia

Ass es sment Sy stem

Analytic Geometry Formula Sheet

Below are the formulas you may find useful as you take the test. However, you may find that you do not need to use all of the formulas. You may refer to this formula sheet as often as needed.

Area of a Circle
A = πr 2

360

rAceLngth = }2πr

}P(B)
Multiplication Rule for Independent Events
P(A and B) = P(A) • P(B)
Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)

Equation of a Circle
(x – h)2 + (y – k)2 = r 2
Circumference of a Circle
C = πd  or  C = 2πr π ≈ 3.14
Arc Length of a Circle

}adj

= opp

tan

hyp

= }adj ;

cos

}hyp

= opp ;

sin

Conditional Probability
P(A|B) = P(A and B)

V = }4 πr 3

3
Statistics Formulas

3
1

V = }3 πr 2h

V = }1 Bh

Pyramid Cone
Sphere

}2}a
Average Rate of Change
The change in the y-value divided by the change in the x-value for two distinct points on a graph.
Geometry Formulas Pythagorean Theorem
a 2 + b 2 = c 2
Trigonometric Relationships

2

x = −b ± √b – 4ac

V = πr 2h

Volume
Cylinder

Quadratic Formula

360

y = ax2 + bx + c y = a(x – h)2 + k

Standard Form:
Vertex Form:

Area of a Sector of a Circle
Area of Sector  = }πr 2

Quadratic Formulas
Quadratic Equations

You can find mathematics formula sheets on the Georgia Milestones webpage at
http://www .gadoe .org/Curriculum-Instruction-and-Assessment/Assessment/Pages/Georgia-Milestones- EOC-Resources .aspx .
Page  190 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

77°

52°

Selected-Response Look at the triangle .

51°

77°

Which triangle is similar to the given triangle?

A .

66°

70°

B .

25°

77°

C .

D .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 191 of 230

Constructed-Response
The diagrams of two shapes are shown .

12

10 inches

inches	12 inches

5 inches
5 inches

How many times the volume, in cubic inches, of the pyramid is the volume, in cubic inches, of the cone? Round your answer to the nearest tenth . Explain how you found your answer or show your work . Write your answer in the space provided .

Page  192 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Extended Constructed-Response
The diagram shows right triangle ABC with altitude BD .
B

A	D	C

Prove △ABC is similar to △BDC . Write your answer in the space provided .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 193 of 230

Selected-Response Which equation is true?
A .   sin 40°   =   tan 50°
B .   cos 40° =   cos 50°
C .   sin 40°   =   sin 50°
D .   cos 40° =   sin 50°

Item 5
Selected-Response
Which point is NOT on a circle with a center of (0, 0) and a radius of 10?

A . (0, 5)
B . (10, 0)
C . (0, −10)
D . (−8, 6)

Page  194 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Study the triangle .

x

5 in.

3 in.

What is sin x? Explain how you determined your answer . Write your answer in the space provided .

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 195 of 230

Explain why the formula for the area of a sector of a circle is A =

πr2θ

360

, where r is the radius of the

circle and θ is the measure in degrees of the central angle of the sector . Write your answer in the space provided .

Page  196 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points A, B, C, D, and E are located on circle O, as shown in this figure .

A

O	2x°

D

C

E	B

The measure of XCD is 80° . What is the value of x ? A . 50
B .  40
C .  35
D .  25

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 197 of 230

A pyramid and a rectangular prism have congruent bases and equal heights . Write a statement comparing the volume of the figures, and explain your reasoning . Write your answer in the space provided .

Page  198 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

—	}
Which expression is equivalent to –4Ï28x ∙ Ï7x3 ?
A .  –56x2

}

B .   4x2Ï}7
C .   –4xÏ196
D . –28x

Item 11
Selected-Response

}

Which value is an irrational number? A . 4 + Ï7
— }
B . Ï2 Ï8

— }

5

C . Ï3 Ï12

—	}
D . Ï3 – Ï3

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 199 of 230

Part A	Explain how you could rewrite the expression 3x + 2(x2 – 4x + 1) + 5x – 2 to write it with the fewest number of terms . Write your answer in the space provided .

Part B	When the expression from Part A is rewritten with the fewest number of terms, how many nonzero terms does it contain? Write your answer in the space provided .

Part B  	

Part A

Page  200 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Select THREE equations that are true .

A . (3x2 + 7x – 4) + (x2 + x + 3) = 4x2 + 8x – 1
B . (5x2  – 6x + 2) – (2x2  – 4x + 1) = 3x2  – 2x + 1  C .  (3x2  + 5x + 4) + (x2  + 2x – 5) = 3x2  + 7x – 1   D .  (6x2  – 3x – 8) – (3x2  + 5x – 4) = 3x2  – 2x – 4 E . (x + 8)(x – 9) = x2 + x – 72
F . (x – 3)(x – 7) = x2 – 10x + 21

Item 14
Selected-Response
A weather balloon is 10 yards in diameter . It is in the shape of a sphere . What is the volume of the weather balloon to the nearest cubic yard?

A . 59  cubic  yards B .  105 cubic yards C . 294 cubic yards  D . 523 cubic yards

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 201 of 230

The coordinate grid shows the graph of the quadratic function f(x) .
y

6
5
4
3
2
1

6 –5 –4 –3 –2 –1 0

–1
–2
–3
–4
–5

1 2 3 4 5

–6

6

–

x

The equation of f(x) can be written in the form f(x) = a(x – h)2 + k, where a, h, and k are rational numbers .

Part A
What are the values of h and k?

A . h = –2 and k = 3  B . h = –2 and k = 6  C .  h = 2 and k = 3  D . h = 2 and k = 4

Part B
Which statement describes the value of a for the function f(x)?

A . The value of a is a number less than –1 .
B .   The value of a is a number between –1 and 0 .   C .   The value of a is a number between 0 and 1 .   D . The value of a is a number greater than 1 .

Page  202 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Part A
Which situation could be modeled by a function with a domain of all positive integers?

A .   the distance a runner has moved during a race as a function of time since   the race started
B .   the amount of fish food required in a fish tank as a function of the number of fish       in the tank
C . the amount of power required to operate a computer as a function of the length of time the computer is on
D . the amount of water required by an animal as a function of the mass of the animal

Part B
Select the situation that could be modeled by a function with a domain that includes positive and negative real numbers .

A . the height of a plant as a function of time since the seed was planted
B . the elevation of a hiker as a function of the number of steps taken
C . the temperature as a function of the time of day
D .   the amount of time required to read a book as a function of the number of words    in the book
E . the amount of precipitation as a function of the outdoor temperature

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 203 of 230

The table defines a quadratic function .

x	y	
–1	5	
0	1	
1	–1	
3	1	
What is the average rate of change between x = –1 and x = 1? A . undefined

3

B .  – 1

C .	–3

D .	–4

Page  204 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Part A	What are the zeros of the function f(x) = x2 – 6x + 8 ? Explain how you determined your answer . Write your answer in the space provided .

Part B	Explain how you know that the function g(x) = x2 – 6x + 10 has a minimum value and not a maximum value . Find the minimum value of the function . Write your answer in the space provided .

Part B

Part A

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 205 of 230

Study this equation of a circle .

x2  – 6x + y2  + 2y + 6 = 0

Which of these represents the center and radius of the circle?

A . center:  (3,  –1),  radius:  4
B . center:  (–3,  1),  radius:  4
C . center:  (3,  –1),  radius:  2
D .   center: (–3, 1), radius:  2

Item 20
Selected-Response
One bag of lawn fertilizer can cover approximately 5,000 square feet . Mike’s lawn is about 500 square feet . Mike fertilizes his lawn an average of 4 times per year . How many full years will he be able to fertilize his lawn with one bag of fertilizer?

A .   2 years
B .   3 years
C .   9 years
D . 10 years

Page  206 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

How many zeros does the quadratic function f(x) have? Explain how you determined your answer . Write your answer in the space provided .

f(x) = x2  + 15x + 56

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 207 of 230

There are 52 cards in a deck . Each card in the deck is numbered with the whole numbers 1 through 13 . Each number appears on 4 cards in the deck . A student draws a card from the deck and then draws another card without replacing the first .

Part A	What is the probability of picking a 1 on the first draw and then a 7 on the second draw?
Write your answer in the space provided .

Part B	Explain why picking a 1 first and a 7 second are NOT independent events . Write your answer in the space provided .

Part B

Part A   	

Page  208 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

When rolling a number cube with sides labeled 1 through 6, what is the probability of rolling an even number or a number less than 3?
A .  5
6
B .  2
3
C .  1
2
D .  1
3

Item 24
Selected-Response
What is the probability of having rolled a 5 on a number cube with sides labeled 1 through 6 if you know that you rolled an odd number?
A .  1
6
B .  1
3
C .  1
2
D .  2
3

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 209 of 230

ADDITIONAL PRACTICE ITEMS ANSWER KEY

Item	Standard/ Element	DOK
Level	Correct Answer	Explanation	

1	

MGSE9-12 .G .SRT .3	

1	

A	The correct answer is choice (A) . The missing angle  of the triangle in choice (A) is 52°, making it similar   to the triangle given . Choices (B), (C), and (D) are incorrect because they have angle measures that are different than the original triangle .	
2	MGSE9-12 .G .GMD .3	2	N/A	See scoring rubric and exemplar responses beginning on page 214 .	
3	MGSE9-12 .G .SRT .5	3	N/A	See scoring rubric and exemplar responses beginning on page 216 .	

4	

MGSE9-12 .G .SRT .7	

1	

D	The correct answer is choice (D) cos 40° = sin 50° .   The sine of an angle is equal to the cosine of the angle’s complement . Choices (A), (B), and (C) are incorrect because they do not correspond to any trigonometric identities .	

5	
MGSE9-12 .G .GPE .4	
2	
A	The correct answer is choice (A) (0, 5) . The point    is only 5 units away from the center of the circle .
Choices (B), (C), and (D) are incorrect because they are 10 units away from the center of the circle .	
6	MGSE9-12 .G .SRT .8	2	N/A	See scoring rubric and exemplar responses on page 218 .	
7	MGSE9-12 .G .C .5	2	N/A	See scoring rubric and exemplar responses beginning on page 219 .	

8	

MGSE9-12 .G .C .2	

2	

D	The correct answer is choice (D) 25 . The measure  of arc CD is 80° and the measure of angle DAC is 40° . Since the angles in a triangle add to 180°, the
measure of angle AOB is 50°; if 2x = 50, then x = 25 .
Choice (A) is incorrect because it is the measure of angle AOB . Choice (B) is incorrect because the answer is true only if the triangle is isosceles .
Choice (C) is incorrect because a computation error was made when determining the value of x .	
9	MGSE9-12 .G .GMD .1	2	N/A	See scoring rubric and exemplar responses beginning on page 221 .	

10	

MGSE9-12 .N .RN .2	

2	

A	The correct answer is choice (A) –56x2 . The given
}
expression can be rewritten as –4 ∙ Ï28 ∙ 7 ∙ x4,
which simplifies to –56x2 . Choices (B), (C), and (D)  are incorrect because the perfect squares are not factored out of the radical correctly .	

Page  210 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Item	Standard/ Element	DOK
Level	Correct Answer	Explanation	

11	

MGSE9-12 .N .RN .3	

1	

A	}
The correct answer is choice (A) 4 + Ï7 . Choice (B)
is incorrect because it simplifies to 4 . Choice (C) is incorrect because it simplifies to 6 . Choice (D) is
5
incorrect because it simplifies to 0 .	
12	MGSE9-12 .A .APR .1	2	N/A	See scoring rubric and exemplar responses beginning on page 223 .	

13	

MGSE9-12 .A .APR .1	

2	

A/B/F	The correct answers are choices (A), (B), and (F) . Choices (A) and (B) are correct because they combine like terms correctly . Choice (F) is correct because
the binomials  are  multiplied  correctly .  Choices  (C) and (D) are incorrect  because  they  do  not  combine like terms correctly . Choice (E) is incorrect  because after the binomials are multiplied, the x terms are not combined correctly .	

14	

MGSE9-12 .G .GMD .3	

2	

D	The correct answer is choice (D) 523 cubic yards . Choice (A) is incorrect because the value is squared and not cubed . Choice (B) is incorrect because the
radius is squared instead of cubed . Choice (C) is incorrect because 3 was used instead of 4 .
4	3	

15	

MGSE9-12 .F .IF .7a	

2	

Part A: D
Part B: B	Part A: The correct answer is choice (D) h = 2 and k = 4 . The maximum of the function is at point  (2, 4), so the value of h is 2 and the value of k is   4 . Choices (A), (B), and (C) are incorrect because
they all give incorrect values for the maximum of the function, (h, k) .
Part B: The correct answer is choice (B) The value   of a is a number between −1 and 0 . The function must have a negative number for a, and the shape of the graph indicates that it will be between −1  and 0 . Choice (A) is incorrect because the graph with a value of a less than –1 would be a less wide parabola . Choices (C) and (D) are incorrect because the graph would have a minimum and an increasing end behavior .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 211 of 230

Item	Standard/ Element	DOK
Level	Correct Answer	Explanation	

16	

MGSE9-12 .F .IF .5	

2	

Part A: B
Part B: E	Part A: The correct answer is choice (B) the amount   of fish food required in a fish tank as a function of   the number of fish in the tank . It is the only choice that models the domain (the number of fish in a tank) as using only positive integers .
Part B: The correct answer is choice (E) the amount of precipitation as a function of the outdoor temperature . It is the only choice that models the domain (the outdoor temperature) as using positive and negative real numbers .	

17	

MGSE9-12 .F .IF .6	

2	

C	The correct answer is choice (C) –3 . Rate of change is found by finding the slope of the line containing the indicated points . Choice (A) is incorrect because it incorrectly computes the slopes . Choice (B) is incorrect because it reverses the numerator and denominator in the slope formula . Choice (D) is incorrect because it incorrectly computes the slope .	
18	MGSE9-12 .A .SSE .3a MGSE9-12 .A .SSE .3b	3	N/A	See scoring rubric and exemplar responses beginning on page 225 .	

19	

MGSE9-12 .G .GPE .1	

2	

C	The correct answer  is  choice  (C)  center:  (3,  –1), radius: 2 . When the equation is changed to standard form using completing the square, the h- and k-values are 3 and –1 and r2 = 4, so r = 2 . Choices (A) and
(B) are incorrect because the radius comes from taking the square root of the constant in standard form . Choice (D) is incorrect because the signs of the center are opposite .	

20	

MGSE9-12 .G .MG .2	

2	

A	The correct answer is choice (A) 2 years . The fertilizer will run out halfway into the third year, lasting only
2 full years . Choice (B) is incorrect because the fertilizer will not last through 3 years . Choice (C) is incorrect because the number of times the fertilizer is applied is subtracted from the total amount instead  of divided . Choice (D) is incorrect because the  number of times a year the fertilizer is applied is not divided by the total amount of fertilizer .	
21	MGSE9-12 .F .IF .8a	2	N/A	See scoring rubric and exemplar responses on page 227 .	
22	MGSE9-12 .S .CP .5	1	N/A	See scoring rubric and exemplar responses beginning on page 228 .	

Page  212 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Item	Standard/ Element	DOK
Level	Correct Answer	Explanation	

23	

MGSE9-12 .S .CP .7	

1	

B	The correct answer is choice (B) 2 . An even number
3
or a number less than 3 includes the outcomes of 1,
2, 4, and 6 and there are 6 outcomes in the sample space; 4 simplifies to 2 . Choice (A) is incorrect
6	3
because the probability of rolling a 1 and the
probability of rolling a number less than 3 were added together without subtracting the overlap . Choice (C) is incorrect because it is the probability of an even number only . Choice (D) is incorrect because it is the probability of a number less than 3 only .	

24	

MGSE9-12 .S .CP .3	

2	

B	The correct answer is choice (B) 1 . With the
3
conditional probability we assume that an odd
number was rolled, which reduces our sample space to 1, 3, and 5 . Out of those possibilities, the probability of rolling a 5 is 1, 1 successful outcome
3
out of 3 total outcomes . Choice (A) is incorrect
because it is the probability of rolling 5 without knowing an odd number was rolled . Choice (C) is incorrect because it is the probability of rolling an odd number . Choice (D) is incorrect because it is the complement of the correct answer .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 213 of 230

ADDITIONAL PRACTICE ITEMS SCORING RUBRICS AND EXEMPLAR RESPONSES
Item 2
Scoring Rubric

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of using volume formulas to solve problems .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of using volume formulas to solve problems .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of using volume formulas to solve problems .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Page  214 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Exemplar Response

Points Awarded	Sample Response	

2	The volume of the cone is about 3 .1 times the volume of the pyramid .
AND
I found the volume of the cone and then divided it by the volume of the pyramid to find the answer . Or other valid explanation .	

1	The volume of the cone is about 3 .1 times the volume of the pyramid with no explanation or an incorrect explanation .
OR
an explanation that contains a computation error but contains the correct process	
0	Response is irrelevant, inappropriate, or not provided.	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 215 of 230

Scoring Rubric

Points	Rationale	

4	The response achieves the following:
•	The response demonstrates a complete understanding of using congruence and similarity criteria for triangles to solve problems and prove relationships in geometric figures .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

3	The response achieves the following:
•	The response demonstrates a nearly complete understanding of using congruence and similarity criteria for triangles to solve problems and prove relationships in geometric figures .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

2	The response achieves the following:
•	The response demonstrates a partial understanding of using congruence and similarity criteria for triangles to solve problems and prove relationships in geometric figures .
•	The response is only partially correct .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

1	The response achieves the following:
•	The response demonstrates a minimal understanding of using congruence and similarity criteria for triangles to solve problems and prove relationships in geometric figures .
•	The response is only minimally correct .
•	The response shows the incomplete or inaccurate application of a relevant strategy .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of using congruence and similarity criteria for triangles to solve problems and prove relationships in geometric figures .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Page  216 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Exemplar Response

Points Awarded	Sample Response	

4	/BDC is a right angle, by the definition of an altitude .
AND
That means /ABC is congruent to /BDC since they are both right angles .
AND
The Reflexive Property of Congruence means that /BCD is congruent to /ACB since they are the same angle .
AND
That proves Angle-Angle Similarity, so △ABC is similar to △BDC . Or other valid explanation .	
3	The student correctly answers three of the four parts .	
2	The student correctly answers two of the four parts .	
1	The student correctly answers one of the four parts .	
0	Response is irrelevant, inappropriate, or not provided.	
Note: If a student makes an error in one part that is carried through to subsequent parts, then the student is not penalized again for the same error.

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 217 of 230

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of finding the sine of an angle in a triangle .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of finding the sine of an angle in a triangle .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of finding the sine of an angle in a triangle .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Exemplar Response

Points Awarded	Sample Response	

2	The sine of x is equal to the ratio of 4 to 5 . Or other valid response .
AND
Use the Pythagorean Theorem to find that the length of the third side of the triangle is 4  inches . The sine of an angle is the ratio of the length of the opposite leg compared to the length of the hypotenuse . Or other valid explanation .	

1	The sine of x is equal to the ratio of 4 to 5 with no explanation or an incorrect explanation .
OR
an explanation that contains a computation error but contains the correct process	
0	Response is irrelevant, inappropriate, or not provided.	

Page  218 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of the equation for the area of a sector .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of the equation for the area of a sector .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of the equation for the area of a sector .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 219 of 230

Points Awarded	Sample Response	

2	I split the formula for the area into two parts, πr² and θ . The first part of the formula
360
shows pi times the radius squared, which is the area of the entire circle . Or other valid
explanation .
AND
The second part of the formula shows the degree measure of the central angle of the sector divided by the degree measure of the entire circle, so it represents the fraction of the circle     that consists of the sector . The product of  these  two  parts  represents  the  fraction  of  the area of the circle that is included in the sector . Or other valid explanation .	

1	I split the formula for the area into two parts, πr² and θ . The first part of the formula
360
shows pi times the radius squared, which is the area of the entire circle . Or other valid
explanation .
OR
The second part of the formula shows the degree measure of the central angle of the sector divided by the degree measure of the entire circle, so it represents the fraction of the circle     that consists of the sector . The product of  these  two  parts  represents  the  fraction  of  the area of the circle that is included in the sector . Or other valid explanation .	
0	Response is irrelevant, inappropriate, or not provided.	

Page  220 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of the formulas for the volume of pyramids and rectangular prisms .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of the formulas for the volume of pyramids and rectangular prisms .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of the formulas for the volume of pyramids and rectangular prisms .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 221 of 230

Points Awarded	Sample Response	

2	The volume of a rectangular prism is equal to the area of its base times its height, and the volume of a pyramid is equal to one-third times the area of its base times its height . Since the pyramid and the rectangular prism have congruent bases, their bases have the same area . Or other valid explanation .
AND
Since the heights are congruent, they also have the same height . So the volume of the pyramid is one-third the volume of the rectangular prism . Or other valid explanation .	

1	The volume of a rectangular prism is equal to the area of its base times its height, and the volume of a pyramid is equal to one-third times the area of its base times its height . Since the pyramid and the rectangular prism have congruent bases, their bases have the same area . Or other valid explanation .
OR
Since the heights are congruent, they also have the same height . So the volume of the pyramid is one-third the volume of the rectangular prism . Or other valid explanation .	
0	Response is irrelevant, inappropriate, or not provided.	

Page  222 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of adding, subtracting, and multiplying polynomials .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of adding, subtracting, and multiplying polynomials .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of adding, subtracting, and multiplying polynomials .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 223 of 230

Points Awarded	Sample Response	

2	Part A: First, distribute the factor 2 to each  term  of  the  expression  in  parentheses . Then, combine the like terms by adding or subtracting their coefficients . Or other valid explanation .
AND
Part B: There is only one nonzero term . Or other valid explanation .	

1	Part A: First, distribute the factor 2 to each  term  of  the  expression  in  parentheses . Then, combine the like terms by adding or subtracting their coefficients . Or other valid explanation .
OR
Part B: There is only one nonzero term . Or other valid explanation .	
0	Response is irrelevant, inappropriate, or not provided.	

Page  224 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points	Rationale	

4	The response achieves the following:
•	The response demonstrates a complete understanding of rewriting a quadratic function to find different properties .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

3	The response achieves the following:
•	The response demonstrates a nearly complete understanding of rewriting a quadratic function to find different properties .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

2	The response achieves the following:
•	The response demonstrates a partial understanding rewriting a quadratic function to find different properties .
•	The response is only partially correct .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

1	The response achieves the following:
•	The response demonstrates a minimal understanding rewriting a quadratic function to   find different properties .
•	The response is only minimally correct .
•	The response shows the incomplete or inaccurate application of a relevant strategy .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of rewriting a quadratic function to find different properties .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 225 of 230

Points Awarded	Sample Response	

4	Part A: The zeros are 2 and 4 .
AND
To find the zeros, I set the value of the function equal to 0 . Then I factored the quadratic expression on the right side of the equation . Next, I used the Zero Product Property to set      each factor equal to 0 . Then I solved each of the resulting equations for x . These values of x     are the zeros of the function . Or other valid explanation.
AND
Part B: The coefficient of the x2 is positive, so the function opens up, which means it has a minimum value instead of a maximum . Or other valid explanation.
AND
The minimum value of the function is 1 .	
3	The student correctly answers three of the four parts .	
2	The student correctly answers two of the four parts .	
1	The student correctly answers one of the four parts .	
0	Response is irrelevant, inappropriate, or not provided.	

Note: If a student makes an error in one part that is carried through to subsequent parts, then the student is not penalized again for the same error.

Page  226 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of finding the zeros of a quadratic function .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of finding the zeros of a quadratic function .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of finding the zeros of a quadratic function .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Exemplar Response

Points Awarded	Sample Response	

2	There are two zeros for this function .
AND
I set the expression equal to zero and factored the expression . Since there are two factors, I   know there are two zeros . Or other valid explanation .	

1	There are two zeros for this function with no explanation or an incorrect explanation .
OR
an explanation that contains a computation error but contains the correct process	
0	Response is irrelevant, inappropriate, or not provided.	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 227 of 230

Points	Rationale	

2	The response achieves the following:
•	The response demonstrates a complete understanding of independent events .
•	The response is correct and complete .
•	The response shows the application of a reasonable and relevant strategy .
•	Mathematical ideas are expressed coherently in the response, which is clear, complete, logical, and fully developed .	

1	The response achieves the following:
•	The response demonstrates a partial understanding of independent events .
•	The response is mostly correct but contains either a computation error or an unclear or incomplete explanation .
•	The response shows the application of a relevant strategy, though the strategy may be only partially applied or may remain unexplained .
•	Mathematical ideas are expressed only partially in the response .	

0	The response achieves the following:
•	The response demonstrates limited to no understanding of independent events .
•	The response is incorrect .
•	The response shows no application of a strategy .
•	Mathematical ideas cannot be interpreted or lack sufficient evidence to support even a limited understanding .	

Page  228 of 230	Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents

Points Awarded	Sample Response	

2	Part A: 4 or other valid response
663
AND
Part B: When the first card is drawn from the deck, there are 52 cards to choose from . Because the first card is not replaced, there are only 51 cards to choose from when the second card is drawn . So the result of picking a 1 on the first draw affects the sample     space and, therefore, the probability of picking a 7 on the second draw . This means the events are not independent . Or other valid explanation .	

1	Part A: 4 or other valid response
663
OR
Part B: When the first card is drawn from the deck, there are 52 cards to choose from . Because the first card is not replaced, there are only 51 cards to choose from when the second card is drawn . So the result of picking a 1 on the first draw affects the sample     space and, therefore, the probability of picking a 7 on the second draw . This means the events are not independent . Or other valid explanation .	
0	Response is irrelevant, inappropriate, or not provided.	

Georgia Milestones Analytic Geometry EOC Study/Resource Guide for Students and Parents	Page 229 of 230

Study/Resource Guide for Students and Parents Analytic Geometry
End-of-Course