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VOLUME 1

ALGEBRA 2 WITH PROBABILITY SOUTH CAROLINA

ISBN: 979-8-89353-913-4 Math Nation AGA was originally developed by Illustrative Mathematics®, and is copyright 2019 by Illustrative Mathematics. It is licensed under Creative Commons Attribution 4.0 International License (CCBY 4.0). The online curriculum resource platform oﬀered at "mathnation.com", including the videos and practice questions, as well as custom curriculum adaptations, are additions to the original Illustrative Mathematics content is copyright 2024 by Accelerate Learning, Inc. The trademark "Math Nation" is owned by Accelerate Learning, Inc. All other trademarks and product names referred to in the Math Nation AGA curriculum are the property of their respective owners and used solely for educational purposes. Unless otherwise stated, Math Nation has no relationship with any of the companies or brands mentioned in the curriculum and does not endorse or have a preference for any of those companies or brands. This curriculum includes public domain images or openly licensed images that are copyright by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.

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Table of Contents Unit 1: Function Foundations.......................................................... 1 Lesson 1: Exploring Parent Functions........................................3 Lesson 2: Average Rate of Change......................................... 15 Lesson 3: Digging Deeper into Average Rate of Change............. 23 Lesson 4: Systems of Linear Inequalities................................. 29 Lesson 5: Introduction to Linear Programming......................... 39 Lesson 6: Using Linear Programming to Solve Problems............ 51 Lesson 7: Matching Up to Data.............................................. 59 Lesson 8: Translating Functions.............................................. 69 Lesson 9: More Translations.................................................. 77 Lesson 10: Reflecting Functions............................................. 85 Lesson 11: Expressing Transformations of Functions Algebraically......................................... 93 Lesson 12: Scaling the Outputs............................................ 103 Lesson 13: Scaling the Inputs.............................................. 111 Lesson 14: Inverse Functions............................................... 119 Lesson 15: Finding Inverses and Verifying Graphically............. 127 Lesson 16: Finding and Interpreting Inverse Functions............ 135 Unit 2: Polynomial Functions...................................................... 143 Lesson 1: Let’s Make a Box................................................. 145 Lesson 2: Funding the Future............................................... 151 Lesson 3: Introducing Polynomials........................................ 157

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Lesson 4: Connecting Factors and Zeros................................ 163 Lesson 5: Different Forms of Polynomials.............................. 167 Lesson 6: Using Factors and Zeros....................................... 173 Lesson 7: End Behavior...................................................... 179 Lesson 8: Graphing Polynomial Functions.............................. 187 Lesson 9: Multiplicity.......................................................... 197 Lesson 10: Key Features of Polynomial Functions.................... 207 Lesson 11: Polynomial Division – Part 1................................ 219 Lesson 12: Polynomial Division – Part 2................................ 227 Lesson 13: Connecting Polynomial Division and Graphing........ 235 Lesson 14: The Remainder and Factor Theorems.................... 241 Lesson 15: The Difference of Squares................................... 251 Lesson 16: Factoring Sums and Differences of Cubes.............. 259 Lesson 17: Relating Perfect Squares and Cubes to Graphs of Polynomials................................................... 265 Unit 3: Complex Numbers and Polynomials................................ 273 Lesson 1: A New Kind of Number......................................... 275 Lesson 2: Introducing the Number ....................................... 281 Lesson 3: Adding and Subtracting Complex Numbers.............. 287 Lesson 4: Multiplying Complex Numbers............................... 295 Lesson 5: Operations with Complex Numbers......................... 301 Lesson 6: Working Backward............................................... 307 Lesson 7: Solving Quadratic Equations.................................. 311 Lesson 8: Completing the Square and Complex Solutions........ 319 Lesson 9: The Quadratic Formula and Complex Solutions......... 327 Lesson 10: Graphs of Quadratic Functions and the Types of Solutions.............................................. 333 iv | Table of Contents

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Lesson 11: Real and Non-Real Solutions................................ 343 Lesson 12: Determining Zeros within a Real-World Context...... 349 Lesson 13: Solving Polynomials for Real and Complex Zeros......... 359 Lesson 14: Factoring Polynomials from a .............................. 369 Real-World Context Lesson 15: Solving One-Variable Quadratic Inequalities........... 375 Lesson 16: Solving Two-Variable Quadratic Inequalities........... 383 Lesson 17: Solving Polynomial Inequalities............................ 393 Lesson 18: Solving Polynomial Inequalities in a Real-World Context............................................................ 401 Unit 4: Rational Functions.......................................................... 409 Lesson 1: Rewriting Rational Expressions.............................. 411 Lesson 2: Graphs of Rational Functions – Part 1..................... 417 Lesson 3: Graphs of Rational Functions – Part 2..................... 425 Lesson 4: Key Features of Rational Functions......................... 431 Lesson 5: Graphs of Rational Functions with Removeable Discontinuities..................................................... 441 Lesson 6: Multiplying and Dividing Rational Expressions.......... 451 Lesson 7: Adding and Subtracting Rational Expressions with Like Denominators........................................ 461 Lesson 8: Adding and Subtracting Rational Expressions with Unlike Denominators..................................... 469 Lesson 9: Rational Equations – Part 1................................... 475 Lesson 10: Rational Equations – Part 2................................. 481 Lesson 11: Solving Rational Equations.................................. 490 Unit 5: Radical Functions............................................................ 497 Lesson 1: Exploring Rational Exponents................................ 499 Lesson 2: Positive and Negative Rational Exponents................ 505

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Lesson 3: Squares and Square Roots.................................... 517 Lesson 4: Key Features of Square Root Functions................... 523 Lesson 5: Cubes and Cube Roots.......................................... 533 Lesson 6: Key Features of Cube Root Functions...................... 541 Lesson 7: Graphing Square Root and Cube Root Functions Given a Written Description................................... 549 Lesson 8: Solving Real-World Problems Using Square Root and Cube Root Functions...................................... 557 Lesson 9: Inequivalent Equations........................................5569 Lesson 10: Solving Radical Equations – Part 1........................ 577 Lesson 11: Solving Radical Equations – Part 2........................ 583

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Unit 1: Function Foundations

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Unit 1, Lesson 1: Exploring Parent Functions

Warm-Up: Which One Doesn’t Belong? The graphs of 4 functions are shown. Function A

Function B

Function C

Function D

1. Which function doesn’t belong?

2. Explain your reasoning.

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Collaborative Activity: Representations of Linear, Quadratic, and Exponential Functions Work with your partner to complete the following. 1. The graphs of three linear functions are shown. Linear Function

Graph

Table of Values a. Match each table of values to the graph it corresponds with by writing the number of the table in the empty row of the previous table. Table of Values 1

𝒙

𝒚

Table of Values 2

𝒙

𝒚

Table of Values 3

𝒙

𝒚

−3

−0.8

−1

5 2

−1

−5

−1

−1.6

7 2

−2

−1.2

−2

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b. Each linear function can also be represented by linear equations in different forms. Match each graph to a set of equations by writing the letter of the graph in the empty column below. Standard Form

Slope-Intercept Form

Point-Slope Form

3𝑥 − 𝑦 = 2

𝑦 = 3𝑥 − 2

(𝑦 − 4) = 3(𝑥 − 2)

𝑥 − 2𝑦 = −6

2𝑥 + 5𝑦 = −10

𝑦 = 1 𝑥 + 3 2

𝑦 = − 2 𝑥 − 2 5

Graph

(𝑦 − 4) = 1 (𝑥 − 2) 2

(𝑦 + 1.6) = −2 (𝑥 + 1) 5

c. On each graph, draw an X on the coordinate that was used to create the point-slope form of the equation in the table. Then, circle the coordinates of the point in the table of values. d. Plot a point at the-intercept of each graph. Then, underline the value of the 𝑦-intercept in the equation of the line that reveals it.

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2. The graphs of three quadratic functions are shown. Quadratic Function

Graph

Table of Values a. Match each table of values to the parabola it corresponds with by writing the number of the table in the empty row of the previous table. Table of Values 4

𝒙

𝒚

Table of Values 5

𝒙

𝒚

Table of Values 6

𝒙

𝒚

−3

−8

−1

−8

−1

−6

−1

−4

−2

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b. Each quadratic function can also be represented by equations in different forms. Match each graph to a set of equations by writing the letter of the graph in the empty column below. Graph

Equations

Form

�(𝑥) = 𝑥2 + 𝑥 − 6

Standard

�(𝑥) = (𝑥 − 2)(𝑥 + 3)

Factored

�(𝑥) = �𝑥 + 1 � − 6.25

Vertex

�(𝑥) = −2𝑥2 − 4𝑥 − 2

Standard

�(𝑥) = −2(𝑥 + 1)(𝑥 + 1)

Factored

�(𝑥) = −2(𝑥 + 1)2

Vertex

�(𝑥) = 2𝑥2 + 4𝑥 − 6

Standard

�(𝑥) = 2(𝑥 + 3)(𝑥 − 1)

Factored

�(𝑥) = 2(𝑥 + 1)2 − 8

Vertex

c. Draw an X on each graph at the vertex of the parabola. Then, underline the values that can be used to determine the vertex in the equation that reveals it. d. Plot a point at the 𝑥-intercept(s) of each graph. Then, circle the form of equations that can be used to reveal them. e. Underline the value of the 𝑦-intercept in the equation that reveals it.

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3. The graphs of two exponential functions are shown. Exponential Function

Graph

Table of Values a. Match each table of values to the exponential graph it corresponds with by writing the number in the empty row of the previous table. Table of Values 7

𝒙

𝒚

Table of Values 8

𝒙

𝒚

−3

−1

−2

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4.5

6.75

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b. Each exponential function can also be represented by equations in different forms. Match each graph to a corresponding set of equations by writing the letter of the graph in the last column. Exponential Functions

Graph

�(𝑥) = 3(1.5)𝑥

�(𝑥) = 3�1 + 1 �

𝑥

�(𝑥) = 2(0.5)𝑥

�(𝑥) = 2�1 − 1 �

𝑥

2 2

c. Plot a point at the 𝑦-intercept of each graph. Then, underline the value of the 𝑦-intercept in each equation.

d. Both exponential functions have a horizontal asymptote at 𝑦 = 0. Explain what this means in terms of the graph.

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Collaborative Activity: Parent Functions Card Sort 1. Work with your partner to sort the cards to match the equation, graph, table of values, and key features of the parent function for each function type. The equation of each parent function is matched to the corresponding function type in the table. Complete the table by writing the letter of the graph, table of values, and set of key features for each parent function. Function Type Linear Quadratic Cubic Exponential growth Exponential decay Absolute value Square root Cube root

Parent Function

Graph

Table of Values

Key Features

𝑦 = 𝑥

𝑦 = 𝑥2

𝑦 = 𝑥3

𝑦 = (2)𝑥

𝑦 = �1 � 2

𝑥

𝑦 = |𝑥|

𝑦 = √𝑥 3

𝑦 = √𝑥

Lesson Summary This lesson is the introduction to the study of functions in this course. A family of functions share the same characteristics and general shape. The most basic of these is called the parent function. Transformations may be applied to the parent function to create other members of the family of functions sharing similar characteristics. Knowing information about a parent function is helpful for identifying transformations of that parent function.

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Throughout this course, you will explore a variety of parent functions and their graphs. Some of the parent functions should already be familiar, such as linear, quadratic, absolute value, and exponential functions. The remaining parent functions explored in this lesson (and some additional functions) will be studied in detail in future lessons.

Practice Problems 1. Representations of transformations of different parent functions are shown in the table. Determine which family of functions each transformation belongs to. Representation of Transformation

Equation of Parent Function

□ □ □ □

□

𝒙

𝒚

−2 3

−1 4

2 7

□ □ □ □ □

𝑦 = |𝑥 + 5|

□ □ □

𝒙

𝒚

−1 1 4

1 2

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□ □ □

𝑦 = 𝑥 𝑦 = 𝑥 𝑦 = 𝑥

2 3

𝑦 = |𝑥|

𝑦 = 𝑥

𝑦 = 𝑥2 𝑦 = 𝑥

𝑦 = |𝑥|

𝑦 = 𝑥

𝑦 = 𝑥2 𝑦 = 𝑥3

𝑦 = |𝑥| 𝑦 = 𝑥

𝑦 = 𝑥2 𝑦 = 𝑥

𝑦 = |𝑥|

□ □ □

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑥

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2. Representations of transformations of different parent functions are shown in the table. Determine which family of functions each transformation belongs to. Representation of Transformation

Equation of Parent Function □

𝒙

𝒚

−2 6

−1 3

2 6

□ □ □

□ □ □ □

□ 𝑥

�(𝑥) = �1 � − 5 2

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□ □ □

𝑦 = 𝑥 𝑦 = 𝑥 𝑦 = 𝑥

2 3

𝑦 = |𝑥|

𝑦 = 𝑥 𝑦 = 𝑥 𝑦 = 𝑥

2 3

𝑦 = |𝑥| 𝑦 = 𝑥

𝑦 = 𝑥2 𝑦 = 𝑥3

𝑦 = |𝑥|

□ □ □

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑦 = √𝑥 𝑦 = 2𝑥

𝑦 = �1 � 2

𝑥

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3. The graph of a transformed parent function is shown. Daniel identified the function family of the graph as quadratic, but he made an error. a. Explain why Daniel is incorrect.

b. Identify the correct parent function for the graph.

Review Problem 4. Consider the linear function 𝑦 = 1 𝑥 − 2. 3

a. Determine the slope and 𝑦-intercept of the linear function. b. Graph the linear function on the coordinate plane.

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Unit 1, Lesson 2: Average Rate of Change

Warm-Up: Evaluating Quotients Evaluate each of the expressions when 𝑥 = 2 and 𝑦 = − 2. 1. 𝑦 + 3 𝑥 + 5

1−𝑦

2. 𝑥 − 6

Guided Activity: Average Rate of Change 1. The graph of a function is shown. The coordinates of 3 points on the graph are labeled.

20 15

a. Discuss with your partner how a secant line can be used to find the average rate of change between any 2 points on the graph. b. What is the average rate of change between (5.06, 4.47) and (16.3, 0.74)? c. What is the average rate of change between (5.06, 4.47) and (35.78, − 3.77)? © Accelerate Learning Inc. - All Rights Reserved

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10 (5.06, 4.47)

(16.3, 0.74) -5

0 -5

x 30

(35.78, −3.77)

-10

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2. A table of values for function 𝑔 is shown. 𝒙

𝒚

−3

−78

−2 0

−1 28

−8

154

a. Determine the average rate of change over the interval [ −2, 3].

b. Determine the average rate of change over the interval [ −1, 1].

c. Determine the average rate of change over the interval [2, 5].

3. Determine the average rate of change over the interval [ −1.5, 2.4] for the function 𝑓(𝑥) = 3(𝑥)2 − 4𝑥 + 2.

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Exploration Activity: Comparing Average Rate of Change 1. Three companies’ total sales over 10 weeks, in thousands of dollars, are shown. Company A

Company B

Company C

a. Complete the table by determining the average rate of change for each of the functions over the given intervals. Round to the nearest tenth. Interval

Company A

Company B

Company C

[2, 4] [4, 6] [6, 8] [2, 8] b. Discuss with your partner similarities between the 3 companies’ sales over the 10 weeks. c. Why is the average rate of change different for companies A and C than for company B?

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2. Which function will have a constant average rate of change? A. 𝑦 = 3𝑥2 + 2𝑥 − 5

B. 𝑦 = 4|𝑥 − 2| + 3 C. 𝑦 = − 2(3)𝑥 + 5 D. 3𝑥 − 5𝑦 = 4

Lesson Summary A line can be drawn to connect any 2 points on a graph. This line is called a secant line and can be used to determine the average rate of change between any 2 points on a graph. The average rate of change of a function 𝑓 between inputs 𝑎 and 𝑏 is the change in the outputs divided by the change in the inputs: 𝑓(𝑏)−𝑓(𝑎) . It is the slope of the line joining (𝑎, 𝑓(𝑎)) and 𝑏−𝑎

(𝑏, 𝑓(𝑏)) on the graph.

The average rate of change can be used to summarize the overall change for a function over a specified interval. The average rate of change is different between any 2 points on a curve for all functions except a linear function. Linear functions always have a constant average rate of change, which is equivalent to the slope of the line.

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Three different representations of 𝑓(𝑥) are shown. Graph

Table

Equation

𝒙

𝒚

−1.5

−5

−4

0 1

−3.2

𝑦 = 4 (𝑥 − 1)(𝑥 + 4) 5

The average rate of change of 𝑓(𝑥) over the interval [−1.5, 1] is 𝑓(𝑏)−𝑓(𝑎) = 𝑓(1)−𝑓(−1.5) = 0+5 = 2. 2.5

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𝑏−𝑎

1−(−1.5)

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Practice Problems 1. Six functions are shown in the table, represented in different ways.

𝒙

𝒚

−2 9

Function A

Function B

𝑦 = 3𝑥 − 5

𝑦 = 𝑥2 + 4

Function C

Function D

Function E

Function F

−3

−7

−15

−27

𝒙

𝒚

−3

−30

−2

−11

−2

2 5

Select all of the functions that have a constant average rate of change. □ Function A □ Function B □ Function C □ Function D □ Function E □ Function F 20 | Unit 1

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2. Shirana and Madeline were riding their bikes to practice for a race. Graphs are shown that represent the distance, in miles (mi.), that each biker rode for the first 10 minutes (min.) of their ride. Shirana

Madeline

a. Complete the table by determining the average rate of change for each biker over each given interval. Round to the nearest hundredth, if necessary. Interval

Shirana

Madeline

[1, 2] [2, 5] [5, 6] [8, 9] b. Explain what the average rates of change indicate about the 2 bikers.

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3. Determine the average rate of change for the function 𝑓(𝑥) = 2𝑥2 − 𝑥 + 1 over each interval. a. �0, 1 � 2

b. �− 1, 2� c. �−3.1, −3�

Review Problems 4. Graph the quadratic function 𝑦 = − 2(𝑥 + 1)(𝑥 − 3) on the coordinate plane.

5. The system of equations � 2𝑥 + 𝑦 = 5 is graphed −4𝑥 + 3𝑦 = 5 on the coordinate plane. Use the graph to determine the solution to the system of equations.

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Unit 1, Lesson 3: Digging Deeper into Average Rate of Change

Warm-Up: Average Rate of Change Consider a function with an equation of 𝑦 = −𝑥2 − 3𝑥.

1. What is the average rate of change over the interval [ − 2.1, 1.6]?

2. Which interval has a positive rate of change? A. [1, 2]

B. [ − 3, 0] C. [ − 2, 0]

D. [ − 2, − 4]

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Guided Activity: Interpreting Average Rate of Change in Real-World Contexts 1. Over the course of the 20th century, intense protection of the southern white rhino has led to a significant increase in their population. Over the last few years, however, increases in poaching rates have led to another decline in their population. The table shows the estimated population size of the southern white rhino from 1929 to 2021. Years Since 1929 Population Size

150

3,800

5,565

6,760

17,470

21,316

20,375

15,940

a. Describe the pattern of the population size of the rhino from 1929 to 2021.

b. Find the average rate of change for the time period of 1991 to 2007.

c. Interpret the average rate of change found in part B in terms of the context.

d. Discuss with your partner why the average rate of change for the time period of 1991 to 2007 does not accurately describe the population change of the rhino from 1929 to 2021.

e. Find the average rate of change for the time period of 2012 to 2021.

f. Interpret the average rate of change found in part E in terms of the context.

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Collaborative Activity: Average Rate of Change 1. The function 𝑃(𝑥) = 2,639(1.1075)𝑥 represents the growth of the population of Queen Creek, Arizona, from 1990 to 2021, where 𝑥 is the number of years since 1990.

a. Discuss with your partner how to use the function to determine the average rate of change over a specified interval. b. Complete the statements. To calculate the average rate of change given a function, first determine the 𝑥-values

for the interval based on the information given. Substitute the

𝑥-values

into the function to determine the corresponding

𝑦-values 𝑦-values

of the 2 distinct points used to calculate the average rate of change.

𝑥-values

𝑦-values

c. What is the average rate of change for the time period of 1990 to 2002? Include units.

d. What is the average rate of change for the time period of 1998 to 2017? Include units.

e. Discuss with your partner what the slopes tell you about whether the graph is increasing or decreasing over any given interval. Summarize your discussion.

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2. In the first half of a hockey season, a player scored a total of 34 goals in the 15 games he played. In the second half of the season, the same player scored a total of 54 goals in the 36 games he played. What is the player’s average rate of change in number of goals per game from the first half of the season to the second half? Round to the nearest hundredth.

3. The average high temperature in degrees Fahrenheit (℉), by month, in a city in South Carolina can be modeled by the function 𝑡(𝑥) = − 0.9𝑥2 + 13𝑥 + 43, where 𝑥 is the month of the year. a. What is the average rate of change of the city’s high temperature for the interval between March and May?

b. What is the average rate of change of the city’s high temperature for the interval between August and November?

c. Interpret the average rates of change found in parts A and B in this context.

Lesson Summary The average rate of change can be determined using a graph, a table, or an equation modeling real-world situations. The slope between any 2 points on a function can be used to determine the average rate of change over that interval. In real-world situations, for example, the average rate of change is often determined between the beginning and end of a time period, such as a year or a decade. The average rate of change of a linear function is constant and is often referred to as simply the rate of change, or the constant rate of change, because the value will be the same between any 2 points on the line. The rate of change of a linear function is the slope. 26 | Unit 1

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Practice Problems The use of mobile phones as a form of communication has grown exponentially since 1976. The table shows the number of mobile phone subscriptions, in millions, from 1976 to 2020. 1. Consider the interval from 1976 to 1979. a. Find the average rate of change over this interval.

b. Interpret the average rate of change for this interval in terms of the context. Include units.

2. What is the average rate of change, in millions of mobile phone subscriptions per year, for the period from 1992 to 2003?

3. What is the average rate of change, in millions of mobile phone subscriptions per year, for the period from 2003 to 2018?

4. Why do you think the average rate of change from 2003 to 2018 is higher than the average rate of change from 1992 to 2003?

5. Interpret the average rate of change for the interval 2003 to 2018 in terms of the context.

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Years since 1976

Mobile Phone Subscriptions (in millions)

0.023

2.55

0.149

11.21

145.04

1,140

35 39 42 44

23.25

738.16

5,890 7,180 7,850 8,270

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Review Problem 6. Sergio graphed the linear inequality 𝑦 < − 2 𝑥 + 1, but he made an error. His graph 3 is shown. a. Explain Sergio’s error.

b. Graph the linear inequality 𝑦 < − 2 𝑥 + 1 3

on the coordinate plane.

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Unit 1, Lesson 4: Systems of Linear Inequalities

Warm-Up: A Silly Riddle Here is a riddle: “I am thinking of two numbers that add up to 5.678. The difference between them is 9.876. What are the two numbers?” 1. Name any pair of numbers whose sum is 5.678.

2. Name any pair of numbers whose difference is 9.876. 3. The riddle can be represented with two equations. Write the equations.

4. Solve the riddle. Explain or show your reasoning.

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Guided Activity: A Quilting Project To make a quilt, a quilter is buying fabric in two colors, light and dark. He needs at least 9.5 yards (yd.) of fabric in total. The light color costs $9 a yd. The dark color costs $13 a yd. The quilter can spend up to $110 on fabric. Here are two graphs that represent the two constraints.

1. Write an inequality to represent the length constraint. Let 𝑥 represent the yards of light fabric and 𝑦 represent the yards of dark fabric. 2. Select all the pairs that satisfy the length constraint. □ (5, 5)

□ (2.5, 4.5) □ (7.5, 3.5) □ (12, 10) 30 | Unit 1

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3. Write an inequality to represent the cost constraint.

4. Select all the pairs that satisfy the cost constraint. □ (1, 1) □ (4, 5) □ (8, 3)

□ (10, 1) 5. Explain why (2, 2) satisfies the cost constraint, but not the length constraint.

6. Find at least one pair of numbers that satisfies both constraints. Be prepared to explain how you know.

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7. What does the pair of numbers represent in this situation?

Exploration Activity: Focusing on the Details Here are the graphs of the inequalities in this system. �

𝑥 < 𝑦 𝑦 ≥ −2𝑥 − 6

Decide whether each point is a solution to the system. Be prepared to explain how you know. 1. (3, −5) 2. (0, 5) 3. (−6, 6) 4. (3, 3) 5. (−2, −2)

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Collaborative Activity: Scavenger Hunt Members of a high school math club are doing a scavenger hunt. Three items are hidden in the park, which is a rectangle that measures 50 meters (m) by 20 m. • The clues are written as systems of inequalities. One system has no solutions. • The locations of the items can be narrowed down by solving the systems. A coordinate plane can be used to describe the solutions. Can you find the hidden items? Sketch a graph to show where each item could be hidden.

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Lesson Summary This lesson explored pairs of two-variable linear inequalities used to represent the constraints in a situation. Each pair of inequalities forms a system of inequalities. Two or more inequalities that represent the constraints in the same situation form a system of inequalities. A solution to a system of linear inequalities is any (𝑥, 𝑦) pair that makes both inequalities true, or any pair of values that simultaneously meet all the constraints in the situation. The solutions to a system of inequalities are best represented by a region on a graph. All pairs of values that make the inequalities in a system true are solutions to the system. The solutions to a system of inequalities can be represented by the points in the region where the graphs of the two inequalities overlap. The graphs of 2 inequalities, 𝑦 > 2𝑥 and 𝑥 + 𝑦 < 10, are shown on 2 coordinate planes.

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𝑦 > 2𝑥 , graph 𝑥 + 𝑦 < 10 both inequalities on the same coordinate plane, as shown. The solution set to the system of inequalities is represented by the region where the 2 graphs overlap. To solve the system of inequalities �

Practice Problems 1. Two inequalities are graphed on the same coordinate plane. Which region represents the solution to the system of the two inequalities?

2. Select all the pairs of 𝑥 and 𝑦 that are solutions to the system of inequalities. □ 𝑥 = 0, 𝑦 = 0

�

𝑦 ≤ − 2𝑥 + 6 𝑥 − 𝑦 < 6

□ 𝑥 = − 5, 𝑦 = − 15 □ 𝑥 = 4, 𝑦 = − 2

□ 𝑥 = 3, 𝑦 = 0

□ 𝑥 = 10, 𝑦 = 0

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3. Jada has $200 to spend on flowers for a school celebration. She decides that the only flowers that she wants to buy are roses and carnations. Roses cost $1.45 each and carnations cost $0.65 each. Jada buys enough roses so that each of the 75 people attending the event can take home at least one rose.

a. Write an inequality to represent the constraint that every person takes home at least one rose.

b. Write an inequality to represent the cost constraint.

4. Here are the graphs of the equations 3𝑥 + 𝑦 = 9 and 3𝑥 − 𝑦 = 9 on the same coordinate plane. a. Label each graph with the equation it represents.

b. Identify the region that represents the solution set to 3𝑥 + 𝑦 < 9. Is the boundary line a part of the solution? Use a colored pencil or crosshatching to shade the region.

c. Identify the region that represents the solution set to 3𝑥 − 𝑦 < 9. Is the boundary line a part of the solution? Use a different colored pencil or cross-hatching to shade the region.

d. Identify a point that is a solution to both 3𝑥 + 𝑦 < 9 and 3𝑥 − 𝑦 < 9.

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Review Problem 5. Which coordinate pair is a solution to the inequality 4𝑥 − 2𝑦 < 22? A. (4, − 3) B. (4, 3)

C. (8, − 3) D. (8, 3)

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Unit 1, Lesson 5: Introduction to Linear Programming

Warm-Up: Which One Doesn’t Belong: Graphs of Solutions Which one doesn’t belong?

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Guided Activity: Introduction to Linear Programming Businesses often try to maximize their profits or minimize their costs. These situations involve several constraints beyond the variables of interest. Using systems of inequalities to optimize, or determine maximum or minimum values, within several linear inequalities’ constraints, is called linear programming. 1. Consider the system of linear inequalities shown.

�

𝑦 ≤ 𝑥 + 2

𝑦 ≤ −1 𝑥 + 5 2

𝑥 ≥ 0 𝑦 ≥ 0

a. Graph the system of linear inequalities on the coordinate plane.

The region that satisfies all 4 linear inequalities’ constraints is called the feasible solution.

b. Using all the constraints as 1 system of linear inequalities, a polygon is formed. How many vertices does the polygon have?

c. Identify the coordinate points of the vertices of the feasible solution represented by the polygon.

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The objective function is a function that defines the quantity of interest that is looking to be minimized or maximized. The objective function is written as 𝑓(𝑥, 𝑦), read as “𝑓 of 𝑥 and 𝑦”, where the function is dependent on both the 𝑥- and 𝑦-values. d. Evaluate the objective function 𝑓(𝑥,𝑦) = 3𝑥 − 4𝑦 for each of the vertices of the feasible solution. Vertex (𝒙, 𝒚)

𝒇(𝒙, 𝒚)

e. What is the maximum value of the objective function?

f. What is the minimum value of the objective function?

2. Caesar has an online business where he sells his digital photographs and animated posters. It takes Caesar 2 hours (hr.) to create 1 animated poster and 30 minutes (min.) to create 1 digital photograph. Caesar spends at most 14 hr. a week creating animated posters and digital photographs. To ensure he doesn’t have too much stock, Caesar creates no more than 10 items per week. a. Define the variables in this scenario.

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b. Write the constraint inequalities for this scenario.

c. Graph the system of linear inequalities defined in part B.

d. Identify the coordinate points of the vertices of the feasible solution represented by the polygon.

Caesar makes a profit of $10 per photograph and $20 per animated poster. e. Write the objective function that can be used to answer the question.

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f. Evaluate the objective function 𝑓(𝑥,𝑦) for each of the vertices of the feasible solution. Vertex (𝒙, 𝒚)

𝒇(𝒙, 𝒚)

g. How many photographs and posters should Caesar make each week to maximize his profit?

Collaborative Activity: Linear Programming 1. Each week, Layla can make 10 to 25 bracelets and 15 to 30 pairs of earrings to sell at a local craft fair. Layla brings at most 45 pieces of jewelry to sell at the craft fair every week. She earns $6 for each bracelet and $4 for each pair of earrings sold. How can Layla maximize her profits? a. Define the variables in this scenario.

b. Write the constraint inequalities for this scenario.

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c. Graph the system of linear inequalities defined in part B.

d. Identify the coordinate points of the vertices of the feasible solution represented by the polygon.

e. Write the objective function that can be used to answer the question. Vertex (𝒙, 𝒚)

𝒇(𝒙, 𝒚)

f. Evaluate the objective function 𝑓(𝑥,𝑦) for each of the vertices of the feasible solution.

g. Explain how Layla can maximize her profits.

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2. The graph of a system of linear inequalities is shown. a. Identify the coordinate points of the vertices of the feasible solution represented by the polygon.

b. Evaluate the objective function 𝑓(𝑥,𝑦) for each of the vertices of the feasible solution. Vertex (𝒙, 𝒚)

𝒇(𝒙, 𝒚) = 𝟑𝒚 − 𝒙

c. Use the objective function 𝑓(𝑥, 𝑦) = 3𝑦 − 𝑥 to determine the maximum and minimum values within the constraints.

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Lesson Summary To optimize a function means to determine the maximum or minimum value of the objective function based on the defined constraints. A constraint is a limitation on the possible values of variables in a model, often expressed by an equation or inequality or by specifying that the value must be an integer. To optimize using linear programming, apply the following process. 1. Define the variables. 2. Write a system of inequalities to represent the constraints. 3. Graph the system of inequalities. 4. Find the coordinates of the vertices of the feasible region. 5. Write a linear function, an objective function, to be maximized or minimized. 6. Substitute the coordinates of the vertices into the objective function. 7. Identify the maximum or minimum result, based on the question. Use the result to answer the question in context. To determine the maximum or minimum value of the objective function, use the vertices of the feasible region to evaluate the function. The result of substituting these 𝑥- and 𝑦-values into the objective function is the value of the function within the defined constraints.

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Practice Problems 1. A system of linear inequalities is shown. 𝑥 ≥ 5 𝑥 ≤ 20 � 𝑦 ≥ 0 𝑥 + 𝑦 ≤ 40

a. Graph the system of linear inequalities on the coordinate plane. b. Use the objective function 𝑓(𝑥, 𝑦) = 10𝑥 + 12𝑦 to determine the maximum and minimum values within the constraints.

2. The graph of the constraints of a system of linear inequalities is shown. a. Determine the vertices of the feasible region.

b. Use the objective function 𝐶 = 2𝑥 + 3𝑦 to determine the maximum and minimum values.

c. Complete the statements. The maximum value of the objective function is ________ at the point (_____, _____). The minimum value of the objective function is ________ at the point (_____, _____). © Accelerate Learning Inc. - All Rights Reserved

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Review Problems 3. A gardener is buying some topsoil and compost to fill his garden. His budget is $70. Topsoil costs $1.89 per cubic foot (cu. ft.), and compost costs $4.59 per cu. ft. Select all statements or representations that correctly describe the gardener’s constraints in this situation. Let 𝑡 represent the cubic feet of topsoil and 𝑐 the cubic feet of compost. □ The combination of 7.5 cu. ft. of topsoil and 12 cu. ft. of compost is within the gardener’s budget. □ If the line represents the equation 1.89𝑡 + 4.59𝑐 = 70, this graph represents the solutions to the gardener’s budget constraint.

□ 1.89𝑡 + 4.59𝑐 ≥ 70

□ The combination of 5 cu. ft. of topsoil and 20 cu. ft. of compost is within the gardener’s budget. □ 1.89𝑡 + 4.59𝑐 ≤ 70

□ If the line represents the equation 1.89𝑡 + 4.59𝑐 = 70, this graph represents the solutions to the gardener’s budget constraint.

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4. Here is a riddle: • The sum of two numbers is less than 10.

• If we subtract the second number from the first, the difference is greater than 3.

Write a system of inequalities that represents this situation. Let 𝑓 represent the first number and 𝑠 represent the second number.

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Unit 1, Lesson 6: Using Linear Programming to Solve Problems

Warm-Up: Feasible Region Complete each inequality with a digit from 1 to 5, such that the point (4, 2) is within the feasible region created by all the constraints. Use each digit only once. 𝑥 ≥ ______

𝑦 ≥ ______

𝑦 ≤ ______𝑥 + ______ 𝑦 ≥ 1 𝑥

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Guided Activity: Optimization Problems Using Linear Programming 1. A bakery sells loaves of bread and packs of muffins at a local farmers’ market. Each batch of bread takes 4 hours (hr.) to prepare and 1 hr. to bake. Each batch of muffins takes 2 hr. to prepare and 2 hr. to bake. The staff at the bakery can do baking prep for a maximum of 16 hr. and bake for a maximum of 10 hr. a. Define the variables.

b. Write the constraints as a system of inequalities.

c. Graph the system of inequalities on the coordinate plane.

7 6

d. Identify the vertices of the feasible region.

5 4 3 2

e. The bakery makes $45 per batch of bread sold and $25 per batch of muffins sold. How many batches of bread and muffins should be sold at the farmers’ market to maximize the bakery’s profits?

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1 -3

-2

-1

x 1

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Collaborative Activity: Solving Linear Programming Problems 1. The graph of the constraints of a system of linear inequalities is shown.

6 5

a. Determine the vertices of the feasible region.

4 (4, 3)

3 2 (–2, 0) -5

b. Use the objective function 𝐶 = 2𝑥 + 3𝑦 to determine the maximum and minimum values.

-4

-3

-2

-1

(4, 0)

0 -1

-2

(0, –2)

-3 -4 -5 -6 -7

c. Complete the statements. The maximum value of the objective function is ________ at the point (_____, _____). The minimum value of the objective function is ________ at the point (_____, _____).

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2. A local food truck sells nachos and pretzels. To make a profit, the food truck must sell at least 30 plates of nachos, but they cannot prepare more than 70 each day. The food truck must also sell at least 10 pretzels, but they cannot prepare more than 40 each day. The food truck cannot prepare more than a total of 90 plates of and pretzels all together. a. The graph of the constraint equations is shown. Shade in the feasible region on the graph. Label the vertices of the feasible region.

70 60 50 40 30 20 10 0 -10

x 10

-20

b. The food truck makes a profit of $2.40 per order of nachos and $1.75 per order of a pretzel. Write the equation of the objective profit function.

c. How many orders of nachos and pretzels should the food truck sell to make the maximum profit?

3. The manager of a travel agency is printing brochures and fliers to advertise special discounts on vacations during the winter months. Each brochure costs $0.08 to print, and each flier costs $0.04 to print. A brochure is printed on 3 pages, and a flier is printed on 2 pages. The manager doesn’t want to use more than 600 pages, but she needs to print at least 50 brochures and 150 fliers. How many of each kind of advertisement should she print to minimize the cost?

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Lesson Summary An optimization problem involves using a system of inequalities to determine the maximum and minimum values of an objective function, dependent on the (𝑥,𝑦) coordinates within the feasible region. This process is known as linear programming. Consider the system of linear inequalities, its graph, and the feasible region marked on the graph shown. System of Inequalities

Graph of System of Inequalities

6 4

𝑦 ≤ 1 𝑥 + 4

2 � 𝑦 ≤ −52 𝑥 + 4

𝑦 ≥ −6 𝑥 ≥ −4

�(𝑥, 𝑦) = 2𝑥 + 3𝑦

x -8

-6

-4

-2

-2 -4 -6 -8

To solve the optimization problem, evaluate each vertex of the feasible region using the objective function. In this example, the coordinate point (0, 4) creates a maximum value of 12 for the objective function. The coordinate point (−4, −6) creates a minimum value of −26 for the objective function.

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Practice Problems 1. The graph of the constraints of a system of linear inequalities is shown.

a. Determine the vertices of the feasible region.

7 6 5 4 3

b. Use the objective function 𝐶 = − 𝑥 + 𝑦 to determine the maximum and minimum values.

2 1 -1

0 -1

x 1

-2 -3 -4

c. Complete the statements. The maximum value of the objective function is ________ at the point (_____, _____). The minimum value of the objective function is ________ at the point (_____, _____).

2. The math club at a high school wants to sell hats and T-shirts to raise money for a competition. The company they’re using to make the hats and T-shirts requires the club to order at least 5 hats and at least 5 T-shirts. In the past, the club has found that the hats do not sell as well as the T-shirts, so they want to order a maximum of 30 hats. The club can only afford to order at most 70 total hats and T-shirts. a. Define the variables.

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b. Write the constraints as a system of inequalities.

c. Graph the system of inequalities on the coordinate plane.

70 60

d. Determine the vertices of the feasible region.

50 40 30 20 10

e. The club makes a profit of $8.50 for each hat sold and $7 for each T-shirt sold. Write the objective function.

0 -10

x 5

f. How many hats and T-shirts does the club need to sell to maximize its profit?

3. Tiana is solving a linear programming problem. She graphs the constraints and finds that the vertices of the feasible region are (2, 7), (12, 7), (2, 22), (12, 10). The objective function is 𝐶 = 4𝑥 + 12𝑦. Use the objective function to determine the maximum and minimum values.

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Review Problems 4. Solve the absolute value equation |𝑥 − 3| = 7.

5. Find the average rate of change of the function 𝑓(𝑥) = 𝑥2 − 6 over the interval [−2, −1].

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Unit 1, Lesson 7: Matching Up to Data

Warm-Up: Notice and Wonder: Cooling Down What do you notice? What do you wonder?

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Exploration Activity: Which Function? These activities require the use of an applet, so please make your way over to the digital platform to find the link. A bottle of soda water is left outside on a cold day. The scatter plot shows the temperature 𝑇, in degrees Fahrenheit, of the bottle ℎ hours after it was left outside. Here are 2 functions you can use to model the temperature as a function of time. 𝑓(ℎ) = 45 +

20 ℎ + 0.5

𝑔 (ℎ) = 45 + 33(0.5)ℎ + 0.5

1. Which function better fits the shape of the data? Explain your reasoning.

Use the applet to zoom out on the graphs.

2. Does this change your opinion about which function fits better?

3. Where do you see the 45 in the expression for each function on the graph? 4. For the function you thought didn’t fit the shape of the data as well, how would you change it to fit better?

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Collaborative Activity: What Happened to the Graph? Your teacher will give you a card. Take turns describing the transformation of the graph on your card for your partner to draw and drawing the transformed graph from your partner’s description. 1.

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Lesson Summary The data on the graph shows the temperature, 𝑇, in degrees Fahrenheit (℉), of a can of soda ℎ hours after it was put into the refrigerator.

To build a function that fits the data well, start with a simpler function that has the same general shape as the data when graphed, and then transform it. An exponential decay function may best model the data. The shape of the data can be modeled using a simpler equation like 𝑇 = (0.7)ℎ, but the graph doesn’t fit where the data is, as shown. The graph of the function given by 𝑇 = 36 + 45(0.7)ℎ isn’t represented by a simple equation, but it does fit the data. The transformations of multiplying by 45 and adding 36 to the graph allow the model to fit the data. © Accelerate Learning Inc. - All Rights Reserved

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The remainder of this unit will focus on how to translate, reflect, and stretch graphs to fit real-world situations.

Practice Problems 1. Describe a transformation that gives the graph representing 𝑔 from the graph representing 𝑓.

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2. Describe a way to transform each graph so that it goes through the labeled points.

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3. Describe a way to transform each graph so that it better matches the data.

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4. Does the function 𝑓 or the function 𝑔 fit the data better? Explain your reasoning.

Review Problems 5. Select all solutions to the system of linear inequalities � 𝑦 ≤ −2𝑥 + 5 . 𝑦 + 3 > 2𝑥 □ (0, 0) □ (0, 5) □ (2, 1) □ (4, 0)

□ (0, −5) □ (−2, 2)

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6. A table of values for a function, ℎ(𝑥), is shown.

What is the average rate of change of ℎ(𝑥) over the interval [−2, 1]?

𝒙

𝒉(𝒙)

−1

−4

−2

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−3

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Unit 1, Lesson 8: Translating Functions

Warm-Up: What Happened to the Equation? Graph each function using technology. Describe how to transform 𝑓(𝑥) = 𝑥2(𝑥 − 2) to get to the functions shown here in terms of both the graph and the equation. 1. ℎ(𝑥) = 𝑥2(𝑥 − 2) − 5

2. 𝑔(𝑥) = (𝑥 − 4)2(𝑥 − 6)

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Exploration Activity: Writing Equations for Vertical Translations The graph of function 𝑔 is a vertical translation of the graph of polynomial 𝑓.

1. Complete the 𝑔(𝑥) column of the table. 2. If 𝑓(0) = − 0.86 , what is 𝑔(0)? Explain how you know.

3. Write an equation for 𝑔 (𝑥) in terms of 𝑓(𝑥) for any input 𝑥.

4. The function ℎ can be written in terms of 𝑓 as ℎ(𝑥) = 𝑓(𝑥) − 2.5. Complete the ℎ(𝑥) column of the table.

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𝒙

𝒇(𝒙)

−3

−5.8

1.2

−3.3

−4

−0.7

𝒈(𝒙)

𝒉(𝒙) = 𝒇(𝒙) − 𝟐.𝟓

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5. Sketch the graph of function ℎ.

6. Write an equation for 𝑔 (𝑥) in terms of ℎ(𝑥) for any input 𝑥.

Guided Activity: Heating the Kitchen A bakery kitchen has a thermostat set to 65 degrees Fahrenheit (℉). Starting at 5:00 a.m., the temperature in the kitchen rises to 85℉ when the ovens and other kitchen equipment are turned on to bake the daily breads and pastries. The ovens are turned off at 10:00 a.m. when the baking finishes. 1. Sketch a graph of the function 𝐻 that gives the temperature in the kitchen 𝐻(𝑥), in ℉, 𝑥 hours (hr.) after midnight.

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2. The bakery owner decides to change the shop hours to start and end 2 hr. earlier. This means the daily baking schedule will also start and end 2 hr. earlier. Sketch a graph of the new function 𝐺, which gives the temperature in the kitchen as a function of time.

3. Explain what 𝐻(10.25) = 80 means in this situation. Why is this reasonable?

4. If 𝐻(10.25) = 80, then what would the corresponding point on the graph of 𝐺 be? Use function notation to describe the point on the graph of 𝐺.

5. Write an equation for 𝐺 in terms of 𝐻. Explain why your equation makes sense.

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Lesson Summary A pumpkin catapult is used to launch a pumpkin vertically into the air from a starting height of 0 feet (ft.). The function ℎ gives the pumpkin’s height, ℎ(𝑡), in ft. above the ground, 𝑡 seconds (sec.) after launch.

Now consider what would happen if the pumpkin had been launched at the same time but from a platform 30 ft. above the ground. Let function 𝑔 represent this pumpkin’s height, 𝑔(𝑡), in ft. How would the graphs of ℎ and 𝑔 compare?

• Since the height of the second pumpkin is 30 ft. greater than the first pumpkin at all times 𝑡, the graph of function 𝑔 is translated up 30 ft. from the graph of function ℎ. For example, the point (2, 66) on the graph of ℎ shows that ℎ(2) = 66, so the original pumpkin was 66 ft. high after 2 sec. The new pumpkin would be 30 ft. higher than that, so 𝑔(2) = 96. Since all the outputs of 𝑔 are 30 more than the corresponding outputs of ℎ, the function 𝑔(𝑡) can be expressed in terms of ℎ(𝑡) using function notation as 𝑔(𝑡) = ℎ(𝑡) + 30.

• Now suppose instead that the pumpkin launched 5 sec. later, from a height of 0 ft. Let function 𝑘 represent this pumpkin’s height, 𝑘(𝑡), in ft. The graph of 𝑘 is translated right 5 sec. from the graph of ℎ. The output values of 𝑘 are the same as the output values of ℎ 5 sec. earlier. For example, 𝑘(7) = 66 and ℎ(7 − 5) = ℎ(2) = 66. This means 𝑘(𝑡) can be expressed in terms of ℎ(𝑡) as 𝑘(𝑡) = ℎ(𝑡 − 5).

The graphs of all 3 functions are shown on the coordinate plane.

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Practice Problems 1. The thermostat in an empty apartment is set to 65℉ from 4:00 a.m. to 5:00 p.m. and to 50℉ from 5:00 p.m. until 4:00 a.m. Here is a graph of the function 𝐻 that gives the temperature 𝐻(𝑥), in ℉, in the apartment 𝑥 hr. after midnight.

a. The owner of the apartment decides to change to a new schedule and they set the thermostat to change 3 hr. later in the morning and the evening. On the same axes, sketch a graph of the new function, 𝐺, giving the temperature as a function of time. b. Explain what 𝐻(6.5) = 65 means in this context. Why is this a reasonable value for the function?

c. If 𝐻(6.5) = 65, then what is the corresponding point on the graph of 𝐺? Use function notation to describe the point on the graph of 𝐺. d. Write an expression for 𝐺 in terms of 𝐻.

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2. A pumpkin pie recipe says to bake the pie at 425℉ for 15 minutes (min.), and then to adjust the temperature down to 350℉ for 45 additional min. The function 𝑃 gives the oven temperature setting 𝑃(𝑡), in ℉, 𝑡 min. after the pie is placed in the oven. a. Explain what 𝑃(30) = 350 means in this context.

b. Diego discovers that the temperature inside the oven is always 25℉ warmer than the oven’s temperature setting. The function 𝐵 gives the actual temperature of Diego’s oven. If 𝑃(30) = 350, then what is the corresponding point on the function 𝐵? c. Write an expression for 𝐵 in terms of 𝑃.

3. Here is the graph of 𝑦 = 𝑓(𝑥) for a function 𝑓. a. On the same axes, sketch a graph of 𝑔(𝑥) = 𝑓(𝑥) + 2.

b. On the same axes, sketch a graph of ℎ(𝑥) = 𝑓(𝑥 + 2).

c. How do the graphs of 𝑔 and ℎ compare to 𝑓?

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4. The graph of a line and two points are shown on the coordinate plane. a. Describe a horizontal translation of the line to a line that contains the two labeled points.

b. Describe a vertical translation of the line to a line that contains the two labeled points.

Review Problem 5. Does the function 𝑓 or the function 𝑔 fit the data better? Explain your reasoning.

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Unit 1, Lesson 9: More Translations

Warm-Up: Moving a Graph How can we translate the graph of 𝐴 to match one of the other graphs?

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Exploration Activity: New Hours for the Kitchen Remember the bakery with the thermostat set to 65 degrees Fahrenheit, (℉)? At 5:00 a.m., the temperature in the kitchen rises to 85℉ due to the ovens and other kitchen equipment being used until they are turned off at 10:00 a.m. When the owner decided to open 2 hours (hr.) earlier, the baking schedule changed to match.

1. Andre thinks, “When the bakery starts baking 2 hr. earlier, that means I need to subtract 2 from 𝑥 and that 𝐺(𝑥) = 𝐻(𝑥 − 2).” How could you help Andre understand the error in his thinking?

2. The graph of 𝐹 shows the temperatures after a change to the thermostat settings. What did the owner do?

3. Write an expression for 𝐹 in terms of the original baking schedule, 𝐻.

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Guided Activity: Thawing Meat A piece of meat is taken out of the freezer to thaw. The data points show the temperature of the meat, 𝑀, in ℉, 𝑡 hr. after it was taken out of the freezer.

The graph 𝑀 = 𝐺(𝑡), where 𝐺(𝑡) = − 62(0.85)𝑡, models the shape of the data but is in the wrong position. 𝒕

𝑴

13.1

1.84

0.41

22.9

2.37

36.1

3.53

38.8

2.95 3.74 4.17 4.92

36.8 40

42.2 45.8

Brooke thinks changing the equation to 𝐵(𝑡) = 𝐺(𝑡) + 75.1 creates a good model for the data. Nikhil thinks 𝑁(𝑡) = (𝑡 + 1) + 68 is better.

1. Without graphing, describe how Brooke and Nikhil each transformed the graph of 𝐺 to make their new function fit the data.

2. Use technology to graph the data, 𝐵, and 𝑁, on the same axes. 3. Whose function do you think best fits the data? Be prepared to explain your reasoning.

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Lesson Summary Remember the pumpkin catapult? The function ℎ gives the height, ℎ(𝑡), in feet (ft.), of the pumpkin above the ground 𝑡 seconds (sec.) after launch. Now, let 𝑘 represent the pumpkin’s height, 𝑘(𝑡), in ft., if it were launched 5 sec. later. • Graphed on the same axes, 𝑘 and ℎ look identical, but the graph of 𝑘 is translated 5 units to the right of the graph of 𝑔, as shown. • Since the pumpkin’s height 𝑘(𝑡) at time 𝑡 is the same as the height ℎ(𝑡) of the original pumpkin at time 𝑡 − 5, 𝑘 can be written in terms of ℎ as 𝑘(𝑡) = ℎ(𝑡 − 5).

Suppose there were a third function, 𝑣, where 𝑣(𝑡) = ℎ(𝑡 + 4). Even without graphing 𝑣, it can be determined that the graph reaches a maximum height of 66 ft. To evaluate 𝑣(𝑡), evaluate ℎ at the input 𝑡 + 4, which is 0 when 𝑡 = − 4. The graph of 𝑣 is translated 4 sec. to the left of the graph of ℎ. This means that 𝑣(𝑡) is the height, in ft., of a pumpkin launched from the catapult 4 sec. earlier.

Practice Problems 1. Here is a graph of 𝑓 and a graph of 𝑔. Express 𝑔 in terms of 𝑓 using function notation.

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2. Tyler leaves his house at 7:00 a.m. to go to school. He walks for 20 minutes (min.) until he reaches his school, 1 mile (mi.) from his house. The function 𝑑 gives the distance 𝑑(𝑡), in mi., of Tyler from his house 𝑡 min. after 7:00 a.m. a. Explain what 𝑑(5) = 0.25 means in this context.

b. On snowy days, Tyler’s school has a 2 hr. delayed start time (120 min.). The function 𝑠 gives Tyler’s distance 𝑠(𝑡), in mi., from home 𝑡 min. after 7:00 a.m. with a 120 min. delayed start time. If 𝑑(5) = 0.25, then what is the corresponding point on the function 𝑠?

c. Write an expression for 𝑠 for in terms of 𝑑.

d. A new function, 𝑛, is defined as 𝑛(𝑡) = 𝑑(𝑡 + 60). Explain what this means in terms of Tyler’s distance from school.

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3. Technology required. Here are the data for the population 𝑓, in thousands, of a city 𝑑 decades after 1960 along with the graph of the function given by 𝑓(𝑑) = 25 ⋅ (1.19)𝑑. Elena thinks that shifting the graph of 𝑓 up by 50 will match the data. Han thinks that shifting the graph of 𝑓 up by 60 and then right by 1 will match the data. a. What functions define Elena’s and Han’s graphs?

b. Use graphing technology to graph Elena’s and Han’s proposed functions along with 𝑓.

c. Which graph do you think fits the data better? Explain your reasoning.

4. Here is a graph of 𝑦 = 𝑓(𝑥 + 2) − 1 for a function 𝑓. Sketch the graph of 𝑦 = 𝑓(𝑥).

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Review Problems 5. Describe how to transform the graph of 𝑓 to the graph of 𝑔: a. using only translations

b. using a reflection and a translation

6. Here is a graph of function 𝑓 and a graph of function 𝑔. Express 𝑔 in terms of 𝑓 using function notation.

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Unit 1, Lesson 10: Reflecting Functions

Warm-Up: Notice and Wonder: Reflections What do you notice? What do you wonder?

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Exploration Activity: Reflecting Across Here is the graph of function 𝑓 and a table of values.

1. Let 𝑔 be the function defined by 𝑔(𝑥) = − 𝑓(𝑥). Complete the table.

2. Sketch the graph of 𝑔 on the same axes as the graph of 𝑓 but in a different color.

3. Describe how to transform the graph of 𝑓 into the graph of 𝑔. Explain how the equation produces this transformation.

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𝒙

𝒇(𝒙)

−1.5

−4.3

−1.8

2.6

3.9

−3

−1

0.6

𝒈(𝒙) = − 𝒇(𝒙)

−4

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Exploration Activity: Reflecting Across a Different Way Here is another copy of the graph of 𝑓 from the earlier activity. This time, let ℎ be the function defined by ℎ(𝑥) = 𝑓(−𝑥). 1. Use the definition of ℎ to find ℎ(0). Does your answer agree with your prediction?

2. What does your prediction tell you about ℎ(−0.6)? Does your answer agree with the definition of ℎ?

𝒙

𝒇(𝒙)

3. Complete the tables. The values for 𝑥 will not be the same for the two tables.

−1.5

−4.3

4. Sketch the graph of ℎ on the same axes as the graph of 𝑓 but in a different color.

−1.8

2.6

3.9

5. Describe what happened to the graph of 𝑓 to transform it into the graph of ℎ. Explain how the equation produces this transformation.

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−3

−1

0.6

𝒙

𝒉(𝒙) = 𝒇(−𝒙)

−4

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Lesson Summary The graphs of the functions 𝑓, 𝑔, and ℎ, where 𝑔(𝑥) = − 𝑓(𝑥) and ℎ(𝑥) = 𝑓(−𝑥), are shown. Function

𝒇(𝒙)

𝒈(𝒙)

𝒉(𝒙)

None

𝑔(𝑥) = − 𝑓(𝑥)

ℎ(𝑥) = 𝑓(−𝑥)

Graph

Transformation from 𝒇

• For the equation 𝑔(𝑥) = − (𝑥), notice that for the same input 𝑥, the value of 𝑔(𝑥) will be the opposite of the value of 𝑓(𝑥). For example, since 𝑓(0) = 1, 𝑔(0) = − 𝑓(0) = − 1. This relationship is visible in the graph of (𝑥), where 𝑔 is the reflection of 𝑓 across the 𝑥-axis.

• The equation ℎ(𝑥) = 𝑓(−𝑥) shows that the 2 functions have the same output for opposite inputs. For example, 1 and − 1 are opposites, so the equations ℎ(1) = 𝑓(−1) and ℎ(−1) = 𝑓(1) are also true. The graph of ℎ(𝑥) demonstrates this relationship, where ℎ is the reflection of 𝑓 across the 𝑦-axis.

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Practice Problems 1. The dashed function is the graph of 𝑓 and the solid function is the graph of 𝑔. Express 𝑔 in terms of 𝑓.

2. The table gives some values of functions 𝑓 and 𝑔. 𝒙

𝒇(𝒙)

𝒈(𝒙)

−1

1 2

−2

0 1

1 4

1 2 1 4

2 4

Which of these equations could be true for all values of 𝑥? A. 𝑓(𝑥) = −𝑔(𝑥) B. 𝑓(𝑥) = 𝑔(−𝑥)

C. 𝑓(𝑥) = −𝑔(−𝑥) D. 𝑓(𝑥) = 𝑔(𝑥)

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3. Here is the graph of a function 𝑓.

a. On the same axis, sketch a graph of 𝑓 reflected over the 𝑦-axis and then translate it 3 units up. b. Write an equation (in terms of 𝑓) for a function 𝑔 that has the graph that you drew.

Review Problems 4. Describe a transformation of the line that contains the two labeled points.

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5. The thermostat in an apartment is set to 75 degrees Fahrenheit (℉) while the owner is awake and to 60℉ while the owner is sleeping. The function 𝑊 gives the temperature 𝑊(𝑥), in ℉, in the apartment 𝑥 hours (hr.) after midnight. When it is hot outside, the owner changes the settings to be exactly 10 degrees warmer than 𝑊 to save energy. The function 𝐻 gives the temperature 𝐻(𝑥), in ℉, 𝑥 hr. after midnight when it is hot outside. a. If 𝑊(6.5) = 75, then what is the corresponding point on 𝐻? Use function notation to describe the point on 𝐻. b. If 𝑊(2) = 60, then what is the corresponding point on 𝐻? Use function notation to describe the point on 𝐻. c. Write an expression for 𝐻 in terms of 𝑊. 6. A ball is hit in the air. Its height, ℎ, in feet (ft.), 𝑡 seconds (sec.) after being hit is modeled by the function ℎ(𝑡) = 4 + 50𝑡 − 32𝑡2. Which function models the height of an identical ball that follows the same path but is hit 2 sec. after the first ball? A. 𝐴(𝑡) = ℎ(𝑡) + 2 B. 𝐴(𝑡) = ℎ(𝑡) − 2

C. 𝐴(𝑡) = ℎ(𝑡 + 2) D. 𝐴(𝑡) = ℎ(𝑡 − 2)

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Unit 1, Lesson 11: Expressing Transformations of Functions Algebraically

Warm-Up: Describing Translations Let 𝑔(𝑥) = |𝑥|. Complete the table. Be prepared to explain your reasoning. Verbal Description (The graph of 𝒚 = 𝒈(𝒙) is . . .) . . . translated left 5 units.

Function Notation

Expression

𝑔(𝑥 + 5)

. . . translated left 5 units and down 3 units.

|𝑥 + 5| − 3 𝑔(−𝑥)

| − 𝑥|

. . . translated left 5 units, and then down 3 units, and then reflected across the 𝑦-axis.

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Guided Activity: Translating Vertex Form Let 𝑓 be the function given by 𝑓(𝑥) = (𝑥)2.

1. Write an equation for the function 𝑔 whose graph is the graph of 𝑓 translated 3 units left and up 5 units. 2. What is the vertex of the graph of 𝑔? Explain how you know.

3. Write an equation for a quadratic function ℎ whose graph has a vertex at (1.5, 2.6).

4. Write an equation for a quadratic function 𝑘 whose graph opens downward and has a vertex at (3.2, − 4.7).

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Collaborative Activity: Transformations – Horizontal and Vertical Shifts 1. The graph and table of values for the function 𝑓(𝑥) are shown. Original Function 𝒇(𝒙)

Graph

𝒙

3 2 1 -5 -4 -3 -2 -1 0 -1

(1, 1)

x 1

-2

(−5, −3) (−1, −3)

Table of Values

-3

(4, −2)

-4 -5

𝒚

−5

−3

−1

−3

−2

a. Discuss with your partner how the graph of the original function will change if 2 is subtracted from each 𝑥-value and 5 is added to each 𝑦-value of the original function. b. Complete the table of values for a new function, 𝑅(𝑥), by subtracting 2 from each 𝑥-value and adding 5 to each 𝑦-value of the original function, 𝑓(𝑥). Then, graph the new function using the new ordered pairs.

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New Function 𝑹(𝒙)

Graph

Table of Values

𝒙

𝒚

c. The transformation can be represented by 𝑓(𝑥 + 2) + 5. Discuss this representation with your partner. Summarize your discussion.

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2. A new function, 𝐾(𝑥), was created by adding 5 to each 𝑥value and 6 to each 𝑦-value of the original function, 𝑓(𝑥). a. Complete the table of values for 𝐾(𝑥). Then, graph the function. New Function 𝑲(𝒙)

Graph

Table of Values

𝒙

𝒚

b. The function notation for the transformation is 𝐾(𝑥) = 𝑓(𝑥 − 5) + 6. Explain how the notation is related to the operation that was used on the 𝑥-values. c. Explain how the notation 𝐾(𝑥) = 𝑓(𝑥 − 5) + 6 is related to the operation that was used on the 𝑦-values.

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3. A new function, 𝑊(𝑥), was created by adding 3 to each 𝑥value and subtracting 4 from each 𝑦-value of the original function, 𝑓(𝑥). a. Complete the table of values for 𝑊(𝑥). Then, graph the function. New Function 𝑾(𝒙)

Graph

Table of Values

𝒙

𝒚

b. Complete the function notation of the transformation. 𝑊(𝑥) = ______________

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Lesson Summary The equation of a parent function can be used to write an equation for its transformation. For example, let 𝑓(𝑥) = (𝑥)2.

The function 𝑔(𝑥) can be built from 𝑓(𝑥) by applying a series of transformations such as a reflection of 𝑓 across the 𝑥-axis, a translation up 10 units, and a translation left 3 units. The new function 𝑔 is a transformation of 𝑓, where 𝑔(𝑥) = − 𝑓(𝑥 + 3) + 10. Each transformation is shown in the equation.

When an equation of a function is given with transformations already applied, use the same knowledge of transformations to graph and analyze key features of the transformed function based on what you know about its parent function. For example, 𝐻(𝑡) = 10 − (1.2)𝑡 + 5 is a function that has several transformations applied to its parent function. The parent function, 𝑓(𝑡), is an exponential function with the equation 𝑓(𝑡) = (1.2)𝑡. The transformations applied to (𝑡) are adding 5 to the input to get (1.2)𝑡 + 5, multiplying the output by − 1 to get −(1.2)𝑡 + 5, and adding 10 to the output to get 10 − (1.2)𝑡 + 5. The graph of 𝐻 can be obtained from the graph of 𝑓(𝑡) = 1.2𝑡 by translating left 5 units, reflecting across the 𝑥-axis, and translating up 10 units. Consider the point (0, 1) on the graph of 𝑓. After translating, reflecting, and translating again, it becomes the point (−5, 9) on the graph of 𝐻, as shown.

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Practice Problems 1. Here is a graph of 𝑓(𝑥) = 𝑒𝑥 and a graph of 𝑔, which is a transformation of 𝑓. Write an equation for the function 𝑔.

2. Describe the transformation(s) that take(s) the graph of function 𝑓 to the graph of function 𝑔. a. 𝑓(𝑥) = 3𝑥 and 𝑔(𝑥) = −3𝑥 + 2.7

b. 𝑓(𝑥) = 𝑥5 and 𝑔(𝑥) = (−𝑥 + 3.1)5 + 1

c. 𝑓(𝑥) = |𝑥| and 𝑔(𝑥) = |𝑥| − 26

d. 𝑓(𝑥) = √𝑥 and 𝑔(𝑥) = −√𝑥 − 0.004

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3. Write the equation of a parabola with the features described. a. Vertex at (1, 4) and opens upward

b. Vertex at (2, −3) and opens downward

Review Problem 4. Here are graphs of functions 𝑓 and 𝑔.

Which sequences of transformations take the graph of 𝑓 to the graph of 𝑔? Select all that apply. □ reflection over the 𝑦-axis, then translation up by 2 □ reflection over the 𝑥-axis, then translation up by 2 □ translation up 2, then reflection over the 𝑦-axis

□ translation up 2, then reflection over the 𝑥-axis □ translation up 2, and then translation left by 5

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Unit 1, Lesson 12: Scaling the Outputs

Warm-Up: Notice and Wonder: Arch You Glad to See Me? What do you notice? What do you wonder?

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Exploration Activity: The Hulme Arch Bridge The Hulme Arch Bridge in Manchester, England is shaped like a parabola. The ends of the arch are 52 meters (m) apart, and it is 25 m high.

1. Use the description to help you label the 3 coordinate pairs on the graph.

2. Han wants to model the shape of the arch with the graph of a function, and he chooses 𝐻(𝑥) = 𝑥(52 − 𝑥), where 𝐻(𝑥) is the height in meters above a point 𝑥 m along the base of the arch from the left end. a. For the 𝑥-coordinates of the three points, what are the corresponding points on the graph of 𝐻?

b. What aspects of the shape does Han’s function model well, and what parts does it not model well?

c. Compare the height of Han’s graph with the height of the Hulme Arch Bridge. How can you change the outputs of 𝐻 to make it fit better? What would the revised version of 𝐻(𝑥) be?

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Collaborative Activity: Transformations by a Factor The graph and table of values for the function 𝑓(𝑥) are shown.

Original Function 𝒇(𝒙)

Graph

Table of Values

𝒙

4 3 2 1 -5 -4 -3 -2 -1 0 -1

(1, 1)

x 1

-2

(−5, −3) (−1, −3)

-3 -4 -5

1. A new function, 𝑔(𝑥), was created by multiplying each 𝑦-value of 𝑓(𝑥) by 2. a. Complete the table of values for a new function, 𝑔(𝑥). Then, graph the function.

(4, −2)

𝒚

−5

−3

−1

−3

−2

Original Function 𝒈(𝒙)

Graph

Table of Values

𝒙

𝒚

b. Describe what happened to the original graph when each 𝑦-value was multiplied by 2.

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c. The function notation for the transformation is 𝑔(𝑥) = 2𝑓(𝑥). Explain how the notation is related to the operation that was used on the 𝑦-values.

2. A new function, ℎ(𝑥), is shown.

Original Function 𝒉(𝒙)

a. Complete the table of values.

Graph

Table of Values

𝒙

4 3

b. Describe how the original graph of 𝑓(𝑥) was changed to create ℎ(𝑥).

𝒚

2 1 -5 -4 -3 -2 -1 0 -1

x 1

-2 -3 -4 -5

c. Discuss with your partner how this transformation can be determined from the table of values. Summarize your discussion.

d. Complete the function notation of the transformation. ℎ(𝑥) = _________

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Lesson Summary As explored in previous lessons, a parent function can be translated up, down, left, or right or reflected to create new functions. Sometimes, an additional transformation may be needed. For example, the towers of the Golden Gate Bridge in San Francisco are 1,280 m apart, and the tops of the towers are 152 m above the roadway. If the 𝑥-axis can be placed at the top of the towers, the parabolic shape of the suspension cable between the towers can be modeled by a quadratic function with zeros at 0 and 1,280, given by 𝑓(𝑥) = 𝑥(𝑥 − 1280). Unfortunately, the model should have a value of − 152 at 𝑥 = 640, which is halfway between the towers, and instead the value for this function is 𝑓(640) = 640(640 − 1280) = − (640)2 = − 409,600. No number of translations or reflections of the current model will create a graph that matches the shape of the suspension cables and also keeps the zeros at 0 and 1,280. So instead of adding to or subtracting from the output, a scale factor, 𝑘, can be multiplied by the output instead. Using multiplication means the zeros stay where they are while scaling all the other output values. To convert − 409,600 to the needed value of − 152, multiply all the output values by a very small scale factor. Using the scale factor 𝑘 =

152 = 0.000371, the graph of 𝑦 = 𝑘𝑓(𝑥) = 0.000371𝑓(𝑥) has the 409,600

same shape as the bridge. This graph can also be represented by the equation 𝑦 = 0.000371𝑥(𝑥 − 1280)

Multiplying by a scale factor less than 1 makes all the output values, 𝑦, closer to the zeros than in the parent function, which compresses the graph vertically. In cases where the goal is to stretch the graph vertically, use a scale factor greater than 1. © Accelerate Learning Inc. - All Rights Reserved

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Practice Problems 1. In each pair of graphs shown here, the values of function 𝑔 are the values of function 𝑓 multiplied by a scale factor. Express 𝑔 in terms of 𝑓 using function notation.

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2. Here is the graph of 𝑦 = 𝑓(𝑥) for a cubic function 𝑓.

a. Will scaling the outputs of 𝑓 change the 𝑥-intercepts of the graph? Explain how you know.

b. Will scaling the outputs of 𝑓 change the 𝑦-intercept of the graph? Explain how you know.

3. The function 𝑓 is given by 𝑓(𝑥) = 2𝑥, while the function 𝑔 is given by 𝑔(𝑥) = 4 ⋅ 2𝑥. Kiran says that the graph of 𝑔 is a vertical scaling of the graph of 𝑓. Mai says that the graph of 𝑔 is a horizontal shift of the graph of 𝑓. Do you agree with either of them? Explain your reasoning.

Review Problems 4. The dashed function is the graph of 𝑓 and the solid function is the graph of 𝑔. Express 𝑔 in terms of 𝑓.

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5. Here is a graph of 𝑓(𝑥) = (𝑥)3 and a graph of 𝑔, which is a transformation of 𝑓. Write an equation for the function 𝑔.

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Unit 1, Lesson 13: Scaling the Inputs

Warm-Up: Out and Back Every weekend, Elena takes a walk along the straight road in front of her house for 2 miles (mi.), then turns around and comes back home. Let’s assume Elena walks at a constant speed. Here is a graph of the function 𝑓 that gives her distance 𝑓(𝑡), in mi., from home as a function of time 𝑡 if she walks 2 miles per hour (mph).

1. Sketch a graph of the function 𝑔 that gives her distance 𝑔(𝑡), in mi., from home as a function of time 𝑡 if she walks 4 mph.

2. Write an equation for 𝑔 in terms of 𝑓. Be prepared to explain why your equation makes sense.

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Exploration Activity: A New Set of Wheels Cal is visiting an amusement park with a large Ferris wheel. Cal records his height above the ground 𝑡 seconds (sec.) after getting on the ride. The table shows Cal’s elevation, in feet (ft.), on the Ferris wheel at different times, 𝑡. Cal got on the ride 80 sec. ago. Right now, at time 0 sec., he is at the top of the ride.

1. Assuming the Ferris wheel, 𝐹, moves at a constant speed for the next 80 sec., complete the table. Time (sec.) Height (ft.)

−80 0

−60 31

−40 106

−20

181

212

Cal plans to try out 2 other Ferris wheels. • The first wheel is twice the height of 𝐹 and rotates at the same speed. The function 𝑔 gives Cal’s height, 𝑔(𝑡), in ft., 𝑡 sec. after starting his descent from the top.

• The second wheel is the same height as 𝐹 but rotates at half the speed. The function ℎ gives Cal’s height, ℎ(𝑡), in ft., 𝑡 sec. after starting his descent from the top. 2. Complete the table for functions 𝐹 and 𝑔.

𝒕

3. Explain why there is not enough information to find the exact values for ℎ(20) and ℎ(60). 4. Complete as much of the table as you can for the function ℎ, modeling Cal’s height on the second Ferris wheel.

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𝑭(𝒕)

𝒈(𝒕)

𝒉(𝒕)

20 40 60 80

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5. Express 𝑔 and ℎ in terms of 𝐹. Be prepared to explain your reasoning.

Collaborative Activity: The Many Transformations of a Function 𝑷

Function 𝑘 is a transformation of function 𝑃 due to a scale factor. 1. Write an equation for 𝑘 in terms of 𝑃.

2. On the same axes, graph the function 𝑚 where 𝑚(𝑥) = 𝑃(0.75𝑥).

3. The highest point on the graph of 𝑃 is (1, 2). What is the highest point on the graph of a function 𝑛 where 𝑛(𝑥) = 𝑃(5𝑥)? Explain or show your reasoning.

4. The point farthest to the right on the graph of 𝑃 is (4, 0). If the point farthest to the right on the graph of a function 𝑞 is (18, 0), write a possible equation for 𝑞 in terms of 𝑃.

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Lesson Summary Two graphs show the distance traveled by 2 trains at 𝑡 hours (hr.) into their journeys.

Train A traveled 25 mi. in 1 hr., and train B traveled 25 mi. in half the time. Similarly, train A traveled 150 mi. in 4 hr. while train B traveled 150 mi. in only 2 hr. Train B is traveling at twice the speed of train A. A train traveling twice the speed will get to any particular point along the track in half the time, so the graph for train B is compressed horizontally by a factor of 1 compared 2

to the graph of train A. If the function 𝑓(𝑡) represents the distance that train A travels in 𝑡 hr., then 𝑓(2𝑡) represents the distance that train B travels in 𝑡 hr., because train B travels as far in 𝑡 hr. as train A travels in 2𝑡 hr.

If a different train, train C, were going one-fourth the speed of train A, then its motion would be represented by 𝑠 = 𝑓(0.25𝑡), and the graph would be stretched horizontally by a factor of 4, since it would take 4 times as long for train C to travel the same distance as train A.

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Practice Problems 1. Here are graphs of functions 𝑓 and 𝑔. For each, determine the value of 𝑘 so that 𝑔(𝑥) = 𝑓(𝑘𝑥).

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Use the table of values for the function 𝑓(𝑥) to answer questions 2 and 3.

𝒙

−5

−3

−1 4 2. A new function, 𝑟(𝑥), was created by multiplying each 𝑥-value of 𝑓(𝑥) by 2. a. Complete the table of values for 𝑟(𝑥). Then, graph the function.

𝒚

−3 −2

Original Function 𝒓(𝒙)

Graph

Table of Values

𝒙

𝒚

b. Describe what happened to the original graph when each 𝑥-value was multiplied by 2.

c. The function notation for the transformation is 𝑟(𝑥) = 𝑓 �1 𝑥�. Explain how the 2

notation is related to the operation that was used on the 𝑥-values.

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3. A new function, 𝑡(𝑥), was created by multiplying each 𝑥-value of 𝑓(𝑥) by 3. a. Complete the table of values for 𝑡(𝑥). Then, graph the function. Original Function 𝒕(𝒙)

Graph 10

Table of Values

𝒙

𝒚

x -15

-10

-5

-10

b. Complete the function notation of the transformation. 𝑡(𝑥) = _________

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Review Problem 4. The height of a hot air balloon, in ft., 𝑚 min. after takeoff is modeled by the function 𝑓(𝑚) = 16𝑚. a. How many min. does it take for the balloon to reach 200 ft.?

b. Another balloon takes off 5 min. later and rises at the same speed. Write an equation for the function 𝑔, where 𝑔(𝑡) is the height, in ft., of this balloon in terms of 𝑚. Explain your reasoning.

c. Sketch graphs of the two functions 𝑓 and 𝑔.

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Unit 1, Lesson 14: Inverse Functions

Warm-Up: What Does It Say? Here is an encoded message, a message that has been converted into a code.

WRGDB LV D JRRG GDB.

Can you figure out what it says in English? How was the original message encoded?

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Exploration Activity: Deciphering Codes with Functions 1. Now it’s your turn to write a secret code! a. Write a short and friendly message with 3–4 words.

b. Pick a number from 1 to 10. Then, encode your message by shifting each letter that many steps forward or backward in the alphabet, wrapping around from Z to A as needed. Consider using this table to create a key for your cipher.

Plain text

A B C D E F G H I

J K L M N O P Q R S T U V W X Y Z

Cipher text c. Give your encoded message to a partner to decode. If requested, give the number you used. d. Decode the message from your partner. Ask for their number, if needed.

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2. Suppose 𝑚 and 𝑐 each represent the position number of a letter in the alphabet, but 𝑚 represents the letters in the original message and 𝑐 the letters in your secret code. a. Complete the table. Letter in message 𝑚 𝑐

Letter in code b. Use 𝑚 and 𝑐 to write an equation that can be used to encode an original message into your secret code.

c. Use 𝑚 and 𝑐 to write an equation that can be used to decode your secret code into the original message.

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Guided Activity: U.S. Dollars and Mexican Pesos An American traveler who is heading to Mexico exchanges some U.S. dollars for Mexican pesos. At the time of his travel, 1 dollar can be exchanged for 19.32 pesos.

At the same time, a Mexican businesswoman who is in the United States is exchanging some Mexican pesos for U.S. dollars at the same exchange rate. 1. Find the amount of money in pesos that the American traveler would get if he exchanged: a. 100 dollars b. 500 dollars 2. Write an equation that gives the amount of money in pesos, 𝑝, as a function of the dollar amount, 𝑑, being exchanged.

3. Find the amount that the Mexican businesswoman would get if she exchanged: a. 1,000 pesos b. 5,000 pesos 4. Explain why it might be helpful to write the inverse of the function you wrote earlier. Then, write an equation that defines the inverse function.

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Lesson Summary Sometimes it’s useful to reverse a function so that the original output is now the input. Suppose Drew lives 400 meters (m) from school and walks to school. A linear function gives Drew’s distance to school, 𝐷, in m, after he has walked 𝑤 m from home, and is defined by the function 𝐷 = 400 − 𝑤.

With this equation, if it’s given how far Drew has walked from home, 𝑤, his remaining distance to school, 𝐷, can be found using the inverse function. In the inverse function, 𝑤 is the input and 𝐷 is the output. When Drew’s remaining distance to school, 𝐷, is known and the distance he has already walked, 𝑤, is unknown, the inverse function can be used. The inverse function can be found by solving the original function 𝐷 = 400 − 𝑤 for 𝑤, as shown.

The equation 𝑤 = 400 − 𝐷 represents the inverse of the original function. With this equation, the distance that Drew has walked from home can be found when his remaining distance to school is given.

𝐷

= 400 − 𝑤

𝑤

= 400 − 𝐷

𝐷 + 𝑤

= 400

In general, if a function takes 𝑎 as its input and gives 𝑏 as its output, its inverse function takes 𝑏 as the input and 𝑎 as the output.

Practice Problems 1. Noah’s cousin is exactly 7 years younger than Noah. Let 𝐶 represent Noah’s cousin’s age and 𝑁 represent Noah’s age. Ages are measured in years. a. Write a function that defines the cousin’s age as a function of Noah’s age. What are the input and output of this function?

b. Write the inverse of the function you wrote. What are the input and output of this inverse function?

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2. Noah’s cousin is exactly 7 years younger than Noah. Let 𝑀 represent Noah’s cousin’s age in months and 𝑁 represent Noah’s age in years. a. If Noah is 15 years old, how old is his cousin, in months?

b. When Noah’s cousin is 132 months old, how old is Noah, in years?

c. Write a function that gives the age of Noah’s cousin in months, as a function of Noah’s age in years.

d. Write the inverse of the function you wrote. What are the input and the output of this inverse function?

3. The number of years, 𝑦, is a function of the number of months, 𝑚. The number of months, 𝑚, is also a function of the number of years, 𝑦. a. Write two equations, one to represent each function.

b. Explain why the two functions are inverses.

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4. Each equation represents a function. For each, find the inverse function. a. 𝑐 = 𝑤 + 3 b. 𝑦 = 𝑥 − 2 c. 𝑦 = 5𝑥 d. 𝑤 = 𝑑 7

Review Problem 5. Here are the graphs of 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑓(𝑥 − 1) for a function 𝑓. Which graph corresponds to each equation? Explain how you know.

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Unit 1, Lesson 15: Finding Inverses and Verifying Graphically

Warm-Up: Shopping for Cookbooks Lin is comparing the cost of buying cookbooks at different online stores. • Store A sells them at $9 each and offers free shipping.

• Store B sells them at $9 each and charges $5 for shipping.

• Store C sells them at 𝑝 dollars and charges $5 for shipping.

• Store D sells them at 𝑝 dollars and charges 𝑓 dollars for shipping. 1. Write an equation to represent the total cost, 𝑇, in dollars as a function of 𝑛 cookbooks bought at each store.

2. Write an equation to find the number of books, 𝑛, that Lin could buy if she spent 𝑇 dollars at each store.

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Exploration Activity: From Celsius to Fahrenheit If we know the temperature in degrees Celsius, 𝐶, we can find the temperature in degrees Fahrenheit, 𝐹, using the equation 𝐹 = 9 𝐶 + 32. 5

1. Complete the table with temperatures in degrees Fahrenheit or degrees Celsius. 𝑪

𝑭

100

104

62.6

2. The equation 𝐹 = 9 𝐶 + 32 represents a function. Write an equation to represent the 5

inverse function. Be prepared to explain your reasoning.

3. The equation 𝑅 = 9 (𝐶 + 273.15) defines the temperature in degrees Rankine as a 5

function of the temperature in degrees Celsius.

Show that the equation 𝐶 = (𝑅 − 491.67) ∙ 5 defines the inverse of that function. 9

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Guided Activity: Digging Deeper into Inverse Functions 1. Consider the linear functions 𝑓(𝑥) = 1 𝑥 + 3 and 𝑔(𝑥) = 2𝑥 − 6. 2

a. Complete the tables of values for 𝑓(𝑥) and 𝑔(𝑥). 𝒙

−4

−2

−1

𝒙

2.5

𝒇(𝒙)

𝒈(𝒙)

3.5

b. What do you notice about the coordinate pairs of 𝑓(𝑥) and 𝑔(𝑥)?

c. Graph 𝑓(𝑥) and 𝑔(𝑥) on the coordinate plane.

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d. Graph the line 𝑦 = 𝑥 on the same coordinate plane. e. Complete the statements.

◯

The inputs of 𝑓(𝑥) correspond to the correspond to the

onto 𝑔 (𝑥) by a

◯

◯ ◯

◯

𝑥-values

𝑦-values

translation reflection

◯

and 𝑔(𝑥) are inverse functions.

𝑦-values

of 𝑔(𝑥). The outputs of 𝑓(𝑥)

of 𝑔(𝑥). When graphed, 𝑓(𝑥) can be mapped ◯

◯

rotation

𝑥-values

◯

over 𝑦 = 𝑥.

up 1 and right 1.

180° about the origin.

Therefore, 𝑓(𝑥)

An inverse function is a function, 𝑓−1(𝑥), that reverses the original function, 𝑓(𝑥), such that all input-output pairs in the original function are mapped to output-input pairs in the inverse function. Graphically, inverse functions are reflections of each other over the line 𝑦 = 𝑥. 1. The graph of the function ℎ(𝑥) = 𝑥3 − 2 is shown on the coordinate plane.

a. Complete the tables of values for ℎ(𝑥) and ℎ−1(𝑥). Use the values of ℎ(𝑥) to complete the table for ℎ−1(𝑥). 𝒙

−2

𝒉(𝒙)

𝒙

𝒉−1(𝒙)

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b. Graph ℎ−1(𝑥) on the same coordinate plane.

c. To determine the equation of the inverse of ℎ(𝑥), Amir began by rewriting ℎ(𝑥) in terms of 𝑦 and then switching the 𝑥 and 𝑦 in the equation. Amir’s partial work is shown. Complete Amir’s work to determine the equation of ℎ−1(𝑥). 𝑦 = 𝑥3 − 2

𝑥 = 𝑦3 − 2

d. Verify whether the equation of ℎ−1(𝑥) matches the table of values and the graph that you created in parts A and B.

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Lesson Summary It’s helpful to interpret the inverse of a function in terms of a situation and the quantities it represents. Suppose a linear function gives the cost, 𝐶, in dollars, of renting some equipment for 𝑛 hours (hr.). The function is defined by the equation 𝐶 = 8.25𝑛 + 30. If the number of hr. of rental, 𝑛, is given, it can be substituted into the expression 8.25𝑛 + 30 and evaluated to find the cost, 𝐶.

What is the inverse of this function, and what information about the length and cost of rental can it provide? To find the inverse of the function 𝐶 = 8.25𝑛 + 30, solve for 𝑛 as shown. 8.25𝑛 + 30 = 𝐶

8.25𝑛 = 𝐶 − 30 𝑛 = 𝐶 − 30 8.25

If the cost of the rental, 𝐶, is known, it can be substituted into the expression 𝐶 − 30 to 8.25 find the length of the rental, 𝑛, in hr.

Notice that the equation defining the inverse function is found by reversing the process that defines the original function. • Using the order of operations, the original rule, 𝐶 = 8.25𝑛 + 30, multiplies the input, 𝑛, by 8.25 and adds 30 to the result to get the output, 𝐶.

• The rule of the inverse function, 𝐶 − 30 , subtracts 30 from the input, 𝐶, and then 8.25

divides the result by 8.25 to get the output, 𝑛.

Since the inverse functions reverse the operations of solving, their equations are the reflections of each other over the line 𝑦 = 𝑥. One way to verify whether 2 functions are inverses of each other is to graph each function and check whether each point (𝑥, 𝑓(𝑥)) on 1 function corresponds to (𝑓(𝑥)𝑥,) on the other function.

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Practice Problems 1. Find the inverse of 𝑓(𝑥) = 5 − 3𝑥 . 2

2. Function 𝑟(𝑥) is shown on the coordinate plane. Graph the function 𝑟−1(𝑥).

3. A table of values for 𝑤(𝑥) is shown. 𝒙

𝒘(𝒙)

−3

− 10

−2

−7

−1

−4

−1

Use the values of 𝑤(𝑥) to complete the table for 𝑤−1(𝑥). 𝒙

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Review Problem 4. Describe how to move the graph so that it better matches the data.

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Unit 1, Lesson 16: Finding and Interpreting Inverse Functions

Warm-Up: Water in a Tank A tank contained some water. The function 𝑤 represents the relationship between 𝑡, time in minutes, and the amount of water in the tank in liters. The equation 𝑤(𝑡) = 80 − 2.5𝑡 defines this function. 1. Discuss with a partner:

a. How is the water in the tank changing? Be as specific as possible.

b. What does 𝑤(𝑡) represent? Is 𝑤(𝑡) the input or the output of this function?

2. Sketch a graph of the function. Be sure to label the axes.

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Exploration Activity: Phones in Homes In 2004, less than 5% of the homes in the U.S. relied only on cell phones. Since then, the percentage of homes that use only cell phones has increased. The table and graph show the percentages of homes with only cell phones from 2004 to 2009. Years Since 2004

Percentages

4.4

9.6

6.7

13.6

22.7

17.5

The linear function, 𝑃(𝑡) = 3.7𝑡 + 3.3 can be used to model the percentage of homes with only cell phones as a function of 𝑡 years since 2004. 1. Find the value of 𝑃(6). Then, explain what it means in this situation. 2. Solve 𝑃(𝑡) = 30 for 𝑡. What does the solution represent? 3. Assuming the trend continues and the function remains valid, what equation could be used to determine the years that correspond to 50%, 75%, or 100% of homes relying only on cell phones? Explain or show your reasoning.

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Collaborative Activity: Writing Inverses of Functions The graphs and equations of functions 𝑓(𝑥) and 𝑔(𝑥) are shown on the coordinate plane. 𝒇(𝒙) = − 𝟑𝒙 + 𝟐

𝒈(𝒙) = 𝒙𝟑 + 𝟏

1 -6 -5 -4 -3 -2 -1 0 -1

x 1

-6 -5 -4 -3 -2 -1 0 -1

-2

-3

-4

-5

-6

x 1

1. Graph 𝑓−1(𝑥) and 𝑔−1(𝑥) on the same coordinate plane as their inverses. 2. Determine the equations of 𝑓−1(𝑥) and 𝑔−1(𝑥). 3. Evaluate each of the following. a. 𝑓(3) = _____

b. 𝑔−1(2) = _____

c. 𝑓−1(2) = _____

d. 𝑔(1) = _____

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4. For each given pair of functions, verify graphically whether the functions are inverses. a. 𝑝(𝑥) = 4𝑥 − 5 and 𝑛(𝑥) = 𝑥 + 5 4

7 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1

x 1

-2 -3 -4 -5 -6 -7 -8

3 b. 𝑤(𝑥) = 1 𝑥3 and 𝑡(𝑥) = 2 √𝑥

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Lesson Summary A function is a relation where every 𝑥-value can map to exactly 1 𝑦-value. The line 𝑦 = 𝑥 is sometimes called the identity function because every coordinate point (𝑥, 𝑦) includes the same value for 𝑥 and 𝑦. A function 𝑓(𝑥) and its inverse, 𝑓−1(𝑥), have a unique graphical relationship where the function is reflected over the identity function to create the graph of the inverse function. To graph the inverse function, 𝑓−1(𝑥), plot the relationship (𝑦, 𝑥) to represent 𝑓(𝑥) reflected over the line 𝑦 = 𝑥. Three examples are shown of functions, their inverses, and the identity function graphed on the same coordinate plane. y

Notice that some functions have an inverse that is not a function because there are 𝑥-values mapped to multiple 𝑦-values. You will explore such relationships more deeply in future courses. To verify graphically whether 2 graphs are inverses, select several coordinate points on 1 graph, (𝑥, 𝑦), and verify that the reflected points, (𝑦, 𝑥), exist on the graph of the inverse. For example, the graphs of 𝑓(𝑥) and 𝑓−1(𝑥) are shown on the coordinate plane.

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Notice that the inverse of any coordinate point (𝑥, 𝑦) on the blue graph can be found on the red graph as (𝑦, 𝑥). For example, (2, − 2) is a point on the blue graph, and the reflected point (−2, 2) exists on the red graph, which may mean that the 2 graphs are inverses of each other.

• Select 2–3 points (𝑥, 𝑦) on 1 graph to check whether the reflected point (𝑦, 𝑥) exists on the other graph to verify that the 2 graphs are inverses of each other.

• Another way to verify there is an inverse relationship between the 2 graphs is to graph the identity function, 𝑦 = 𝑥, and verify that it acts as a line of reflection for the 2 graphs.

Practice Problems 1. Functions 𝑓 and 𝑔 are inverses, and 𝑓(−2) = 3. Is the point (3, − 2) on the graph of 𝑓, on the graph of 𝑔, or neither? 2. Here are two equations that relate two quantities, 𝑝 and 𝑄: 𝑄 = 7𝑝 + 1,999 𝑝 = 𝑄 − 1,999

Select all statements that are true about 𝑝 and 𝑄.

□ 𝑄 = 7𝑝 + 1,999 could represent a function, but 𝑝 = 𝑄 − 1,999 could not. □ Each equation could represent a function.

□ 𝑝 = 𝑄 − 1,999 could represent a function, but 𝑄 = 7𝑝 + 1,999 could not. 7

□ The two equations represent two functions that are inverses of one another. □ If 𝑄 = 7𝑝 + 1,999 represents a function, then the inverse function can be defined by 𝑝 = 7𝑄 − 1,999.

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3. Match each function to its inverse. A. 𝑦 = 2𝑥 − 3

1. 𝑥 = 𝑦 + 2 3

B. 𝑦 = 3𝑥

2. 𝑥 = 𝑦 + 3

D. 𝑦 = 𝑥 − 2

4. 𝑥 = 𝑦 + 2

C. 𝑦 = 3𝑥 − 2 E. 𝑦 = 𝑥 + 2 F. 𝑦 = 𝑥−2 3

3. 𝑥 = 3𝑦 + 2

5. 𝑥 = 𝑦 3

6. 𝑥 = 𝑦 − 2

4. Functions ℎ and 𝑗 are inverses. When 𝑥 is − 10, the value of ℎ(𝑥) is 7, or ℎ(−10 ) = 7. a. What is the value of 𝑗(7)?

b. Determine if each point is on the graph of ℎ, on the graph of 𝑗, or neither. Explain your reasoning. i.

(−10, 7)

ii.

(7, −10)

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Review Problem 5. The function 𝑓(𝑥) is shown on the coordinate plane. Graph the function 𝑔(𝑥) = 𝑓(𝑥 + 2) − 3.

7 6 5 4 3 2 1 -8 -7 -6 -5 -4 -3 -2 -1 0 -1

x 1 2

-2 -3 -4 -5 -6 -7 -8

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Unit 2: Polynomial Functions

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Unit 2, Lesson 1: Let’s Make a Box

Warm-Up: Which One Doesn’t Belong: Boxes Which one doesn’t belong? A. length: 4 centimeters (cm) width: 8 cm height: 10 cm

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Exploration Activity: Building Boxes Your teacher will give you some supplies. 1. Construct an open-top box from a sheet of paper by cutting out a square from each corner and then folding up the sides.

2. Calculate the volume of your box, and complete the table with your information.

Side Length of Square Cutout (inches (in.))

Length (in.)

Width (in.)

Height (in.)

Volume of Box (cu. in.)

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Collaborative Activity: Building the Biggest Box 1. The volume 𝑉(𝑥) in cubic inches (cu. in.) of the open-top box is a function of the side length 𝑥 in. of the square cutouts. Make a plan to figure out how to construct the box with the largest volume.

Pause here so your teacher can review your plan.

2. Write an expression for 𝑉(𝑥).

3. Use graphing technology to create a graph representing 𝑉(𝑥). Approximate the value of 𝑥 that would allow you to construct an open-top box with the largest volume possible from one piece of paper.

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Lesson Summary Polynomials can be used to model a variety of situations. A polynomial is the sum or difference of terms which have variables raised to non-negative integer powers and which have coefficients that may be real or complex. A polynomial can be used to model the volume of a box created by removing squares from each corner of a rectangular piece of paper.

• Let 𝑉(𝑥) be the volume of the box in cu. in., where 𝑥 is the side length, in inches, of each square removed from the 4 corners. • To define 𝑉 using a polynomial expression, use the fact that the volume of a cube is lengthwidthheight. If the original piece of paper has dimensions 3 in. by 8 in., then 𝑉(𝑥) = (3 − 2𝑥)(8 − 2𝑥)(𝑥).

What are some reasonable values for 𝑥? Cutting out squares with side lengths less than 0 in. doesn’t make sense, and similarly, it is impossible to cut out squares larger than 1.5 in., since the short side of the paper is only 3 in. (and 3 − 1.5 ⋅ 2 = 0). You may remember that the name for the set of all input values that make sense to use with a function is the domain. For this context, a reasonable domain is larger than 0 in. but less than 1.5 in. By graphing this function, it is possible to find the maximum value within a specific domain. The graph of 𝑦 = 𝑉(𝑥) over this domain is shown. It looks like the largest possible volume of a box made this way from a 3 in. by 8 in. piece of paper is about 7.4 cu. in.

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Practice Problems 1. A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150 meters (m) of fencing for the other three sides. The area 𝐴(𝑥), in square meters (sq. m), of the schoolyard is a function of the length 𝑥, in m, of each of the sides perpendicular to the school wall. a. Write an expression for 𝐴(𝑥).

b. What is the area of the schoolyard when 𝑥 = 40?

c. What is a reasonable domain for 𝐴 in this context?

2. Noah finds an expression for 𝑉(𝑥) that gives the volume of an open-top box, in cu. in., in terms of the length 𝑥, in inches, of the cutout squares used to make it. This is the graph Noah gets if he allows 𝑥 to take on any value between −1 and 5. a. What would be a more appropriate domain for Noah to use instead?

b. What is the approximate maximum volume for his box?

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3. Mai wants to make an open-top box by cutting out corners of a square piece of cardboard and folding up the sides. The cardboard is 10 cm by 10 cm. The volume 𝑉(𝑥), in cubic centimeters (cu. cm), of the open-top box is a function of the side length 𝑥, in cm, of the square cutouts. a. Write an expression for 𝑉(𝑥).

b. What is the volume of the box when 𝑥 = 3? Review Problems 4. Here is the graph of 𝑓.

a. Graph the function 𝑔 given by 𝑔𝑥 = − 𝑓(𝑥).

b. Graph the function ℎ given by ℎ(𝑥) = 𝑓(− 𝑥).

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Unit 2, Lesson 2: Funding the Future

Warm-Up: Notice and Wonder: Writing Numbers What do you notice? What do you wonder?

300 + 20 + 9

3 100s, 2 10s, 9 1s

3(102) + 2(101) + 9(100)

Exploration Activity: Polynomials in the Integers Consider the polynomial function 𝑝 given by 𝑝(𝑥) = 5𝑥3 + 6𝑥2 + 4𝑥.

1. Evaluate the function at 𝑥 = − 5 and 𝑥 = 15.

2. How does knowing that 5,000 + 600 + 40 = 5,640 help you solve the equation 5𝑥3 + 6𝑥2 + 4𝑥 = 5,640?

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Collaborative Activity: A Yearly Gift At the end of 12th grade, Clare’s aunt started investing money for her to use after graduating from college four years later. The first deposit was $300. If 𝑟 is the annual interest rate of the account, then at the end of each school year the balance in the account is multiplied by a growth factor of 𝑥 = 1 + 𝑟. 1. After one year, the total value is 300𝑥. After two years, the total value is 300𝑥 ⋅ 𝑥 = 300𝑥2. Write an expression for the total value after graduation in terms of 𝑥.

2. If Clare’s aunt had invested another $500 at the end of her freshman year, what would the expression be for the total value after graduation in terms of 𝑥?

Pause here for a whole-class discussion.

3. Suppose that $250 was invested at the end of sophomore year, and $400 at the end of junior year in addition to the original $300 and the $500 invested at the end of freshman year. Write an expression for the total value after graduation in terms of 𝑥. 4. The total amount 𝑦, in dollars, after four years is a function 𝑦 = 𝐶(𝑥) of the growth factor 𝑥. If the total Clare receives after graduation is 𝐶(𝑥) = 1,580, use a graph to find the interest rate that the account earned.

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Lesson Summary Consider an account in which $200 is invested at an annual interest rate of 𝑟. This means that at the end of each year, the balance in the account is multiplied by a growth factor of 𝑥 = 1 + 𝑟. After the first year, the amount in the account can be expressed as 200𝑥, which is a polynomial. Similarly, the amount will be 200𝑥2 after 2 years, 200𝑥3 after 3 years, etc.

• If an additional $350 is invested at the end of the first year, the polynomial expression can be revised accordingly. The amount of money in the account after 1 year is the same, but the amount of money after 2 years is now 200𝑥 + 350𝑥 = 200𝑥2 + 350𝑥. • If $400 more is invested at the end of the second year, and $150 more is invested at the end of the third year, the polynomial expression would be 200𝑥4 + 350𝑥3 + 400𝑥2 + 150𝑥.

Let 𝐷(𝑥) be the amount of money, in dollars, in the account after 4 years, and let 𝑥 be the growth factor, where 𝐷(𝑥) = 200𝑥4 + 350𝑥3 + 400𝑥2 + 150𝑥. A graph of 𝑦 = 𝐷(𝑥) helps visualize how the amount in the account after 4 years depends on different values of 𝑥. This polynomial model can be used to examine the effect of different annual interest rates, or to estimate what the annual interest rate needs to be to achieve a specific quantity at the end of the 4 years. • For example, point A is at (1.04, 𝐷(1.04), or approximately (1.04, 1216). This point represents that the amount in the account after 4 years with an interest rate of 4% each year is approximately $1,216. • Similarly, if the goal is for the account to have $2,000 after 4 years, that corresponds to point B, and at that point, the growth rate is approximately 1.25 each year, since (1.25, 𝐷(1.25)) ≈ (1.25, 2000). Therefore, an interest rate of 25% will produce a total of $2,000 in the account after 4 years, though it is unlikely that a bank would offer such a rate. • Note also that the values of 𝑥 < 1 correspond to negative rates, which would represent a decrease in the total amount in the account. Polynomial models are adaptable to a variety of situations, even as they grow in complexity. © Accelerate Learning Inc. - All Rights Reserved

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Practice Problems 1. Select all polynomial expressions that are equivalent to 6𝑥4 + 4𝑥3 − 7𝑥2 + 5𝑥 + 8.

□ 16𝑥

□ 6𝑥 + 4𝑥 − 7𝑥 + 5𝑥 + 8𝑥 5

□ 6𝑥 + 4𝑥 − 7𝑥 + 5𝑥 + 8 4

□ 8 + 5𝑥 + 7𝑥 − 4𝑥 + 6𝑥 2

□ 8 + 5𝑥 − 7𝑥 + 4𝑥 + 6𝑥 2

2. Each year a certain amount of money is deposited in an account which pays an annual interest rate of 𝑟 so that at the end of each year the balance in the account is multiplied by a growth factor of 𝑥 = 1 + 𝑟. $500 is deposited at the start of the first year, an additional $200 is deposited at the start of the next year, and $600 at the start of the following year. a. Write an expression for the value of the account at the end of three years in terms of the growth factor 𝑥. b. What is the amount (to the nearest cent) in the account at the end of three years if the interest rate is 2%?

3. Consider the polynomial function 𝑝 given by 𝑝(𝑥) = 5𝑥3 + 8𝑥2 − 3𝑥 + 1. Evaluate the function at 𝑥 = −2.

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Review Problem 4. Tyler finds an expression for 𝑉(𝑥) that gives the volume of an open-top box, in cubic inches (cu. in.), in terms of the length 𝑥, in inches (in.), of the square cutouts used to make it. This is the graph Tyler gets if he allows 𝑥 to take on any value between − 1 and 7. a. What would be a more appropriate domain for Tyler to use instead?

b. What is the approximate maximum volume for his box?

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Unit 2, Lesson 3: Introducing Polynomials

Warm-Up: Which One Doesn’t Belong: What Are Polynomials? Which one doesn’t belong? A. 4 − 𝑥2 + 𝑥3 − 4𝑥

B. 2𝑥4 + 𝑥2 − 5.7𝑥 + 2 C. 𝑥2 + 7𝑥 − 𝑥1/3 + 2

D. 𝑥5 + 8.36𝑥3 − 2.4𝑥2 + 0.32𝑥

Collaborative Activity: Card Sort: Equations and Graphs Your teacher will give you a set of cards. Group them into pairs that represent the same polynomial function. Be prepared to explain your reasoning.

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Guided Activity: Let’s Make Some Curves Use graphing technology to write equations for polynomial functions whose graphs have the characteristics listed when plotted on the coordinate plane. 1. A 1st degree polynomial function whose graph intercepts the vertical axis at 8.

2. A 2nd degree polynomial function whose graph has only positive 𝑦-values. 3. A 2nd degree polynomial function whose graph contains the point (0, − 9).

4. A 3rd degree polynomial function whose graph crosses the horizontal axis more than once.

5. A 4th degree or higher polynomial function whose graph never crosses the horizontal axis.

Lesson Summary A polynomial function of 𝑥 is a function given by a sum of terms, each of which is a coefficient multiplied by a whole-number power of 𝑥. Polynomials are often classified by degree, according to the term with the highest exponent on the independent variable. The degree of a polynomial in 𝑥 is the highest exponent occurring on when you write the polynomial out as a sum of non-zero constants times powers of 𝑥 (with like terms collected). For example, a quadratic function, like 𝑔(𝑡) = 10 + 96𝑡 − 16𝑡2, is considered a 2nd degree polynomial because the highest exponent on 𝑡 is 2. Similarly, a linear function, like 𝑓(𝑥) = 3𝑥 − 10, is considered a 1st degree polynomial. In Lesson 1, you worked with the 158 | Unit 2

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function 𝑉(𝑥) = (11 − 2𝑥)(8.5 − 2𝑥)(𝑥). This is an example of a 3rd degree polynomial, which is easier to see after applying the distributive property to rewrite the function in standard form as 𝑉(𝑥) = 4𝑥3 − 39𝑥2 + 93.5𝑥.

Graphs of polynomials have a variety of appearances. Three graphs are shown of different polynomials with degrees 1, 3, and 6, respectively.

Graphs of polynomials can curve up and down multiple times, so they can have points that are higher or lower than the rest of the points around them. These points are called relative maximums and relative minimums. The relative maximum is a point on the graph of a function that is higher than any of the points around it.

The relative minimum is a point on the graph of a function that is lower than any of the points around it.

In the second graph, there is a relative maximum at about (−3, 18) and a relative minimum at (2, 0). The word relative is used because while these are maximums and minimums relative to surrounding points, there are other points that are higher or lower. Future lessons will explore connections between equations and graphs of polynomials.

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Practice Problems 1. Select all points where relative minimum values occur on this graph of a polynomial function.

□ Point 𝐴

□ Point 𝐵 □ Point 𝐶

□ Point 𝐷 □ Point 𝐸 □ Point 𝐹

□ Point 𝐺

□ Point 𝐻 2. Add one term to the polynomial expression 14𝑥19 − 9𝑥15 + 11𝑥4 + 5𝑥2 + 3 to make it into a 22nd degree polynomial.

3. Identify the degree, leading coefficient, and constant value of each of the following polynomials. a. 𝑓(𝑥) = 𝑥3 − 8𝑥2 − 𝑥 + 8 Degree: _______

Leading coefficient: _______

b. ℎ(𝑥) = 2𝑥4 + 𝑥3 − 3𝑥2 − 𝑥 + 1 Degree: _______

Leading coefficient: _______

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Constant value: _______

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c. 𝑔(𝑥) = 13.2𝑥3 + 3𝑥4 − 𝑥 − 4.4 Degree: _______

Leading coefficient: _______

Constant value: _______

Review Problems 4. We want to make an open-top box by cutting out corners of a square piece of cardboard and folding up the sides. The cardboard is a 9 in. by 9 in. square. The volume 𝑉(𝑥) in cu. in. of the open-top box is a function of the side length 𝑥 in in. of the square cutouts. a. Write an expression for 𝑉(𝑥). b. What is the volume of the box when 𝑥 = 1? c. What is a reasonable domain for 𝑉 in this context? 5. Consider the polynomial function 𝑝 given by 𝑝(𝑥) = 7𝑥3 − 2𝑥2 + 3𝑥 + 10. Evaluate the function at 𝑥 = −3.

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Unit 2, Lesson 4: Connecting Factors and Zeros

Warm-Up:Factored Form What do you notice? What do you wonder? 𝑓(𝑥) = (𝑥 + 5)(𝑥 + 1)(𝑥 − 3)

𝑔(𝑥) = (𝑥 + 5)(𝑥 + 1)(𝑥 − 2)

ℎ(𝑥) = (𝑥 + 4)(𝑥 + 1)(𝑥 − 2)

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Guided Activity: What Values Make the Equations True? Find all values of 𝑥 that make the equation true. 1. (𝑥 + 4)(𝑥 + 2)(𝑥 − 1) = 0

2. (𝑥 + 4)(𝑥 + 2)(𝑥 − 1)(𝑥 − 3) = 0

3. (𝑥 + 4)2(𝑥 + 2)2 = 0

4. −2(𝑥 − 4)(𝑥 − 2)(𝑥 + 1)(𝑥 + 3) = 0

5. (2𝑥 + 8)(7𝑥 − 3)(𝑥 − 10) = 0

6. 𝑥2 + 3𝑥 − 4 = 0

7. 𝑥(3 − 𝑥)(𝑥 − 1)(𝑥 + 0.75) = 0

8. (𝑥2 − 4)(𝑥 + 9) = 0

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Collaborative Activity: Factors, Intercepts, and Graphs Your teacher will give you a set of cards. Match each equation to either a graph or a description. Take turns with your partner to match an equation with a graph or a description of a graph.

1. For each match that you find, explain to your partner how you know it’s a match.

2. For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement.

Lesson Summary A polynomial that is written as a product of linear factors, called factored form, can be used to identify several facts about it. For example, the factored form of the polynomial shown in the graph is 𝑃(𝑥) = 0.5(𝑥 − 3)(𝑥 − 2)(𝑥 + 1).

• The graph of 𝑃(𝑥) has 𝑥-intercepts at 𝑥 = 3, 𝑥 = 2, and 𝑥 = −1. • Each of these 𝑥-values makes one of the factors in the expression 0.5(𝑥 − 3)(𝑥 − 2)(𝑥 + 1) equal to zero, so each makes the equation 𝑃(𝑥) = 0 true.

• The values 3, 2, and − 1 are known as the zeros of the function. When a polynomial is not written as the product of linear factors, identifying the zeros from the expression for the polynomial can be more challenging. Future lessons will explore different strategies for identifying the zeros of polynomials.

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Practice Problems 1. What is the value of 4(𝑥 − 2)(𝑥 − 3) + 7(𝑥 − 2)(𝑥 − 5) − 6(𝑥 − 3)(𝑥 − 5) when 𝑥 = 5?

2. Which polynomial function has zeros when 𝑥 = − 2, 3/4, 5? A. 𝑓(𝑥) = (𝑥 − 2)(3𝑥 + 4)(𝑥 + 5) B. 𝑓(𝑥) = (𝑥 − 2)(4𝑥 + 3)(𝑥 + 5) C. 𝑓(𝑥) = (𝑥 + 2)(3𝑥 − 4)(𝑥 + 5)

D. 𝑓(𝑥) = (𝑥 + 2)(4𝑥 − 3)(𝑥 − 5)

3. The graph of a polynomial 𝑓(𝑥) = (2𝑥 − 3)(𝑥 − 4)(𝑥 + 3) has 𝑥-intercepts at 3 𝑥-values. What are they?

Review Problem 4. Han is multiplying 10𝑥4 by 0.5𝑥3 and gets 5𝑥7. He says that 0.5𝑥3 is not a polynomial because 0.5 is not an integer. What is the error in Han’s thinking? Explain your reasoning.

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Unit 2, Lesson 5: Different Forms of Polynomials

Warm-Up: Which One Doesn’t Belong: Small Differences Which one doesn’t belong? A. 𝑦 = (𝑥 + 4)(𝑥 − 6) B. 𝑦 = 2𝑥2 − 8𝑥 − 24

C. 𝑦 = 𝑥2 + 5𝑥 − 25

D. 𝑦 = 𝑥3 + 3𝑥2 − 10𝑥 − 24

Exploration Activity: The Return of the Box Earlier, we learned we can make a box from a piece of paper by cutting squares of side length 𝑥 from each corner and then folding up the sides. Let’s say we now have a piece of paper that is 8.5 inches (in.) by 14 in. The volume 𝑉, in cubic inches (cu. in.), of the box is a function of the side length 𝑥, where 𝑉(𝑥) = (14 − 2𝑥)(8.5 − 2𝑥)(𝑥). 1. Identify the degree and leading term of the polynomial. Explain or show your reasoning.

2. Without graphing, what can you say about the horizontal and vertical intercepts of the graph of 𝑉? Do these points make sense in this situation?

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Guided Activity: Using Diagrams to Multiply 1. Use the distributive property to show that each pair of expressions is equivalent. a. (𝑥 + 2)(𝑥 + 4) and 𝑥2 + 6𝑥 + 8 b. (𝑥 + 6)(𝑥 + − 5) and 𝑥2 + 𝑥 − 30 c. (𝑥2 + 10𝑥 + 7)(2𝑥 − 1) and 2𝑥3 + 19𝑥2 + 4𝑥 − 7 d. (4𝑥3 − 8)(𝑥2 + 3) and 4𝑥5 + 12𝑥3 − 8𝑥2 − 24 2. Write a pair of expressions that each have 2 or 3 terms, and trade them with your partner. Multiply the expressions they gave you.

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Collaborative Activity: Spot the Differences This activity requires the use of an applet, so please make your way over to the digital platform to find the link. Let 𝑓(𝑥) = (𝑥 − 2)(𝑥 + 3)(𝑥 − 7) and 𝑔(𝑥) = 1 /2 (𝑥 − 2)(𝑥 + 3)(𝑥 − 7).

1. Use the applet to explore both functions in the same window of −10 ≤ 𝑥 ≤ 10 and −100 ≤ 𝑦 ≤ 100 by using slider 𝑎 to graph 𝑔(𝑥). Describe how the 2 graphs are similar and how they are different.

2. Identify the degree of these functions. Use the expressions and their graphs to explain how you determined their degree. Consider rewriting each function in standard form to check your answer.

3. Let 𝑚(𝑥) = (3𝑥 − 6)(𝑥 + 3)(𝑥 − 7). What do you think the graph of 𝑦 = 𝑚(𝑥) will look like compared to 𝑦 = 𝑓(𝑥)?

Use the applet to check your prediction.

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Lesson Summary Polynomials can be expressed in different, equivalent algebraic forms. Each of these forms reveals different information about features of the polynomial and its graph. For example, equivalent expressions for polynomial 𝑃(𝑥) in factored form and standard form are shown with the key features of the polynomial that can be identified from each. 𝑃(𝑥) = 0.5(𝑥 − 1)2(𝑥 + 2)(2𝑥 − 3)

From factored form, the zeros can be identified.

𝑃(𝑥) = 𝑥4 − 1.5𝑥3 − 3𝑥2 + 6.5𝑥 − 3

• For this function, the zeros are 1, −2, and 3/2 . The graph of the function will have 𝑥-intercepts at 𝑥 = 1, 𝑥 = − 2, and 𝑥 = 3/2. From standard form, 2 key features of the polynomial function can be identified: the constant term and the degree. • The constant term, which is − 3 in this example, shows the value of the function when 𝑥 = 0. In a graph of the function, this point is known as the 𝑦-intercept.

• The degree, which is 4 in this example, reveals information about the general shape of the graph of the polynomial, which will be explored in detail in future lessons.

Practice Problems 1. 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 4) and 𝑔(𝑥) = 1/3(𝑥 + 3)(𝑥 − 4). The graphs of each are shown here. Which graph represents which polynomial function? Explain how you know.

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2. For each polynomial function, rewrite the polynomial in standard form. Then state its degree and constant term. a. 𝑓(𝑥) = (𝑥 + 1)(𝑥 + 3)(𝑥 − 4)

Standard form: _________________________________ Degree: ______ Constant term: ______

b. 𝑔(𝑥) = 3(𝑥 + 1)(𝑥 + 3)(𝑥 − 4)

Standard form: _________________________________ Degree: ______ Constant term: ______

3. Tyler incorrectly says that the constant term of (𝑥 + 4)(𝑥 − 4) is zero. a. What is the correct constant term?

b. What is Tyler’s mistake? Explain your reasoning.

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Review Problems 4. Select all polynomial expressions that are equivalent to 5𝑥3 + 7𝑥 − 4𝑥2 + 5.

□ 13𝑥

□ 5𝑥 − 4𝑥 + 7𝑥 + 5 3

□ 5𝑥 + 4𝑥 ⋅ 2 + 7𝑥 + 5 3

□ 5 + 4𝑥 − 7𝑥 + 5𝑥 2

□ 5 + 7𝑥 − 4𝑥 + 5𝑥 2

5. Select all the points which are relative minimums of this graph of a polynomial function.

□ Point 𝐴

□ Point 𝐵 □ Point 𝐶

□ Point 𝐷 □ Point 𝐸 □ Point 𝐹

□ Point 𝐺 6. What are the 𝑥-intercepts of the graph of 𝑦 = (3𝑥 + 8)(5𝑥 − 3)(𝑥 − 1)?

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Unit 2, Lesson 6: Using Factors and Zeros

Warm-Up: More Than Factors 𝑀 and 𝐾 are both polynomial functions of 𝑥 where 𝑀(𝑥) = (𝑥 + 3)(2𝑥 − 5) and 𝐾(𝑥) = 3(𝑥 + 3)(2𝑥 − 5).

1. How are the two functions alike? How are they different?

2. If a graphing window of −5 ≤ 𝑥 ≤ 5 and −20 ≤ 𝑦 ≤ 20 shows all intercepts of a graph of 𝑦 = 𝑀(𝑥), what graphing window would show all intercepts of 𝑦 = 𝐾(𝑥)?

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Exploration Activity: Choosing Windows Mai graphs the function 𝑝 given by 𝑝(𝑥) = (𝑥 + 1)(𝑥 − 2)(𝑥 + 15) and sees this graph. She says, “This graph looks like a parabola, so it must be a quadratic.”

1. Is Mai correct? Use graphing technology to check.

2. Explain how you could select a viewing window before graphing an expression like 𝑝(𝑥) that would show the main features of a graph.

3. Using your explanation, what viewing window would you choose for graphing 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 1)(𝑥 − 2)(𝑥 − 28)?

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Collaborative Activity: What’s the Equation? Write a possible equation for a polynomial whose graph has the following horizontal intercepts. Check your equation using graphing technology. 1. (4, 0)

2. (0, 0) and (4, 0)

3. (−2, 0), (0, 0), and (4, 0)

4. (−4, 0), (0, 0), and (2, 0)

5. (−5, 0), (1/2, 0), and (3, 0)

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Lesson Summary The zeros of a polynomial function can be used to determine what an expression for the polynomial might be. For example, suppose polynomial function 𝑍 satisfies 𝑍(𝑥) = 0 when 𝑥 is −1, 2, or 4. One way to write a polynomial expression is as a product of linear factors. To write a possible expression for 𝑍(𝑥), multiply together a factor that is zero when 𝑥 = −1, a factor that is zero when 𝑥 = 2, and a factor that is zero when 𝑥 = 4. There are many possible expressions for 𝑍(𝑥). Using linear factors, one possibility is 𝑍(𝑥) = (𝑥 + 1)(𝑥 − 2)(𝑥 − 4).

Another possibility is 𝑍(𝑥) = 2(𝑥 + 1)(𝑥 − 2) (𝑥 − 4), since the 2 (or any other rational number) does not change the values of 𝑥 that make the function equal to zero. To check that these expressions satisfy the given information about 𝑍, test the values −1, 2, and 4 to make sure that 𝑍(𝑥) = 0 for each of the 𝑥-intercepts. Alternatively, graph both possible versions of 𝑍 to show that the graphs intersect the horizontal axis at −1, 2, and 4, as shown here.

Practice Problems 1. Diego wrote 𝑓(𝑥) = (𝑥 + 2)(𝑥 − 4) as an example of a function whose graph has 𝑥-intercepts at 𝑥 = − 4, 2. What was his mistake?

2. Write a possible equation for a polynomial whose graph has horizontal intercepts at 𝑥 = 2, −1/2, −3.

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3. Which polynomial function’s graph is shown here? A. 𝑓(𝑥) = (𝑥 + 1)(𝑥 + 3)(𝑥 + 4)

B. 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 3)(𝑥 + 4) C. 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 3)(𝑥 − 4) D. 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 3)(𝑥 − 4)

Review Problems 4. Which expression is equivalent to (3𝑥 + 2)(3𝑥 − 5)? A. 6𝑥 − 3

B. 9𝑥2 − 10

C. 9𝑥2 − 3𝑥 − 10

D. 9𝑥2 − 9𝑥 − 10 5. What is the value of 6(𝑥 − 2)(𝑥 − 3) + 4(𝑥 − 2)(𝑥 − 5) when 𝑥 = −3?

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6. Match each polynomial function with its leading coefficient. A. 𝑃(𝑥) = (𝑥 + 2)(2𝑥 − 3)(4𝑥 + 7) B. 𝑃(𝑥) = 1 /2 (𝑥 − 2)(2𝑥 − 3)(4𝑥 + 7)

1. 2.

C. 𝑃(𝑥) = 5(𝑥 − 2)(2𝑥 − 3)(4𝑥 + 7)

E. 𝑃(𝑥) = 1 /4 (𝑥 + 2)(2𝑥 − 3)(4𝑥 + 7)

D. 𝑃(𝑥) = − (𝑥 − 2)(2𝑥 − 3)(4𝑥 + 7)

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4 2

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Unit 2, Lesson 7: End Behavior

Warm-Up: Notice and Wonder: A Different View What do you notice? What do you wonder? 𝑦 = 𝑥3 + 4𝑥2 − 𝑥 − 4

𝑦 = 𝑥4 − 10𝑥2 + 9

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Exploration Activity: Polynomial End Behavior 1. For your assigned polynomial, complete the column for the different values of 𝑥. Discuss with your group what you notice. 𝒙

−1,000

𝒚 = 𝒙𝟐 + 𝟏

𝒚 = 𝒙𝟑 + 𝟏

𝒚 = 𝒙𝟒 + 𝟏

𝒚 = 𝒙𝟓 + 𝟏

−100 −10 −1 1 10 100 1,000 2. Sketch what you think the end behavior of your polynomial looks like, then check your work using graphing technology.

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Collaborative Activity: Understanding End Behavior The graphs of 2 different polynomial functions are shown on the coordinate plane.

𝒇(𝒙) = −𝟖𝒙𝟑 + 𝟏𝟎𝒙𝟐 + 𝟐𝟖𝒙 − 50

𝟏𝟓 𝟐

𝒈(𝒙) = 𝟏𝟐𝒙𝟒 + 𝟐𝟔𝒙𝟑 − 𝟑𝒙𝟐 − 𝟐𝟐.𝟓𝒙 50

x -5

-4

-3

-2

-1

-5

-4

-3

-2

-1

-10

-20

-30

-40

-50

The end behavior of a function refers to the appearance of the function’s graph at the “ends,” as the 𝑥-values get infinitely smaller or infinitely larger. 1. Use the description of end behavior to complete each statement.

a. For the function 𝒇(𝒙), as the 𝑥-values get infinitely smaller, the ends of the graph of 𝒇(𝒙) approach ¡ ¡

−∞. ∞.

b. For the function 𝒇(𝒙), as the 𝑥-values get infinitely larger, the ends of the graph of 𝒇(𝒙) approach −∞. ¡ ∞. ¡

c. For the function 𝒈(𝒙), as the 𝑥-values get infinitely smaller, the ends of the graph of 𝒈(𝒙) approach ¡ −∞. ¡ ∞.

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d. For the function 𝒈(𝒙), as the 𝑥-values get infinitely larger, the ends of the graph of 𝒈(𝒙) approach −∞. ¡ ∞. ¡

2. Complete the table for the 2 polynomial functions.

Function

Degree

Leading Coefficient

𝒇(𝒙)

𝒈(𝒙)

End Behavior As 𝑥→ − ∞, 𝑓(𝑥)→______. As 𝑥→∞, 𝑓(𝑥)→______.

As 𝑥→ − ∞, 𝑓(𝑥)→______. As 𝑥→∞, 𝑓(𝑥)→______.

3. Make a conjecture about the relationship between the degree, leading coefficient, and end behavior of a polynomial function.

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Lesson Summary As explored in previous lessons, when the expression for a polynomial function 𝑓 is written in factored form with a factor (𝑥 − 𝑎), then 𝑎 is a zero of 𝑓, which means 𝑓(𝑎) = 0 and the point 𝑎, 0 is on the graph of the function. But what about other values of 𝑥? In particular, as the values of 𝑥 approach either end of the graph, what happens to the values of 𝑓(𝑥)? The answer to this question depends on the degree of the polynomial and its leading coefficient, because any negative real number raised to an even power results in a positive number, while any negative real number raised to an odd power results in a negative number. The graphs of the functions 𝑦 = 𝑥2, 𝑦 = 𝑥3, and 𝑦 = 𝑥4 are shown. 𝒚 = 𝒙𝟐

𝒚 = 𝒙𝟑

𝒚 = 𝒙𝟒

For both 𝑦 = 𝑥2 and 𝑦 = 𝑥4, as 𝑥 approaches either −∞ or ∞, the values of 𝑦 approach ∞. But for 𝑦 = 𝑥3, as 𝑥 approaches −∞ or ∞, the values of 𝑦 approach opposite directions. One key feature used to describe functions is related to what value the dependent variable approaches as the independent variable approaches the ends of the graph. This is called the end behavior of the function. The end behavior is how the outputs of a function change as we look at input values farther and farther from .

The value of the leading term and the degree of a polynomial are used together to determine the end behavior of the function. The table summarizes the end behavior patterns of polynomial functions.

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Degree

Sign of Leading Coefficient

End Behavior

Even

Positive

Odd

Positive

As 𝑥→ − ∞, 𝑦→∞. As 𝑥→∞, 𝑦→∞.

Even

Negative

Odd

Negative

As 𝑥→ − ∞, 𝑦→ − ∞. As 𝑥→∞, 𝑦→∞.

As 𝑥→ − ∞, 𝑦→ − ∞. As 𝑥→∞, 𝑦→ − ∞. As 𝑥→ − ∞, 𝑦→∞. As 𝑥→∞, 𝑦→ − ∞.

Practice Problems 1. Match each polynomial with its end behavior. Some end behavior options may be used more than once, or not at all. A. 𝑓(𝑥) = 2𝑥3 + 3𝑥4 + 𝑥2 − 1

1. As 𝑥→ − ∞, 𝑓(𝑥)→∞. As 𝑥→∞, 𝑓(𝑥)→∞.

B. 𝑓(𝑥) = 1 − 3𝑥 + 𝑥2

2. As 𝑥→ − ∞, 𝑓(𝑥)→ − ∞. As 𝑥→∞, 𝑓(𝑥)→∞.

D. 𝑓(𝑥) = 2𝑥 + 5

4. As 𝑥→ − ∞, 𝑓(𝑥) − ∞. As 𝑥→∞, 𝑓(𝑥)→ − ∞.

C. 𝑓(𝑥) = 9 + 𝑥4

3. As 𝑥→ − ∞, 𝑓(𝑥)→∞. As 𝑥→∞, 𝑓(𝑥)→ − ∞.

2. Which polynomial function has the end behavior described?

A. 𝑓(𝑥) = 5𝑥2 − 2𝑥 + 1

As 𝑥 → −∞, 𝑓(𝑥) → −∞.

B. 𝑓(𝑥) = 6𝑥3 + 4𝑥2 − 15𝑥 + 32

C. 𝑓(𝑥) = 7𝑥4 − 2𝑥3 + 3𝑥2 + 8𝑥 − 10 D. 𝑓(𝑥) = 8𝑥6 + 1

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3. The graph of a polynomial function 𝑓 is shown. Which statement about the polynomial is true? A. The degree of the polynomial is even. B. The degree of the polynomial is odd. C. The constant term of the polynomial is even. D. The constant term of the polynomial is odd.

Review Problems 4. Andre wants to make an open-top box by cutting out corners of a 22 inch (in.) by 28 in. piece of poster board and then folding up the sides. The volume 𝑉(𝑥) in cubic inches (cu. in.) of the open-top box is a function of the side length 𝑥 in in. of the square cutouts. a. Write an expression for 𝑉(𝑥). b. What is the volume of the box when 𝑥 = 6? c. What is a reasonable domain for 𝑉 in this context? 5. Kiran wrote 𝑓(𝑥) = (𝑥 − 3)(𝑥 − 7) as an example of a function whose graph has 𝑥-intercepts at 𝑥 = −3, −7. What was his mistake?

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Unit 2, Lesson 8: Graphing Polynomial Functions

Warm-Up: It’s a Cover-Up Match each of the graphs to the polynomial equation it represents. For the graph without a matching equation, write down what must be true about the polynomial equation. 1. 𝑦 = 𝑥(𝑥 + 3)(2 − 𝑥)

2. 𝑦 = 1 − 3𝑥 + 5𝑥4

3. 𝑦 = 5(𝑥 + 3) − 5𝑥

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Collaborative Activity: The Case of Unexpected End Behavior 1. Write an equation for a polynomial with the properties listed. • It has even degree. • It has at least 2 terms. • As 𝑥 → ∞, 𝑦 → −∞, and as 𝑥 → −∞, 𝑦 → −∞.

Pause here so your teacher can review your work.

2. Write an equation for a polynomial with the properties listed. • It has odd degree. • It has at least 2 terms. • As 𝑥 → −∞, 𝑦 → ∞, and as 𝑥 → ∞, 𝑦 → −∞.

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Guided Activity: Sketching Graphs of Polynomials For each function, use the given equation and graph to complete the following. 1. 𝑦 = 𝑥(𝑥 + 2)(𝑥 − 2)

y 5

a. Find the zeros of the function. x -5

b. Mark the zeros on the graph.

-5

c. What is the degree of the polynomial?

d. What is the sign of the leading coefficient?

e. Complete the description of the end behavior of the function. As 𝑥 → −∞, 𝑦 → ______. As 𝑥 → ∞, 𝑦 → ______. 2. 𝑦 = 0.2(𝑥 + 3)(𝑥 − 4)(𝑥 + 1)(𝑥 − 2)

y 5

a. Find the zeros of the function. x -5

b. Mark the zeros on the graph.

-5

c. What is the degree of the polynomial?

d. What is the sign of the leading coefficient?

e. Complete the description of the end behavior of the function. As 𝑥 → −∞, 𝑦 → ______. As 𝑥 → ∞, 𝑦 → ______.

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3. Complete the statements. The zero(s) of a polynomial function

zero(s) can be determined from

¡ is the 𝑦-intercept

¡ are the 𝑥-intercepts

¡ factored form

of the function. The

using the zero-product

¡ standard form property. The

¡ 𝑦-intercept

¡ 𝑥-intercepts

function is written in

of the function is the constant term when the

¡ factored form.

¡ standard form.

4. Complete the table to summarize how to use the degree of a polynomial function and the sign of the leading coefficient to determine its end behavior. Degree

Sign of Leading Coefficient

Left End Behavior

Right End Behavior

Even

Positive (+)

As 𝑥 → −∞, 𝑦 → ______.

As 𝑥 → ∞, 𝑦 → ______.

Positive (+)

As 𝑥 → −∞, 𝑦 → ______.

As 𝑥 → ∞, 𝑦 → ______.

Odd

Negative (−) Negative (−)

As 𝑥 → −∞, 𝑦 → ______.

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As 𝑥 → ∞, 𝑦 → ______.

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5. Consider the polynomial function 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)(𝑥 + 5). a. Complete the table of key features.

Key Features Sign of leading coefficient

𝑥-intercept(s) 𝑦-intercept

Left end behavior

As 𝑥 → −∞, 𝑦 → ______.

Right end behavior

As 𝑥 → ∞, 𝑦 → ______.

(0, −15)

Degree

b. Sketch the graph of 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)(𝑥 + 5) using the key features.

x -15

-10

-5

-10

-15

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6. Consider the polynomial function ℎ(𝑥) = − 0.5𝑥(𝑥 − 4)(𝑥 + 1)(𝑥 + 6). a. Complete the table of key features for the polynomial. Key Features 𝑥-intercept(s) 𝑦-intercept Degree

Sign of leading coefficient Left end behavior

As 𝑥 → −∞, 𝑦 →______.

Right end behavior

As 𝑥 → ∞, 𝑦 → ______.

b. Sketch the graph of ℎ(𝑥) = − 0.5𝑥(𝑥 − 4)(𝑥 + 1)(𝑥 + 6) using the key features.

x -6

-4

-2

-4

-6

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Lesson Summary Key features of functions can be used to create rough graphs, or sketches, of the functions. Some of the key features that can be useful in sketching the graph of a polynomial function 𝑓 are shown. Key Feature 𝑥-intercept(s) 𝑦-intercept Degree Sign of leading coefficient

Impact on Graph Determine 𝑥-value(s) of 𝑓(𝑥) = 0. The zero(s) is (are) the 𝑥-intercept(s), where the polynomial intersects the 𝑥-axis.

Determine the 𝑦-value of 𝑓(0). This value is the 𝑦-intercept of the function.

Determine the highest exponent of all the terms. An even degree polynomial means both end behaviors approach the same direction. An odd degree polynomial means the end behaviors approach opposite directions. The sign of the leading coefficient determines the end behavior of the polynomial function as 𝑥 → ∞.

Consider the function 𝑓(𝑥) = −2(𝑥 + 1)(2𝑥 − 5)(𝑥 − 3). The key features and graph of the function 𝑓(𝑥) are shown. Key Features

𝑥-intercept(s)

Graph

𝑦-intercept

(0, − 30)

Sign of leading coefficient

Negative

Degree

𝑥 = − 1, 𝑥 = 5/2 , 𝑥 = 3 3

15 10 5 x -3

-2

-1

0 -5

-10 -15 -20

Left end behavior

Right end behavior

As 𝑥 → −∞, 𝑦 → ∞. As 𝑥 → ∞, 𝑦 → −∞.

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Practice Problems 1. Sketch the graph of 𝑇(𝑥) = − 0.1𝑥(𝑥 + 10)(𝑥 + 8). y

x -10

-5

-10

2. Match each polynomial with its end behavior. Some end behavior options may be used more than once, or not at all. A. 𝑓(𝑥) = − 2𝑥 + 3

1. As 𝑥 → −∞, 𝑓(𝑥) → ∞. As 𝑥 → ∞, 𝑓(𝑥) → ∞.

B. 𝑓(𝑥) = 𝑥2 − 6𝑥 + 3

2. As 𝑥 → −∞, 𝑓𝑥 → −∞. As 𝑥 → ∞, 𝑓(𝑥) → ∞.

D. 𝑓(𝑥) = 7 − 𝑥4

4. As 𝑥 → −∞, 𝑓𝑥 → −∞. As 𝑥 → ∞, 𝑓(𝑥) → −∞.

C. 𝑓(𝑥) = 1 − 𝑥2 + 2𝑥3

3. As 𝑥 → −∞, 𝑓(𝑥) → ∞. As 𝑥 →∞, 𝑓(𝑥) → −∞.

3. State the degree and end behavior of 𝑓(𝑥) = −𝑥3 + 5𝑥2 + 6𝑥 + 1. Explain or show your reasoning.

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4. The graph of a polynomial function 𝑓 is shown. Select all the true statements about the polynomial.

□ The degree of the polynomial is even. □ The degree of the polynomial is odd. □ The leading coefficient is positive. □ The leading coefficient is negative.

□ The constant term of the polynomial is positive. □ The constant term of the polynomial is negative. Review Problem 5. Select all the polynomial functions whose graphs have 𝑥-intercepts at 𝑥 = 4, −1/4, −2.

□ (𝑥 + 4)(4𝑥 − 1)(𝑥 − 2) □ (𝑥 − 4)(4𝑥 + 1)(𝑥 + 2) □ (𝑥 − 4)(4𝑥 − 1)(𝑥 − 2)

□ (𝑥 + 4)(4𝑥 + 1)(𝑥 + 2)

□ (2𝑥 + 4)(4𝑥 − 1)(𝑥 − 2)

□ (4𝑥 − 16)(4𝑥 + 1)(𝑥 + 2)

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Unit 2, Lesson 9: Multiplicity

Warm-Up: Notice and Wonder: Duplicate Factors What do you notice? What do you wonder? 𝑦 = (𝑥 − 3)2

𝑦 = (𝑥 + 1)(𝑥 − 3)2

𝑦 = (𝑥 − 3)3

𝑦 = (𝑥 − 6)(𝑥 − 3)2

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Guided Activity: Sketching Graphs of Polynomials 1. Consider the information in the table.

y 10

a. Sketch a polynomial using the information from the table.

Degree Zeros Leading Coefficient 𝒚-intercept

𝑥 = − 5 𝑥 = 4 with a multiplicity of 2 Negative −2

-10

-5

-10

b. Compare your sketch to your partner’s sketch. Note any differences or similarities.

c. How did you determine how high or how low to draw the maximums or minimums?

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2. Consider the polynomial 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)(𝑥 + 1)(𝑥 − 1).

a. Complete the table of key features for the polynomial. Key Features Degree 𝑥-intercept(s)

Type of polynomial Sign of leading coefficient in standard form

𝑦-intercept

Opens up/down

Left end behavior

As 𝑥→_____, 𝑦→_____.

Right end behavior

As 𝑥→_____, 𝑦→_____.

b. Sketch the graph of 𝑓(𝑥) = (𝑥 + 3)(𝑥 − 1)(𝑥 + 1)(𝑥 − 1) using the key features.

y 10

-10

-5

-10

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3. Consider the polynomial 𝑓(𝑥) = − 0.5𝑥(2𝑥 − 5)(𝑥 − 1)(𝑥 − 3)(3𝑥 + 4). a. Complete the table of key features for the polynomial. Key Features Degree 𝑥-intercept(s) 𝑦-intercept

Opens up/down

Type of polynomial Sign of leading coefficient in standard form Left end behavior

As 𝑥 → _____, 𝑦 → _____

Right end behavior

As 𝑥 → _____, 𝑦 → _____

b. Sketch the graph of 𝑓(𝑥 )= − 0.5𝑥(2𝑥 − 5)(𝑥 − 1)(𝑥 − 3)(3𝑥 + 4) using the key features.

-10

-5

-20

-40

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Collaborative Activity: Sketching Polynomials 1. For polynomials 𝐴–𝐹:

a. Write the degree, all zeros, and complete the sentence about the end behavior. b. Sketch a possible graph. c. Check your sketch using graphing technology. 𝐴(𝑥) = (𝑥 + 2)(𝑥 − 2)(𝑥 − 8)

𝐵𝑥 = − (𝑥 + 2)(𝑥 − 2)2

End behavior: As 𝑥→ − ∞, 𝑦→ ______

𝐶(𝑥) = (𝑥 + 6)(𝑥 + 2)2

𝐷(𝑥) = − (𝑥 + 6)2(𝑥 + 2)

Degree: _____

End behavior: As 𝑥→ − ∞, 𝑦→ ______

Degree: _____

Zeros: _________

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Zeros: _________

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𝐸(𝑥) = (𝑥 + 4)(𝑥 − 2)3

𝐹(𝑥) = 𝑥3(𝑥 + 4)(𝑥 − 3)2

End behavior: As 𝑥→ − ∞, 𝑦→ ______

Degree: _____

Zeros: _________

Degree: _____

Zeros: _________

Pause here for your teacher to check your work.

2. Create your own polynomial for your partner to figure out. a. Create a polynomial with degree greater than 2 and less than 8 and write the equation in the space given. b. Trade papers with a partner, then fill out the information about their polynomial and complete a sketch. c. Trade papers back. Check your partner’s sketch using graphing technology. Your polynomial: ________________________________ Degree: _____

Zeros: _________

End behavior: As 𝑥→ − ∞, 𝑦→ ______

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Lesson Summary While the zeros, or 𝑥-intercepts, of a polynomial function can be determined using a factored expression, the number of times that a factor is repeated also provides important information about how a zero interacts with the horizontal axis. For example, 𝑦 = (𝑥 + 3)(𝑥 − 1)(𝑥 − 4) has 3 factors with no duplicates. This results in a graph where each zero crosses the 𝑥-axis as shown, at 𝑥 = − 3, 𝑥 = 1, and 𝑥 = 4.

In this function, each factor, 𝑥 + 3, 𝑥 − 1, and 𝑥 − 4, has a multiplicity of 1.

Multiplicity is the power to which a factor occurs in the factored form of a polynomial.

For the function 𝑦 = (𝑥 + 3)2(𝑥 − 4), there are still 3 factors, but 2 of them are (𝑥 + 3). This results in a graph that has a relative maximum at 𝑥 = − 3 and crosses the 𝑥-axis at 𝑥 = 4. The factor (𝑥 + 3) has a multiplicity of 2, while the factor (𝑥 − 4) has a multiplicity of 1.

Combining all information about factors, degree, end behavior, the sign of the leading coefficient, and multiplicities of polynomials grants the ability to sketch polynomials written in factored form. For example, consider what the graph of 𝑦 = (𝑥 + 3)(𝑥 − 4)3 would look like. • The factors can be used to identify that the function has zeros at 𝑥 = − 3 and 𝑥 = 4.

• Since (𝑥 + 3) has a multiplicity of 1 and (𝑥 − 4) has a multiplicity of 3, the graph of the polynomial crosses the 𝑥-axis at −3 and crosses the 𝑥-axis again as a cubic curve at 4. © Accelerate Learning Inc. - All Rights Reserved

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• Since this is a 4th degree polynomial with a positive leading coefficient, the right and left end behaviors both approach ∞. The graph of 𝑦 = (𝑥 + 3)(𝑥 − 4)3 is shown.

Practice Problems 1. Draw a rough sketch of the graph of 𝑔(𝑥) = (𝑥 − 3)(𝑥 + 1)(7𝑥 − 2). y

-10

-5

-10

2. Draw a rough sketch of the graph of 𝑓(𝑥) = (𝑥 + 1)2(𝑥 − 4). y

-10

-5

-10

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3. Technology required. Predict the end behavior of each polynomial function, then check your prediction using technology. a. 𝐴(𝑥) = (𝑥 + 3)(𝑥 − 4)(3𝑥 − 7)(4𝑥 − 3) b. 𝐵(𝑥) = (3 − 𝑥)2(6 − 𝑥) c. 𝐶(𝑥) = - (4 − 3𝑥)(𝑥4) d. 𝐷(𝑥) = (6 − 𝑥)6 Review Problems 4. 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 6) and 𝑔(𝑥) = 2(𝑥 + 1)(𝑥 − 6). The graphs of each are shown. Which graph represents which polynomial function? Explain how you know.

5. State the degree and end behavior of 𝑓(𝑥) = 8𝑥3 + 2𝑥4 − 5𝑥2 + 9. Explain or show your reasoning.

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Unit 2, Lesson 10: Key Features of Polynomial Functions

Warm-Up: Determining Zeros Identify the zeros of each polynomial function. 1. 𝑓(𝑥) = 𝑥(𝑥 − 4)(𝑥 − 3) 2. 𝑔(𝑥) = (4𝑥2 + 1)(2𝑥 − 1) 3. ℎ(𝑥) = (𝑥 + 2)(𝑥 − 2)(𝑥 + 1)(𝑥 − 1)

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Guided Activity: Key Features of Polynomial Functions 1. Function 𝑓 is represented as 𝑓(𝑥) = (𝑥 + 2)2(2𝑥 − 1).

a. Explain which part(s) of the equation can be used to determine the end behavior of 𝑓(𝑥). b. Determine the 𝑥-intercepts of 𝑓(𝑥) and their multiplicities. c. Determine the 𝑦-intercept of 𝑓(𝑥).

d. Sketch a graph of 𝑓(𝑥) on the coordinate plane.

y 5 4 3 2 1 -5 -4 -3 -2 -1 0 -1

-2 -3 -4 -5

e. Compare your sketch to your partner’s. Discuss any differences between your graphs.

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The zeros of polynomial functions may also be referred to as roots. f. Complete the table of key features for 𝑓(𝑥), using interval notation where applicable. Key Features Positive interval(s) Negative interval(s)

Root(s) 𝑦-intercept Domain

Left end behavior

Range As 𝑥→_____, 𝑦→_____.

Right end behavior

As 𝑥→_____, 𝑦→_____.

Relative maximum and relative minimum points are turning points in a polynomial where the function changes from increasing to decreasing or vice versa.

g. Function 𝑓 has a relative maximum of (−2, 0) and a relative minimum of (−1/3, −49/27). Use the values of the relative maximum and relative minimum to determine the increasing and decreasing intervals of the function in interval notation and set notation. Increasing interval(s): _______________________________ _______________________________

Decreasing interval(s): _______________________________ _______________________________

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2. A graph of 𝑔(𝑥) is shown.

y (−2, 6.4)

(−1, 3.6)

(−3, 0)

7 6 5 4 3 2 1

‐4 ‐3 ‐2 ‐1 0 ‐1 ‐2 ‐3 ‐4 ‐5

(0, 0) (2, 0)

1 2 3 4 (0.75, −0.879) (3, −3.6)

a. Explain which points can be used to determine positive and negative intervals of 𝑔(𝑥).

b. Explain which points can be used to determine increasing and decreasing intervals of 𝑔(𝑥).

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c. Complete the table of key features of 𝑔(𝑥), using set notation where applicable. Key Features Increasing interval(s)

𝑥-intercept(s)

Decreasing interval(s)

𝑦-intercept Domain

Positive interval(s)

Range

Negative interval(s)

Left end behavior

As 𝑥→_____, 𝑦→_____

Right end behavior

As 𝑥→_____, 𝑦→_____

3. Consider the function 𝑣(𝑥) = 5𝑥3(𝑥 − 3)2(𝑥 + 1). a. Identify the zeros and their multiplicities.

b. Complete the statements of end behavior for the function. As 𝑥→_____, 𝑦→_____.

As 𝑥→_____, 𝑦→_____.

c. Determine the positive and negative intervals of 𝑣(𝑥).

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4. The graph of ℎ(𝑥) is shown on the coordinate plane.

a. Select all of the key features that are useful for writing the equation of the polynomial function.

□ Zeros □ Range □ Domain □ 𝑦-intercept

□ End behavior □ Positive interval □ Increasing interval

b. The leading coefficient of ℎ(𝑥) is −0.02. Write the equation of the polynomial function. ℎ(𝑥) = ______________________________

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Collaborative Activity: Card Sort 1. Your teacher will give you 16 cards, each featuring either the graph of a polynomial function or a list of key features. Match each graph card to a key feature card. 2. Complete the table by identifying the graph card and the key feature card that match each of the given equations. One set of cards will not have an equation that matches. Equation

Graph Card

Key Feature Card

𝑦 = − 0.5(𝑥 + 4)(𝑥 + 2)(𝑥 − 2) 𝑦 = 0.4𝑥(𝑥 + 4)(𝑥 + 2)(𝑥 − 2) 𝑦 = 0.6(𝑥 + 2)(𝑥 − 2)2

𝑦 = −0.4(𝑥 + 2)2(2𝑥 − 2)2 𝑦 = 0.2(𝑥 − 2)(𝑥 + 2)2

𝑦 = − 2𝑥(𝑥 + 2)(𝑥 − 2)

𝑦 = − 1.3𝑥2(𝑥 + 2)(𝑥 − 2)

𝑦 = − 0.2(𝑥 + 4)(𝑥 + 2)2(𝑥 − 2) 3. Identify the equation that does not have a matching set of cards.

4. Rewrite the equation so that it matches the remaining graph and key feature cards.

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Lesson Summary Key features of polynomial functions can give clues about the graph and equation of the function. Key features can also be determined from the equation and the graph of a given function. The graph of function 𝑓(𝑥) = 2𝑥(𝑥 + 2)2(𝑥 − 2)3 is shown.

7 60 (−1, 54) 50 40 30 20 (−2, 0)

The table summarizes the key features of the function with explanations of how to determine them. Several of the key features are identified using both interval notation and set notation.

Key Feature

Zero(s)

𝑦-intercept Domain

𝑥 = 0

𝑥 = − 2 multiplicity of 2 𝑥 = 2 multiplicity of 3 (0, 0)

( − ∞, ∞) {𝑥|𝑥 ∋ ℝ}

Range

[− 22.475, ∞)

{𝑥|𝑥≥ − 22.475}

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‐3 ‐2 ‐1 0 ‐10 ‐20 ‐30

(0, 0) 0) (2, 0) x (0, 1 2 3 (0.667, −22.475)

Explanation Set each factor equal to 0, and solve for 𝑥. Note the multiplicity of each zero. Remember that 𝑥-intercepts may be referred to as zeros or roots. Evaluate the function when 𝑥 = 0. This is the point where the function crosses the 𝑦-axis. Determine the interval of all possible 𝑥-values for the function.

Determine the interval of all possible 𝑦-values for the function. Use the graph of the function to determine maximum or minimum values of the function.

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Key Feature Increasing interval(s)

Decreasing interval(s)

Positive interval(s)

Negative interval(s)

Left end behavior

Right end behavior

(−2, −1) ∪ (0.667, ∞)

{𝑥| −2 < 𝑥 < − 1 ∪ 𝑥 > 0.667} (−∞, −2) ∪ (−1, 0.667)

{𝑥 | 𝑥 < −2 ∪ −1 < 𝑥 < 0.667} (−∞, −2) ∪ (−2, 0) ∪ (2, ∞)

{𝑥 | 𝑥 < −2 ∪ −2 < 𝑥 < 0 ∪ 𝑥 > 2} (0, 2)

{𝑥 | 0 < 𝑥 < 2} As 𝑥→ − ∞, 𝑦→∞

As 𝑥→∞, 𝑦→∞

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Explanation Increasing intervals are intervals over which as the 𝑥-values increase, the 𝑦-values increase.

Decreasing intervals are intervals over which as the 𝑥-values increase, the 𝑦-values decrease. Relative maximum and minimum points help determine these intervals. Positive intervals are intervals of 𝑥-values over which the 𝑦-values of the function are positive. Negative intervals are intervals of 𝑥-values over which the 𝑦-values of the function are negative. Zeros of the function help determine these intervals. Left end behavior identifies what the 𝑦-values of the function approach as the 𝑥-values approach − ∞.

Right end behavior identifies what the 𝑦-values of the function approach as the 𝑥-values approach ∞.

End behavior is affected by the degree of the polynomial and the leading coefficient.

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Practice Problems 1. Consider the function 𝑤(𝑥) = 𝑥(𝑥 − 2)(𝑥 + 2). a. Identify the zeros of 𝑤(𝑥).

b. Identify the end behavior of 𝑤(𝑥). As 𝑥→_____, 𝑦→_____.

As 𝑥→_____, 𝑦→_____.

c. Determine the positive and negative intervals of the function.

2. The graph of ℎ(𝑥) = − 2(𝑥 + 3)2(𝑥 − 1)2 is shown. 1 5

(1, 0) x 0 ‐5 ‐4 ‐3 ‐2 ‐1 1 2 3 ‐5 ‐10 ‐15 (−2, −18) (0, −18) ‐20 ‐25 (−1, −32) ‐30 ‐35 (−3, 0)

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Complete the table of key features for the function. Key Features Increasing interval(s)

Zero(s)

Decreasing interval(s)

𝑦-intercept Domain

Positive interval(s)

Range

Negative interval(s)

Left end behavior

As 𝑥→_____, 𝑦→_____.

Right end behavior

As 𝑥→_____, 𝑦→_____.

Review Problem 3. State the degree and end behavior of 𝑓(𝑥) = 2𝑥3 − 3𝑥5 − 𝑥2 + 1. Explain or show your reasoning.

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Unit 2, Lesson 11: Polynomial Division – Part 1

Warm-Up: Notice and Wonder: A Different Use for Diagrams 1. What do you notice? What do you wonder? a. (𝑥 − 3)(𝑥 + 5) = 𝑥2 + 2𝑥 − 15 𝑥

𝑥

𝑥2

−3

b. (𝑥 − 1) (𝑥2 + 3𝑥 − 4) = 𝑥3 + 2𝑥2 − 7𝑥 + 4 𝑥

𝑥2

𝑥3

− 𝑥2

𝑥

𝑥3

−2

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5𝑥

− 3𝑥

−1

c. (𝑥 − 2)( ?) = (𝑥3 − 𝑥2 − 4𝑥 + 4)

− 15

3𝑥

3𝑥2

− 3𝑥

−4

− 4𝑥 +4

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Exploration Activity: Factoring with Diagrams Priya wants to sketch a graph of the polynomial 𝑓 defined by 𝑓(𝑥) = 𝑥3 + 5𝑥2 + 2𝑥 − 8. She knows 𝑓(1) = 0, so she suspects that (𝑥 − 1) could be a factor of 𝑥3 + 5𝑥2 + 2𝑥 − 8 and writes (𝑥3 + 5𝑥2 + 2𝑥 − 8) = (𝑥 − 1)(?𝑥2 + ?𝑥 + ?) and draws a diagram.

𝑥

1. Finish Priya’s diagram.

−1

𝑥3

2. Write 𝑓(𝑥) as the product of (𝑥 − 1) and another factor.

3. Write 𝑓(𝑥) as the product of three linear factors.

4. Make a sketch of 𝑦 = 𝑓(𝑥).

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Collaborative Activity: More Factoring with Diagrams Here are some polynomial functions with known factors. Rewrite each polynomial as a product of linear factors. Note: you may not need to use all the columns in each diagram. For some problems, you may need to make another diagram. 1. 𝐴(𝑥) = 𝑥3 − 7𝑥2 − 16𝑥 + 112, (𝑥 − 7)

𝑥

−7

𝑥2

𝑥3

− 7𝑥2

2. 𝐵(𝑥) = 2𝑥3 − 𝑥2 − 27𝑥 + 36, �𝑥 − 2 �

𝑥

− 2

2𝑥2

2𝑥3

− 3𝑥2

2𝑥2

3. 𝐶(𝑥) = 𝑥3 − 3𝑥2 − 13𝑥 + 15, (𝑥 + 3)

𝑥

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4. 𝐷(𝑥) = 𝑥4 − 13𝑥2 + 36, (𝑥 − 2), (𝑥 + 2)

(Hint: 𝑥4 − 13𝑥2 + 36 = 𝑥4 + 0𝑥3 − 13𝑥2 + 0𝑥 + 36)

5. 𝐹(𝑥) = 4𝑥4 − 15𝑥3 − 48𝑥2 + 109𝑥 + 30, (𝑥 − 5), (𝑥 − 2), (𝑥 + 3)

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Lesson Summary Factors and zeros of polynomials can be used to take polynomial functions given in standard form and rewrite them in factored form. For example, consider the polynomial 𝑝(𝑥) = 𝑥3 − 5𝑥2 − 2𝑥 + 24, where 𝑝 (− 2) = 0. If 𝑝( − 2) = 0, then (𝑥 + 2) is a factor of the polynomial. To factor the polynomial, one strategy is to use an area model. One of the dimensions of the rectangle is the known factor of (𝑥 + 2), and the other dimension is unknown. Since the degree of 𝑝(𝑥) is 3 and the known factor has a degree of 1, the remaining expression, 𝑞(𝑥), must be 2nd degree. A partially completed area model is shown.

𝑥

𝑥2 𝑥3

• To set up an area model, first fill in the known factor, (𝑥 + 2), as one dimension, and fill in the leading term of 𝑝(𝑥), 𝑥3, as one of the partial products. • From here, it can be concluded that the leading term of 𝑞(𝑥) must be 𝑥2.

• Complete the rest of the area model using similar thinking and paying close attention to the signs of each term. For example, 2𝑥2 belongs in the bottom left cell of the partial products because it’s the product of 2 and 𝑥2. However, since the 𝑥2 term in 𝑝(𝑥) is − 5, the remaining cell with 𝑥2, the middle cell of the middle row, must have a value of − 7𝑥2. The completed area model is shown.

𝑥

𝑥2 𝑥3

2𝑥2

− 7𝑥

− 7𝑥2

− 14𝑥

12𝑥 24

The area model shows that 𝑝(𝑥) = 𝑥3 − 5𝑥2 − 2𝑥 + 24 = (𝑥 + 2)(𝑥2 − 7𝑥 + 12). With a bit more factoring, it can be found that 𝑝(𝑥) = (𝑥 + 2)(𝑥 − 3)(𝑥 − 4).

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Practice Problems 1. The polynomial function 𝑝(𝑥) = 𝑥3 − 3𝑥2 − 10𝑥 + 24 has a known factor of (𝑥 − 4). a. Rewrite 𝑝(𝑥) as the product of linear factors.

b. Draw a rough sketch of the graph of the function.

2. Tyler thinks he knows one of the linear factors of 𝑃(𝑥) = 𝑥3 − 9𝑥2 + 23𝑥 − 15. After finding that 𝑃(1) = 0, he suspects that 𝑥 − 1 is a factor of 𝑃(𝑥). Here is the diagram he made to check if he’s right, but he set it up incorrectly. What went wrong?

𝑥

𝑥2

𝑥3

𝑥2

−8𝑥

−8𝑥2 −8𝑥

−15

−15𝑥 −15

3. The polynomial function 𝑞(𝑥) = 2𝑥4 − 9𝑥3 − 12𝑥2 + 29𝑥 + 30 has known factors (𝑥 − 2) and (𝑥 + 1). Which expression represents 𝑞(𝑥) as the product of linear factors? A. (2𝑥 − 5)(𝑥 + 3)(𝑥 − 2)(𝑥 + 1)

B. (2𝑥 + 3)(𝑥 − 5)(𝑥 − 2)(𝑥 + 1)

C. (2𝑥 + 15)(𝑥 − 1)(𝑥 − 2)(𝑥 + 1) D. (2𝑥 − 15)(𝑥 + 1)(𝑥 − 2)(𝑥 + 1)

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Review Problems 4. Each year a certain amount of money is deposited in an account which pays an annual interest rate of 𝑟 so that at the end of each year the balance in the account is multiplied by a growth factor of 𝑥 = 1 + 𝑟. $1,000 is deposited at the start of the first year, an additional $300 is deposited at the start of the next year, and $500 at the start of the following year. a. Write an expression for the value of the account at the end of three years in terms of the growth factor 𝑥. b. Determine (to the nearest cent) the amount in the account at the end of three years if the interest rate is 4%. 5. State the degree and end behavior of 𝑓(𝑥) = 5 + 7𝑥 − 9𝑥2 + 4𝑥3. Explain or show your reasoning.

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Unit 2, Lesson 12: Polynomial Division – Part 2

Warm-Up: Notice and Wonder: Different Divisions What do you notice? What do you wonder?

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Exploration Activity: Polynomial Long Division 1. Diego used the long division shown here to figure out that 6𝑥2 − 7𝑥 − 5 = (2𝑥 + 1)(3𝑥 − 5). Show what it would look like if he had used a diagram.

2𝑥 1

6𝑥2

Pause here for a whole-class discussion.

2. (𝑥 − 2) is a factor of 2𝑥3 − 7𝑥2 + 𝑥 + 10, which means there is some other factor 𝐴 where 2𝑥3 − 7𝑥2 + 𝑥 + 10 = (𝑥 − 2)(𝐴). Finish the division started here to find the value of 𝐴.

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3. Jada used the diagram shown here to figure out that 2𝑥3 + 13𝑥2 + 16𝑥 + 5 = (2𝑥 + 1)(𝑥2 + 6𝑥 + 5). Show what it would look like if she had used long division.

2𝑥 1

𝑥2

2𝑥3 𝑥2

6𝑥

12𝑥2 6𝑥

𝟓

10𝑥 5

Guided Activity: More Long Division Here are some polynomial functions with known factors. Rewrite each polynomial as a product of linear factors using long division. 1. 𝐴(𝑥) = 𝑥3 − 7𝑥2 − 16𝑥 + 112, (𝑥 − 7)

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2. 𝐶(𝑥) = 𝑥3 − 3𝑥2 − 13𝑥 + 15, (𝑥 + 3)

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Collaborative Activity: Missing Numbers Here are pairs of equivalent expressions, one in standard form and the other in factored form. Find the missing numbers. 1. 𝑥2 + 9𝑥 + 14 and (𝑥 + 2)(𝑥 + ________)

2. 𝑥2 − 9𝑥 + 20 and (𝑥 − ________)(𝑥 − ________)

3. 2𝑥2 + 2𝑥 − 24 and 2(𝑥 + ________)(𝑥 − 3)

4. ________𝑥3 + 11𝑥2− 17𝑥 + 6 and ( − 𝑥 + 3)(2𝑥 − 1)(𝑥 − 2) 5. 6𝑥3 + 2𝑥2 − 16𝑥 + 8 and (𝑥 − 1)(2𝑥 + 4)(________ 𝑥 − 2)

6. 2𝑥3 + 7𝑥2 − 7𝑥 − 12 and (2𝑥 − 3)(𝑥 + ________)(𝑥 + ________)

7. 𝑥3 + 6𝑥2 + ________ 𝑥 − 10 and (𝑥 + 2)(𝑥 − 1)(𝑥 + ________)

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Lesson Summary In earlier grades, you learned how to add, subtract, and multiply numbers. You also learned that one way to divide numbers, such as 1,573 divided by 11, is by using long division. The process for dividing 1,573 by 11 using long division is shown in 3 stages, leading to the quotient, 1,573÷11 = 143.

1 11 1573 −11

14 11 1573 −11

143 11 1573 −11

47 −44

33 −33 0

Polynomials form a system similar to integers. They can be added, subtracted, and multiplied, as explored in Algebra 1. To divide polynomials, long division can be applied. Organize the terms in the polynomial from highest degree to lowest degree, similar to how the digits in a whole-number dividend are organized by place value. The division of <APS_NEWLINE>𝑥<APS_NEWLINE>3<APS_NEWLINE> + 5<APS_ NEWLINE>𝑥<APS_NEWLINE>2<APS_NEWLINE> + 7𝑥 + 3 by 𝑥 + 1 is shown in 3 stages, like the previous example of 1,573÷11. Notice how the terms of the same degree are arranged in the same columns, similar to how digits with the same place value are organized in long division of whole numbers.

x2 x + 1 x3 + 5x2 + 7x + 3 −x3 − x2 4x2 + 7x

x2 + 4x x + 1 x3 + 5x2 + 7x + 3 −x3 − x2 4x2 + 7x −4x2 + 4x

3x + 3 At each stage, the focus is only on the term with the largest exponent that remains. At the conclusion of the division, (𝑥3 + 5𝑥2 + 7𝑥 + 3) ÷ (𝑥 + 1) = 𝑥2 + 4𝑥 + 3, which means 𝑥3 + 5𝑥2 + 7𝑥 + 3 = (𝑥 + 1)(𝑥2 + 4𝑥 + 3).

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Practice Problems 1. The polynomial function 𝐵(𝑥) = 𝑥3 − 21𝑥 + 20 has a known factor of (𝑥 − 4). Rewrite 𝐵(𝑥) as a product of linear factors.

2. Let the function 𝑃 be defined by 𝑃(𝑥) = 𝑥3 + 7𝑥2 − 26𝑥 − 72, where (𝑥 + 9) is a factor. To rewrite the function as the product of two factors, long division was used but an error was made. How can we tell by looking at the remainder that an error was made somewhere?

3. For the polynomial function 𝐴(𝑥) = 𝑥4 − 2𝑥3 − 21𝑥2 + 22𝑥 + 40 we know (𝑥 − 5) is a factor. Select all the other linear factors of 𝐴(𝑥).

□ (𝑥 + 1) □ (𝑥 − 1)

□ (𝑥 + 2) □ (𝑥 − 2)

□ (𝑥 + 4) □ (𝑥 − 4)

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Review Problems 4. Match the polynomial function with its constant term. A. 𝑃(𝑥) = (𝑥 − 2)(𝑥 − 3)(𝑥 + 7)

B. 𝑃(𝑥) = (𝑥 + 2)(𝑥 − 3)(𝑥 + 7) 1

−210

ii. −42

C. 𝑃(𝑥) = 2 (𝑥 − 2)(𝑥 − 3)(𝑥 + 7)

iii. 21

E. 𝑃(𝑥) = − 5(𝑥 − 2)(𝑥 − 3)(𝑥 + 7)

v. 210

D. 𝑃(𝑥) = 5(𝑥 − 2)(𝑥 − 3)(𝑥 + 7)

iv. 42

5. What are the solutions to the equation (𝑥 − 2)(𝑥 − 4) = 8?

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Unit 2, Lesson 13: Connecting Polynomial Division and Graphing

Warm-Up: What Else Is True? 𝐺(𝑥) is a polynomial. Here are some things we know about it: • It has degree 3.

• Both 𝑥 and (𝑥 + 4) are factors of 𝐺.

• It has 2 horizontal intercepts, but only 1 is negative. • Its leading coefficient is negative.

What else do we know is true about 𝐺(𝑥)?

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Exploration Activity: Info Gap: More Polynomials Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner. If your teacher gives you the data card: 1. Silently read the information on your card. 2. Ask your partner, “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!) 3. Before telling your partner the information, ask, “Why do you need to know (that piece of information)?” 4. Read the problem card, and solve the problem independently. 5. Share the data card, and discuss your reasoning.

If your teacher gives you the problem card: 1. Silently read your card and think about what information you need to answer the question. 2. Ask your partner for the specific information that you need. 3. Explain to your partner how you are using the information to solve the problem. 4. When you have enough information, share the problem card with your partner, and solve the problem independently. 5. Read the data card, and discuss your reasoning.

Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.

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Collaborative Activity: Even More Polynomials 1. Without letting your partner see, do the following: a. write a polynomial of degree 3 or 4 in factored form b. sketch the graph of your polynomial c. rewrite its expression in standard form

2. On a separate slip of paper, write the standard form of your polynomial along with 1 of the factors (or 2 factors, if the polynomial has degree 4). Trade slips with your partner.

3. Use the information your partner gave you about their polynomial to: a. rewrite their polynomial in factored form

b. sketch a graph of their polynomial showing all horizontal intercepts

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4. Once you and your partner have finished graphing, check your factored form and graph with your partner and discuss any differences.

Lesson Summary The equation of a polynomial can be represented in several different forms, including standard form or factored form. For example, the function 𝑃(𝑥) = 𝑥3 − 7𝑥 + 6 is written in standard form, but it can also be written in factored form as 𝑃(𝑥) = (𝑥 − 2)(𝑥 + 3)(𝑥 − 1). The graph of 𝑃(𝑥) is shown. Each representation of 𝑃(𝑥) provides different information about the polynomial function.

• For example, to quickly approximate the value of 𝑃(𝑥) for a specific value of 𝑥, the graph of the function may be most useful. • If the graph of 𝑃(𝑥) is not given or available, the factored form of the equation can be used to determine the zeros of the polynomial and to sketch a graph of 𝑃(𝑥). The standard form of 𝑃(𝑥) can be used to determine the end behavior and degree of the function. To determine the zeros of 𝑃(𝑥) when the function is given in standard form, strategies for dividing polynomials can be used to determine the factors of the function. To write a factored polynomial function in standard form, distribute the factors, and write the terms in descending degree order.

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Practice Problems 1. We know these things about a polynomial function, 𝑓(𝑥): it has exactly one relative maximum and one relative minimum, it has exactly three zeros, and it has a known factor of (𝑥 − 4). Sketch a graph of 𝑓(𝑥) given this information.

2. Mai graphs a polynomial function, 𝑓(𝑥), that has three linear factors (𝑥 + 6), (𝑥 + 2), and (𝑥 − 1). But she makes a mistake. What is her mistake?

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3. Here is the graph of a polynomial function with degree 4.

Select all of the statements that are true about the function.

□ The leading coefficient is positive. □ The constant term is negative. □ It has 2 relative maximums. □ It has 4 linear factors.

□ One of the factors is (𝑥 − 1). □ One of the zeros is 𝑥 = 2.

□ There is a relative minimum between 𝑥 = 1 and 𝑥 = 3. Review Problems 4. Kiran thinks he knows one of the linear factors of 𝑃(𝑥) = 𝑥3 + 𝑥2 − 17𝑥 + 15. After finding that 𝑃(3) = 0, Kiran suspects that 𝑥 − 3 is a factor of 𝑃(𝑥), so he sets up a diagram to check. Here is the diagram he made to check his reasoning, but he set it up incorrectly. What went wrong?

𝑥

𝑥2 𝑥3

3𝑥2

4𝑥

4𝑥2

12𝑥

5. The polynomial function 𝐵(𝑥) = 𝑥3 + 8𝑥2 + 5𝑥 − 14 has a known factor of (𝑥 + 2). Rewrite 𝐵(𝑥) as a product of linear factors.

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Unit 2, Lesson 14: The Remainder and Factor Theorems

Warm-Up: Notice and Wonder: Division Leftovers What do you notice? What do you wonder?

A. 330 = 33(10) + 0

B. 330 = 4(82) + 2

C. 330 = 5(66) + 0

Exploration Activity: The Unknown Coefficient 1. Consider the polynomial function 𝑓(𝑥) = 𝑥4 − 𝑢𝑥3 + 24𝑥2 − 32𝑥 + 16, where 𝑢 is an unknown real number. a. If 𝑥 − 2 is a factor of the polynomial, what is the value of 𝑢?

b. Discuss with your partner what the value of 𝑢 is, and explain how you determined the value. Summarize your discussion.

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2. Consider the third-degree polynomial 𝑃(𝑥) = 2𝑥3 + 5𝑥2 + 𝑥 + 10.

One diagram illustrating 𝑃(𝑥) divided by a linear factor and another diagram illustrating 𝑃(𝑥) divided by a quadratic factor are shown.

2x ²

𝑷(𝒙) 𝒙 + 𝟑

𝑷(𝒙) 𝒙𝟐 − 𝒙

–x

2x ³

–x ²

6x ²

–3x

x²

2x ³

7x ²

–x

–2x ²

–7x

a. What is the remainder of each of these divisions?

b. For each division, how does the degree of the remainder compare to the degree of the divisor?

c. Discuss with your partner whether the remainder could ever have the same degree of the divisor or a higher degree. Give an example to show that this is possible, or explain why it is not possible.

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Collaborative Activity: A Study of Remainders 1. Several polynomial functions are shown. 𝐴(𝑥) = 6𝑥2 − 7𝑥 − 5

𝐵(𝑥) = 3𝑥2 + 15𝑥 − 42

𝐶(𝑥) = 2(𝑥)3 + 13(𝑥)2 + 16𝑥 + 5

𝐷(𝑥) = 3(𝑥)3 − 2𝑥2 − 15𝑥 + 14

𝐸(𝑥) = 8(𝑥)4 − 41(𝑥)3 − 18(𝑥)2 + 101𝑥 + 70

𝐹(𝑥) = 𝑥4 + 5𝑥3 − 27𝑥2 − 101𝑥 − 70

Determine which of the polynomials could have (𝑥 − 2) as a factor. Could Have (𝒙 − 𝟐) as a Factor

Could Not Have 𝒙 − 𝟐 as a Factor

2. Select 1 of the polynomials that could not have (𝑥 − 2) as a factor. a. Explain why (𝑥 − 2) is not a factor.

b. Divide the polynomial by (𝑥 − 2). What is the remainder? 3. List the remainders for each of the polynomials when divided by (𝑥 − 2).

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4. How do these values compare to the value of the functions at 𝑥 = 2?

5. Find the remainder of the quotient of 𝑀(𝑥) and 𝑁(𝑥), where 𝑀(𝑥) = (𝑥)3 − 7𝑥 − 6 and 𝑁(𝑥) = 𝑥 − 4.

6. Evaluate 𝑀(4).

7. Discuss with your partner what connections there are between 𝑀(4) and the 𝑀(𝑥) remainder of 𝑁(𝑥) . Summarize your discussion.

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Guided Activity: The Remainder Theorem 1. Consider 𝑓(𝑥) = 2(𝑥)3 − 3(𝑥)2 − 59𝑥 + 30. a. Determine if (𝑥 − 5) is a factor.

b. Determine 𝑓(5).

c. Explain the relationship between parts A and B.

2. Select all of the polynomial expressions that have (𝑥 + 6) as a factor. o 2𝑥3 − 19𝑥2 − 7𝑥 + 294 o 4𝑥3 + 11𝑥2 − 75𝑥 + 18

o 3𝑥3 + 34(𝑥)2 + 61𝑥 − 210 o (𝑥)3 + 6(𝑥)2 + 11𝑥 + 6

o 10(𝑥)3 + 53(𝑥)2 − 41𝑥 + 6

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2. Let 𝑝 be a polynomial function defined by 𝑝(𝑥) = 2(𝑥)3 + 5(𝑥)2 − 6𝑥 − 9.

a. Use the information to rewrite the expression for 𝑝(𝑥) as the product of linear factors. 𝑝( − 4) = − 33

𝑝( − 3) = 0

𝑝( − 2) = 7

𝑝( − 1) = 0

b. With your partner, compare linear factors. If there are differences, verify together if there is more than 1 correct answer.

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Lesson Summary When using long division to divide 1,573 by 12, the quotient includes a remainder of 1, so 1,573 = 12(131) + 1. When 1,573 is divided by 11 instead, the quotient includes a remainder of 0, so 1,573 = 11(143). A remainder of 0 means that 11 is a factor of 1,573. The same is true for polynomials. While (𝑥)3 + 5(𝑥)2 + 7𝑥 + 3÷(𝑥 + 2) results in a remainder that is not 0, the same polynomial divided by (𝑥 + 1) results in a remainder of 0. This means that (𝑥 + 1) is a factor of (𝑥3 + 5(𝑥)2 + 7𝑥 + 3), while (𝑥 + 2) is not. The long division of (𝑥)3 + 5(𝑥)2 + 7𝑥 + 3) ÷ (𝑥 + 1) is shown.

• The factor theorem states that 𝑘 is a zero of 𝑓 if and only if (𝑥 − 𝑘) is a factor where 𝑓(𝑘) = 0. Since (𝑥 + 1) is a factor of 𝑓(𝑥) = (𝑥)3 + 5𝑥2 + 7𝑥 + 3, as shown by the long division, 𝑓(− 1) = 0. Therefore, the point (− 1, 0) is a point on the graph of 𝑓(𝑥). • The remainder theorem states that when the polynomial 𝑓(𝑥) is divided by (𝑥 − 𝑘), the remainder equals 𝑓(𝑘). If 𝑘 is not a zero of 𝑓, the remainder of the division is useful in helping to determine the point (𝑘, 𝑓(𝑘)), where the remainder of dividing 𝑓(𝑥) by (𝑥 − 𝑘) is the value of 𝑓(𝑘).

Practice Problems 1. For the polynomial function 𝑓(𝑥) = 𝑥3 − 2𝑥2 − 5𝑥 + 6, we have 𝑓(0) = 6,𝑓(2) = − 4, 𝑓(− 2) = 0,𝑓(3) = 0,𝑓(− 1) = 8,𝑓(1) = 0. Rewrite 𝑓(𝑥) as a product of linear factors. 2. Select all the polynomials that have (𝑥 − 4) as a factor. o 𝑥3 − 13𝑥 − 12

o 𝑥3 + 8𝑥2 + 19𝑥 + 12 o 𝑥3 + 6𝑥 + 5𝑥 − 12

o 𝑥3 − 𝑥2 − 10𝑥 − 8

o 𝑥2 − 4

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3. Write a polynomial function, 𝑝(𝑥), with degree 3 that has 𝑝(7) = 0.

4. Long division was used here to divide the polynomial function 𝑝(𝑥) = 𝑥3 + 7𝑥2 − 20𝑥 − 110 by (𝑥 − 5) and to divide it by (𝑥 + 5).

𝑥𝑥² + 12𝑥𝑥 + 40 𝑥𝑥 − 5 𝑥𝑥³ + 7𝑥𝑥² − 20𝑥𝑥 − 110 ―𝑥𝑥³ + 5𝑥𝑥² 12𝑥𝑥² − 20𝑥𝑥 ―12𝑥𝑥² + 60𝑥𝑥 40𝑥𝑥 − 110 ―40𝑥𝑥 + 200 90

𝑥𝑥² + 2𝑥𝑥 − 30 𝑥𝑥 + 5 𝑥𝑥³ + 7𝑥𝑥² − 20𝑥𝑥 − 110 ―𝑥𝑥³ − 5𝑥𝑥² 2𝑥𝑥² − 20𝑥𝑥 ―2𝑥𝑥² − 10𝑥𝑥 ―30𝑥𝑥 − 110 30𝑥𝑥 + 150 40

c. What is 𝑝( − 5)? d. What is 𝑝(5)? Review Problems 2

5. Which polynomial function has zeros when 𝑥 = 5, 3 , − 7? A. 𝑓(𝑥) = (𝑥 + 5) (2𝑥 + 3) (𝑥 − 7)

A. 𝑓(𝑥) = (𝑥 + 5) (3𝑥 + 2) (𝑥 − 7) B. 𝑓(𝑥) = (𝑥 − 5) (2𝑥 − 3) (𝑥 + 7) C. 𝑓(𝑥) = (𝑥 − 5) (3𝑥 − 2) (𝑥 + 7)

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6. The polynomial function 𝑞(𝑥) = 3𝑥4 + 8𝑥3 − 13𝑥2 − 22𝑥 + 24 has known factors (𝑥 + 3) and (𝑥 + 2). Rewrite 𝑞(𝑥) as the product of linear factors.

7. We know these things about a polynomial function 𝑓(𝑥): it has degree 3, the leading coefficient is negative, and it has zeros at 𝑥 = − 5, − 1, 3. Sketch a graph of 𝑓(𝑥) given this information.

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Unit 2, Lesson 15: The Difference of Squares

Warm-Up: Factoring Use the digits 1 to 6 no more than once each to fill in the blanks and make the mathematical statement true. 𝑥2 + _______𝑥 + ________ = (𝑥 + _______)(𝑥 + _______)

Exploration Activity: What Is a Difference of Squares? The table shows 2 algebra tile models and their corresponding expressions. 𝟒𝒙𝟐 − 𝟏𝟐𝒙 + 𝟗 x²

–x –x –x

x²

–x –x –x

–x

–x –x

x²

𝟒𝒙𝟐 − 𝟗 x²

x²

1 1 1

–x

–1 –1 –1

–x

1 1 1

–x

–1 –1 –1

–x

1 1 1

–x

–1 –1 –1

The quadratic expression 4𝑥2 − 12𝑥 + 9 is a perfect square trinomial.

1. Discuss with your partner why the quadratic expression 4𝑥2 − 9 is not a perfect square trinomial, even though the algebra tile model is a square. Summarize your discussion.

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2. Use the algebra tile models to rewrite each quadratic expression in factored form. 𝟒𝒙𝟐 − 𝟏𝟐𝒙 + 𝟗

𝟒𝒙𝟐 − 𝟗

3. What do you notice about the two factored forms?

4. The quadratic expression, 4𝑥2 − 9, is a difference of squares. Discuss with your partner why 4𝑥2 − 9 is called a “difference.”

5. The squares shown represent each term of the binomial 4𝑥2 − 9. 4𝑥2

What is the side length of each square? 𝟒𝒙𝟐

𝟗

Side Length = ________

The squares are a geometric representation of each term. The quadratic expression 4𝑥2 − 9 can be rewritten with each term as a monomial squared, as shown. (2𝑥)2 − (3)2

6. Discuss with your partner how the factored form of 4𝑥2 − 9 reflects the expression (2𝑥)2 − (3)2. Summarize your discussion.

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7. Use the table to determine if each expression is a difference of squares. a. Complete the table.

Quadratic Expression

− 100𝑔4 + 1 16𝑦2 − 100𝑦 − 𝑏2 + 25 25𝑤2 − 40𝑤 + 16 9𝑥2 − 4𝑦6 𝑞2 − 2𝑞𝑧 + 𝑧2 49𝑔2 − 81ℎ2

The Expression Is a Binomial that Can Be Written as a Difference

The Absolute Value of the First Term Is a Perfect Square

The Absolute Value of the Second Term Is a Perfect Square

Yes No

b. Circle the expressions in the table that are considered a difference of squares, based on your check marks.

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Guided Activity: Factoring a Difference of Squares 1. Consider the expression 49𝑔2 − 81ℎ2.

a. Rewrite 49𝑔2 − 81ℎ2 as a difference of two squares. (__________)2 − (__________)2

b. Factor 49𝑔2 − 81ℎ2.

2. Consider the expression 9𝑥2 − 4𝑦6.

a. Rewrite 9𝑥2 − 4𝑦6 as a difference of squares.

(__________)2 − (__________)2

b. Factor 9𝑥2 − 4𝑦6.

3. Both − 121𝑦2 + 81 and − 1 + 16𝑥2 are differences of squares. Discuss with your partner how to rewrite them as a difference of squares. Then, factor each. − 𝟏𝟐𝟏𝒚𝟐 + 𝟖𝟏

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− 𝟏 + 𝟏𝟔𝒙𝟐

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4. Consider the expression 2𝑥2 − 50𝑦2.

a. Rewrite 2𝑥2 − 50𝑦2 with a greatest common factor (GCF). __________ (__________ − __________)

b. Rewrite 2𝑥2 − 50𝑦2 as a difference of squares.

c. Factor 2𝑥2 − 50𝑦2.

_________ ((__________)2 − (__________)2)

Collaborative Activity: Factoring Algebraic Expressions Your teacher will give you a set of cards featuring algebraic expressions in standard and factored forms. Match each expression in standard form to its equivalent factored form. One expression is not factorable. Be prepared to explain your reasoning for the matches you create.

Lesson Summary In previous courses, you learned strategies for factoring algebraic expressions. Sometimes, the structure of the polynomial is helpful in determining how to factor it. • A difference of perfect squares is represented by a binomial of the form (𝑎)2 − (𝑏)2, which can be factored into the form (𝑎 + 𝑏)(𝑎 − 𝑏), where 𝑎 and 𝑏 can be monomial expressions or any nonzero values. • In some expressions, a greatest common factor (GCF) may need to be factored out to find a binomial in the form of a difference of perfect squares. For example, the expression 36(𝑥)5 − 100𝑥3 is not in the form of a difference of perfect squares. However, after factoring out the GCF of 4𝑥3, the expression is 4𝑥3(9𝑥2 − 25). In such a case, the remaining expression can be factored as a difference of perfect squares. The fully factored form of 36𝑥5 − 100𝑥3 is 4𝑥3(3𝑥 − 5)(3𝑥 + 5). © Accelerate Learning Inc. - All Rights Reserved

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Practice Problems 1. Factor each expression. a. 1 − 25𝑘2

b. 49𝑧4 − 16

2. Factor each expression. a. 48𝑥4 − 27𝑥2

b. 144𝑥 − 25𝑥3

3. Select all of the expressions that are equivalent to 12𝑥3 − 108𝑥.

□ 12𝑥 − 9𝑥 3

□ 12𝑥(𝑥 − 3) □ 3𝑥(2𝑥 + 6)

2 2

□ 3(4𝑥 − 6)(4𝑥 + 6) □ 12𝑥(𝑥 − 3)(𝑥 + 3)

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4. Carolina factored the expression 12𝑥6 − 75, but she made an error. Her work is shown. 12𝑥6 − 75 = 3(4𝑥2 − 25) = 3(2𝑥 + 5)(2𝑥 − 5)

Explain the error Carolina made in her work.

Review Problems 5. Describe the end behavior of 𝑓(𝑥) = 1 + 7𝑥 + 9(𝑥)3 + 6𝑥4 − 2𝑥5.

6. Which polynomial function’s graph is shown? A. 𝑓(𝑥) = (𝑥 + 1)(𝑥 + 2)(𝑥 + 4) B. 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 2)(𝑥 + 4) C. 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 2)(𝑥 − 4) D. 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 4)

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Unit 2, Lesson 16: Factoring Sums and Differences of Cubes

Warm-Up: Which One Doesn’t Belong? 1. Which one doesn’t belong? Be prepared to explain your reasoning. A. 𝑥2 − 81

B. 3𝑥2 − 9

C. 4𝑥2 + 5

D. 𝑥2 + 5𝑥 + 4

Exploration Activity: Factoring a Sum of 2 Cubes Consider 2 cubes. The first cube has side lengths that measure 𝑎 units, and the second cube has side lengths that measure 𝑏 units.

The volume of the first cube is 𝑉 = 𝑎 ∙ 𝑎 ∙ 𝑎 = 𝑎3, and the volume of the second cube is 𝑉 = 𝑏 ∙ 𝑏 ∙ 𝑏 = 𝑏3.

The sum of the volumes of the cubes is 𝑎3 + 𝑏3. Enclosing the space around cube 𝑏 helps to determine the factored form of 𝑎3 + 𝑏3.

1. The volume of the new prism is

¡ 𝑎2 (𝑎 + 𝑏).

¡ 𝑎2 (𝑎 − 𝑏).

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The top of the prism can be decomposed to find the volume of the empty space outside the 2 given cubes.

Prism 2

Prism 1

2. Determine the dimensions and volume of prism 1 and prism 2. A list of possible dimensions is provided.

Possible Dimensions 𝑎

𝑏

______, ______, and _________

Prism

𝑎2

Dimensions

𝑏2

(𝑎 + 𝑏)

Volume

(𝑎 − 𝑏)

¡ 𝑏2(𝑎 − 𝑏) ¡ 𝑎𝑏(𝑎 − 𝑏) ¡ 𝑏2(𝑎 − 𝑏) ¡ 𝑎𝑏(𝑎 − 𝑏)

3. Complete the steps to find the factored form of 𝑎3 + 𝑏3. Step 1

Expression

Write an expression representing the combined volumes of prism 1 and prism 2. Factor out (𝑎 − 𝑏) from both expressions.

Factor out the GCF from the second binomial.

Subtract the combined volume of prisms 1 and 2 from the volume of the entire rectangular prism found in problem 1.

(𝑎 − 𝑏)(________________) ______(𝑎 − 𝑏)(____________) 𝑎2 (𝑎 + 𝑏) − 𝑏(𝑎 − 𝑏)(𝑎 + 𝑏)

Factor out (𝑎 + 𝑏) from both expressions.

(𝑎 + 𝑏)(_____________________________)

Rewrite the expression after distribution.

(𝑎 + 𝑏)(_________________________)

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Exploration Activity: Factoring a Difference of 2 Cubes Consider the same 2 cubes with side lengths that measure 𝑎 units and 𝑏 units, respectively.

The difference of the volumes of the 2 cubes is 𝑎3 − 𝑏3. The prism can be decomposed to find the total volume of 𝑎3 − 𝑏3 in factored form. Each prism in the decomposition is labeled as prism 1, prism 2, or prism 3.

1. Work with your partner to determine the volume of each prism. Complete the statements.

a. The volume of prism 1 is

b. The volume of prism 2 is

c. The volume of prism 3 is

¡ 𝑏2 (𝑎 − 𝑏). ¡ 𝑎2 (𝑎 − 𝑏). ¡ 𝑎𝑏 (𝑎 − 𝑏). ¡ 𝑏2 (𝑎 − 𝑏). ¡ 𝑎2 (𝑎 − 𝑏). ¡ 𝑎𝑏 (𝑎 − 𝑏). ¡ 𝑏2 (𝑎 − 𝑏). ¡ 𝑎2 (𝑎 − 𝑏). ¡ 𝑎𝑏 (𝑎 − 𝑏).

2. Use the volumes of prisms 1–3 to rewrite 𝑎3 − 𝑏3, the volume of the entire prism, in factored form. a. First, write an expression for the sum of the volumes of prisms 1–3. This expression represents the volume of the entire prism. b. Determine the GCF, and then factor the expression completely.

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Guided Activity: Factoring a Sum or Difference of 2 Cubes 1. Completely factor each expression. a. 𝑥3 + 216

b. 27𝑎3 − 64

c. 2𝑐3 − 250

d. 4𝑟3 + 108

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Lesson Summary Some algebraic expressions can be factored using existing patterns, such as perfect squares and perfect cubes. A summary of some of the factoring you have seen over the past 2 lessons is shown in the table. Difference of perfect squares Sum of 2 cubes Difference of 2 cubes

𝑎2 − 𝑏2 = (𝑎 − 𝑏) (𝑎 + 𝑏)

𝑎3 + 𝑏3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏2)

𝑎3 − 𝑏3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏2)

In many cases in this course and future courses, multiple factoring strategies may need to be applied to an expression to factor it completely. For example, consider the expression 3𝑥6 − 192.

• To completely factor the expression 3𝑥6 − 192, first factor out the GCF to rewrite the expression as 3(𝑥6 − 64).

• The second factor in the resulting expression is a difference of 2 cubes, so the expression can be rewritten as 3(𝑥2 − 4)(𝑥4 + 4𝑥2 + 16). • The second factor in this expression can be factored as a difference of perfect squares, resulting in 3𝑥6 − 192 = 3(𝑥 + 2)(𝑥 − 2)(𝑥4 + 4𝑥2 + 16).

Practice Problems 1. Which of these expressions is equivalent to 𝑥6 + 27? A. (𝑥2 − 3) (𝑥2 + 3𝑥 + 9)

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2. Completely factor each expression. a. 𝑥3 − 1000 b. 64𝑎3 + 343 3. Completely factor each expression. a. 2𝑑4 − 1024𝑑 b. 3𝑐3 + 24 Review Problems 4. A polynomial function, 𝑓(𝑥), has 𝑥-intercepts at ( − 6, 0) and (2, 0). What is one possible factor of 𝑓(𝑥)?

5. What are the 𝑥-intercepts of the graph of 𝑦 = (5𝑥 + 7)(2𝑥 − 1)(𝑥 − 4)? A. − 7 , − 1 ,4 5

B. 5 , 1 ,4 7

C. − 7 , 1 ,4 5

D. 5 ,2, 4

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Unit 2, Lesson 17: Relating Perfect Squares and Cubes to Graphs of Polynomials

Warm-Up: Let’s Find Some Differences 1. Calculate the following differences: a. 302 − 292

b. 412 − 402

c. 182 − 172

2. What do you notice about these calculations?

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Guided Activity: Graphing Perfect Squares and Cubes y

1. Polynomial function 𝑝(𝑥) is defined by the equation 𝑝(𝑥) = 𝑥2 − 16. The graph of the parent function 𝑓(𝑥) = 𝑥2 is shown on the coordinate plane. a. Describe a sequence of transformations that can be applied to the parent function to create 𝑝(𝑥).

b. What information about the graph of 𝑝(𝑥) can be determined from the equation?

c. Use what you know about a difference of perfect squares to rewrite 𝑝(𝑥) in factored form.

d. What information about the graph of 𝑝(𝑥) can be determined from the factored form?

e. Complete the key features table for 𝑝(𝑥) = 𝑥2 − 16. Key Features

Sign of leading coefficient

𝑥-intercept(s)

Left end behavior

Degree

Right end behavior

𝑦-intercept

As 𝑥→ − ∞, 𝑦→______. As 𝑥→∞, 𝑦→______.

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2. Polynomial function 𝑔(𝑥) is defined by the equation 𝑔(𝑥) = (𝑥 − 2)(𝑥2 + 2𝑥 + 4). a. Write the equation of the parent function 𝑓(𝑥).

b. Use the factorization of squares and cubes to rewrite the equation of 𝑔(𝑥). c. Write the sequence of transformation(s) applied to the parent function 𝑓(𝑥) to create 𝑔(𝑥). d. Complete the key features table for 𝑔(𝑥). Key Features Sign of leading coefficient

𝑥-intercept(s)

Left end behavior

Degree

Right end behavior

𝑦-intercept

As 𝑥→ − ∞, 𝑦→______. As 𝑥→∞, 𝑦→______. y

e. Graph 𝑔(𝑥) on the coordinate plane.

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Collaborative Activity: Graphs of Polynomial Functions 1. Consider the polynomial function 𝑟(𝑥) = 27 − 𝑥3.

a. Identify the degree and sign of the leading coefficient of 𝑟(𝑥). Degree: _______

Leading coefficient: _______

b. Describe the transformations applied to the parent function 𝑓(𝑥) = 𝑥3 to create 𝑟(𝑥). c. Determine the 𝑥-intercept(s) of 𝑟(𝑥). d. Graph 𝑟(𝑥) on the coordinate plane.

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2. The graphs of 4 functions are shown. Graph A

Graph B y

Graph C

Graph D

x 0

a. Complete the table by rewriting each given equation in the form indicated and matching it to the graph of the function. Standard Form

𝑦 = 16 − 𝑥4 𝑦 = 𝑥3 − 64

Factored Form

Graph

𝑦 = (2𝑥 + 3)(2𝑥 − 3)

𝑦 = (𝑥 + 1)(𝑥2 − 𝑥 + 1)

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b. Explain how you determined which graphs matched the equations in part A.

c. Compare your answers to your partner’s, and adjust your responses as needed.

Lesson Summary Each form of the equation of a polynomial function provides information about what the graph of the function may look like. • The standard form of a polynomial function reveals the degree, end behavior, and 𝑦-intercept of the function, where the 𝑦-intercept is the constant value in the polynomial equation. • The factored form of a polynomial function reveals the zero(s), or 𝑥-intercept(s), of the graph and can be used to determine intervals over which the function is positive or negative. When polynomial functions have a degree higher than 2, factoring becomes more complex. Some ways to factor such functions include using the greatest common factor, an area model, long division, and rules of factoring perfect squares and perfect cubes. In many cases, a polynomial function may be factored using multiple strategies. For example, to graph the polynomial function 𝑦 = (𝑥 − 4)(𝑥2 + 4𝑥 + 16), identify any key features that can be determined from the factored form; in this case, one such feature is an 𝑥-intercept at (4, 0). Use the rules of factoring perfect cubes to rewrite the polynomial function as 𝑦 = 𝑥3 − 43, or 𝑦 = 𝑥3 − 64. The standard form of the function reveals that the graph is the parent function 𝑦 = 𝑥3 shifted down 64 units, such that the 𝑦-intercept is (0, − 64).

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Practice Problems 1. Consider the function 𝑔(𝑥) = 81 − 16𝑥4.

a. Describe the sequence of transformations applied to the parent function 𝑓(𝑥) = 𝑥4 to create 𝑔(𝑥).

b. Complete the key features table for 𝑔(𝑥).

Key Features Sign of leading coefficient

𝑥-intercept(s)

Left end behavior

Degree

Right end behavior

𝑦-intercept

2. Consider the function ℎ(𝑥) = 27 − 8𝑥3.

As 𝑥→ − ∞, 𝑦→______. As 𝑥→∞, 𝑦→______. y

a. Describe the sequence of transformations applied to the parent function 𝑓(𝑥) = 𝑥3 to create ℎ(𝑥).

b. Graph the function ℎ(𝑥).

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Review Problems 3. For the polynomial function 𝐴(𝑥) = 2𝑥3 + 5𝑥2 − 28𝑥 − 15, we know (𝑥 + 5) is a factor. Rewrite 𝐴(𝑥) as a product of linear factors.

4. Which point is a relative minimum? A. A B. B C. C D. D

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Unit 3: Complex Numbers and Polynomials

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Unit 3, Lesson 1: A New Kind of Number

Warm-Up: Numbers Are Inventions Jada was helping her cousin with his math homework. He was supposed to solve the equation 8 + 𝑥 = 5. He said, “If I subtract 8 from both sides, I get 𝑥 = 5 − 8. This doesn’t make sense. You can’t subtract a bigger number from a smaller number. If I have 5 grapes, I can’t eat 8 of them!” What do you think Jada could say to her cousin to help him understand why 5 − 8 actually does make sense?

Collaborative Activity: The Square Root of Negative One Numbers on the number line are often called real numbers.

1. The equation 𝑥2 = 9 has 2 real solutions. How can you see this on the graph of 𝑦 = 𝑥2? Draw points on this real number line to represent these 2 solutions.

2. How many real solutions does 𝑥2 = 0 have? Explain how you can see this on the graph of 𝑦 = 𝑥2. Draw the solution(s) on a real number line. 3. How many real solutions does 𝑥2 = −1 have? Explain how you can see this on the graph of 𝑦 = 𝑥2. Draw the solution(s) on a real number line.

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Exploration Activity: The Imaginary Number, 𝑖

The square of a number, 𝑛, is the number that can be multiplied by itself to result in 𝑛. ─, for any value such that 𝑛 < 0, is not a real 1. Discuss with your partner why �𝑛 number. Summarize your discussion.

Outside of the real number system exists the imaginary number, 𝑖, defined as �–1.

2. Given 𝑖 = �–1, complete the table. Expression

Expanded Form

Simplified

𝑖1

�–1

𝑖

𝑖2

𝑖3

𝑖4 𝑖5

(�–1)2

−1

𝑖2 ∙ 𝑖2 = −1 ∙ −1

𝑖2 ∙ 𝑖 = −1 ∙ 𝑖

−𝑖

𝑖6 𝑖7 𝑖8

3. What patterns do you notice in the table?

4. How can the patterns be used to find an equivalent value of 𝑖𝑛?

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5. Evaluate the values in the table.

𝑖15 =

𝑖56 =

𝑖24 =

𝑖74 =

𝑖41 =

𝑖101 =

6. Anju and Rupa each evaluated 𝑖40. Their work is shown. Anju

I know that 𝑖4 = 1. Since 40 = 10, this 4 represents the value of the exponent. If I multiply 1 by itself 10 times, I get 1. Therefore, 𝑖40 = 1. Explain who is correct.

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Rupa

(�–1)40 = (�–1)20 (�–1)2

(�–1)20 = 1 ; (�–1)2 = −1 (1 )(−1) = −1 𝑖40 = −1

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Lesson Summary Sometimes the number line is referred to as the real number line.

All real numbers are either positive, negative, or 0, and they can be plotted on this line. Real numbers are the set of all rational and irrational numbers.

All the numbers you have used in math class before this lesson have been real numbers. Several facts that are true for all real numbers are shown. • A positive number multiplied by a positive number is always positive. So, when positive number is squared, the result will always be positive. • A negative number multiplied by a negative number is always positive. So, when a negative number is squared, the result will always be positive. •

0 squared is 0.

Squaring a real number never results in a negative number. Therefore, it can be concluded that the equation 𝑥2 = −1 does not have any real number solutions. In other words, none of the numbers on the real number line can satisfy this equation and make it true. Mathematicians developed a new number that cannot be plotted on the real number line. This new number comes from the solution to the equation 𝑥2 = −1. This new number, �–1, is a solution to the equation 𝑥2 = −1, so (�–1)2 = −1. All numbers including �–1 are called imaginary numbers. An imaginary number is a number on the imaginary number line. It can be written as 𝑏𝑖, where 𝑏 is a real number and 𝑖2 = −1.

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Practice Problems 1. Select all the true statements.

□ �–1 is an imaginary number.

□ There are no real numbers that satisfy the equation 𝑥 = �–1. □ Because �–1 is imaginary, no one does math with it. □ The equation 𝑥2 = −1 has real solutions. □ �–1 = −1 because −1 ⋅ −1 = −1.

2. Plot each number on the real number line, or explain why the number is not on the real number line.

a. �4

b. −�4 c. �–4

d. �8

e. −�8 f. �–8

3. Explain why (𝑥 − 4)2 = −9 has no real solutions.

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Review Problems 4. The graph of a polynomial function 𝑓 is shown. Select all the true statements about the polynomial.

□ The degree of the polynomial is even. □ The degree of the polynomial is odd. □ The leading coefficient is positive. □ The leading coefficient is negative. □ The constant term of the polynomial is positive.

□ The constant term of the polynomial is negative.

5. Which term can be added to the polynomial expression 5𝑥7 − 6𝑥6 + 4𝑥4 − 4𝑥2 to make it into a 10th degree polynomial? A. 10

B. 5𝑥3

C. 5𝑥7 D. 𝑥10

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Unit 3, Lesson 2: Introducing the Number 𝑖 Warm-Up: Math Talk: Squared 1. Find the value of each expression mentally. (2�3)2 2

�1 �3� 2

(2�–1)2 2

�1 �–1� 2

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Guided Activity: Simplifying Square Roots Using Imaginary Numbers 1. Simplify �50 within the real number system. 2. Simplify �−50 within the real number system. Taking the square root of a negative value results in no real solution. Imaginary numbers allow for negative values under a square root to be simplified within the complex number system. 3. Complete the final step in simplifying �−50 within the complex number system. √-50 = √ -1 √ 50 = i √50 =

i√

4. Simplify each square root within the complex number system. a. �−16

b. �−242

c. �−27

d. −�−96

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Collaborative Activity: An Imaginary Puzzle Complete the table by simplifying the expressions within the complex number system. Match the fun facts to the correct state using the table on the next page. Expression 1. 𝑖38 2. �−36 3. �−20 4. 𝑖27 5. −�−100 6. �−1331 7. 𝑖149 8. −�−153

Simplified Expression

Fun Fact

State

This state has the most roads, totaling to 679,917 miles of lanes.

This state has the longest coastline, measuring 6,640 miles.

This state has the longest Main Street, measuring 33 miles. This state has only 2 escalators in the entire state. This state has the highest mean elevation of 6,800 feet.

This state has the lowest mean elevation of 60 feet.

This state has the most miles of interstate per square miles. This state has the largest mall in the US, occupying 4.87 million square feet.

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Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri 284 | Unit 3

6𝑖

−6𝑖

Montana Nebraska Nevada

10𝑖

New Hampshire

−10𝑖

New Mexico

11𝑖�11

North Carolina

2�−5

Ohio

2𝑖�5

Oregon

3𝑖�17

New Jersey

5𝑖

New York

10�𝑖

−10𝑖�10

North Dakota

Oklahoma

−3�17𝑖

Pennsylvania

11�11𝑖

South Carolina

𝑖

−3�17 36𝑖

Rhode Island

South Dakota Tennessee

100𝑖

Texas

2𝑖

Vermont

−17𝑖�3 −2𝑖

Utah

Virginia

11�−11

Washington

5𝑖�2

Wisconsin

−3𝑖�17 10

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West Virginia

Wyoming

𝑖�11 𝑖�5

4𝑖�5 3𝑖

11𝑖

2𝑖�10 9𝑖�17

10𝑖�2

2𝑖�10

121𝑖�11 2�5𝑖

−2𝑖�10 −100𝑖

−3�−17 6

10𝑖�10 −10�𝑖 −1

−9𝑖�17 −2�5𝑖

−11�11𝑖 3�17 −36𝑖

3�17𝑖 −𝑖

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Lesson Summary Outside of the real number system exists the imaginary number, 𝑖, defined as �−1.

The solutions to the equation 𝑥2 = 𝑎 could be real or imaginary numbers. Every positive real number has 2 real square roots. For example, the solutions to the equation 𝑥2 = 35 are �35 and −�35 because those are the 2 numbers that square to make 35. Remember, while typically the � symbol is defined to indicate the principal square root, there are actually 2 square roots of any positive real number, 1 positive and 1 negative. The principal square root is the positive square root of a positive real number.

For this reason, equations in the form 𝑥2 = 𝑝 have 2 solutions, ±�𝑝. Using the previous example, the solutions to 𝑥2 = 35 are ±�35 because (�35)2 = 35 and −�352 = 35.

Similarly, every negative real number has 2 imaginary square roots. The 2 square roots of −1 are written as 𝑖 and −𝑖. That means that 𝑖2 = −1 and (−𝑖)2 = −1.

Practice Problems 1. Write these imaginary numbers using the number 𝑖. a. �−36

b. �−10 c. −�−100 d. −�−17

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2. Match each expression to an equivalent expression. A. 2𝑖 ∙ 8 B. 16𝑖3 C. (2𝑖)4 D. 2𝑖 ⋅ 8𝑖

1. 2. 3. 4.

−16 16 −16𝑖 16𝑖

3. a. Diego squared a number and got 4. Andre squared a different number and got 4. What were the numbers that Diego and Andre squared?

b. Jada squared a number and got −4. Elena squared a different number and got −4. What were the numbers that Jada and Elena squared? Review Problems 4. Find all solutions to each equation. a. 𝑎2 = 1

b. 𝑏2 = 13

c. 𝑐2 = −9

d. 𝑑2 = −5 5. Explain how you know that �−1 is not a negative number.

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Unit 3, Lesson 3: Adding and Subtracting Complex Numbers

Warm-Up: Math Talk: Telescoping Sums 1. Find the value of these expressions mentally. 2 − 2 + 20 − 20 + 200 − 200 100 − 50 + 10 − 10 + 50 − 100 3+2+1+0−1−2−3 1 + 2 + 4 + 8 + 16 + 32 − 16 − 8 − 4 − 2 − 1

Guided Activity: The Complex Number System A complex number is a number that can be written in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. In the form 𝑎 + 𝑏𝑖, 𝑎 is considered the real part of the complex number, and 𝑏𝑖 is the imaginary part of the complex number. Complex numbers where 𝑎 = 0 are often referred to as pure imaginary numbers. 1. A list of numbers is shown. 3,

∛3 , 2

−2𝑖,

4 + �6𝑖,

−1 , 6

𝜋 − 𝑖,

1 (1 − �−4), 2

Sort the numbers into the categories shown in the table.

Real Numbers (𝒃 = 𝟎)

∛−8,

5𝑖2, 𝑖

Complex Numbers (𝒂 + 𝒃𝒊)

Pure Imaginary Numbers Other Complex Numbers (𝒂 = 𝟎, 𝒃 ≠ 𝟎) (𝒂 ≠ 𝟎, 𝒃 ≠ 𝟎)

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The number system includes categories, or subsets, of numbers. With the introduction of additional number types, the system grows to represent the unique characteristics of each number. A Venn diagram representing the relationship between different number types within the complex number system is shown. Complex Numbers Real Numbers Rational Numbers

Irrational Numbers

Integers Whole Numbers Natural Numbers

Pure Imaginary Numbers

2. Classify each real number expression by writing them in the appropriate sections of the Venn diagram. �49, 2 , −2𝜋, −8 , 3

√70 , 2

4 − �4,

1.03

3. Two new subsets of numbers have been introduced in this unit. Discuss with your partner whether an imaginary number is also a complex number.

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Exploration Activity: Adding and Subtracting Complex Numbers 1. For each complex number, place a box around the real part, and circle the imaginary part. 5 + 2𝑖 2 − 3𝑖 −5𝑖 + 23 10 + 2𝑖 2. Louis was asked to add and subtract 2 complex numbers, 2 + 5𝑖 and 3 − 8𝑖. His work is shown. Add Subtract

(2 + 5𝑖) + (3 − 8𝑖) = 5 − 3𝑖

(2 + 5𝑖) − (3 − 8𝑖) = 2 + 5𝑖 − 3 + 8𝑖 = −1 + 13𝑖

With your partner, explain what Louis did to add and subtract the 2 expressions.

3. Perform the given operation for the complex numbers. a. (3 + 5𝑖) + (7 − 2𝑖) b. (3 − 8𝑖) − (6 + 12𝑖)

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Collaborative Activity: Adding and Subtracting Complex Numbers Maze 1. Add or subtract each complex expression. Shade in the correct path to complete the maze from start to end. Start (−3 + 2i) + (4 − 5i)

(3 − 4i) − (−2 + 9i)

(7 + 9i) − (5 − 3i)

2 + 6i

26 + 4i

5 − 13i

5i + 1

12 + 6i

1 + 7i

1 − 3i

(37 − 5i) + (11 + 9i)

−6 + 14i

21 + 2i

48 − 4i

−10 − 16i

(−8 − i) − (2 − 15i)

12 + 12i

48 + 4i

(22 − i) + (−1 + i)

−10 + 14i

End

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Lesson Summary When a real number and an imaginary number are combined, the result is a complex number. A complex number is a number in the complex plane. It can be written as 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers and 𝑖2 = −1. A complex number has 2 parts: the real part, 𝑎, and the imaginary part, 𝑏𝑖. The number system can be organized using a Venn diagram as shown in the lesson or as a hierarchy where each subset of numbers is shown below the broader number type. The complex number system’s hierarchy is shown in the image. Notice that real numbers and pure imaginary numbers are both subsets of complex numbers.

Complex numbers Pure imaginary numbers

Real numbers

Rational numbers

Ratio of integers

• Real numbers are complex numbers where 𝑏 = 0.

• Pure imaginary numbers are complex numbers where 𝑎 = 0.

Irrational numbers

Integers

Negative integers

Whole numbers

Zero

Natural numbers

The process of adding and subtracting complex numbers is similar to that of adding and subtracting polynomials. With polynomials, like terms are combined. With complex numbers, like parts are combined. The sum and difference of the complex numbers 2 + 3𝑖 and 4 + 5𝑖 are shown. (2 + 3𝑖) + (4 + 5𝑖) = (2 + 4) + (3𝑖 + 5𝑖) = 6 + 8𝑖

(2 + 3𝑖) − (4 + 5𝑖) = (2 − 4) + (3𝑖 − 5𝑖) = −2 − 2𝑖

In general, the sum (or difference) of complex numbers results in the sum (or difference) of the coefficients of the real parts and the coefficients of the imaginary parts of the complex numbers. (𝑎 + 𝑏𝑖) + (𝑐 + 𝑑𝑖) = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖

(𝑎 + 𝑏𝑖) − (𝑐 + 𝑑𝑖) = (𝑎 − 𝑐) + (𝑏 − 𝑑)𝑖

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Practice Problems 1. Which expression is equivalent to (3 + 9𝑖) − (5 − 3𝑖)? A. −2 − 12𝑖 B. −2 + 12𝑖 C. 15 + 27𝑖 D. 15 − 27𝑖 2. What are 𝑎 and 𝑏 when you write �−16 in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers? A. 𝑎 = 0, 𝑏 = −4 B. 𝑎 = 0, 𝑏 = 4

C. 𝑎 = −4, 𝑏 = 0 D. 𝑎 = 4, 𝑏 = 0

3. Fill in the boxes to make a true statement: (

− 3𝑖) − (15 +

𝑖) = 7 − 12𝑖

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Review Problems 4. Plot each number on the real number line, or explain why the number is not on the real number line.

a. �16

b. −�16 c. �−16

d. 561 2

e. −561 2

f. (−56)2 5. Which expression is equivalent to �−4? A. −2𝑖 B. −4𝑖 C. 2𝑖 D. 4𝑖

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Unit 3, Lesson 4: Multiplying Complex Numbers

Warm-Up: 𝑖 Squared

Write each expression in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. 1. 4𝑖 ⋅ 3𝑖

2. 4𝑖 ⋅ −3𝑖 3. −2𝑖 ⋅ −5𝑖 4. −5𝑖 ⋅ 5𝑖 5. (−5𝑖)2

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Collaborative Activity: Equivalent Expressions with Imaginary Numbers 1. Take turns with your partner to match an expression in column A with an equivalent expression in column B. Some expressions in column B may not be used, and some may be used more than once. a. Draw an arrow to show each match. b. For each match that you make, explain to your partner how you know it is a match. c. For each match that your partner finds, listen carefully to their explanation. If you disagree, discuss your thinking, and work to reach an agreement.

5 ∙ 7𝑖

−9

5𝑖 ∙ 7𝑖 3𝑖2

−35

8𝑖3

(3𝑖)2

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𝑖4

−3

(−𝑖)2

−8𝑖

−𝑖2

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Guided Activity: Multiplying Complex Numbers 1. Multiply the complex numbers by following the steps described. Original Problem

(2 + 3𝑖)(4 + 5𝑖)

Apply Distributive Property

8 + 10𝑖 + 12𝑖 + 15𝑖2

Combine Like Terms Simplify

2. Determine the product of each pair of factors in the two-way table. Check the box in the cells where multiplying the factors results in a real number product.

𝟗 − 𝟑𝒊

𝟗 + 𝟑𝒊

𝟕𝒊

−𝟑

𝟕𝒊

−𝟑 3. Determine each product. Write the answer in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. a. (−3 + 9𝑖)(5𝑖) b. (8 + 𝑖)(−5 + 3𝑖) c. (3 + 2𝑖)2 d. (3 + 2𝑖)(3 − 2𝑖) © Accelerate Learning Inc. - All Rights Reserved

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Lesson Summary 4𝑖 ⋅ 4𝑖 ⋅ 4𝑖 When an imaginary number is raised to a power, 2 use the fact that 𝑖 = −1 to write the result in the form 𝑎 + 𝑏𝑖. For example, (4𝑖)3 = 4𝑖 ⋅ 4𝑖 ⋅ 4𝑖. To rewrite expressions with powers of 𝑖, group the 𝑖 factors, and use laws of exponents to rewrite in simplest possible form as shown.

= = = =

(4 ⋅ 4 ⋅ 4) ⋅ (𝑖 ⋅ 𝑖 ⋅ 𝑖) 64 ⋅ (𝑖2 ⋅ 𝑖) 64 ⋅ −1 ⋅ 𝑖 −64𝑖

The complex number system follows the same properties of operations as the real number system, including associative, commutative, and distributive properties. The distributive property can be applied to multiply complex numbers. To multiply complex numbers, use the distributive property, and apply the property that 𝑖2 = −1 to rewrite the product in 𝑎 + 𝑏𝑖 form. The multiplication of (2 + 3𝑖)(4 + 5𝑖) is shown. (2 + 3𝑖)(4 + 5𝑖)

= = =

8 + 10𝑖 + 12𝑖 + 15𝑖2 8 + 10𝑖 + 12𝑖 – 15 −7 + 22𝑖

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Practice Problems 1. Determine each product. Write the answer in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. a. 3𝑖(6 − 5𝑖)

b. (7 − 2𝑖)(2 + 3𝑖) + (3 − 5𝑖)(9 + 5𝑖)

c. (2 − 8𝑖)(1 + 3𝑖)

d. −2(7 − 2𝑖)(2 + 3𝑖)

2. Lin says, “When you add or multiply two complex numbers, you will always get an answer you can write in 𝑎 + 𝑏𝑖 form.” Noah says, “I don’t think so. Here are some exceptions I found:” (7 + 2𝑖) + (3 − 2𝑖) = 10 (2 + 2𝑖)(2 + 2𝑖) = 8𝑖

a. Check Noah’s arithmetic. Is it correct?

b. Can Noah’s answers be written in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers? Explain or show your reasoning.

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3. Which expression is equivalent to 2𝑖(5 + 3𝑖)? A. −6 + 10𝑖 B. 6 + 10𝑖

C. −10 + 6𝑖 D. 10 + 6𝑖

4. Explain to someone who missed class how you would write (3 − 5𝑖)(−2 + 4𝑖) in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. Review Problems 5. Select all the expressions that are equivalent to (3𝑥 + 2)(𝑥 − 4) for all real values of 𝑥.

□ 3𝑥 − 12 2

□ 3𝑥 − 10𝑥 − 8 2

□ 3(𝑥 + 2𝑥 − 4) 2

□ 3(𝑥 − 3𝑥) − (𝑥 + 8) 2

□ 3𝑥(𝑥 − 3) − 2(5𝑥 + 4) □ 3𝑥(𝑥 − 4) + 2(𝑥 − 4)

6. Write an expression for two of the graphs in terms of 𝑓(𝑥).

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Unit 3, Lesson 5: Operations with Complex Numbers

Warm-Up: Which One Doesn’t Belong: Complex Expressions Which one doesn’t belong? A. 𝑖2

B. (1 + 𝑖) + (1 − 𝑖)

C. (1 + 𝑖)2

D. (1 + 𝑖)(1 − 𝑖)

Exploration Activity: Powers of 𝑖

1. Write each power of 𝑖 in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. If 𝑎 or 𝑏 is zero, you can ignore that part of the number. For example, 0 + 3𝑖 can simply be expressed as 3𝑖. 𝑖0 = _____________

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2. What is 𝑖100? Explain your reasoning.

3. What is 𝑖38? Explain your reasoning.

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Collaborative Activity: Add ‘Em Up (or Subtract or Multiply) For each row, your partner and you will each rewrite an expression so it has the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. You and your partner should get the same answer. If you disagree, work to reach agreement.

Partner A

Partner B

(7 + 9𝑖) + (3 − 4𝑖)

5𝑖(1 − 2𝑖)

2𝑖(3 + 4𝑖)

(1 + 2𝑖) − (9 − 4𝑖)

(4 − 3𝑖)(4 + 3𝑖)

(5 + 𝑖) + (20 − 𝑖)

(2𝑖)4

(3 + 𝑖�7)(3 − 𝑖�7)

(1 + 𝑖�5) − (−7 − 𝑖�5)

(−2𝑖)(−�5 + 4𝑖)

(1 𝑖) (1 𝑖) (3 𝑖)

(1 𝑖)3

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Lesson Summary A complex number that includes an imaginary term must have 𝑖 raised to a power of 1. If an expression has an imaginary term with an exponent that is greater than 1, use the laws of exponents to rewrite the expression into an equivalent form. For example, (3 + 5𝑖)(7 − 2𝑖) can be multiplied (3 + 5𝑖)(7 − 2𝑖) using the distributive property and written as a complex number in 𝑎 + 𝑏𝑖 form, as shown.

= = = =

21 − 6𝑖 + 35𝑖 − 10𝑖2 21 + 29𝑖 − 10(−1) 21 + 29𝑖 + 10 31 + 29𝑖

Keeping track of the negative signs is especially important when multiplying because it is easy to mix up the fact that 𝑖2 = −1 with the fact that −𝑖2 = −(−1) = 1. For example, to write the difference (−6 + 3𝑖) − (2 − 4𝑖) as a single complex number in the form 𝑎 + 𝑏𝑖, consider distributing the negative and combining like terms, as shown.

(−6 + 3𝑖) − (2 − 4𝑖)

= = =

−6 + 3𝑖 − 2 − (−4𝑖) −8 + 3𝑖 + 4𝑖 −8 + 7𝑖

Precision with negative signs is also important when adding and subtracting terms. In the example, it is a common mistake to subtract 4𝑖 rather than subtracting −4𝑖.

Practice Problems 1. Select all expressions that are equivalent to 8 + 16𝑖.

□ 2(4 + 8𝑖)

□ 2𝑖(8 − 4𝑖) □ 4(2𝑖 − 4)

□ 4𝑖(4 − 2𝑖)

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2. Which expression is equivalent to (−4 + 3𝑖)(2 − 7𝑖)? A. −29 − 22𝑖 B. −29 + 34𝑖 C. 13 − 22𝑖 D. 13 + 34𝑖 3. Match the equivalent expressions. A. 𝑖2(3 + 𝑖)

i. (3 + 5𝑖) − (10 + 4𝑖)

B. −4𝑖 ⋅ 5𝑖

ii. (2 + 4𝑖)(2 − 4𝑖)

D. (1 + 2𝑖)(−1 + 3𝑖)

iv. (−6 + 12𝑖) − (−21 − 8𝑖)

C. 5𝑖(4 − 3𝑖)

iii. (1 − 4𝑖) + (−4 + 3𝑖)

Review Problems 4. Write each expression in 𝑎 + 𝑏𝑖 form. a. (−8 + 3𝑖) − (2 + 5𝑖) b. 7𝑖(4 − 𝑖) c. (3𝑖)3 d. (3 + 5𝑖)(4 + 3𝑖) e. (3𝑖)(−2𝑖)(4𝑖)

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5. Write each expression in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. a. 4(3 − 𝑖)

b. (4 + 2𝑖) + (8 − 2𝑖) c. (1 + 3𝑖)(4 + 𝑖) d. 𝑖(3 + 5𝑖) e. 2𝑖 ⋅ 7𝑖

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Unit 3, Lesson 6: Working Backward

Warm-Up: What’s Missing? Here are some complex numbers with an unknown difference: (10 + 4𝑖) − (___ + ____𝑖) = ? 1. If the result of this subtraction is a real number, what could the second complex number be?

2. If the result of this subtraction is an imaginary number, what could the second complex number be?

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Collaborative Activity: Info Gap: What Was Multiplied? Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner. If your teacher gives you the data card:

If your teacher gives you the problem card:

1. Silently read the information on your card.

1. Silently read your card and think about what information you need to answer the question.

2. Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!) 3. Before telling your partner the information, ask “Why do you need to know (that piece of information)?” 4. Read the problem card, and solve the problem independently. 5. Share the data card, and discuss your reasoning.

2. Ask your partner for the specific information that you need. 3. Explain to your partner how you are using the information to solve the problem. 4. When you have enough information, share the problem card with your partner, and solve the problem independently. 5. Read the data card, and discuss your reasoning.

Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.

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Lesson Summary When complex numbers are multiplied, each part of the first complex number is distributed to the other complex number. This means that there will always be the same pattern in multiplying 2 complex numbers. (𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖) = 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑏𝑐𝑖 + 𝑏𝑑𝑖2

Use the fact that 𝑖2 = −1 to rearrange the product and make it easier to identify the real and imaginary parts of the resulting product so that (𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖) = (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖.

Every time complex numbers are multiplied, the result is not only a complex number but also a specific complex number that comes from combining the parts of the numbers in a specific way. If 𝑎 and 𝑐 are the real parts of the complex numbers and 𝑏𝑖 and 𝑑𝑖 are the imaginary parts, then the result will always have 𝑎𝑐 − 𝑏𝑑 as a real part and (𝑎𝑑 + 𝑏𝑐)𝑖 as an imaginary part.

Practice Problems 1. Select all the expressions that are equivalent to (3 − 5𝑖)(−8 + 2𝑖).

□ −24 + 6𝑖 − 40𝑖 + 10𝑖

□ −24 + 46𝑖 − 10

□ −24 + 6𝑖 + 40𝑖 − 10𝑖

□ −14 − 34𝑖 □ −34 − 34𝑖

□ −24 + 46𝑖 + 10 □ 46𝑖 − 14

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3. Without going through all the trouble of writing the left side in the form 𝑎 + 𝑏𝑖, how could you tell that this equation is false? (−9 + 2𝑖)(10 − 13𝑖) = −68 − 97𝑖 4. Andre spilled something on his math notebook, and some parts of the problems he was working on were erased. Here is one of the problems: (________ − 2𝑖)( ________ + 2𝑖) = ________ −10𝑖

a. What could go in the blanks?

b. Could other numbers work, or is this the only possibility? Explain your reasoning.

Review Problem 5. Find the exact solution(s) to each of these equations, or explain why there is no solution. a. 𝑥2 = 49 b. 𝑥3 = 49 c. 𝑥2 = −49 d. 𝑥3 = −49

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Unit 3, Lesson 7: Solving Quadratic Equations

Warm-Up: Find the Perfect Squares The expression 𝑥2 + 8𝑥 + 16 is equivalent to (𝑥 + 4)2. Select all the expressions that are equivalent to (𝑥 + 𝑛)2 for some number 𝑛.

□ 𝑥2 + 10𝑥 + 25 □ 𝑥2 + 10𝑥 + 29 □ 𝑥2 − 6𝑥 + 8 □ 𝑥2 − 6𝑥 + 9 □ 𝑥2 − 6𝑥 − 9

□ 𝑥2 − 16𝑥 − 64 □ 𝑥2 + 16𝑥 + 64 □ 𝑥2 − 16𝑥 + 64 Collaborative Activity: Card Sort 1. Work with your partner to match each graph with its corresponding equation and 𝑥-intercepts. Record the letters for each match in the table. Graph

Equation

𝒙-intercepts

K L M

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2. Factor each given expression. Given Expression

6𝑥2 + 26𝑥 – 20

6𝑥2 + 19𝑥 – 20

6𝑥2 − 22𝑥 + 20

6𝑥2 + 7𝑥 – 20

Factored Form

3. Work with your partner to match each expression to an equation and corresponding 𝑥-intercepts. Expression

Equation in Factored Form

𝒙-intercepts

B C D

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Exploration Activity: Different Ways to Solve It Elena and Han solved the equation 𝑥2 − 6𝑥 + 7 = 0 in different ways. Elena said, “First I added 2 to each side: 𝑥2 − 6𝑥 + 7 + 2 = 2

So that tells me:

(𝑥 − 3)2 = 2

I can find the square roots of both sides: 𝑥 − 3 = ±�2

Which is the same as:

𝑥 = 3 ± �2

So the two solutions are 𝑥 = 3 + �2 and 𝑥 = 3 − �2.”

Han said, “I used the quadratic formula: 𝑥 = −𝑏 ± �𝑏 − 4 ⋅ 𝑎 ⋅ 𝑐 2

2⋅𝑎

Since 𝑥2 − 6𝑥 + 7 = 0, that means 𝑎 = 1, 𝑏 = − 6, and 𝑐 = 7. I know: 𝑥 = 6 ± �36 − 4 ⋅ 1 ⋅ 7 2⋅1

𝑥 = 6 ± �8 2

So:

𝑥 = 3 ± √8 2

I think the solutions are 𝑥 = 3 + √8 and 2 𝑥 = 3 − √8 .” 2

Do you agree with either of them? Explain your reasoning.

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Collaborative Activity: Solve These Ones Solve each quadratic equation with the method of your choice. Be prepared to compare your approach with a partner‘s. 1. 𝑥2 = 100 2. 𝑥2 = 38 3. 𝑥2 − 10𝑥 + 25 = 0 4. 𝑥2 + 14𝑥 + 40 = 0 5. 𝑥2 + 14𝑥 + 39 = 0 6. 3𝑥2 − 5𝑥 − 11 = 0

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Lesson Summary Quadratic equations that are factorable can be rewritten from a given form into factored form such as (𝑥 + 3)(𝑥 – 5) = 0 or (2𝑥 − 7)(𝑥 + 1) = 0. Using the zero product property, each factor can be set equal to 0 to determine the 𝑥-intercepts, or zeros, of the quadratic equation. The zero product property says that if the product of two numbers is 0, then one of the numbers must be 0. Quadratic equations that are not factorable are best solved by completing the square or using the quadratic formula. To complete the square, note that the perfect square (𝑥 + 𝑛)2 is equal to 𝑥2 + (2𝑛)𝑥 + 𝑛2. Consider the quadratic equation 𝑥2 − 5𝑥 = 25.

• Compare the coefficients of 𝑥 in 𝑥2 + (2𝑛)𝑥 + 𝑛2 to the expression 𝑥2 − 5𝑥 to see that 2𝑛 = 5, or just 𝑛 = − 5 . 2

• This means the perfect square �𝑥 − 5 � is equal to 2

𝑥2 − 5𝑥 + 25 , so adding 25 to each side of our equation 4

will result in a perfect square.

−√125, so 𝑥 − 5 = ± √125, which means the 2 solutions 2

2 are 𝑥 = 5 ± √125. 2 2

𝑥2 − 5𝑥 + 25 =

• The 2 numbers that square to make 125 are √125 and 4

𝑥2 − 5𝑥 = 4

�𝑥 − 5 � 2

25 + 25 4

100 + 25 4 4 125 4

In another example, to solve the quadratic equation 3𝑥2 − 2𝑥 = 9, the familiar strategies of factoring or completing the square may be too complex or not possible. Each side of the equation can be divided by 3, and then we can complete the square like before; however, the equation would get even messier, and the chance of making a mistake might be higher. With messier equations like this, it is often most efficient to use the quadratic formula. The quadratic formula 𝑥 = −𝑏±�𝑏 −4𝑎𝑐 gives the solutions of the 2

2𝑎

quadratic equation 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0, where 𝑎 is not 0.

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To use this formula, first rewrite the equation in standard form, and identify 𝑎, 𝑏, and 𝑐.

Rearranging the equation 3𝑥2 − 2𝑥 = 9 into standard form results in 3𝑥2 − 2𝑥 − 9 = 0. So 𝑎 = 3, 𝑏 = − 2, and 𝑐 = − 9. Pay close attention to the negative signs. Using the quadratic formula results in 𝑥 = −(−2)±� (−2) −4(3)(−9) = 2±�4+(12)(−9) . 2

2(3)

The exact solutions of this quadratic equation are 𝑥 = 1+2�7 and 𝑥 = 1−2�7 . The 3

approximate solutions of this quadratic equation are 𝑥 ≈ 2.097 and 𝑥 ≈ −1.431.

Practice Problems 1. What number should be added to the expression 𝑥2 − 15𝑥 to result in an expression equivalent to a perfect square? A. −7.5 B. 7.5

C. −56.25 D. 56.25

2. Noah uses the quadratic formula to solve the equation 2𝑥2 + 3𝑥 − 5 = 4. He finds 𝑥 = −2.5 or 1. But, when he checks his answer, he finds that neither −2.5 nor 1 are solutions to the equation. Here are his steps. 𝑎 = 2, 𝑏 = 3, 𝑐 = −5 a. Explain what Noah’s mistake was.

𝑥 = −3 ± �3 − 4 ⋅ 2 ⋅ −5 2

2⋅2

b. Solve the equation correctly.

𝑥 = −3 ±4 �49

𝑥 = −2.5 or 1

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3. Solve each quadratic equation with the method of your choice. a. 𝑥2 − 2𝑥 = −1 b. 𝑥2 + 8𝑥 + 14 = 23 c. 𝑥2 − 15 = 0 d. 7𝑥2 − 2𝑥 − 5 = 0 e. 2𝑥2 + 12𝑥 = 8 4. What are the solutions to the equation 𝑥2 − 4𝑥 = −3? a. 4 ± �16 − 4 ⋅ 0 ⋅ −3 2⋅0

b. 4 ± �16 − 4 ⋅ 1 ⋅ −3 2⋅1

c. 4 ± �16 − 4 ⋅ 1 ⋅ 3 2⋅1

d. −4 ± �16 − 4 ⋅ 1 ⋅ 3 2⋅1

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Review Problems 5. Which expression is equivalent to �−23? a. −23𝑖 b. 23𝑖

c. −𝑖�23

d. 𝑖�23

6. Write each expression in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. a. 5𝑖2

b. 𝑖2 ⋅ 𝑖2 c. (−3𝑖)2 d. 7 ⋅ 4𝑖 e. (5 + 4𝑖) − (−3 + 2𝑖)

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Unit 3, Lesson 8: Completing the Square and Complex Solutions

Warm-Up: Creating Quadratic Equations Match each equation in standard form to its factored form and its solutions. 1. 𝑥2 − 25 = 0

• (𝑥 − 5𝑖)(𝑥 + 5𝑖) = 0

3. 𝑥2 + 25 = 0

• (𝑥 − �5)(𝑥 + �5) = 0

2. 𝑥2 − 5 = 0

• (𝑥 − 5)(𝑥 + 5) = 0

• �5, −�5

• 5, −5

• 5𝑖, −5𝑖

Exploration Activity: Sometimes the Solutions Aren’t Real Numbers What are the solutions to these equations? 1. (𝑥 − 5)2 = 0

2. (𝑥 − 5)2 = 1

3. (𝑥 − 5)2 = −1

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Guided Activity: Finding Complex Solutions Solve these equations by completing the square. 1. 𝑥2 − 8𝑥 + 13 = 0

2. 𝑥2 − 8𝑥 + 19 = 0

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Collaborative Activity: Can You See the Solutions on a Graph? 1. How many real solutions does each equation have? How many non-real solutions? a. 𝑥2 − 8𝑥 + 13 = 0

b. 𝑥2 − 8𝑥 + 16 = 0

c. 𝑥2 − 8𝑥 + 19 = 0

2. How do the graphs of these functions help us answer the previous question? a. 𝑓(𝑥) = 𝑥2 − 8𝑥 + 13

b. 𝑔(𝑥) = 𝑥2 − 8𝑥 + 16

c. ℎ(𝑥) = 𝑥2 − 8𝑥 + 19

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Lesson Summary Sometimes quadratic equations have real solutions, and sometimes they do not. Consider a quadratic equation where 𝑥2 is equal to a negative number, 𝑥2 = −𝑘 (assume 𝑘 is positive). This equation will have imaginary solutions 𝑖�𝑘 and −𝑖�𝑘. By similar reasoning, an equation of the form (𝑥 − ℎ)2 = −𝑘 will have non-real solutions if 𝑘 is positive. In this case, the solutions are ℎ + 𝑖�𝑘 and ℎ − 𝑖�𝑘.

It isn’t always clear just by looking at a quadratic equation whether or not the solutions will be real. For example, the solutions to the equation 𝑥2 − 12𝑥 + 41 = 0 will require factoring, using square roots, completing the square, or using the quadratic formula to solve. All quadratic equations can be solved by completing the square. The solution to the equation 𝑥2 − 12𝑥 + 41 = 0 is shown by completing the square. 𝑥2 − 12𝑥 + 36 + 5

(𝑥 – 6)2

−5

(𝑥 – 6)2 + 5

𝑥 − 6 𝑥

±𝑖�5 6 ± 𝑖�5

This equation has 2 non-real, complex solutions, 𝑥 = 6 + 𝑖�5 and 𝑥 = 6 − 𝑖�5.

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Practice Problems 1. Find the solution or solutions to each equation. a. 𝑥2 + 0.5𝑥 − 14 = 0

b. 𝑥2 + 12𝑥 + 36 = 0

c. 𝑥2 − 3𝑥 + 8 = 0

d. 𝑥2 + 4 = 0

2. Which describes the solutions to the equation 𝑥2 + 7 = 0? A. One real solution

B. Two real solutions C. One non-real solution D. Two non-real solutions

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3. Determine the number of real solutions and non-real solutions to each equation. Use the graphs; don’t do any calculations to find the solutions.

𝑦 = 𝑥2 − 6𝑥 + 7

a. 𝑥2 − 6𝑥 + 7 = 0 b. 3𝑥2 + 2𝑥 + 1 = 0 c. −𝑥2 − 3𝑥 + 2 = 0 d. 𝑥2 − 6𝑥 + 7 = −2

𝑦 = −𝑥2 − 3𝑥 + 2

e. −𝑥2 − 3𝑥 + 2 = 6 f. 3𝑥2 + 2𝑥 + 1 = 2 𝑦 = 3𝑥2 + 2𝑥 + 1

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Review Problems 4. a. Write (5 − 5𝑖)2 in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers.

b. Write (5 − 5𝑖)4 in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers.

5. What number 𝑛 makes this equation true? 11

A. 4

𝑥2 + 11𝑥 + 121 = (𝑥 + 𝑛)2 4

B. 2

C. 11

121

D. 4

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Unit 3, Lesson 9: The Quadratic Formula and Complex Solutions

Warm-Up: Math Talk: Real or Not? Mentally decide whether the solutions to each equation are real numbers or not. 𝑤2 = −367 𝑥2 + 25 = 0 (𝑦 + 5)2 = 0 (𝑧 + 5)2 = −367

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Exploration Activity: Using the Discriminant Kiran was using the quadratic formula to solve the equation 𝑥2 − 12𝑥 + 41 = 0. He wrote the equation shown. 𝑥 = 12 ±√144 – 144 2

After noticing that the number under the square root is negative, he said, “This equation doesn’t have any solutions.” 1. Explain whether you agree with Kiran.

2. Write �–20 as an imaginary number.

3. Solve the equation 3𝑥2 − 10𝑥 + 50 = 0.

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Collaborative Activity: Solving All Kinds of Quadratics For each row, you and your partner will each solve a quadratic equation. You should each get the same answer. If you disagree, work to reach agreement. Partner A

Partner B

𝑥2 − 4𝑥 − 4 = 0

(𝑥 − 2)2 = 8

(𝑦 − 2)2 = −8

𝑦2 − 4𝑦 + 12 = 0

2𝑧2 + 6𝑧 = −19

(𝑧 + 32 ) = −294 𝑤2 + 3𝑤 = 5

4𝑡2 − 20𝑡 + 25 = 0

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(𝑤 + 32 ) = 294 2

4(𝑡2 − 5𝑡) = −25

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Lesson Summary 2 The quadratic formula, 𝑥 = −𝑏±√𝑏2𝑎– 4𝑎𝑐 , can always be used to determine solutions of quadratic equations in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0. The discriminant is the value under the square root in the quadratic formula, 𝑏2 − 4𝑎𝑐. The value of the discriminant provides information about the type of solutions of the quadratic equation.

• When the discriminant is positive, the quadratic equation has 2 real solutions. • When the discriminant is negative, the quadratic equation has 2 complex solutions. • When the discriminant is zero, the quadratic equation has 1 real solution with a multiplicity of 2.

When the discriminant of a quadratic equation is negative, the square root will produce an imaginary number and include 𝑖 in the solution(s) to the quadratic.

For example, to solve the equation 5𝑥2 + 𝑥 + 10 = 0 using the quadratic formula produces the results shown. 𝑥 =

𝑥 =

−1±√12 − 4⋅5⋅10 2⋅5 −1±√− 199 10

There are 2 complex solutions to the equation 5𝑥2 + 𝑥 + 10 = 0.

�199 �199 𝑥 = − 1 + 10 𝑖 and 𝑥 = − 1 − 10 𝑖. 10 10

Since the equation 5𝑥2 + 𝑥 + 10 = 0 has no real solutions, the graph of the function 𝑦 = 5𝑥2 + 𝑥 + 10 does not intersect the 𝑥-axis.

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Practice Problems 1. Clare solves the quadratic equation 4𝑥2 + 12𝑥 + 58 = 0, but when she checks her answer, she realizes she made a mistake. Explain what Clare’s mistake was. 𝑥 =

−12±√122 − 4 ⋅ 4 ⋅58 2⋅4

𝑥 =

− 12±√− 784 8

𝑥 = 𝑥 =

− 12±√144 − 928 8 − 12±28𝑖 8

𝑥 = −1.5±28𝑖

2. Write in the form 𝑎 + 𝑏𝑖, where 𝑎 and 𝑏 are real numbers: a. 5±√− 4 2

b. 10±√− 16 2

c. − 3±√− 144 6

3. Priya is using the quadratic formula to solve two different quadratic equations. For the first equation, she writes 𝑥 = 4±√16 − 72 . 12

For the second equation, she writes 𝑥 = 8±√64 − 24 . 6

Which equation(s) will have real solutions? Which equation(s) will have non-real solutions? Explain how you know.

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Review Problems 4. Find the exact solution(s) to each of these equations, or explain why there is no solution. a. 𝑥2 = 25 b. 𝑥3 = 27 c. 𝑥2 = 12 d. 𝑥3 = 12

5. Kiran is solving the equation √𝑥 + 2 − 5 = 11 and decides to start by squaring both sides. Which equation results if Kiran squares both sides as his first step? A. 𝑥 + 2 − 25 = 121 B. 𝑥 + 2 + 25 = 121

C. 𝑥 + 2 − 10√𝑥 + 2 + 25 = 121

D. 𝑥 + 2 + 10√𝑥 + 2 + 25 = 121

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Unit 3, Lesson 10: Graphs of Quadratic Functions and the Types of Solutions

Warm-Up: Solving Quadratic Equations 1. Solve the following quadratic equation using the square root method. 13𝑥2 + 6 = 58

2. Solve the following quadratic equation by factoring. 𝑥2 − 4𝑥 = 12

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Collaborative Activity: Factoring the Sum of 2 Squares 1. Leah and Giselle are factoring 2𝑥2 + 8. Both Leah and Giselle have found the GCF and are deciding what the next step in factoring should be. Leah’s Explanation I think that the polynomial factors to 2(𝑥2 + 4).

I know that 𝑥2 + 4 is a sum of squares and must be factored completely using complex numbers.

Giselle’s Explanation I think that the GCF is 2, so the polynomial factors to 2(𝑥2 + 4).

I know that the square root of 4 is 2. This tells me that this is a sum of squares.

a. Which explanation(s) do you agree with, Leah’s, Giselle’s, both, or neither?

b. Explain your choice.

Leah and Giselle discussed the next steps.

Leah Says

Since 𝑥2 + 4 is a sum of squares, the problem can be written as 𝑥2 + 0𝑥 + 4.

However, there are no real numbers that can multiply to get 4 and add to get 0.

Giselle Says

I also think that 𝑥2 + 4 is a sum of squares.

However, the rule for the sum of 2 squares is the same as the difference of 2 squares. Therefore, 𝑥2 + 4 factors into (𝑥 + 2)(𝑥 − 2).

2. With your partner, discuss Leah and Giselle’s conversation, and record your reactions to the application of the sum of 2 squares.

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3. Complete the statement about factoring 𝑥2 + 4. Since there are no 2

real complex

numbers with a product of 4 and a sum of 0,

the polynomial must be factored using

real complex

numbers.

4. Work with your partner to factor 𝑥2 + 4.

5. Complete the factorization of 2𝑥2 + 8.

(

)(

)

Guided Activity: Graphing Quadratic Functions 1. Consider the function 𝑓(𝑥) = 𝑥2 − 6𝑥 + 5.

a. Explain whether the graph of 𝑓(𝑥) opens up or down. b. Select all of the possible methods for determining the roots of 𝑓(𝑥).

□ Factoring

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c. Determine the root(s) of 𝑓(𝑥). d. Determine the vertex of 𝑓(𝑥). e. Complete the key features table for 𝑓(𝑥). Key Features Root(s)

Domain

𝑦-intercept

Range Opens up/ down

Vertex Left end behavior

As 𝑥 → _____, 𝑦 → _____.

Right end behavior

As 𝑥 → _____, 𝑦 → _____.

f. Graph 𝑓(𝑥) on the coordinate plane.

y 8 6 4 2 x -8

-6

-4

-2

-2 -4 -6 -8

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2. A table of key features for the quadratic function 𝑔(𝑥) is shown. Key Features 𝑥-intercept(s)

(−3, 0)

Domain

(−∞, ∞)

𝑦-intercept

(0, −9)

Range

(−∞, 0]

Vertex

(−3, 0)

Left end behavior

Opens up/ down

Down

Right end behavior

As 𝑥 → −∞, 𝑦 → −∞

As 𝑥 → ∞, 𝑦 → −∞

Use the key features to graph 𝑔(𝑥) on the coordinate plane. y

10 8 6 4 2

x -10

-8

-6

-4

-2

-2 -4 -6 -8 -10

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3. Quadratic function ℎ(𝑥) is represented by the equation ℎ(𝑥) = 𝑥2 + 4𝑥 + 6. a. Determine the zero(s) of ℎ(𝑥).

2 real b. The function ℎ𝑥 = 𝑥 + 4𝑥 + 6 has 1 real 2 complex

solution(s), which means

that the graph of 𝑓(𝑥) will intersect the 𝑥-axis

0 1 2

times.

c. Determine the vertex of ℎ(𝑥). d. Graph ℎ(𝑥) on the coordinate plane.

y 8 6 4 2 x

-8

-6

-4

-2

-2 -4 -6 -8

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4. Explain how complex zeros of a quadratic function relate to its graph.

Lesson Summary In Algebra 1, the study of quadratic functions was limited to real number solutions. When quadratic equations were not factorable or the quadratic formula produced a solution with a negative under the square root, the solutions were described as “no real solutions” because the complex number system had not yet been introduced. It is not that there were no solutions, but rather, it was that there were no real solutions. Now that you have been introduced to the complex number system, you have the knowledge and understanding to determine the solutions of any quadratic equation where the combinations of solutions could be 2 real solutions, 2 complex (or imaginary) solutions, or 1 real solution with a multiplicity of 2. The types of solutions are dependent on the value of 𝑏2 − 4𝑎𝑐, the discriminant, and how it compares to 0.

• If 𝑏2 − 4𝑎𝑐 > 0, the quadratic equation will have 2 real solutions, and the graph of the quadratic function will intersect the 𝑥-axis at 2 unique points.

• If 𝑏2 − 4𝑎𝑐 = 0, the quadratic equation will have 1 real solution with a multiplicity of 2, and the graph of the quadratic function will have its vertex on the 𝑥-axis at the unique solution.

• If 𝑏2 − 4𝑎𝑐 < 0, the quadratic equation will have 2 complex solutions, and the graph of the quadratic function will not intersect the 𝑥-axis.

Recall that solutions to quadratic equations may also be referred to as zeros, roots, and 𝑥-intercepts. Typically, 𝑥-intercepts are expressed as points, (𝑥, 0), and zeros and roots are expressed in the form 𝑥 = ____.

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Practice Problems 1. Complete the statements for each quadratic function. a. The function 𝑓(𝑥) = 2𝑥2 + 5𝑥 + 12 has

2 real 1 real 2 complex

that the graph of 𝑓(𝑥) will intersect the 𝑥-axis b. The function 𝑓(𝑥) = 9𝑥2 − 12𝑥 + 4 has

0 1 2

2 real 1 real 2 complex

that the graph of 𝑓(𝑥) will intersect the 𝑥-axis

0 1 2

solutions, which means

times.

solutions, which means

times.

2. Graph the quadratic function 𝑚(𝑥) = 𝑥2 − 4𝑥 + 7 on the coordinate plane.

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3. Select all the functions that have 2 complex solutions.

□ 𝑘(𝑥) = 𝑥2 + 2𝑥 + 5

□ 𝑔(𝑥) = 2𝑥2 + 7𝑥 − 15 □ 𝑝(𝑥) = 𝑥2 − 4𝑥 + 13

□ 𝑓(𝑥) = 𝑥2 − 10𝑥 + 26

□ ℎ(𝑥) = 3𝑥2 − 6𝑥 − 24 Review Problems 4. Solve the quadratic equation 5𝑥2 + 2𝑥 + 7 = 8. 5. Complete the statements for each quadratic function.

a. The function 𝑓(𝑥) = 𝑥 − 6𝑥 + 9 has a discriminant that is 2

Therefore, the function 𝑓(𝑥) has

0 real roots. 1 real root. 2 real roots.

b. The function ℎ(𝑥) = 2𝑥2 − 7𝑥 + 3 has a discriminant that is Therefore, the function ℎ(𝑥)s has

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zero. positive. negative.

0 real solutions. 1 real solution. 2 real solutions.

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Unit 3, Lesson 11: Real and Non-Real Solutions

Warm-Up: Notice and Wonder: Where Is It 0? What do you notice? What do you wonder? 𝑓(𝑥) = 2𝑥2 + 4𝑥

𝑓(𝑥) = 0 when 𝑥 = 0, −2.

𝑔(𝑥) = 2𝑥2 + 4𝑥 + 2

𝑔(𝑥) = 0 when 𝑥 = −1.

ℎ(𝑥) = 2𝑥2 + 4𝑥 + 4

ℎ(𝑥) = 0 when 𝑥 = −1 + 𝑖, −1 − 𝑖.

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Exploration Activity: Real or Not? Here are some equations. Equation

Prediction

𝑥2 − 6𝑥 + 5 = 0

𝑥2 − 6𝑥 + 13 = 0

−𝑥2 + 6𝑥 − 9 = 0 −𝑥2 − 9 = 0

1. Which equations will have real solutions, and which ones will not? Write your prediction in the table.

Pause here for discussion.

2. What advice would you give to someone who is trying to figure out whether a quadratic equation has real solutions?

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Collaborative Activity: Make Your Own 1. Create three different quadratic equations. At least one should have solutions with an imaginary part, and at least one should have solutions with only real parts.

2. Solve your equations. Write down your equations and their solutions in the table. Equation

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Solutions

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Lesson Summary The graph of a quadratic function can help determine whether or not it has real zeros. The real zeros are the 𝑥-coordinates of the points where the graph touches the 𝑥-axis. So, if the graph doesn’t touch the 𝑥-axis at all, then the function does not have real zeros. In such cases, the graph cannot be used to identify the exact complex zeros of the function, but the zeros can be determined algebraically by setting the quadratic function equal to 0 and solving.

The quadratic formula can also be used to determine whether 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 has real zeros. Imaginary numbers are the square roots of negative numbers, so it is helpful to look at the part of the quadratic formula that involves a square root, which is �𝑏2 − 4𝑎𝑐. If 4𝑎𝑐 is greater than 𝑏2, then the number under the radical will be negative, which means that the zeros of the quadratic function will have an imaginary part. For example, consider the quadratic function 𝑦 = 𝑥2 + 2𝑥 + 2. The graph of the function is shown. Since its graph does not intersect the 𝑥-axis, this function doesn’t have any real zeros.

The equation can also be used to verify whether 4𝑎𝑐 is greater than 𝑏2. Since 𝑎 = 1, 𝑏 = 2, and 𝑐 = 2, 𝑏2 − 4𝑎𝑐 = −4. Therefore, the zeros of this function have an imaginary part. The exact values of complex zeros of quadratic functions can be determined by using the quadratic formula or completing the square.

Practice Problems 1. The graph shows the equation 𝑦 = 2𝑥2 + 0.5𝑥 − 4. Based on the graph, what number could you put in the box to create an equation that has no real solutions? 2𝑥2 + 0.5𝑥 − 4 =

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2. Without calculating the solutions, determine whether each equation has real solutions or not.

𝑦 = 𝑥2 − 4𝑥 + 7

a. −0.5𝑥2 + 3𝑥 = 0 b. 𝑥2 − 4𝑥 + 7 = 0 c. 2𝑥2 − 2𝑥 − 1 = 0 d. −0.5𝑥2 + 3𝑥 = 3

𝑦 = 2𝑥2 − 2𝑥 − 1

e. 𝑥2 − 4𝑥 + 7 = 5 f. 2𝑥2 − 2𝑥 − 1 = −4

𝑦 = −0.5𝑥2 + 3𝑥

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3. Write a quadratic equation that has two non-real solutions. How did you decide what equation to write?

Review Problems 4. Find the solution or solutions to each equation. a. −2𝑥2 + 2𝑥 = 2.5

b. 4.5𝑥2 + 3𝑥 + 1 = 0 2

c. 1 𝑥2 + 5𝑥 = −14 2

d. −𝑥2 − 1.5𝑥 + 5 = 7

5. Elena and Kiran were solving the equation 2𝑥2 − 4𝑥 + 3 = 0 and they got different answers. Elena wrote 1 ± 𝑖�0.5 , and Kiran wrote 1 ± 𝑖√8 . Are their answers 4 equivalent? Say how you know.

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Unit 3, Lesson 12: Determining Zeros within a Real-World Context

Warm-Up: Solving Quadratics Using Different Methods 1. Complete the table by identifying the solution method you would use for each quadratic equation based on its structure. Be prepared to explain your reasoning. Quadratic Equation

Solution Method ¡ Factoring

𝑥2 + 5𝑥 = 7𝑥 − 2

¡ Quadratic formula ¡ Completing the square ¡ Factoring

4𝑥2 + 5 = 12𝑥

¡ Quadratic formula ¡ Completing the square ¡ Factoring

8𝑥 + 𝑥2 = −4

¡ Quadratic formula ¡ Completing the square

2. Solve 1 of the equations in the table using the method you selected. Show your work.

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Collaborative Activity: Using Quadratics to Model Real-World Contexts 1. Organizers of a music festival charge $60 for admission. The organizers polled the attendees to determine how increasing the price would affect attendance. The equation 𝑃(𝑥) = −20,000𝑥2 − 20,000𝑥 + 600,000 models profit as a function of 𝑥, where 𝑥 is each $1 increase in the admission fee. a. Discuss with your partner which factors would affect your attendance at a music festival. List your top 3 factors.

b. What admission fee would you consider to be too high?

c. If 𝑥 = 2, what is the admission fee to the festival?

d. Solve −20,000𝑥2 − 20,000𝑥 + 600,000 = 0 by factoring. Show your work.

e. Interpret the solutions in terms of the context.

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f. Ask 2 other classmates for their interpretations of the solutions to −20,000𝑥2 − 20,000𝑥 + 600,000 = 0. Write a summary of each of their interpretations of the solutions. You are the only person who should write in the first column of the table. Have each person initial next to your summary, indicating that your summary is accurate. My Summary of Their Interpretation of the Solutions in Terms of the Context

Initials

g. Review your classmates’ interpretations with your partner. If necessary, adjust your interpretation.

2. The compact disc (CD) was invented in 1979, many years before online music existed. At the time, it was the most sophisticated way to store and play music. In 1982, CDs were made available to the public. CDs quickly became the most efficient way of storing music and replaced vinyl records and cassette tapes. 3. The function 𝐶(𝑥) = −2𝑥2 + 38𝑥 + 40 models the sales, in hundreds of millions of dollars, of CDs in the years since 1990. a. Solve − 2𝑥2 + 38𝑥 + 40 = 0 by factoring.

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b. Interpret the year that each zero represents. Zero

Year

c. Discuss the interpretation of each zero with your partner. Then, determine whether the interpretation makes sense. Summarize your discussion.

d. In 2001, buying songs online and storing them on portable music players or computers became an option for the first time. By 2008, music streaming sites began to launch. Discuss with your partner why one of the zeros shows that online access to music by these companies and others may have led to CD sales eventually reaching zero. Summarize your discussion.

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Guided Activity: Finding Zeros 1. A landscape designer is adding a garden to a client’s backyard. One side of the garden will be alongside an existing fence. The client has 180 feet (ft.) of fencing left over from another project. The client would like the garden to use all of the fencing.

existing fence

width

The equation 𝐴 = 180𝑤 − 2𝑤2 can be used to find the area of the garden, where 𝑤 is the garden’s width in feet.

length

a. Find the zeros of the equation 𝐴 = 180𝑤 − 2𝑤2.

b. Interpret both of the zeros in terms of the context.

2. A statistician gathers data over several years, starting in 1995, on the average number of trips a household makes to run errands. The statistician uses the data to find the equation, 𝐸(𝑥) = −2𝑥2 + 68𝑥 + 480, which models the average annual number of trips for errands per household for years since 1995. The graph of the equation is shown.

y 1200 1100

(12,1008)

1000 900 800 700 600 500 400

a. Interpret 𝐸(12). Include units.

300 200 100 -10 -5 0 -100

10 15 20 25 30 35 40 45

-200

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b. Determine the zeros of the function. Show your work.

c. Interpret both of the zeros in terms of the context.

d. Discuss with your partner whether both zeros are viable. If a zero is nonviable, explain why.

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Lesson Summary Quadratic functions are often used to model real-world phenomena such as the path of an object’s height with respect to time as it is thrown in the air, the area of a rectangle with variable dimensions, and profit and loss trends over time. In previous lessons, you have learned how to solve quadratic functions using various strategies. While most equations can be solved using more than 1 method, the table summarizes when some methods may be more efficient than others.

Method

When to Use

Types of Solutions that Can Be Determined

Graphing

Use when the quadratic is given in a form that is easy to graph, such as vertex form.

Real solutions only

Factoring

Use if the equation includes factors that can be easily determined.

Real and complex solutions

Using square roots Completing the square Using the quadratic formula

Use if the equation only includes a squared term and a constant term, such as 𝑥2 − 7 = 0 or (𝑥 − 2)2 = 7.

Real and complex solutions

Use for any equation. The resulting equation shows vertex form of a quadratic function, which is also easier to graph.

Real and complex solutions

Use for any equation where other methods may not be possible or efficient.

Real and complex solutions

After solving quadratic equations that model a real-world context, always check the solutions to determine if they are viable in context. In some contexts, viable solutions may be constrained to only positive solutions or only integers, and in others only solutions within a reasonable domain are viable. Note that complex solutions will often not be included in the reasonable solution(s) of a real-world problem.

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Practice Problems 1. The equation 𝐻(𝑥) = −16𝑥2 + 64𝑥 + 192 models the height, 𝐻, in feet (ft.), of a toy rocket 𝑥 seconds (sec.) after it is launched from the top of a 16 story building. a. Determine the zeros of the function. Show your work.

b. Interpret both of the zeros in terms of the context.

c. Determine whether both zeros are viable. If a zero is nonviable, explain why.

2. The area, 𝐴, in square feet (sq. ft.) of a projected image on a screen is modeled by the equation 𝐴 = 0.36𝑤2, where 𝑤 is the distance, in ft., the projector is from the screen. How far from the screen should the projector be placed so that the area of projected image is 100 sq. ft.?

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3. A group of engineers is designing a section of a road whose height will be below sea level. Taking the local terrain into account, the road’s curve in elevation can be modeled by the equation 𝑦 = 0.00003𝑥2 − 0.06𝑥, where 𝑥 is the horizontal distance, in ft., between the 2 points where the road is at sea level. The engineers need to place warning road signs where the road will be below sea level. How far apart will the warning signs be located?

Review Problems 4. Find the solution(s) to each equation, or explain why there is no solution. a. 2𝑥2 − 2 = 5 1 3

b. (𝑥 + 1)2 = 81

c. 3𝑥2 + 14 = 12

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Unit 3, Lesson 13: Solving Polynomials for Real and Complex Zeros

Warm-Up: Which One Doesn’t Belong? 1. Which one doesn’t belong? A. 2𝑥2 − 4𝑥 = 1 B. 𝑥2 + 2 = −𝑥

C. 1 = 4𝑥2

D. 4𝑥2 + 1 = 4𝑥 2. Explain your reasoning.

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Collaborative Activity: Factoring to Find Zeros

y 20

1. The function 𝑓(𝑥) = 𝑥3 − 2𝑥2 − 9𝑥 + 18 is shown on the coordinate grid. a. Use the graph to determine the solutions to 𝑥3 − 2𝑥2 − 9𝑥 + 18 = 0.

-4

-2

b. Work with your partner to solve 𝑥3 − 𝑥2 − 7𝑥 + 18 = 𝑥2 + 2𝑥 by completing the steps shown. 𝑥3 − 𝑥2 − 7𝑥 + 18 = 𝑥2 + 2𝑥 𝑥3 − 2𝑥2 − 9𝑥 + 18 = 0

(𝑥3 − 2𝑥2) + ( − 9𝑥 + 18) = 0

________(𝑥 − 2) + ________(𝑥 − 2) = 0 (𝑥 − 2) (______________) = 0

(𝑥 − 2)(_____________)(_______________) = 0

𝑥 − 2 = 0

𝑥 = 2

___________ = 0 𝑥 = __________

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___________ = 0

𝑥 = __________

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2. Charley and Evie solved 𝑥3 − 4𝑥2 + 4𝑥 − 16 = 0 by factoring. Their work is shown. Charley

Evie

𝑥3 − 4𝑥2 + 4𝑥 − 16 = 0

𝑥2(𝑥 − 4) + 4(𝑥 − 4) = 0

(𝑥 − 4)(𝑥2 + 4) = 0

𝑥 − 4 = 0

(𝑥 − 4)(𝑥2 + 4) = 0

𝑥2 + 4 = 0

𝑥 = 4

(𝑥 − 4)(𝑥 + 2𝑖)(𝑥 − 2𝑖) = 0

𝑥 =−4 2

𝑥 = 4, − 2𝑖, 2𝑖

𝑥 = ±2𝑖

a. How are Charley’s and Evie’s methods similar?

b. How are Charley’s and Evie’s methods different?

c. Discuss with your partner which sketch best represents 𝑓(𝑥) = 𝑥3 − 4𝑥2 + 4𝑥 − 16. Sketch A

Sketch B

y x

Explain your choice.

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3. Consider the equation 81𝑥4 − 16 = 0. a. Solve 81𝑥4 − 16 = 0 by factoring.

b. Li Hua did not solve 81𝑥4 − 16 = 0 by factoring. Her work is shown. 81𝑥4 − 16 = 0 81𝑥4 = 16

81𝑥4 = 16 81 81 𝑥4 = 16 81

√81

𝑥 = ± 4 16 2

𝑥 = ± 3

Compare Li Hua’s solution to the solution found by factoring. If the solutions are not the same, then reconcile the differences with your partner.

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4. Work with your partner to match each type of expression from the first column with its factored form in the second column by drawing an arrow.

(𝑎 − 𝑏)2 (𝑎 + 𝑏)2

Sum of squares 𝑎2 + 𝑏 2

(𝑎 − 𝑏)3

Difference of squares 𝑎2 − 𝑏 2 Sum of cubes 𝑎3 + 𝑏3

Difference of cubes 𝑎3 − 𝑏3

(𝑎 + 𝑏)(𝑎 − 𝑏)

(𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖)

(𝑎 − 𝑏𝑖)(𝑎2 + 𝑎𝑏𝑖 + 𝑏2)

(𝑎 + 𝑏𝑖)(𝑎2 − 𝑎𝑏𝑖 + 𝑏2) (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏2) (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏2)

Guided Activity: Factoring to Find Zeros 1. Find the solutions to 5𝑥4 − 80 = 0.

2. Find the zeros of 𝑔(𝑥) = 125𝑥3 + 8.

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3. Solve 𝑥5 − 2𝑥4 − 𝑥3 = 3𝑥4 + 5𝑥3.

4. Teagan and Cohen solved 2𝑥4 + 7𝑥2 − 3 = 6 − 2𝑥4 + 12𝑥2 using 2 different factoring techniques. Their partial work is shown. Teagan

Cohen

2𝑥4 + 7𝑥2 − 3 = 6 − 2𝑥4 + 12𝑥2

4(𝑥2)2 − 5(𝑥2) − 9 = 0

4𝑥4 + 4𝑥2 − 9𝑥2 − 9 = 0

4𝑥4 − 5𝑥2 − 9 = 0 Let 𝑢 = 𝑥2.

4𝑢2 − 5𝑢 − 9 = 0

4𝑢2 + 4𝑢 − 9𝑢 − 9 = 0

4𝑥4 − 5𝑥2 − 9 = 0

4𝑥2(𝑥2 + 1) − 9(𝑥2 + 1) = 0 (4𝑥2 − 9)(𝑥2 + 1) = 0

4𝑢(𝑢 + 1) − 9(𝑢 + 1) = 0 4𝑢 − 9𝑢 + 1 = 0

(4𝑥2 − 9)(𝑥2 + 1) = 0

a. Discuss with your partner what you notice about what Teagan did to help her factor the polynomial.

b. Finish either Teagan’s or Cohen’s problem to find the zeros.

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c. Use either Teagan’s or Cohen’s method to solve 𝑥4 − 7𝑥2 + 12 = 0.

5. Find the zeros of ℎ(𝑤) = 16𝑤5 − 250𝑤2.

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Lesson Summary The strategies used to solve quadratic equations, such as factoring, using the quadratic formula, completing the square, and graphing, can be used to solve other polynomials with a higher degree than 2. A polynomial function y has the same number of solutions, real or complex, as 16 the degree of the polynomial’s function. 14 The graph of the polynomial function 𝑓(𝑥) = 𝑥3 − 3𝑥2 − 4𝑥 + 12 is shown. Real solutions to the function are the 𝑥-coordinates of the horizontal intercepts of the graph, where 𝑓(𝑥) = 0. The graph can be used to determine the solutions of 𝑓(𝑥) = 0 at 𝑥 = −2, 𝑥 = 2, and 𝑥 = 3.

• The polynomial can also be solved by factoring. First, factor the polynomial by grouping.

12 10 8 6 4 2 x -5

-4

-3

-2

-1

-2 -4 -6

𝑓(𝑥) = 𝑥3 − 3𝑥2 − 4𝑥 + 12

𝑓(𝑥) = 𝑥2(𝑥 − 3) − 4(𝑥 – 3)

𝑓(𝑥) = (𝑥 − 3)(𝑥2 − 4)

• Then, apply the difference of perfect squares, or another factoring technique, to factor the polynomial function to linear factors. 𝑓(𝑥) = (𝑥 − 3)(𝑥2 − 4)

𝑓(𝑥) = (𝑥 − 3)(𝑥 + 2)(𝑥 − 2)

Once the polynomial is expressed as a product of linear factors, or lowest possible degree factors, set each factor equal to zero, and solve for 𝑥. 𝑥 − 3 = 0 𝑥 = 3

𝑥 + 2 = 0 𝑥 = −2

𝑥 − 2 = 0 𝑥 = 2

In some cases, the factors of a polynomial function cannot be rewritten with a degree lower than 2. In these cases, other methods may need to be applied to solve the polynomial.

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The graph of 𝑓(𝑥) = 𝑥3 + 3𝑥2 − 3𝑥 is shown. The zeros, or solutions, of the polynomial function cannot be accurately determined from the graph alone. Although it is possible to estimate that there is a zero at 𝑥 = 0, the other 2 zeros are approximately at 𝑥 ≈ − 3.8 and 𝑥 ≈ 0.8. The exact values can only be determined using the quadratic formula or by completing the square.

14 12 10 8 6 4

To solve the polynomial function algebraically, first determine the greatest common factor if possible.

2 x -5

• In this case, the factors of the polynomial function cannot be factored any further than 𝑓(𝑥) = 𝑥(𝑥2 + 3𝑥 − 3).

-4

-3

-2

-1

-2 -4 -6

• Set each factor equal to 0, so that 𝑥 = 0 and 𝑥2 + 3𝑥 − 3 = 0.

• The quadratic factor cannot be factored into linear factors, but it can be solved using the quadratic formula or by completing the square. • The resulting solutions are 𝑥 = −

3 +√21 3 +√21 and 𝑥 = − . 2 2

Practice Problems 1. Find the solutions of 𝑥3 − 4𝑥2 + 4𝑥 − 16 = 0.

2. Solve 𝑥4 + 2𝑥2 = 35.

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3. Solve 𝑟6 + 3𝑟3 + 2 = 0.

4. Find the zeros of 𝑌(𝑣) = 2𝑣5 − 18𝑣.

Review Problems 5. An open-top box is formed by cutting squares out of a 5 inch (in.) by 7 in. piece of paper and then folding up the sides. The volume 𝑉(𝑥) in cubic inches (cu. in.) of this type of open-top box is a function of the side length 𝑥 in in. of the square cutouts and can be given by 𝑉(𝑥) = (7 − 2𝑥)(5 − 2𝑥)(𝑥). Rewrite this equation by expanding the polynomial.

6. State the degree and end behavior of 𝑓(𝑥) = 4𝑥3 + 3𝑥5 − 𝑥2 − 2. Explain or show your reasoning.

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Unit 3, Lesson 14: Factoring Polynomials from a Real-World Context

Warm-Up: Zeros of Polynomial Equations 1. Select all the zeros of the polynomial equation 9𝑥3 − 18𝑥2 + 4𝑥 − 8 = 0.

□ 𝑥 = 23 □ 𝑥 = 2

□ 𝑥 = 23 𝑖 □ 𝑥 = 2𝑖

□ 𝑥 = − 23 □ 𝑥 = − 2

□ 𝑥 = − 23 𝑖 □ 𝑥 = − 2𝑖

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Collaborative Activity: Polynomials in a Real-World Context In a previous lesson, you explored the volume of a box with dimensions expressed in terms of an unknown height of the box, 𝑥. The volume of the box was modeled using the function 𝑉(𝑥) = (3 − 2𝑥)(8 − 2𝑥)(𝑥). Since the function 𝑉(𝑥) was given in factored form, the zeros of the function were easy to determine and use to create a sketch of the function. Often, functions that model real-world phenomena are presented in standard form. In cases where the polynomial function is not given in a factored form, additional steps must be taken to determine the zeros and other key features of the function. 1. A box is created out of a piece of cardboard, where a square with a length of 𝑥 inches (in.) is cut from each corner. The function 𝑉(𝑥) = (5 − 2𝑥)(10 − 2𝑥)(𝑥) can be used to represent the volume of the box after each cut. Complete the statement. The box will have a volume of 0 cubic inches (cu. in.) when 𝑥 = _______, 𝑥 = ______, or 𝑥 = ________. The box will have a maximum volume when 𝑥 is between ______ in. and _______in. 2. The function 𝑉(𝑥) = 4𝑥3 − 40𝑥2 + 96𝑥 represents the volume, 𝑉, in cu. in. of the box, where 𝑥 represents the cut size, in (1.569, 67.604) inches. The graph of the function 𝑉(𝑥) y is shown. a. Circle the part of the graph that represents the reasonable cut sizes.

(0, 0) 0

(4, 0)

(6, 0) 5

(5.097, –20.193) -50

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b. A table of values for the function 𝑉(𝑥) = 4𝑥3 − 40𝑥2 + 96𝑥 is shown. Cut Size (in.) Volume (cu. in.)

0.5

1.5

2.5

38.5

67.5

52.5

−20

Ryan was looking at the table of values and noticed that a cut of 5 in. has a volume of −20 cu. in. He asked his teacher how this is possible. His teacher asked Ryan, “Why do you think a cut of 4 in. would give a volume of 0 cu. in.?” Help Ryan explain why a cut of 4 in. results in a volume of 0 cu. in.

c. Discuss with your partner why a cut size of 7 in. is not possible, even though the table shows that a cut size of 7 in. results in a volume of 84 cu. in. d. What does the point (1.569, 67.604) represent in the context of the problem?

3. Solve 4𝑥3 − 40𝑥2 + 96𝑥 = 0 by factoring.

4. How do the solutions to 4𝑥3 − 40𝑥2 + 96𝑥 = 0 relate to the context?

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Guided Activity: Polynomials in a Real-World Context 1. The concentration of a certain medication in a patient’s bloodstream, 𝐷, in milligrams (mg), can be modeled by the function 𝐷(𝑥) = −𝑥4 + 5𝑥3 − 10𝑥2 + 50𝑥, where 𝑥 is the amount of the time, in hours, since the medication was administered. a. Find the zeros of 𝐷(𝑥).

b. What do the zeros of 𝐷(𝑥) represent in terms of the context?

Lesson Summary Polynomial functions can be used to model real-world scenarios, including volume of three-dimensional figures, supply and demand business models, remaining medication amount after administration, and others. Apply the same strategies for solving polynomial functions as before, such as factoring, using the quadratic formula, completing the square, and graphing. The algebraic solutions for a polynomial function should also be evaluated for reasonableness in context. Each zero, or 𝑥-intercept, needs to be verified in the scenario to make sure it makes sense. In some cases, a negative solution may not be viable, or a solution greater than or less than a specific value may not make sense based on contextual constraints.

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Practice Problems 1. A learning curve indicates the percentage of a task mastered at a certain time. A mathematician for a company found that the learning curve for new employees for the first 4 days could be modeled by 𝐿(𝑡) = (12𝑡 − 5)(− 0.2𝑡2 + 1), where 𝑡 is the number of days and 𝐿 is the percentage of job tasks learned. a. Find the zeros of 𝐿(𝑡).

b. Determine if the zeros of 𝐿(𝑡) make sense for the context. Justify your decision.

2. A local book publisher determined that the supply function for the books sold is 𝑆(𝑝) = 2𝑥3 − 3𝑥2, and the demand function is 𝐷(𝑝) = 4𝑥 − 6, where 𝑝 is the price of the books sold. The equilibrium function is defined as the difference between the supply and demand functions, 𝐸(𝑝) = (2𝑥3 − 3𝑥2) − (4𝑥 − 6). What should the book price(s) be for the equilibrium function to have a value of 0?

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Review Problems 3. Solve the equation 𝑥3 − 2𝑥2 − 𝑥 + 2 = 0.

4. Solve the equation 𝑥3 − 3𝑥2 + 9𝑥 − 27 = 0.

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Unit 3, Lesson 15: Solving One-Variable Quadratic Inequalities

Warm-Up: Solving Inequalities 1. What is the solution set for −13 ≤ −2𝑥 − 15 < 2?

2. Find the solution set of the inequality −4 < −2𝑥 − 6 < 4.

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Guided Activity: Solving One-Variable Quadratic Inequalities Nora began working through graphing the solution set of the inequality 𝑥2 − 2𝑥 > 8. The first steps of her process are shown in the table. Step

Work 𝑥2 − 2𝑥 > 8

−8−8

𝑥2 − 2𝑥 − 8 > 0 𝑥2 − 2𝑥 − 8 = 0

(𝑥 + 2)(𝑥 − 4) = 0

3 4

𝑥 + 2 = 0

𝑥 − 4 = 0

−2 − 2

+4 +4

𝑥 + 2 = 0

1. What did Nora do from step 1 to step 2?

𝑥 = −2

𝑥 − 4 = 0 𝑥 = 4

2. Plot the solutions to the equation 𝑥2 − 2𝑥 − 8 = 0 with open circles on the number line.

-5

-4

-3

-2

-1

3. Discuss with your partner why the points plotted are open and not closed. 4. What are the 3 intervals created by the solutions of the equation?

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Nora substituted the values − 3, 0, and 5 into the inequality 𝑥2 − 2𝑥 − 8 > 0 to determine if they satisfy the inequality. Her work is shown. 𝒙 −3

Evaluation (− 3)2 − 2(− 3) − 8 > 0 9+6−8>0 7>0

02 − 2(0) − 8 > 0 0−0−8>0 −8 > 0

(5)2 − 2(5) − 8 > 0 25 − 10 − 8 > 0 7>0

Conclusion

When 𝑥 = −3, the inequality is true.

When 𝑥 = 0, the inequality is false.

When 𝑥 = 5, the inequality is true.

5. Explain why Nora chose −3, 0, and 5 as the values to substitute for 𝑥.

6. Shade the number line in problem 2 for the intervals with a true statement.

7. Write the solution set of 𝑥2 − 2𝑥 − 8 > 0 in interval notation. 8. Write the solution set of 𝑥2 − 2𝑥 − 8 > 0 in set notation.

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Collaborative Activity: Solving One-Variable Quadratic Inequalities 1. Consider the inequality 𝑥2 − 8𝑥 + 7 ≥ 0. a. Solve the equation 𝑥2 − 8𝑥 + 7 = 0.

b. Graph the solutions to the equation.

-8

-7

-6

-5

-4

-3

-2

-1

c. Select a number in each interval created by the solutions of the equation, and test them in the inequality 𝑥2 − 8𝑥 + 7 ≥ 0.

d. Complete the graph of the solution set in part B by shading intervals where the inequality is true. e. Write the solution set of 𝑥2 − 8𝑥 + 7 ≥ 0 in interval notation.

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2. Consider −𝑥2 − 3𝑥 + 10 > 0 to complete the following. a. Solve the equation −𝑥2 − 3𝑥 + 10 = 0.

b. Graph the solutions to the equation.

-8

-7

-6

-5

-4

-3

-2

-1

c. Select a number in each interval created by the solutions of the equation, and test each of them in the inequality −𝑥2 − 3𝑥 + 10 > 0.

d. Complete the graph of the solution set in part B by shading intervals where the inequality is true.

e. Write the solution set of −𝑥2 − 3𝑥 + 10 > 0 in interval notation. f. Write the solution set of −𝑥2 − 3𝑥 + 10 > 0 in set notation.

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Lesson Summary To solve a quadratic inequality, begin by applying the same strategies used when solving a quadratic equation, including first rewriting the inequality so that 1 side of the inequality in the statement is 0. Use this statement to determine the values of 𝑥 for which the quadratic expression equals 0. These values can be used to determine possible intervals that are true or false for the given inequality. Select values to substitute into the inequality within each interval to determine which intervals are included in the solution set of the inequality. For example, to solve the inequality 𝑥2 + 4𝑥 < 5, first set the quadratic expression equal to 0, and solve for 𝑥. 𝑥2 + 4𝑥 < 5

𝑥2 + 4𝑥 − 5 < 0 𝑥2 + 4𝑥 − 5 = 0

(𝑥 + 5)(𝑥 − 1) = 0

𝑥 = − 5 and 𝑥 = 1

There are 3 possible solution intervals (values less than −5, values between −5 and 1, and values greater than 1). Use 𝑥-values within each possible solution interval to verify whether the quadratic inequality is true over that interval. Interval

𝑥 < −5

−5 < 𝑥 < 1

𝑥 > 1

𝒙-value −10

Evaluation (−10)2 + 4 (−10) − 5 < 0 100 − 40 − 5 < 0 55 < 0

(0)2 + 4(0) − 5 < 0 0+0−5<0 −5 < 0

(2)2 + 4(2) − 5 < 0 4+8−5<0 7<0

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Conclusion

When 𝑥 < −5, the inequality is false. When −5 < 𝑥 < 1, the inequality is true.

When 𝑥 > 1, the inequality is false.

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The inequality is only true over the interval −5 < 𝑥 < 1. Therefore, that is the solution set of the inequality 𝑥2 + 4𝑥 − 5 < 0. The solution set can be written in interval notation (−5, 1) or set notation {𝑥| −5 < 𝑥 < 1}.

Practice Problems 1. Solve the inequality 𝑥2 + 6𝑥 − 7 > 0. Graph the solution set.

-8

-7

-6

-5

-4

-3

-2

-1

2. Solve the inequality −𝑥2 − 3𝑥 ≥ −18. Graph the solution set.

-8

-7

-6

-5

-4

-3

-2

-1

3. Solve the inequality 𝑥2 + 8𝑥 ≤ 15. Write the solution in interval notation.

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4. Solve the inequality −2𝑥2 + 4𝑥 + 48 ≤ 0. Write the solution in set notation.

Review Problem 5. Solve the polynomial equation 16𝑥4 − 81 = 0.

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Unit 3, Lesson 16: Solving Two-Variable Quadratic Inequalities

Warm-Up: Graphing Quadratic Functions Select all the statements that are true about the function 𝑓(𝑥) = −𝑥2 − 5𝑥 + 6.

□ 𝑓(𝑥) has a 𝑦-intercept at (0, 6).

□ 𝑓(𝑥) has a 𝑦-intercept at (0, −6).

□ 𝑓(𝑥) has zeros at 𝑥 = 2 and 𝑥 = 3.

□ 𝑓(𝑥) has zeros at 𝑥 = −6 and 𝑥 = 1.

□ As 𝑥 → ∞, 𝑓(𝑥) → ∞ and as 𝑥 → −∞, 𝑓(𝑥) → −∞.

□ As 𝑥 → ∞, 𝑓(𝑥) → −∞ and as 𝑥 → −∞, 𝑓(𝑥) → −∞.

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Exploration Activity: Comparing Graphs of Quadratic Equations and Inequalities y

The graph of 𝑦 = (𝑥 + 4)(𝑥 + 2) is shown in red.

The graph of 𝑦 ≥ (𝑥 + 4)(𝑥 + 2) is shown in blue.

1. Discuss with your partner the similarities and differences between the 2 graphs. Summarize your discussion in the table. Similarities between the Red and Blue Graphs

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Differences between the Red and Blue Graphs

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The graph of 𝑦 > (𝑥 + 4)(𝑥 + 2) is shown in green.

2. Discuss with your partner the difference between the green and blue graphs. Summarize your discussion.

The graph of 𝑦 ≤ (𝑥 + 4)(𝑥 + 2) is given in purple.

Consider the shading of the blue, green, and purple graphs to answer the following. 3. Explain why the shading is above the parabola on the blue and green graphs. 0

4. Explain why the shading is below the parabola on the purple graph.

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Guided Activity: Graphing the Solution Set of a Two-Variable Quadratic Inequality 1. Asa graphed 𝑓(𝑥) = 𝑥2 − 4 on the coordinate grid shown.

Discuss the following with your partner, and explain your thinking. a. How can Asa’s graph be helpful in graphing the quadratic inequality 𝑦 < 𝑥2 − 4?

b. Should the graph of 𝑦 < 𝑥2 − 4 have a solid or dashed line? c. How can you determine where the graph of the inequality should be shaded to represent the solution set of the inequality?

2. Graph 𝑦 < 𝑥2 − 4.

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3. Consider the quadratic inequality 𝑦 ≥ −𝑥2 − 4𝑥 − 3. a. Find the 𝑥-intercepts of 𝑦 = −𝑥2 − 4𝑥 − 3.

b. Determine the vertex of 𝑦 = −𝑥2 − 4𝑥 − 3. c. Identify the 𝑦-intercept of 𝑦 = −𝑥2 − 4𝑥 − 3. d. Complete the statement. solid line dashed line

The graph 𝑦 ≥ −𝑥2 − 4𝑥 − 3 will have a and the solution set is shaded

above below

forming a parabola,

the parabola.

e. Graph 𝑦 ≥ −𝑥2 − 4𝑥 − 3. y

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Collaborative Activity: Comparing Graphs 1. Use the inequality 𝑦 < ( −8 − 𝑥)(𝑥 + 2) to complete the following. a. Complete the statement.

solid line dashed line

The graph of the quadratic inequality will have a parabola, and the solution set is shaded

above below

forming a

the parabola.

b. Complete the table by finding each key feature listed. 𝒙-intercept(s)

c. Graph 𝑦 < (−8 − 𝑥)(𝑥 + 2).

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Vertex

𝒚-intercept

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Lesson Summary Inequality symbols are used to compare 2 expressions or to compare when a function will have values that are greater than or less than a specific value. For example, 𝑓(𝑥) = (3𝑥 + 4)(𝑥 − 2) is a quadratic

function with zeros at 𝑥 = − 4 and 𝑥 = 2. From this 3

form, additional key features can be determined to graph 𝑓(𝑥), such as the 𝑦-intercept or other points on the graph. For now, the graph of 𝑓(𝑥) is shown.

To graph the inequality 𝑦 ≤ (3𝑥 + 4)(𝑥 − 2) means to find all the coordinate points that, when substituted into the inequality, will make the statement true.

• Since the comparison symbol is ≤, the curve will be solid (not dashed) to indicate that points along the curve are included in the solution set to the inequality. • Select a point, either above or below the curve, such as (0, 0), to substitute into the inequality, and check whether the point creates a true statement. In this case, 0 ≤ (3⋅0 + 4)(0 − 2) is not true. Therefore, the solution set does not include (0, 0). A similar analysis can be made with the inequality 𝑦 > (3𝑥 + 4)(𝑥 − 2).

• Since the comparison symbol is >, the curve will be dashed to indicate that points along the curve are not included in the solution set of the inequality. • The solution set will be shaded above the graph because, using the same point to verify, 0 < (3⋅0 + 4)(0 − 2) is true. The graphs of 𝑦 ≤ (3𝑥 + 4)(𝑥 − 2) and 𝑦 > (3𝑥 + 4)(𝑥 − 2) are shown.

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𝑦 ≤ (3𝑥 + 4)(𝑥 − 2)

𝑦 > (3𝑥 + 4)(𝑥 − 2)

Practice Problems 1. Graph 𝑦 > 𝑥2 − 4𝑥 + 3.

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2. Graph 𝑦 ≤ −2(𝑥 − 1)(𝑥 + 3).

Review Problems 3. Solve the inequality 𝑥2 − 3𝑥 + 8 > 6, and write the solution set in interval notation.

4. Solve 𝑥2 + 4𝑥 − 5 = −10.

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Unit 3, Lesson 17: Solving Polynomial Inequalities

Warm-Up: Graph Polynomial Function Some features of a polynomial function, 𝑔(𝑥), are listed. • 𝑔(𝑥) is a 4th-degree polynomial.

• The leading coefficient is negative. • The zeros for 𝑔(𝑥) are 𝑥 = 2, 𝑥 = 1, and 𝑥 = −1 with a multiplicity of 2. 1. Sketch a graph of 𝑔(𝑥) on the coordinate plane. y

2. Compare your graph with your partner’s. Identify 1 similarity and 1 difference between your graphs.

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Collaborative Activity: Solving One-Variable Polynomial Inequalities 1. Consider the function 𝑓(𝑥) = 𝑥3 + 𝑥2 − 9𝑥 − 9. a. Determine the zeros of 𝑓(𝑥).

b. Discuss with your partner how you can use the zeros of 𝑓(𝑥) to solve the inequality 𝑥3 + 𝑥2 − 9𝑥 − 9 > 0. c. Select a number in each interval created by the solutions of the equation, and test them in the inequality 𝑥3 + 𝑥2 − 9𝑥 − 9 > 0.

d. Graph the solution set of the inequality 𝑥3 + 𝑥2 − 9𝑥 − 9 > 0 on the number line.

-8

-7

-6

-5

-4

-3

-2

-1

e. Write the solution set of 𝑥3 + 𝑥2 − 9𝑥 − 9 > 0 in interval notation.

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2. Consider the function 𝑔(𝑥) = 𝑥4 + 2𝑥3 − 16𝑥2 − 32𝑥. a. Solve the equation 𝑥4 + 2𝑥3 − 16𝑥2 − 32𝑥 = 0.

b. Select a number in each interval created by the solutions of the equation, and test them in the inequality 𝑥4 + 2𝑥3 − 16𝑥2 − 32𝑥 ≤ 0.

c. Graph the solutions to the inequality on the number line.

-8

-7

-6

-5

-4

-3

-2

-1

d. Write the solution of 𝑥4 + 2𝑥3 − 16𝑥2 − 32𝑥 ≤ 0 in interval notation.

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Guided Activity: Solving Two-Variable Polynomial Inequalities Consider the polynomial function ℎ(𝑥) = (𝑥 + 1)(𝑥 −2)2(2𝑥 − 1).

1. Determine the zeros of the function ℎ(𝑥).

2. Complete the table of key features by identifying the key features of ℎ(𝑥). Key Features Sign of the leading coefficient Degree Left end behavior Right end behavior 𝑦-intercept

3. Some of the ordered pairs on the graph of ℎ(𝑥) include (1, 2), (− 0.5, −6.25), and (2.5, 3.5). Use the key features and provided ordered pairs to graph ℎ(𝑥) on the coordinate plane.

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4. Complete the statement. The graph 𝑦 ≤ (𝑥 + 1)(𝑥 −2)2(2𝑥 − 1) will have a and the solution set is shaded

above below

solid line dashed line

forming a curve,

the curve.

5. Complete the graph in question 3 so that it represents the solution set of the inequality 𝑦 ≤ (𝑥 + 1)(𝑥 − 2)2(2𝑥 − 1).

Lesson Summary To solve polynomial inequalities, use a similar strategy to solving quadratic inequalities. For example, consider the inequality 𝑥3 − 4𝑥2 + 36 ≥ 9𝑥. The steps for how to solve a polynomial inequality and an example solution are shown. Step 1. Rewrite the polynomial inequality as an equation. 2. Rewrite the polynomial equation in standard form so that 1 side is set equal to 0. 3. Factor the polynomial.

4. Determine the zeros of the equation.

5. Test points in each interval created by the zeros to determine when the inequality is true.

6. Write the solution set in interval or set notation.

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Work 𝑥3 − 4𝑥2 + 36 = 9𝑥 𝑥3 − 4𝑥2 − 9𝑥 + 36 = 0 (𝑥 − 4)(𝑥 + 3)(𝑥 − 3) = 0 𝑥 = 4, 𝑥 = 3, 𝑥 = −3

𝑥 < −3 Test: 𝑥 = −5 Inequality: False

−3 < 𝑥 < 3 Test: 𝑥 = 0 Inequality: True 3 < 𝑥 < 4 Test: 𝑥 = 3.5 Inequality: False 𝑥 > 4 Test: 𝑥 = 5 Inequality: True

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Practice Problems 1. Consider the inequality (𝑥 + 1)(𝑥2 − 5𝑥 + 6) > 0.

a. Solve the inequality. Write the solution in interval notation.

b. Graph the solution on the number line.

-8

-7

-6

-5

-4

-3

-2

-1

2. Graph the solution of the polynomial inequality 𝑦 ≤ (𝑥 − 1)(𝑥 + 2)2 on the coordinate plane.

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Review Problems 3. Factor the polynomial 𝑥3 − 𝑥2 + 4𝑥 − 4 completely.

4. Rewrite each expression in the form of 𝑎 + 𝑏𝑖. a. (3 + 4𝑖) + 3(1 − 2𝑖)

b. 2(7 − 5𝑖) − (4 + 9𝑖)

c. (2 + 3𝑖)(4 − 3𝑖)

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Unit 3, Lesson 18: Solving Polynomial Inequalities in a Real-World Context

Warm-Up: Quadratic Inequalities Match The graphs of 4 inequalities are shown. Graph A

Graph B

2 x

-8

-6

-4

-2

x -8

-6

-4

-2

-4

-6

-8

Graph C 8

Graph D

x -8

-6

-4

-2

x -8

-6

-4

-2

-4

-6

-8

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Match each quadratic inequality to its graph. 1. 𝑦 ≤ −𝑥2 + 4𝑥 + 5

_____

3. 𝑦 ≥ 𝑥2 − 4𝑥 − 5

_____

2. 𝑦 ≥ −𝑥2 + 6𝑥 − 5

4. 𝑦 ≤ 𝑥2 + 6𝑥 + 5

_____

Guided Activity: Solving Quadratic Inequalities in Real-World Contexts 1. Rowan plays on the high school baseball team. He hit the ball toward the outfield with an upward velocity of 40 feet per second (ft./sec.) from 3.5 ft. above the plate. Due to gravity, the acceleration of the ball is −16 ft./sec. squared. In relation to time, the equation for the height of the ball is 𝐻(𝑡) = −16𝑡2 + 40𝑡 + 3.5, where 𝑡 is the time in sec. a. Use the quadratic formula to solve 𝐻(𝑡) for 𝑡.

The inequality 0 ≤ −16𝑡2 + 40𝑡 + 3.5 represents the length of time, in sec., that the ball is in the air. b. Write the solution to the inequality in interval notation.

c. If the ball isn’t caught and lands in the outfield, how long was the ball in the air? Include units.

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2. Gustavo is building a fenced-in, rectangular dog run. He wants the length of the run to be 9 ft. longer than the width, 𝑤. To determine the dimensions of the dog run with an area of 220 square feet (sq. ft.) at most, Gustavo uses the inequality shown. 𝑤(9 + 𝑤) ≤ 220

a. Determine the solution set of the inequality.

b. Interpret the solution set of the inequality in context.

c. What are 2 possible dimensions of Gustavo’s dog run? Include units.

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3. Weston and Olivia rock climb together once a month. They use a rope that supports less mass than a traditional climbing rope and is longer. The rope Weston and Olivia use safely supports a mass, 𝑀, in pounds (lb.) modeled by 𝑀 < 4960(𝑟 – 1)2, where 𝑟 is the radius of the rope in inches (in.). a. Graph the solution set of the mass that is supported in relation to the radius of the rope.

6,000

5,500 5,000 4,500 4,000 3,500 3,000 2,500 2,000 1,500 1,000 500

-1

-0.5

0.5

1.5

2.5

b. Discuss with your partner which ordered pairs on the graph of 𝑀 < 4960(𝑟 – 1)2 are not viable for the context. Summarize your discussion.

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Collaborative Activity: Solving Polynomial Inequalities in Real-World Contexts 1. A part of the track of a new roller coaster ride is modeled by the function 𝑟(𝑥), where 𝑟(𝑥) is the height of the roller coaster above the ground, in ft., and 𝑥 is the time, in sec., since the ride began. The graph of 𝑟(𝑥) is shown. a. Identify and interpret the zeros of 𝑟(𝑥).

y 24

(11.9, 20.9)

20 16 12

(44.75, 10.6)

8 4 0 (0, 0) 10 -4

(25, 0)

(35, 0)

(55, 0) 40

(29.8, −3.7)

-8

b. Additional supports must be built when the track of the roller coaster goes below the ground. Shade the polynomial function that represents the solution to 𝑦 ≤ 0, where the roller coaster is below the ground level.

c. Which set represents the solution to the inequality 𝑦 ≤ 0? A. {𝑥| 0 ≤ 𝑥 ≤ 55}

B. {𝑥| 25 ≤ 𝑥 ≤35}

C. {𝑦| 25 ≤ 𝑦 ≤ 35}

D. {𝑦| −3.7 ≤ 𝑦 ≤ 20.9}

d. Interpret the solution to the inequality 𝑦 ≤ 0 in this scenario.

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Lesson Summary The process of solving polynomial inequalities in real-world scenarios is the same as solving them without context. However, after solving, the solution set needs to be considered in context to determine whether there are any nonviable solutions in the set. The stopping distance, in ft., for Kiara’s new car when she is traveling 𝑥 miles per hour (mph) is modeled by the function 𝑠(𝑥) = 1 𝑥2 + 𝑥. Kiara wants to know the speeds she 20 can travel to make sure she has at least 240 ft. of stopping distance in front of her. The inequality 𝑦 ≥ 1 𝑥2 + 𝑥 − 240 can be used to determine the speeds that Kiara can travel 20 and safely stop within 240 ft. •

200

Solve the quadratic equation 0 = 1 𝑥2 + 𝑥 − 240 2

to determine the zeros of the model. The zeros of the quadratic equation are 𝑥 = −80 and 𝑥 = 60.

• Use the inequality 𝑦 ≥ 1 𝑥2 + 𝑥 − 240 to 20

150 100 50

(–80, 0) -200 -150

(60, 0)

-100

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100

150

x 200

150

x 200

-100 -150 -200 -250 -300

• Although this graph represents the solution set to the mathematical inequality, it does not represent the solution set in context. • Since 𝑥 is the speed that Kiara’s car is traveling, it is restricted to 𝑥 ≥ 0. Additionally, 𝑦 ≤ 0 because it cannot take her longer than 240 ft. to stop. Therefore, within the solution set of the inequality 𝑦 ≥ 1 𝑥2 + 𝑥 − 240, only the solutions within 20 the interval 0 ≤ 𝑥 ≤ 60 are viable. As long as Kiara is traveling between 0 and 60 mph, she should have enough space to come to a stop within 240 ft. The graph of the solution set in context is shown.

-50

check whether (0, 0) is a solution to determine the shading of the graph. Since substituting (0, 0) into the inequality results in a true statement, the shading of the solution set should include the origin as shown.

When working with real-world contexts, it is important to verify which part(s) of the solution set is (are) viable.

200

150 100 50 -200 -150

-100

-50

(60, 0) 50

100

-50 -100 -150 -200 -250 -300

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Practice Problems 1. A small-business owner determines that her profit, in dollars, is modeled by 𝑃(𝑥) = −11𝑥2 + 1,080𝑥 − 21,500, where 𝑥 is the number of leather bags she produces. a. Graph 𝑦 < −11𝑥2 + 1,080𝑥 − 21,500. y

5500 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 x -5

-500 -1000

b. Determine the range of leather bags the owner must make to maintain a profit.

2. The volume of a chicken coop in the shape of a rectangular prism can be modeled using the function 𝑉(𝑥) = 3𝑥3 − 14𝑥2 + 15𝑥, where 𝑥 is the width of the coop in yards (yd.). Determine the solution set of possible values for the width of the coop that will be reasonable in this scenario.

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Review Problems 3. Solve 𝑥3 ≤ 4𝑥2. Write the solution set in interval notation.

4. An open-top box is formed by cutting squares out of an 11 inch (in.) by 17 in. piece of paper and then folding up the sides. The volume 𝑉(𝑥) in cubic inches (cu. in.) of this type of open-top box is a function of the side length 𝑥 in inches of the square cutouts and can be given by 𝑉(𝑥) = (17 − 2𝑥)(11 − 2𝑥)(𝑥). Rewrite this equation by expanding the polynomial.

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Unit 4: Rational Functions

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Unit 4, Lesson 1: Rewriting Rational Expressions

Warm-Up: Exploring Domain and Range 1. Find the range of the function 𝑦 = (𝑥 + 3)2 − 5. A. {𝑦|𝑦 ≠ −3} B. {𝑦|𝑦≠ −5}

C. {𝑦|𝑦 ≤ −3} D. {𝑦|𝑦 ≥ −5} E. {𝑦 ∈ 𝑅}

2. What is the domain of the graphed function? A. {0, 1, 2, 3, 4, 5}

B. {0, 1, 2, 3}

C. {𝑥: 0 < 𝑥 < 3} D. {𝑥: 0 < 𝑥 < 5}

E. All real values of 𝑥

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Guided Activity: Factoring Rational Expressions to Generate Equivalent Expressions 1.

Factor the numerator and denominator of the expression

Factors of 𝟓𝒙 + 𝟓

Write the factored form of

The expression 3.

5𝑥 + 5

𝑥 + 6𝑥 + 5 2

5𝑥 + 5

𝑥 + 6𝑥 + 5 2

5𝑥 + 5

𝑥2 + 6𝑥 + 5

Factors of 𝒙𝟐 + 𝟔𝒙 + 𝟓

is a rational expression.

For each 𝑥-value, determine if the rational expressions are equivalent. 𝒙

𝟓𝒙 + 𝟓

𝒙 + 6𝒙 + 𝟓 2

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𝟓

𝒙 + 𝟓

Are they equivalent? Yes No Yes No Yes No Yes No

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Discuss with your partner why the expressions are not always equivalent.

Explain the relationship between the domain and the 𝑥-values in the table where the 2 expressions were not equivalent.

Factor each algebraic expression. 𝟑𝒙 + 𝟗

𝒙𝟐 + 𝟕𝒙 + 𝟏𝟐

3𝑥 + 9

Use the factored expressions in problem 6 to rewrite

Rewrite the factored expression after eliminating the common factor.

Discuss with your partner why the original expression and the factored expression are equivalent but not equal. Summarize your discussion.

10.

What value of 𝑥 makes the expression

11.

Discuss with your partner whether 𝑥 = − 4 is also a restriction on the domain of 3𝑥 + 9 .

𝑥 + 4

𝑥 + 7𝑥 + 12 2

undefined?

𝑥2 + 7𝑥 + 12

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12.

Factor

2𝑥 + 4

2𝑥 + 5𝑥 + 2 2

to find an equivalent rational expression, and determine the

domain restrictions.

Original Expression

2𝑥 + 4 2𝑥2 + 5𝑥 +2

Equivalent Expression

Domain Restrictions

Collaborative Activity: Simplifying Rational Expressions Work with your partner to complete the station activities. Record your solution, and work in the space provided for each station.

Station 1

Station 2

Station 3

Station 4

Station 5

Station 6

Station 7

Station 8

Station 9

Station 10

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Lesson Summary Variables in algebra often represent real numbers. Therefore, operations with rational numbers and rational expressions follow the same rules and properties. A rational expression is a quotient of two polynomials with a non-zero denominator.

Rewriting rational expressions into irreducible forms is similar to rewriting numbers. Rewriting Rational Numbers

Rewriting Rational Expressions

12 = 18

𝑥2 − 5𝑥 + 6 𝑥2 − 4𝑥 + 4

6⋅2 6⋅3

2 3

(𝑥 − 3)(𝑥 − 2) (𝑥 − 2)(𝑥 − 2)

𝑥 − 3 𝑥 − 2

When working with rational expressions, domain restrictions must be included. Any 𝑥-value that, when substituted, results in a 0 in the denominator contributes to 𝑥2 − 5𝑥 + 6 is a necessary domain restriction. For example, the rational expression 2 𝑥 − 4𝑥 + 4 𝑥 − 3 with a domain restriction of 𝑥 ≠ 2. equivalent to 𝑥 − 2

Practice Problems 𝑥2 + 5𝑥 + 4 𝑥 + 4 1. Determine if the rational expressions 2 and (𝑥 + 1)(𝑥 + 2)(𝑥 − 3) 𝑥 − 𝑥 − 6 equivalent when 𝑥 ≠ −1, 𝑥 ≠ −2, and 𝑥 ≠ 3.

2. Rewrite

are

5𝑥 + 25 by factoring, and determine any domain restrictions. 𝑥 + 9𝑥 + 20 2

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Review Problems 3. The graph of a polynomial function 𝑓 is shown. Which statement is true about the end behavior of the polynomial function? A. As 𝑥 → −∞, 𝑦 → ∞, and as 𝑥 → ∞, 𝑦 → −∞.

B. As 𝑥 → −∞, 𝑦 → −∞, and as 𝑥 → ∞, 𝑦 → −∞. C. As 𝑥 → −∞, 𝑦 → ∞, and as 𝑥 → ∞, 𝑦 → ∞.

D. As 𝑥 → −∞, 𝑦 → −∞, and as 𝑥 → ∞, 𝑦 → ∞.

4. The graph shows the height of a tennis ball 𝑡 seconds after it has been hit. The function 𝑓 given by 𝑓(𝑡) = 5 + 30𝑡 − 32𝑡2 models the height of the ball in feet (ft.). a. How high was the ball when it was hit? Where do you see this in the equation?

b. Suppose a second ball follows the same trajectory but is hit from 7 ft. off the ground. Sketch the graph of the height of the second ball on the same axes.

c. Write an equation for a function 𝑔 that defines the height 𝑔(𝑡), in feet, of the second ball hit from 7 ft. off the ground in terms of 𝑓(𝑡).

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Unit 4, Lesson 2: Graphs of Rational Functions – Part 1

Warm-Up: Biking 10 Miles (Part 1) Kiran’s aunt plans to bike 10 miles (mi.). 1. How long will it take if she bikes at an average rate of 8 miles per hour (mph)?

2. How long will it take if she bikes at an average rate of 𝑟 mph?

3. Kiran wants to join his aunt, but he only has 45 minutes (min.) to exercise. What will their average rate need to be for him to finish on time?

4. What will their average rate need to be if they have 𝑡 hours (hr.) to exercise?

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Exploration Activity: Biking 10 Miles (Part 2) Kiran plans to bike 10 mi. 1. Write an equation that gives his time 𝑡, in hr., as a function of his rate 𝑟, in mph.

2. Graph 𝑦 = 𝑡(𝑟).

18 16 14 12

3. What is the meaning of 𝑡(8)? Does this value make sense? Explain your reasoning.

10 8 6 4 2

4. What is the meaning of 𝑡(0)? Does this value make sense? Explain your reasoning.

8 10 12 14 16 18 20

5. As 𝑟 gets closer and closer to 0, what does the behavior of the function tell you about the situation?

6. As 𝑟 gets larger and larger, what does the end behavior of the function tell you about the situation?

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Guided Activity: Rational Functions The equation and parent function for a rational function are shown.

Function Type

Parent Function

Graph 5

4 3 2 1

Rational

𝑓(𝑥) =

1 𝑥

-5 -4 -3 -2 -1 0 -1

x 1

-2 -3 -4 -5

1. Use the parent function to complete the table of values.

𝒙 𝑦

–4

–2

–1

2. Notice that the graph has an asymptote at 𝑥 = 0. Use the equation of the parent function to explain why 𝑥 ≠ 0. 3. Marcus noticed that there was also an asymptote at 𝑦 = 0. He asked his partner, DaiJa, to help him figure out why 𝑦 ≠ 0. DaiJa wrote the following and asked, “Could a fraction ever equal zero?”

Marcus responded, “The only way for a fraction to equal 0 is if the numerator can equal 0.”

Review DaiJa’s work and Marcus’s conclusion with your partner. Use DaiJa’s work and Marcus’s conclusion to write an explanation as to why there is an asymptote at 𝑦 = 0.

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4. Complete the table to describe the key features of the parent function, 𝑓(𝑥) = Key Feature

Description

1 . 𝑥

Domain Range 𝑥-intercept(s) 𝑦-intercept(s) Slope Vertex Axis of symmetry Intervals where the function is increasing or decreasing Intervals where the function is positive or negative End behavior

Collaborative Activity: Card Sort: Graphs of Rational Functions Your teacher will give you a set of cards. Match each rational function with its graphical representation.

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Lesson Summary The distance, 𝑑, that an object moving at constant speed travels is based on the length of time, 𝑡, the object travels and the speed, 𝑟, of the object. Often, this relationship is written as 𝑑 = 𝑟 ⋅ 𝑡. It could also be written as 𝑟 = 𝑑 or 𝑡 = 𝑑 . 𝑡 𝑟

For example, the distance across the English Channel from Dover in England to Calais in France is 33.3 kilometers (km). The time, in hr., it takes for a boat to make this crossing can be modeled by the function 𝑡(𝑟) = 33.3 , where 𝑟 is measured in km per hr. 𝑟 For small values of 𝑟, the journey takes a long time. For larger values of 𝑟 (and a fast boat), the trip is shorter. The graph of the function shows how the travel time decreases as the speed of the boat increases. This observation can also be written as an end behavior: as 𝑟 → ∞, 𝑡 → 0, and as 𝑟 → 0, 𝑡 → ∞. The function 𝑡 is a rational function.

A rational function is a function defined by a fraction with polynomials in the numerator and denominator. Rational functions include polynomials because a polynomial can be written as a fraction with denominator 1.

Unlike the graphs of polynomial functions that look smooth and connected, the graphs of some rational functions can look like separate pieces. For example, the graph of 1 is shown. 𝑓(𝑥) = 𝑥 – 3

The dashed line at 𝑥 = 3 is a representation of a vertical asymptote.

The line x = a is a vertical asymptote for a function f if f is undefined at x = a and its outputs get larger and larger in the negative or positive direction when x gets closer and closer to a on each side of the line.

As 𝑥 gets closer and closer to 3, consider what happens to the value of 𝑓(𝑥). When the numerator, 1, is divided by a negative number with a small absolute value, the result is a negative number with a very large absolute value. This is shown on the left side of the vertical asymptote. When the numerator, 1, is divided by a very small positive number, the result is a very big positive number, which is what is shown on the right © Accelerate Learning Inc. - All Rights Reserved

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side of the vertical asymptote. It is important to note that the drawn-in asymptote is not actually part of the graph of the function. Instead, it is a helpful reminder that the function has no value at 𝑥 = 3 and very large absolute values at inputs very close to 𝑥 = 3.

Practice Problems 1. Jada is planning a kayak trip. She finds an expression for the time, T(𝑠), in hr. it takes her to paddle 10 km upstream in terms of 𝑠, the speed of the current in km per hr. This is the graph Jada gets if she allows 𝑠 to take on any value between 0 and 7.5. a. What would be a more appropriate domain for Jada to use instead?

b. What is the approximate speed of the current if her trip takes 6 hr.?

2. A cylindrical can needs to have a volume of 6 cubic inches (cu. in.). A label is to go 12 gives the area of the label in around the side of the can. The function S(𝑟) = 𝑟 square inches (sq. in.) where 𝑟 is the radius of the can in inches (in.).

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a. As 𝑟 gets closer and closer to 0, what does the behavior of the function tell you about the situation?

b. As 𝑟 gets larger and larger, what does the end behavior of the function tell you about the situation?

3. What is the equation of the vertical asymptote for the graph of the rational function 6 g(𝑥) = ? 𝑥 − 1 A. 𝑥 = 1

B. 𝑥 = −1 C. 𝑥 = 6 D. 𝑥 =

1 6

Review Problem 4. Is this the graph of g(𝑥) = −𝑥2(𝑥 − 2) or ℎ(𝑥) = 𝑥2(𝑥 − 2)? Explain how you know.

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Unit 4, Lesson 3: Graphs of Rational Functions – Part 2

Warm-Up: Rewritten Equations Decide if each of these equations is true or false for 𝑥-values that do not result in a denominator of 0. Be prepared to explain your reasoning. 1.

𝑥 + 7 7 =1+ 𝑥 𝑥

𝑥 𝑥 =1+ 𝑥 + 7 7

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Guided Activity: Publishing a Paperback Let 𝑐 be the function that gives the average cost per book 𝑐(𝑥), in dollars, when using an online store to print 𝑥 copies of a self-published paperback book. Here is a graph of 120 + 4𝑥 𝑐(𝑥) = . 𝑥 1. What is the approximate cost per book when 50 books are printed? 100 books? 50 books: __________ 100 books: __________

2. The author plans to charge $8 per book. About how many should be printed to make a profit?

3. What is the value of 𝑐(𝑥) when 𝑥 =

1 ? How does this relate to the context? 2

4. What does the end behavior of the function say about the context?

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Exploration Activity: Horizontal Asymptotes Here are four graphs of rational functions. 1. Match each function with its graphical representation. a. 𝑎(𝑥) = b. 𝑏(𝑥) =

4 −1 𝑥

1 −4 𝑥

c. 𝑐(𝑥) =

1 + 4𝑥 𝑥

e. 𝑒(𝑥) =

1 − 4𝑥 𝑥

d. 𝑑(𝑥) =

f. 𝑓(𝑥) =

𝑥 + 4 𝑥

4−𝑥 𝑥

g. 𝑔(𝑥) = 1 +

h. ℎ(𝑥) =

4 𝑥

1 +4 𝑥

2. Where do you see the horizontal asymptote of the graph in the expressions for the functions?

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Lesson Summary Consider the rational function 𝑓(𝑥) =

3𝑥 + 1 . Written in this form, the function’s vertical 𝑥

asymptote at 𝑥 = 0 is revealed by using the denominator to determine any value that would cause division by 0. But what can be determined about the value of 𝑓(𝑥) for values of 𝑥 far away from the vertical asymptote? One way to think about these values is to rewrite the expression for 𝑓(𝑥) by breaking 3𝑥 1 1 up the fraction, which results in 𝑓(𝑥) = + , and rewriting it as 𝑓(𝑥) = 3 + . 𝑥 𝑥 𝑥

Written this way, it’s easier to see that as 𝑥 gets larger and larger in either the positive 1 or negative direction, the term will get closer and closer to 0. Because of this, it 𝑥

can be determined that the value of the function will get closer and closer to 𝑦 = 3.

𝑎(𝑥) 𝑟(𝑥) can be rewritten as 𝑔(𝑥) = 𝑐 + , 𝑏(𝑥) 𝑏(𝑥) 𝑟(𝑥) gets where 𝑐 is a constant and 𝑟(𝑥) and 𝑏(𝑥) are polynomial expressions such that 𝑏(𝑥) More generally, if a rational function 𝑔(𝑥) =

closer and closer to 0 as 𝑥 gets larger and larger in both the positive and negative directions, then 𝑔(𝑥) will likewise get closer and closer to 𝑐. Rational functions of this type have a horizontal asymptote at the constant value. The line y = c is a horizontal asymptote of a function if the outputs of the function get closer and closer to c as the inputs gets larger and larger in either positive or negative direction.

Rational function 𝑓(𝑥) with a horizontal asymptote at 𝑦 = 𝑐 has an end behavior that can be described as shown. As 𝑥 → ∞, 𝑦 → 𝑐, and as 𝑥 → −∞, 𝑦 → 𝑐.

Other functions with horizontal asymptotes have a similar end behavior, such as exponential growth and decay functions. In future courses, end behavior as it relates to horizontal asymptotes will be studied in greater detail.

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Practice Problems 1. Rewrite the rational function 𝑔(𝑥) = are constants.

𝑥 − 4 𝑟 in the form 𝑔(𝑥) = 𝑐 + , where 𝑐 and 𝑟 𝑥 𝑥

2. The average cost (in dollars) per mile (mi.) for riding 𝑥 mi. in a cab is 2.5 + 2𝑥 𝑐(𝑥) = . As 𝑥 gets larger and larger, what does the end behavior of the 𝑥 function tell you about the situation?

3. The graphs of two rational functions 𝑓 and 𝑔 are shown. One of them is given by the 2 − 3𝑥 . Which graph is it? Explain how you know. expression 𝑥

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Review Problems 4. Which polynomial function’s graph is shown here? A. 𝑓(𝑥) = (𝑥 + 1)(𝑥 + 2)(𝑥 + 5) B. 𝑓(𝑥) = (𝑥 + 1)(𝑥 − 2)(𝑥 − 5) C. 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 2)(𝑥 + 5) D. 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 5)

5. State the degree and end behavior of 𝑓(𝑥) = 5𝑥3 − 2𝑥4 − 6𝑥2 − 3𝑥 + 7. Explain or show your reasoning.

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Unit 4, Lesson 4: Key Features of Rational Functions

Warm-Up: Vocabulary 1. Match the terms from the word bank with their corresponding descriptions. Word Bank A. Increasing intervals

D. Horizontal asymptotes

B. Decreasing intervals

E. Domain

C. Vertical asymptotes

F. Range

Vocabulary Term

Description Lines that correspond to the zeros of the denominator of a rational function 𝑦-values on a graph that a function approaches but does not actually reach The complete set of possible values of the input of a function Reading a graph from left to right, as the 𝑥-values increase, the 𝑦-values increase

The complete set of possible values of the output of a function Reading a graph from left to right, as the 𝑥-values increase, the 𝑦-values decrease

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Guided Activity: Key Features of the Graph of a Rational Function 1. Use a graphing tool to graph ℎ(𝑥) =

𝑥 . 𝑥 − 3

a. Sketch the graph of ℎ(𝑥) on the coordinate plane.

-5

b. What is occurring at 𝑥 = 3? c. When does the function cross 𝑦 = 1? If it does not, explain why. d. Draw a vertical dashed line on the coordinate plane to represent the vertical asymptote of the function.

e. Draw a horizontal dashed line on the coordinate plane to represent the horizontal asymptote of the function.

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f. With your partner, use the table of values and the graph on your graphing tool to discuss the key features of ℎ(𝑥). Then, in the table, write the key features you are able to determine. Key Features Domain Range Horizontal asymptote Vertical asymptote Increasing interval(s) Decreasing interval(s) Left end behavior Right end behavior

Collaborative Activity: Key Features of the Graph of a Rational Function 1. Consider the function g(𝑥) =

𝑥 − 8 . 𝑥 + 1

a. Discuss with your partner where you think the vertical asymptote of 𝑔(𝑥) will be. Summarize your discussion.

b. Use a graphing tool to graph 𝑔(𝑥) =

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𝑥 − 8 . 𝑥 + 1

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c. Sketch the graph of 𝑔(𝑥) on the coordinate plane. 25

20 15 10 5 -25

-20

-15

-10

-5

-5 -10 -15

d. Compare the vertical asymptote on the graph to your initial assumption. e. Draw a dashed line on the coordinate plane to represent the vertical asymptote. f. Draw a dashed line on the coordinate plane to represent the horizontal asymptote. g. Using the graph and the table of values on your graphing tool, identify the key features for 𝑔(𝑥). Key Features

Increasing interval(s)

𝑥-intercept(s) 𝑦-intercept

Decreasing interval(s)

Domain

Horizontal asymptote

Range

Vertical asymptote

Left end behavior

Right end behavior

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Exploration Activity: Determining Vertical Asymptotes Using Graphs and Equations 1. The vertical asymptote for the function 𝑎(𝑥) =

12𝑥 − 60 is 𝑥 = −7. 𝑥 + 2𝑥 − 35 2

a. Rewrite 𝑎(𝑥) by factoring the numerator and denominator.

b. What do you notice about the factors in the numerator and the denominator?

c. Discuss with your partner how to determine the location of the vertical asymptote without graphing the function. Summarize your discussion.

d. Why do you think 𝑥 = 5 is not a vertical asymptote?

e. Use the graph and the table of values on your graphing tool to discuss with your partner what happens at 𝑥 = 5. Summarize your discussion.

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2. Consider the function 𝑡(𝑥) =

(𝑥 + 3) . (𝑥 − 4)(𝑥 + 3)

a. Hypothesize the vertical asymptote of t(𝑥).

b. Use a graphing tool to graph 𝑡(𝑥) =

(𝑥 + 3) . (𝑥 − 4)(𝑥 + 3)

c. Discuss with your partner if your hypothesis was correct or incorrect. If it was incorrect, discuss what the correct vertical asymptote is and how it can be (𝑥 + 3) . Summarize your discussion. identified from (𝑥 − 4)(𝑥 + 3)

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Collaborative Activity: Determining Horizontal Asymptotes Using Graphs and Equations 1. Use a graphing tool to determine the horizontal asymptotes of 𝑥 𝑥2 𝑎(𝑥) = , 𝑡(𝑥) = , and 𝑥2 + 4 𝑥2 + 4 𝑝(𝑥) =

𝑥2 . 𝑥 + 4

Function 𝑎(𝑥) = 𝑡(𝑥) = 𝑝(𝑥) =

Horizontal Asymptote

𝑥 𝑥 +4 2

𝑥2 𝑥2 + 4 𝑥2 𝑥 + 4

2. Discuss with your partner the differences in the graphs, equations, and horizontal asymptotes. Make note of your observations.

3. Complete the table summarizing how to determine the horizontal asymptote of a rational function. Degree of the Numerator Compared to Degree of the Denominator

Horizontal Asymptote ¡ No horizontal asymptote

Numerator < Denominator

¡ 𝑦=

¡ 𝑦=0

coefficient of numerator coefficient of denominator

¡ No horizontal asymptote

Numerator = Denominator

¡ 𝑦=

¡ 𝑦=0

coefficient of numerator coefficient of denominator

¡ No horizontal asymptote Numerator > Denominator

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¡ 𝑦=

¡ 𝑦=0

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Lesson Summary This lesson explored how to determine key features from expressions of rational functions, including zeros, vertical asymptotes, and horizontal asymptotes. • Zeros, or 𝑥-intercepts, of rational functions occur at the 𝑥-values that make the numerator equal to 0. • Vertical asymptotes of rational functions occur at the 𝑥-values that make the denominator equal to 0. Note that asymptotes are not part of the actual graph of the function but, rather, create boundaries for the graph. For this reason, vertical and horizontal asymptotes are graphed using dashed lines to indicate they are not included. To determine any vertical asymptotes of a rational function, begin by factoring the function. Then, simplify the function. From the simplified equation, set the expression in the denominator equal to 0, and solve for 𝑥.

Any factor that is eliminated when simplifying is represented by an open circle on the graph. The function is not continuous at this point. These values will be explored more closely in the next lesson. An irreducible fraction is a fraction in which the numerator and denominator are integers that have no common factors other than 1. The irreducible form of a 𝑥 + 1 . The rational function is similarly defined. Consider the rational expression 𝑥2 − 1 𝑥 + 1 , where (𝑥 + 1) is a common factor, expression can be rewritten as (𝑥 + 1)(𝑥 − 1) 1 which means the expression is also equivalent. Although the expressions are 𝑥 − 1 𝑥 + 1 1 equivalent, the graphs of functions 𝑓(𝑥) = are not the same. and 𝑓(𝑥) = 2 𝑥 −1 𝑥 − 1 This will be explored in more detail later in the unit. • To determine any horizontal asymptotes of a rational function, compare the degree of the numerator to the degree of the denominator. In some cases, rational functions do not have a horizontal asymptote but, rather, have an oblique asymptote, which will be studied in future math courses.

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Practice Problems 1. Use a graphing tool to complete the table. For each equation, determine whether asymptotes exist. If so, write the equation of the asymptote.

𝒚 =

𝟓 + 𝟐𝒙𝟐 𝟐 − 𝒙 − 𝒙𝟐

𝒚 =

Vertical Asymptote(s)

𝒙 − 𝟏 𝒙𝟐 − 𝟐𝒙 + 𝟏

Vertical Asymptote(s)

¡ None

¡ Yes. Write the equation of all vertical asymptotes.

Horizontal Asymptote(s)

¡ None

¡ Yes. Write the equation of all horizontal asymptotes.

2. Complete the key features table for the function 𝑓(𝑥) = Key Features 𝑥-intercept(s)

−𝑥 . 𝑥+ 3

Increasing interval(s)

𝑦-intercept

Decreasing interval(s)

Domain

Horizontal asymptote

Range

Vertical asymptote

Left end behavior

Right end behavior

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Review Problem 3. The graphs of two rational functions 𝑓 and 𝑔 are shown. Which function must be 10 ? Explain how you know. given by the expression of 𝑥 − 3

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Unit 4, Lesson 5: Graphs of Rational Functions with Removeable Discontinuities

Warm-Up: Factorable Quadratics In the first column, fill in the blanks with integers so that each quadratic expression is factorable. Then, factor each expression.

Expression

Factored Expression

1. 𝑥2 + __________ 𝑥 + 4 2. 𝑥2 + __________ 𝑥 − 12 3. 3𝑥2 +__________ 𝑥 + 8 4. 2𝑥2 + 3𝑥 + __________

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Collaborative Activity: Removable Discontinuities 1. Factor the equation 𝑦 =

𝑥 . Then, simplify. 𝑥2 – 8𝑥

Graph of 𝒚 =

Equation in Factored Form

𝒙 𝒙 − 𝟖𝒙 𝟐

𝑦 = _________

3 2 1

Removable Discontinuity

Simplified Equation

-4

-2

12 x

-1 -2

𝑦 =_________

-3 -4

1 2. Discuss with your partner what is occurring at point (0, − ). Summarize your 8 discussion by describing why you think the graph has a removable discontinuity.

Occasionally, a graph will contain a point, or multiple points, where the function is undefined. This point is indicated on the graph by an open circle. In further studies of mathematics, you will learn why this point is called a removable discontinuity. 𝑥 is a common factor of the numerator and the 𝑥 – 8𝑥 denominator. The zero of this factor is 𝑥 = 0. When 0 is substituted into the 1 1 1 , the result is − . This point (0, − ) represents the hole simplified expression, 𝑥 – 8𝑥 8 8 known as the removable discontinuity. Notice that 𝑥 in the expression

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3. Draw dashed lines to represent the horizontal and vertical asymptotes on the coordinate plane shown. 4. With your partner, use the graph of 𝑦 =

𝑥 to complete the following. 𝑥 – 8𝑥 2

a. Describe the graph of the function to the left of 𝑥 = 0.

b. Describe the graph of the function from 𝑥 = 0 to 𝑥 = 8.

c. Describe the graph of the function to the right of 𝑥 = 8.

When determining the domain and range of a rational function, it is important to consider any removable discontinuities and asymptotes. d. Determine the domain of the function using interval notation.

e. Determine the range of the function using interval notation.

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Guided Activity: Determining Key Features and Graphing a Rational Function from an Equation 1. Complete the table for the function 𝑘(𝑥) =

Factored Form

𝒌𝒙 =

2𝑥 − 1 . 𝑥 + 4

𝟐𝒙 − 𝟏 𝒙 + 𝟒

• Factor completely. • Simplify if possible. Removable Discontinuities • Could the function be simplified? ¡

No. There are no removable discontinuities. Move to the next step. Yes. There are removable discontinuities. To find the point(s) of discontinuity, set the common factor equal to 0, and solve for 𝑥. To find the value of 𝑦, substitute 𝑥 in the simplified equation, and solve for 𝑦.

Express as a point (𝑥, 𝑦). 𝒙-intercept(s)

• Use the simplified equation.

𝒚-intercept

• To find 𝑥-intercept, set 𝑦 = 0, and solve for 𝑥.

• To find the 𝑦-intercept, set 𝑥 = 0, and solve for 𝑦.

Express as a point (𝑥, 0).

Vertical Asymptote(s)

• Use the original equation.

Express as a point (0, 𝑦).

• Use the simplified equation. • Set the expression in the denominator equal to 0. Solve for 𝑥. Write the equation(s) for the vertical asymptote(s), if applicable. 444 | Unit 4

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Horizontal Asymptote

𝒌𝒙 =

𝟐𝒙 − 𝟏 𝒙 + 𝟒

• Compare the degree of the numerator and denominator. ¡

If 𝑁 < 𝐷, then the horizontal asymptote is 𝑦 = 0.

𝐴 If 𝑁 = 𝐷, then the horizontal asymptote is 𝑦 = , where 𝐴 and 𝐵 are the 𝐵 leading coefficients.

Write the equation for the horizontal asymptote, if applicable. Domain

Range

2. Graph 𝑘(𝑥) =

2𝑥 − 1 . 𝑥 + 4

As part of your graph, include any removable discontinuities, 𝑥-intercept(s), 𝑦-intercept, vertical asymptote(s), and horizontal asymptote.

12 11 10 9 8 7 6 5 4 3 2 1

3. Determine the interval(s) where the graph is increasing.

-13 -12 -11 -10 -9

-8

-7

-6

-5

-4

-3

-2

-1

0 -1

x 1

-2 -3 -4 -5 -6 -7

4. Determine the interval(s) where the graph is decreasing.

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Collaborative Activity: Graphing a Rational Function from a Written Description 1. Sketch the graph of a rational function using the following information. • The function is undefined at 𝑥 = −3, 𝑥 = 1, and 𝑦 = 0.

4 3

• The function crosses the horizontal asymptote at (0, 0). • The domain is (−∞, −3) ∪ (−3, 1) ∪ (1, ∞).

2 1 -5

-4

6 • The function has the points: (2, ), 5 1 6 5 ( , − ), (0, 0), (−2, 2), (−5, − ). 2 7 4

-3

-2

-1

-1 -2 -3 -4

2. For what interval(s) is the function increasing?

3. For what interval(s) is the function decreasing?

4. What is the end behavior of the function as 𝑥 → −∞ and as 𝑥 → ∞?

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Lesson Summary Rational functions include several key features you have explored with other functions, such as 𝑥- and 𝑦-intercepts, domain, range, and increasing or decreasing intervals. Some features are unique to rational functions, including vertical and horizontal asymptotes and removeable discontinuities. Steps for identifying different key features of rational functions are listed. • Factor both the numerator and denominator of the rational function. Rewrite the rational function in irreducible form. • Identify any common factors if possible. Write the point(s) of removeable discontinuity by setting any common factor equal to 0 and solving for 𝑥. To find the coordinate point, substitute 𝑥 into the irreducible equation, and solve for 𝑦. • Use the irreducible form to identify the 𝑥-intercept. Use the original expression to identify the 𝑦-intercept. • Use the irreducible form to determine the vertical asymptote(s) by setting the expression in the denominator equal to 0 and solving for 𝑥. • Compare the degree of the numerator and the denominator to determine the equation of the horizontal asymptote if applicable.

• Use the point(s) of removeable discontinuity and asymptote(s) to write the domain and range of the rational function. • Use the graph to determine increasing and decreasing intervals. 𝑥 − 3 . The function can be rewritten in factored form 𝑓(𝑥) = 2𝑥 − 5𝑥 − 3 𝑥 − 3 1 and then rewritten in irreducible form 𝑓(𝑥) = . The key features of (𝑥 − 3)(2𝑥 + 1) 2𝑥 + 1

Consider 𝑓(𝑥) =

𝑓(𝑥) are shown.

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Key Features 𝑥-intercept(s)

None

𝑦-intercept

(0, 1)

Domain

Increasing interval(s)

None

Decreasing interval(s)

1 1 (−∞, − ) ∪ (− , ∞) 2 2

Range Left end behavior

Removable discontinuity

1 {𝑥 | 𝑥 ≠ − } 2

(−∞, 0) ∪ (0, ∞) {𝑥 | 𝑥 ≠ 0}

As 𝑥 → −∞, 𝑦 → 0.

1 {𝑥 | 𝑥 ≠ − } 2 1 (3, ) 7

Asymptote(s) Right end behavior

1 𝑥 = − and 𝑦 = 0 2 As 𝑥 → ∞, 𝑦 →0.

Practice Problems 1. Complete the table of key features for the function 𝑡(𝑥) = Key Features 𝑥-intercept(s) 𝑦-intercept

𝑥 − 1 . 𝑥 − 2𝑥 + 1 2

Increasing interval(s) Decreasing interval(s)

Domain

Removable discontinuity

Range

Asymptote(s)

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2. Sketch the graph of the function 𝑝(𝑥) = 6

5𝑥 − 25 by first determining key features. 𝑥 − 9𝑥 + 20 2

5 4 3 2 1 x

-4

-3

-2

-1 0 -1

-2 -3 -4 -5

Review Problems 3. The polynomial function 𝑝(𝑥) = 𝑥3 + 3𝑥2 − 6𝑥 − 8 has a known factor of (𝑥 + 4). a. Rewrite 𝑝(𝑥) as the product of linear factors.

b. Draw a rough sketch of the graph of the function.

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4. Complete the table by identifying the parent function and the transformation applied to the parent function to create the transformed function.

Transformed Function

Parent Function

Transformation(s)

𝑓(𝑥) = (𝑥 − 3)2 + 1 𝑔(𝑥) = 2𝑥 + 1 − 3 ℎ(𝑥) = −(𝑥 + 1)3

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Unit 4, Lesson 6: Multiplying and Dividing Rational Expressions

Warm-Up: Rewriting Rational Expressions Rewrite each rational expression in irreducible form, and determine any domain restrictions. 1.

𝑥2 + 5𝑥 − 14 𝑥 + 7

(𝑥 − 1)2 2. (𝑥 − 1)(𝑥2 − 1)

(𝑥 − 12)(𝑥 + 3)(𝑥 − 1) 𝑥2 − 13𝑥 + 12

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Guided Activity: Multiplying Rational Expressions 1. Find each product. a.

3 2 ⋅ 5 6

3𝑥2 2𝑦3 b. ⋅ 5𝑦 6𝑥 c. (𝑥 + 3) ⋅

2 𝑥 + 3𝑥 2

2. Discuss with your partner the strategies that you used in problem 1. 3. Complete the table by multiplying each rational expression by Rational Expression

Work

𝑥 + 3 . 2𝑥 + 1

Product

4𝑥 + 2 3𝑥 𝑥 + 2 𝑥 + 6𝑥 + 9 2

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4. Complete the steps for multiplying the expressions first steps are partially completed.

𝑥2 + 𝑥 − 12 𝑥 + 9 ⋅ 2 . The 8𝑥 + 72 𝑥 − 4𝑥 − 32

Step

Work 𝑥2 + 𝑥 − 12

Factor the numerators.

𝑥 + 9

(𝑥 + 4)(𝑥 − 3)

(𝑥 + 9)

Factor the denominators.

8𝑥 + 72

𝑥2 − 4𝑥 − 32

Rewrite the rational expressions in factored form.

(𝑥 + 4)(𝑥 − 3)

⋅

(𝑥 + 9)

Determine the domain restriction(s) by setting the denominator equal to 0. Reduce any common factors that exist in the numerator and denominator. Multiply the remaining numerators, and multiply the remaining denominators.

5. Determine the product of restrictions.

10𝑥 + 15 3𝑥 − 6 and in irreducible form. State the domain 8 − 4𝑥 2𝑥 + 3

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Guided Activity: Dividing Rational Expressions 1. Divide each expression. a.

3 4 ÷ 5 5

9𝑥3 3 b. ⋅ 5𝑥𝑧3 𝑧 c. (𝑥 + 3)2 ÷

𝑥2 + 8𝑥 + 15 2𝑥

2. Ahmed found the quotient of work is shown.

7𝑥 𝑥2 − 6𝑥 ÷ 2 , but he made an error. His 2𝑥 − 10 𝑥 − 11𝑥 + 30

𝑥2 − 6𝑥 7𝑥 7𝑥 𝑥(𝑥 − 6) 7𝑥 𝑥 7𝑥2 ÷ 2 ⋅ = ⋅ = = 𝑥 − 11𝑥 + 30 2𝑥 − 10 2(𝑥 − 5) (𝑥 − 6)(𝑥 − 5) 2(𝑥 − 5) (𝑥 − 5) 2(𝑥 − 5)2 a. Explain to Ahmed the error in his solution.

b. Determine the correct quotient.

c. Identify the domain restrictions for the quotient.

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3. Divide

𝑥 + 2 𝑥 − 2𝑥 − 3 2

𝑥2 − 4 𝑥2 − 9

. State the domain restrictions.

Collaborative Activity: Multiple Operations Complete the given operations. State any domain restrictions. 1.

3𝑥 − 6 4𝑥 − 4

∙

𝑥2 + 2𝑥 − 3 2𝑥 + 12 ÷ 2 𝑥 − 3𝑥 − 10 8𝑥 + 16

2ℎ2 + 10ℎ ℎ2 + 10ℎ + 24 ℎ+4 ÷ 2 ∙ 2 ℎ−1 ℎ + 8ℎ − 9 8ℎ + 12ℎ

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Lesson Summary 45 72 ⋅ , it is often easier to find common factors 24 105 in the numerators and the denominators before multiplying. The expression can be rewritten in an equivalent form to reveal common factors, as shown.

When multiplying fractions such as

5∙9 3∙3∙8 ⋅ 8∙3 3∙7∙5

To write the product in irreducible form, find common factors that are equivalent to Then, express the product using the resulting numerator and denominator.

1 . 1

5∙9 3∙3∙8 5∙8∙3∙3∙9 9 9 ⋅ = =1∙ = 8∙3 3∙7∙5 5∙8∙3∙3∙7 7 7

The same strategies can be used to multiply rational expressions such as 3𝑥 + 18 4𝑥 − 8 ∙ . The product of rational expressions can be rewritten in factored form 4𝑥 − 24 5𝑥 − 10

3(𝑥 + 6) 4(𝑥 − 2) ∙ . Since algebraic expressions represent relationships for all real 4(𝑥 − 6) 5(𝑥 − 2) 3(𝑥 + 6) 4(𝑥 − 2) numbers, a binomial such as (𝑥 − 2) in ∙ is a common factor, 4(𝑥 − 6) 5(𝑥 − 2) whereas (𝑥 + 6) and (𝑥 − 6) are not common factors because (𝑥 + 6) and (𝑥 − 6) do not produce the same value for any value of 𝑥. The resulting product can be expressed 3(𝑥 + 6) as , where 𝑥 ≠ 2 and 𝑥 ≠ 6. 5(𝑥 − 6) 𝑎 𝑐 𝑎 𝑑 𝑎𝑑 ÷ is rewritten as ∙ , or . The same 𝑏 𝑑 𝑏 𝑐 𝑏𝑐 process of rewriting division as multiplication applies to division of rational expressions. 3𝑥2 9𝑥2 − 45𝑥 ÷ . Consider the quotient 𝑥2 − 7𝑥 + 10 𝑥2 − 4𝑥

When dividing fractions, the expression

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• First, rewrite the expression as a product, and then factor. 3𝑥2 9𝑥2 − 45𝑥 3𝑥2 𝑥2 − 7𝑥 + 10 3𝑥2 (𝑥 − 5) (𝑥 − 2) ÷ = ∙ = ∙ 𝑥2 − 7𝑥 + 10 9𝑥2 − 45𝑥 𝑥2 − 4𝑥 𝑥2 − 4𝑥 𝑥(𝑥 − 4) 3(3𝑥2)(𝑥 − 5)

• Before multiplying, rewrite the product using common factors that are equivalent 1 to . 1 3𝑥2 (𝑥 − 5) (𝑥 − 2) ∙ = 𝑥(𝑥 − 4) 3(3𝑥2)(𝑥 − 5)

3𝑥2(𝑥 − 5) 3𝑥2(𝑥 − 5)

∙

(𝑥 − 2) (𝑥 − 2) = 1∙ . 3𝑥(𝑥 − 4) 3𝑥(𝑥 − 4)

• Then, multiply, and leave the answer in factored form, noting any domain restrictions. (𝑥 − 2) , where 𝑥 ≠ 0, 𝑥 ≠ 4, and 𝑥 ≠ 5 3𝑥(𝑥 − 4)

Any product or quotient of rational expressions may involve domain restrictions. Any expression that results in a 0 in the denominator of the original expression needs to be acknowledged as a value that must be excluded from the domain.

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Practice Problems 1. Determine the product of

5𝑥 𝑥2 − 4 . Express the answer in irreducible ∙ 𝑥2 + 5𝑥 + 6 10𝑥

form, and state the domain restrictions.

2. Determine the product of

9𝑣 − 3𝑣2 𝑣2 + 7𝑣 + 12 ∙ . Express the answer in irreducible 𝑣2 − 9 9𝑣 + 36

form, and state the domain restrictions.

3. Determine the quotient of

𝑥2 − 3𝑥 − 40 𝑥2 + 13𝑥 + 40 ÷ . Express the answer in 𝑥2 + 8𝑥 − 20 𝑥2 + 12𝑥 + 20

irreducible form, and state the domain restrictions.

4. Determine the quotient of

𝑥2 𝑥 2 − 𝑦2 4𝑥 𝑦 − 𝑥

. Express the answer in irreducible form.

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Review Problems 5. Which polynomial has (𝑥 + 1) as a factor? A. 𝑥3 + 2𝑥2 − 19𝑥 − 20

B. 𝑥3 − 21𝑥 + 20

C. 𝑥3 + 8𝑥 + 11𝑥 − 20 D. 𝑥3 − 3𝑥2 + 3𝑥 − 1

6. The graph of a polynomial function 𝑓 is shown.

a. What is the degree of 𝑓? b. What is the sign of the leading coefficient of 𝑓?

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Unit 4, Lesson 7: Adding and Subtracting Rational Expressions with Like Denominators

Warm-Up: Patterns in the Sums of Unit Fractions 1. The table shows several expressions and their sums. Expression

Sum

1 1 + 2 3

5 6

1 1 + 3 4

7 12

1 1 + 5 6

11 30

1 1 + 4 5

9 20

a. Describe the pattern between the expressions and the sums.

b. Use the pattern to determine the sum of the expression

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1 1 + . 𝑎 𝑏

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Guided Activity: Adding and Subtracting Rational Expressions with Like Denominators 1. Keilani added the following rational expressions. 5𝑥 2𝑥 7𝑥 + = 𝑥 + 4 𝑥 + 4 2𝑥 + 8

a. Circle the mistake Keilani made.

b. Show the correct addition of the 2 expressions,

5𝑥 2𝑥 + . 𝑥 + 4 𝑥 + 4

2. Fill in the missing information for each set of expressions to generate a single equivalent expression for each.

3. Discuss with your partner which strategy you used to find the missing terms. 4. Penelope and Eugene were subtracting the rational expressions

𝑎 − 6𝑣 𝑎 − 5𝑣 − . 3 24𝑎𝑣 24𝑎𝑣3

Penelope’s Answer

Eugene’s Answer

−11𝑣 24𝑎𝑣3

−𝑣

Each student’s answer contains mistakes. Fix the mistakes Penelope and Eugene made by correctly subtracting 𝑎 − 6𝑣 𝑎 − 5𝑣 − . 3 24𝑎𝑣 24𝑎𝑣3

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Exploration Activity: Equivalent Expressions Mrs. Fowler marked both Efraim’s and Trenton’s answers for adding the 2 rational 𝑥2 − 8𝑥 + 2 5𝑥 − 12 + as correct. expressions 𝑥 + 2 𝑥 + 2 Efraim’s Answer

Trenton’s Answer

(𝑥 − 5)(𝑥 + 2) 𝑥 + 2

𝑥−5

1. Add the rational expressions to get each student’s answer. Then, compare your steps with your partner’s. Work for Efraim’s Answer 𝑥2 − 8𝑥 + 2 𝑥 + 2

5𝑥 − 12 = 𝑥 + 2

Work for Trenton’s Answer 𝑥2 − 8𝑥 + 2 𝑥 + 2

5𝑥 − 12 = 𝑥 + 2

𝑥2 − 8𝑥 + 2 5𝑥 − 12 + , are there any domain restrictions? If so, what 𝑥 + 2 𝑥 + 2 are the restriction(s)?

2. When adding

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Guided Activity: Adding and Subtracting Rational Expressions 1. Add or subtract the following rational expressions. Show your work. Write the answers in irreducible form, and determine the domain restrictions. 7𝑤 + 12 𝑤2 + a. 𝑤 + 3 𝑤 + 3

5𝑥2 − 7𝑥 + 3 4𝑥2 + 𝑥 − 9 + 𝑥2 − 3𝑥 − 18 𝑥2 − 3𝑥 − 18

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2. The steps for adding

4𝑝 − 1 𝑝 + are shown. 3𝑝 − 1 1 − 3𝑝

a. Describe what is occurring at each step. Step

Description

4𝑝 − 1 𝑝 + 3𝑝 − 1 1 − 3𝑝

Given

4𝑝 − 1 (−1)𝑝 + 3𝑝 − 1 (−1)(1 − 3𝑝) 4𝑝 − 1 −𝑝 + 3𝑝 − 1 3𝑝 − 1 4𝑝 − 1 − 𝑝 3𝑝 − 1 3𝑝 − 1 3𝑝 − 1 1

b. Complete the statement. When the denominators of 2 rational expressions are opposites, multiply 1 of the fractions by to get a common denominator.

𝑡2 − 6𝑡 3𝑡 + 2 3. Subtract 2 − . 𝑡 −1 1 − 𝑡2

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Collaborative Activity: Card Sort Complete the card sort with your partner. Use the table to show your work. Expression Number

Work

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Lesson Summary When adding or subtracting fractions, setting like denominators is necessary. The same is true for adding or subtracting rational expressions. Similar strategies are used in both cases. Consider the examples shown. Adding Fractions

Adding Rational Expressions

1 4 + 3 3

𝑥 − 1 2𝑥 + 𝑥 + 3 𝑥 + 3

5 3

3𝑥 − 1 for 𝑥 ≠ −3 𝑥 + 3

1+ 4 3

𝑥 − 1 + 2𝑥 𝑥 + 3

1 4 + , combine the fractions into 1 fraction by adding the numerators, and 3 3 then rewrite the fraction into an irreducible form. The same process can be applied to rational expressions.

To add

When subtracting fractions or rational expressions, pay close attention to the second rational expression. Distribute the negative sign to each term in the second numerator before combining the like terms of the numerators. For example,

4 2𝑥 + 14 4 − (2𝑥 + 14) 4 − 2𝑥 − 14 −2𝑥 − 10 − = = = for 𝑥 ≠ −9. 𝑥 + 9 𝑥 + 9 𝑥 + 9 𝑥 + 9 𝑥 + 9

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Practice Problems 1. Find the sum

4 2𝑥 + 14 − . Determine any domain restrictions. 𝑥 + 9 𝑥 + 9

2. Find the difference restrictions.

3. Find the difference

2𝑛 + 3 𝑛 + 1 − . Determine any domain (𝑛 + 6)(𝑛 + 2) (𝑛 + 6)(𝑛 + 2)

𝑎2 + 4 𝑎2 + 𝑎 − . Determine any domain restrictions. 𝑎2 − 16 𝑎2 − 16

Review Problem 4. Which of these standard form equations is equivalent to (𝑥 + 1)(𝑥 − 2)(𝑥 + 4) (3𝑥 + 7)? A. 𝑥4 + 10𝑥3 + 15𝑥2 − 50𝑥 − 56

B. 𝑥4 + 10𝑥3 + 15𝑥2 − 50𝑥 + 56 C. 3𝑥4 + 16𝑥3 + 3𝑥2 − 66𝑥 − 56

D. 3𝑥4 + 16𝑥3 + 3𝑥2 − 66𝑥 + 56

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Unit 4, Lesson 8: Adding and Subtracting Rational Expressions with Unlike Denominators

Warm-Up: Fraction Operations Rewrite each expression in irreducible form. 1.

1 2 + 3 5 2 − 3𝑥 5𝑥 + 6 6 1 𝑥𝑦 𝑥 ⋅ ÷ 4 3 6

Exploration Activity: Rewriting Fractions with Unlike Denominators In the prior lesson, all the rational expressions had a common denominator. 1. Rewrite

4 2 and as fractions with like denominators. 7 11

2. Discuss with your partner how you wrote equivalent fractions in problem 1. 3. Given that 𝑎, 𝑏, 𝑐, and 𝑑 are rational expressions, rewrite fractions with like denominators.

4. Rewrite the expression

𝑎 𝑐 + as equivalent 𝑏 𝑑

𝑎 𝑐 − as equivalent fractions with like denominators. 𝑏 𝑑

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Guided Activity: Adding and Subtracting Rational Expressions with Unlike Denominators 1. Two rational expressions are shown. 8 3𝑥 and 2 𝑥 − 2𝑥 − 3 𝑥 + 4𝑥 + 3 2

a. Discuss with your partner what strategy could be used to determine the least common denominator (LCD) for the 2 expressions. b. Determine the LCD.

c. Complete the table by rewriting the rational expressions

8 and 𝑥 − 2𝑥 − 3 2

3𝑥 as equivalent rational expressions with a common denominator. 𝑥2 + 4𝑥 + 3 Some steps have been completed. 𝟖 𝒙𝟐 − 𝟐𝒙 − 𝟑

Description of Work

8 (𝑥 − 3)(𝑥 + 1)

Factor each denominator.

(𝑥 + 1)(𝑥 − 3)(𝑥 + 3) 8( ) (𝑥 − 3)(𝑥 + 1)(

)

(𝑥 − 3)(𝑥 + 1)(𝑥 + 3)

Find the LCD. Multiply each denominator by the “missing” factor, and multiply each numerator by the same factor. Simplify the numerators.

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𝟑𝒙 𝒙𝟐 + 𝟒𝒙 + 𝟑

3𝑥 (𝑥 + 3)(𝑥 + 1) (𝑥 + 1)(𝑥 − 3)(𝑥 + 3) 3𝑥 ( ) (𝑥 + 3)(𝑥 + 1)(

)

(𝑥 − 3)(𝑥 + 1)(𝑥 + 3)

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d. Use the equivalent rational expressions with a common denominator to rewrite 8 3𝑥 + 2 as a single rational expression. 2 𝑥 − 2𝑥 − 3 𝑥 + 4𝑥 + 3

Collaborative Activity: Adding and Subtracting Rational Expressions 𝑥2 + 11𝑥 + 8 and 1. Use the expressions 2 𝑥 − 2𝑥 − 8

𝑥 + 3 𝑥 − 4

to complete the following.

a. Determine the LCD of the expressions by factoring each denominator.

𝑥2 + 11𝑥 + 8 b. Rewrite the expressions 2 and 𝑥 − 2𝑥 − 8 expressions with like denominators.

𝑥 + 3 𝑥 − 4

as equivalent rational

𝑥2 + 11𝑥 + 8 c. Use the equivalent expressions from part B to subtract 2 − 𝑥 − 2𝑥 − 8

𝑥 + 3 . 𝑥 − 4

d. Identify the domain restrictions.

2. Use the expressions

𝑥 𝑥 + 4

and

𝑥 − 2 𝑥 + 5

to complete the following.

a. Rewrite the expressions with like denominators.

b. Subtract

𝑥 𝑥 + 4

−

𝑥 − 2 . 𝑥 + 5

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3. Add

2𝑥 + 7 + 3. State the domain restrictions. 10𝑥 − 1

4. A box in the shape of a rectangular prism has dimensions

5 , 𝑥

𝑥 𝑥 − 1 , and . 𝑥 − 1 3

a. Write a single irreducible rational expression that can be used to determine the volume of the box.

b. Write a single irreducible rational expression that can be used to determine the surface area of the box.

Collaborative Activity: Multiple Operations with Rational Expressions 1. Three rational expressions are shown.

a. Rewrite

𝑟 𝑟 + 1

3 𝑟 − 1

𝑟 3 6 2 𝑟 + 1 𝑟 − 1 𝑟 −1 −

6 as a single irreducible rational expression. 𝑟 −1 2

b. State domain restrictions.

3𝑘 2 17𝑘 + 4 + − 2 as a single irreducible rational expression, 𝑘 + 2 𝑘 + 7 𝑘 + 9𝑘 + 14 and identify the domain restrictions.

2. Rewrite

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𝑥 1 1 𝑥 + − − 2 , where 𝑥 ≠ −3 and 𝑥 ≠ 3, as a single 𝑥 + 3 𝑥 − 3 5 𝑥 −9 irreducible rational expression.

3. Rewrite

Lesson Summary Denominators must be the same when adding or subtracting rational numbers. This applies to rational expressions as well. Expressions can be rewritten as equivalent rational expressions with a common denominator before adding or subtracting the expressions. 𝑎 𝑐 and , 𝑏 𝑑 𝑎 𝑐 create a single equivalent expression with a like denominator, such as + = 𝑏 𝑑 To add or subtract rational expressions with unlike denominators, such as

𝑎𝑑 + 𝑐𝑏 𝑎 𝑐 𝑎𝑑 − 𝑐𝑏 or − = , where 𝑏 ≠ 0 and 𝑑 ≠ 0. 𝑏𝑑 𝑏 𝑑 𝑏𝑑

Practice Problems 𝑎2 2 5 − , where 𝑎 ≠ 0 and 𝑦 ≠ 0, as a single irreducible rational − 7 3𝑦 𝑎 expression.

1. Rewrite

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𝑦 − 1 𝑦 + 1 𝑦2 − 5 − + 2 , where 𝑦 ≠ −1 and 𝑦 ≠ 1, as a single irreducible 𝑦 + 1 𝑦 − 1 𝑦 −1 rational expression.

2. Rewrite

3. Rewrite −

2𝑥 4 4(𝑥 + 5) 1 1 − + 2 , where 𝑥 ≠ − , 𝑥 ≠ , and 𝑥 ≠ 2, as a 3𝑥 + 1 2𝑥 − 1 3𝑥 − 5𝑥 − 2 3 2

single irreducible rational expression.

Review Problems 4. Rewrite the expression expression.

3𝑥2 − 17𝑥 − 6 ⋅ 4𝑥2 − 20𝑥 − 24

2𝑥2 − 𝑥 − 3 6𝑥2 − 7𝑥 − 3

as a single irreducible rational

5. Rewrite the expression shown as a single irreducible rational expression. 4𝑥2 − 1 12𝑥2 + 12𝑥 − 9 ÷ 3𝑥3 − 6𝑥2 − 24𝑥 −2𝑥2 + 5𝑥 + 12

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Unit 4, Lesson 9: Rational Equations – Part 1

Warm-Up: Notice and Wonder: Denominators and Solutions 1. What do you notice? What do you wonder? •

2𝑥 + 5 7𝑥 − 5 = , 𝑥 = 2 𝑥 𝑥

• 2𝑥 + 5 = 7𝑥 − 5, 𝑥 = 2 •

2𝑥2 + 3 4𝑥 + 1 = , 𝑥 = 1 𝑥 + 2 𝑥 + 2

• 2𝑥2 + 3 = 4𝑥 + 1, 𝑥 = 1

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Exploration Activity: Rationalizing the Price of T-Shirts The school art club at a large high school is in charge of designing school T-shirts and getting them printed this year. A local business charges $35 to set up their T-shirt printing machine with the design and $4.25 in materials per T-shirt to print. 1. Create an equation to represent the average cost C(𝑥), in dollars, per T-shirt if 𝑥 T-shirts are printed by this business.

2. What is the average cost per shirt to print 25 shirts? 100 shirts?

3. What is the cheapest the average cost per T-shirt will get? Explain or show your reasoning.

4. How many shirts should be printed to have an average cost of $5 or less per shirt? Explain how you know.

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Guided Activity: Batting Averages Tyler is on a school baseball team, and he has had 24 base hits out of 110 at bats this year. 1. What is his current batting average?

2. He wants to raise his batting average to 0.300. How many of the next consecutive at bats need to be base hits to raise his batting average to 0.300? Write and solve an equation to describe this situation using 𝑥 for the number of consecutive base hits. Be prepared to explain how you wrote your equation and each of your solving steps.

3. Unfortunately, Tyler gets no base hits in his next three at bats. Revise your equation and then calculate how many of his next consecutive at bats need to be base hits to raise his batting average to 0.300. Be prepared to explain how you revised your equation and each of your solving steps.

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Lesson Summary Consider a student on a school softball team who wants to raise her batting average to .200. So far this year, she has 20 base hits out of 120 at bats, making her current 20 = .167. batting average .167, since 120 To increase her batting average, she needs to get more base hits, but each base hit means the number of at bats also increases by 1. Since batting average is the number 20 + 𝑥 can be of base hits divided by the number of at bats, the rational expression 120 + 𝑥 used to model how her batting average changes based on the number of consecutive base hits, 𝑥, she gets. Her batting average is the value of this expression to 3 decimal places. The value of 𝑥 that makes this expression equal to .200 will identify how many consecutive base hits she needs to get the batting average she wants. 0.200 = 0.200 ⋅ (120 + 𝑥) =

20 + 𝑥 120 + 𝑥

20 + 𝑥 ⋅ (120 + 𝑥) 120 + 𝑥

0.2(120 + 𝑥) = 20 + 𝑥 24 + 0.2𝑥 = 20 + 𝑥 4 = 0.8𝑥 5 = 𝑥

Even though the original equation included a rational expression on the right side, multiplying each side by (120 + 𝑥) resulted in an equation similar to ones you have solved before. Check the solution 𝑥 = 5 by substituting it into the original expression, 20 + 5 25 = = 0.2. 120 + 5 125

The student needs 5 consecutive base hits to have a batting average of .200.

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Practice Problems 1. A local office supply store charges $18 to set up their business card printing machine with the design and $0.15 in materials per business card to print. Select all equations that could represent an expression for the average cost (𝑥) of printing a batch of 𝐴(𝑥) business cards.

□ 𝐴(𝑥) = □ 𝐴(𝑥) =

18 + 𝑥 0.15

18 + 0.15𝑥 𝑥

□ 𝐴(𝑥) =

0.15 + 18𝑥 𝑥

□ 𝐴(𝑥) =

18 + 0.15𝑥 18 + 𝑥

0.15

□ 𝐴(𝑥) = 18 + 𝑥 18

□ 𝐴(𝑥) = 𝑥 + 0.15 2. The school band is in charge of a new set of uniforms made with a new logo. A local business charges $140 to set up the logo with the design and $0.25 in materials per 140 + 0.25𝑥 logo printed. The function C(𝑥) = represents the average cost per logo if 𝑥 𝑥 uniforms are printed by this business. a. What is the average cost per uniform to get the logo printed on 25 uniforms?

b. What is the average cost per uniform to get the logo printed on 100 uniforms?

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c. How many uniforms should be printed to have an average cost of $1 per logo?

d. What will happen to the price as the number of uniforms printed increases?

3. Two competing sports equipment suppliers sell footballs at different prices. Supplier A charges $85 in shipping and charges $2.59 per football. Supplier B charges $50 shipping and charges $4.29 per football. A school wants to buy 40 balls. Which supplier has the lowest average cost per ball?

Review Problems 1. The graph of a polynomial 𝑓(𝑥) = (5𝑥 − 3)(𝑥 + 4)(𝑥 + 𝑎) has 𝑥-intercepts at −4, 6. What is the value of 𝑎?

3 , and 5

3𝑥 − 4 −22 can be rewritten in the form 𝑓(𝑥) = 3 + . What is 𝑥 + 6 𝑥 + 6 the end behavior of 𝑦 = 𝑓(𝑥)?

2. The function 𝑓(𝑥) =

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Unit 4, Lesson 10: Rational Equations – Part 2

Warm-Up: Solving Proportional Relationships Solve each equation mentally. 1.

𝑥 3 = 2 4 3 1 = 𝑥 6 1 1 = 2 4 𝑥 2 𝑥 = 𝑥 8

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Guided Activity: Solving One-Variable Rational Equations 1. A proportional relationship is shown. 𝑥 33 = 12 20

Vernon and Aaliyah solved the proportional relationship. Their work is shown. a. Discuss the questions shown with your partner.

Vernon

360 (

𝑥 33 )=( ) 360 12 20 30𝑥 = 594

𝑥 = 19.8 • How did Vernon and Aaliyah use the multiplication property of equality to solve the proportional relationship?

Aaliyah 60 (

𝑥 33 )=( )60 12 20 5𝑥 = 99

𝑥 = 19.8

• How might Vernon have decided to use 360? • How might Aaliyah have decided to use 60?

b. Which student used the least common denominator (LCD)?

2. Explain why the expressions 𝑥 + 7 and 𝑥 − 7 do not represent the same value.

3. What is the LCD of the expressions

1 1 and ? 𝑥 + 7 𝑥 − 7

4. A rational equation is shown. 6 4 = 𝑥 + 3 𝑥 − 3 482 | Unit 4

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a. Discuss with your partner how to use the multiplication property of equality to solve the rational equation.

b. Solve

6 4 = . Show your work. 𝑥 + 3 𝑥 − 3

c. Determine whether the value of 𝑥 is a solution to the rational equation using substitution.

5. Consider the following rational equation. 1 1 6 + = 2 𝑥 − 1 𝑥 + 1 𝑥 −1

a. Rewrite the rational equation by factoring 𝑥2 − 1. 1 1 + = 𝑥 − 1 𝑥 + 1

(

6 )(

)

b. Work with your partner to determine which 𝑥-values would result in the equation being undefined.

c. Determine the LCD.

d. Use the multiplication property of equality to write a transformed equation so each partial fraction has the same denominator. The first fraction has been completed.

1 𝑥 − 1

1 𝑥 + 1

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6 )(

)

𝑥 + 1 𝑥 + 1

𝑥 + 1 𝑥2 − 1

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Transformed Equation

𝑥 + 1 𝑥2 − 1

When the equation is multiplied by the LCD, the common factors in the LCD and in the denominator in each fraction reduce to 1. 𝑥 + 1 𝑥 − 1 6 + 2 = 2 2 𝑥 −1 𝑥 −1 𝑥 −1

(𝑥2 − 1)

(𝑥 + 1) + (𝑥 − 1) 6 = 2 2 𝑥 −1 𝑥 −1

((𝑥 + 1) + (𝑥 − 1)) 6 = (𝑥2 − 1) (𝑥2 − 1) (𝑥2 − 1)

e. Discuss with your partner how the resulting equation, (𝑥 + 1) + (𝑥 − 1) = 6, is related to the numerator of the transformed equation. Summarize your discussion.

f. Solve (𝑥 + 1) + (𝑥 − 1) = 6 to find the solution.

g. Is the solution of the transformed equation also a solution of the original 1 1 6 + = 2 ? Justify your answer using mathematics. equation, 𝑥 − 1 𝑥 + 1 𝑥 −1

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6. Two students were solving the equation

3 2 −6 − = 2 . 𝑥 − 1 𝑥 − 4 𝑥 − 5𝑥 + 4

Daniella

(𝑥 − 4)(𝑥 − 1) [

3 2 −6 − = 𝑥 − 1 𝑥 − 4 (𝑥 − 4)(𝑥 − 1)

3 2 −6 − ]=[ ](𝑥 − 4)(𝑥 − 1) 𝑥 − 1 𝑥 − 4 (𝑥 − 4)(𝑥 − 1) 3(𝑥 − 4) − 2(𝑥 − 1) = − 6 Bryson

3 2 −6 − = 𝑥 − 1 𝑥 − 4 (𝑥 − 4)(𝑥 − 1)

3(𝑥 − 4) 2(𝑥 − 1) −6 − = (𝑥 − 4)(𝑥 − 1) (𝑥 − 4)(𝑥 − 1) (𝑥 − 4)(𝑥 − 1) 3(𝑥 − 4) − 2(𝑥 − 1) = − 6

Daniella found the LCD and used the multiplication property of equality. Bryson rewrote the rational expressions by finding a common denominator and then setting the numerators equal. a. Discuss with your partner the 2 different methods shown by Daniella and Bryson. b. Find the solution to

3 2 −6 − = 2 . 𝑥 − 1 𝑥 − 4 𝑥 − 5𝑥 + 4

c. Verify whether the value of 𝑥 is a solution to the rational equation using substitution.

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Collaborative Activity: Rational Expressions, Equations, and Rates 1. Noah likes to go for boat rides along a river with his family. In still water, the boat travels about 8 kilometers per hour (kph). In the river, it takes them the same amount of time, in hours (hr.), 𝑡, to go upstream 5 kilometers (km) as it does to travel downstream 10 km. a. If the speed of the river is 𝑟, write an expression for the time it takes to travel 5 km upstream and an expression for the time it takes to travel 10 km downstream.

b. Use your expressions to calculate the speed of the river. Explain or show your reasoning.

2. Leo can mow a lawn in his neighborhood in 3 hr. It takes Rick 4 hr. to mow the same lawn. a. Complete the statements. It takes Leo 3 hr. to mow the lawn, so he mows _____ of the lawn each hr. It takes Rick 4 hr. to mow the lawn, so he mows _____ of the lawn each hr. b. Complete the table by identifying each person’s work rate and the amount of the lawn each has mowed after 𝑡 hr. Work Rate Leo Rick

Time

Amount of Lawn Mowed (work rate × time)

𝑡 𝑡

c. Write an equation to represent the total amount of time Rick and Leo will spend mowing the lawn altogether.

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d. How long would it take them to mow the lawn together?

Lesson Summary A boat travels about 6 kph in still water. If the boat is on a river that flows at a constant speed of 𝑟 kph, it can travel at a speed of 6 + 𝑟 kph downstream and 6 − 𝑟 kph upstream. If the river current were the same speed as the boat, the boat wouldn’t be able to travel upstream at all. On a particular river, the boat can travel 4 km upstream in the same amount of time it takes to travel 12 km downstream. • Because time is equal to distance divided by speed, the travel time can be 12 4 hr. or hr. If the travel time is not known, an equation expressed as either 6+𝑟 6−𝑟 can be created using the fact that these 2 expressions are equal. The equation can be used to determine the speed of the river. 4 12 = 6−𝑟 6+𝑟

12 4 ⋅ (6 + 𝑟)(6 − 𝑟) = ⋅ (6 + 𝑟)(6 − 𝑟) 6+𝑟 6−𝑟 12(6 − 𝑟) = 4(6 + 𝑟) 72 − 12𝑟 = 24 + 4𝑟 48 = 16𝑟 3=𝑟

• Substituting this value into the original expressions, the resulting expressions are 12 4 4 4 = and = , showing that these 2 expressions are equal when 6+3 3 6−3 3 𝑟 = 3. This means that when the water flow in the river is about 3 kph, it takes the boat 1 hr. and 20 minutes (min.) to travel 4 km upstream and 1 hr. and 20 min. to travel 12 km downstream. • Even though the original equation had a rational expression on each side of the equal sign, multiplying each side by the product of the denominators, (6 + 𝑟)(6 − 𝑟), resulted in a familiar linear equation. Multiplying to get an equation with no variables in the denominators is sometimes referred to as clearing the denominators. © Accelerate Learning Inc. - All Rights Reserved

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In some rational equations, solutions may be mathematically correct without making the original rational equation true. In such cases, the solution is identified as an extraneous solution. When working with rational equations, always substitute each solution back into the original equation to verify whether it is an actual solution or an extraneous solution. The work in the next lesson will address this concept in further detail.

Practice Problems 1. Solve 𝑥 − 1 =

2. Solve

𝑥2 − 4𝑥 + 3 for 𝑥. 𝑥 + 2

4 5 = for 𝑥. 4−𝑥 4+𝑥

3. Kiran jogs at a speed of 6 miles per hour (mph) when there are no hills. He plans to jog up a mountain road, which will cause his speed to decrease by 𝑟 mph. Which expression represents the time, 𝑡, in hr. it will take him to jog 8 miles (mi.) up the mountain road? A. 𝑡 = B. 𝑡 = C. 𝑡 = D. 𝑡 = 488 | Unit 4

8−𝑟 6

8 6+𝑟

6+𝑟 8

8 6−𝑟

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4. Kendra can paint a bedroom in 6 hr., whereas Aliyah can paint the same room in 4 hr. How long does it take for Kenda and Aliyah to paint the room altogether?

Review Problem 5. A softball player has had 8 base hits out of 25 at bats for a current batting average 8 = .320 . How many consecutive base hits does she need if she wants to raise of 25 her batting average to .400? Explain or show your reasoning.

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Unit 4, Lesson 11: Solving Rational Equations

Warm-Up: Notice and Wonder: Thoughtful Multiplication What do you notice? What do you wonder? 3 = 𝑥(𝑥 − 2)

3 ⋅ 𝑥(𝑥 − 2) = 𝑥(𝑥 − 2)

2𝑥 + 1 𝑥 − 2 2𝑥 + 1 𝑥 − 2

3 = 2𝑥2 + 𝑥

⋅ 𝑥(𝑥 − 2)

0 = 2𝑥2 + 𝑥 − 3

Exploration Activity: Rational Solving Jada is working to find values of 𝑥 that make this equation true. 5𝑥 + 5 5 = 𝑥 + 1 𝑥

She says, “If I multiply both sides by 𝑥(𝑥 + 1), I find that the solutions are 𝑥 = 1 and 𝑥 = −1, but when I substitute in 𝑥 = −1, the equation does not make any sense.” 1. Is Jada’s work correct? Explain or show your reasoning.

2. Why does Jada’s method produce an 𝑥-value that does not solve the equation?

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Collaborative Activity: More Rational Solving 1. Several equations are shown. For each one, use what you know about division to identify values of 𝑥 that cannot be solutions to the equation. a.

𝑥2 + 𝑥 − 6 =5 𝑥 − 2

2𝑥 + 1 1 = 𝑥 𝑥 − 2 10 5 = 𝑥 + 8 𝑥 − 8

𝑥2 + 𝑥 + 1 2 = 13 𝑥 − 1

𝑥 + 1 𝑥 − 1 = 4𝑥 3𝑥 1 1 = 𝑥 𝑥(𝑥 + 1)

𝑥 + 2 3 = 𝑥 𝑥 − 2

1 1 = 𝑥 − 3 𝑥(𝑥 − 3)

(𝑥 + 1)(𝑥 + 2) 𝑥 + 2 = 𝑥 + 1 𝑥 + 1

2. Without solving, identify three of the equations that you think would be least difficult to solve and three that you think would be most difficult to solve. Be prepared to explain your reasoning.

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3. Choose three equations to solve. At least one should be from your “least difficult” list and one should be from your “most difficult” list.

Lesson Summary Consider the equation

𝑥 + 2 2 = . The equation can be solved for 𝑥 by 𝑥(𝑥 + 1) (𝑥 + 1)(𝑥 − 1)

multiplying each expression by 𝑥𝑥 + 1𝑥 − 1 to get an equation with no variables in denominators and then rearranging it into an expression that equals 0, as shown. 𝑥 + 2 ⋅ 𝑥(𝑥 + 1)(𝑥 − 1) = 𝑥(𝑥 + 1)

2 ⋅ 𝑥(𝑥 + 1)(𝑥 − 1) (𝑥 + 1)(𝑥 − 1)

(𝑥 + 2)(𝑥 − 1) = 2𝑥

𝑥2 + 𝑥 − 2 = 2𝑥 𝑥2 − 𝑥 − 2 = 0

(𝑥 − 2)(𝑥 + 1) = 0

• Solving the transformed equation, (𝑥 − 2)(𝑥 + 1) = 0, leads to 2 solutions: 𝑥 = 2 and 𝑥 = −1.

2+2 2 = , which • Substituting 𝑥 = 2 into the original equation results in 2(2 + 1) (2 + 1)(2 − 1) is true, since each side is equal to 23. –1 + 2 2 = , –1(–1 + 1) (–1 + 1)(–1 − 1) which isn’t a valid equation because this substitution creates an undefined function with a 0 in the denominator. This means 𝑥 = − 1 is not a solution to this rational equation.

• Substituting 𝑥 = − 1 into the original equation results in

Consider the simpler equation 𝑥 − 5 = 0. This equation has 1 solution, 𝑥 = 5. However, when each side is multiplied by (𝑥 − 1), the result is a new equation, (𝑥 − 1)(𝑥 − 5) = 0, which has solutions 5 and 1. The 1 is a solution to the new equation because when 𝑥 = 1, 𝑥 − 1 = 0. But when 𝑥 = 1 is substituted into the original equation, the resulting equation is 1 − 5 = − 4 = 0, which is not a true equation. Therefore, 𝑥 = 1 is not a solution to the original equation. Because each side of the original equation was multiplied by an expression that has the value 0 when 𝑥 = 1, the 2 sides that were © Accelerate Learning Inc. - All Rights Reserved

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unequal at that specific 𝑥-value, 𝑥 − 5 and 0, are now equal. For this example, 𝑥 = 1 is called an extraneous solution. In the original example, 𝑥 = −1 is an extraneous solution. Although 𝑥 = −1 is a solution to the transformed equation where the original equation was multiplied by 𝑥(𝑥 + 1)(𝑥 − 1) on each side, it is not a solution to the original equation because they are not equivalent. It should be noted that even though we multiplied by 𝑥, (𝑥 + 1), and (𝑥 − 1), only 1 extraneous solution was added. This shows that multiplying by an expression that can equal 0 does not always cause an extraneous solution. When solving a rational equation, each solution must be substituted into the original equation to verify that it is viable and not extraneous.

Practice Problems 1. Identify all values of 𝑥 that make the equation true. a.

2𝑥 + 1 1 = 𝑥 𝑥 − 2

1 2 = 𝑥 + 2 𝑥 − 1

𝑥 + 3 𝑥 + 1 = 1−𝑥 𝑥 + 2

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𝑥 + 2 1 = 𝑥 + 8 𝑥 + 2

2. Kiran is solving

2𝑥 − 3 2 = for 𝑥, and he uses these steps. 𝑥 − 1 𝑥(𝑥 − 1)

(𝑥 − 1)(

2𝑥 − 3 𝑥 − 1

2 = 𝑥(𝑥 − 1)

2𝑥 − 3 2 ) = 𝑥(𝑥 − 1)( ) 𝑥 − 1 𝑥(𝑥 − 1) 2𝑥 − 3 = 2 2𝑥 = 5 𝑥 = 2.5

He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).

3. Identify all values of 𝑥 that make the equation true. a. 𝑥 =

25 𝑥

b. 𝑥 + 2 =

6𝑥 − 3 𝑥

𝑥 3 = 2 𝑥 𝑥

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6𝑥2 + 18𝑥 5 = 3 2𝑥 𝑥

Review Problems 4. Is this the graph of 𝑔(𝑥) = −𝑥4(𝑥 + 3) or ℎ(𝑥) = 𝑥4(𝑥 + 3)? Explain how you know.

5. Elena has a boat that would go 9 miles per hour (mph) in still water. She travels downstream for a certain distance and then back upstream to where she started. Elena notices that it takes her 4 hours (hr.) to travel upstream and 2 hr. to travel downstream. The river’s speed is 𝑟 mph. Write an expression that will help her solve for 𝑟.

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Unit 5: Radical Functions

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Unit 5, Lesson 1: Exploring Rational Exponents

Warm-Up: Which One Doesn’t Belong: Exponents and Equations 1. Which one doesn’t belong? Explain your reasoning. A. 23 = 9 B. 9 = 32 C. 2 ⋅ 2 ⋅ 2 ⋅ 2 = 16 D. 𝑎 ⋅ 20 = 𝑎

Exploration Activity: Fraction of What Exactly? Use the exponent rules and your understanding of roots to find the exact value of each. 1. 251 2

2. 151 2

3. 81 3

4. 21 3

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Collaborative Activity: Exponents and Radicals Match each exponential expression to an equivalent expression. 1 49

• 73

•

• 72

•

• 71

• √7

• 70

• 1/∛7

• 7�1

• ∛7

• 7�2

• 49

• 7�3

• 1/√7

• 72

1 343

• 343 1

• 7�2 1

• 7 • 1

• 73 1

• 7�3

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Guided Activity: You Can Use Any Fraction as an Exponent 1. Use exponent rules to explain why these expressions are equal to each other. �5 1 � 3

(52)3

2. Write 52 using radicals. 3

3. Write 54 using radicals. Show your reasoning using exponent rules. 3

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Lesson Summary 1

Recall from Algebra 1 that the expression 112 can be rewritten into equivalent 1

expressions using exponent rules. When the expression 112 is raised to the power of 2, 1

1 ⋅2

�112 � = 112 , which is simply 11. In other words, when the number 112 is squared, us1

ing exponent rules, the result is 11. That means that 112 must be equal to √11. 1

1 ⋅3

Similarly, 113 must be equal to √11 because �113 � = 113

= 11.

In general, if 𝑎 is any positive number, then 𝑎1 = √𝑎 and 𝑎1 = √𝑎.

Expressions that involve the √ expressions.

or √

symbols are often referred to as radical

Practice Problems 1. Which expression is equivalent to 161 ? 2

A. 1 4

B. 4 C. 8

D. 16.5

2. Select all the expressions equivalent to is equivalent to 410.

□2 ⋅ 2 5

□2

□4 ⋅ 4 4

□4 ⋅ 4 7

6 �3

□ 44

-6

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3. Evaluate 85 . 3

4. Select all expressions that are equal to 643 . 2

□ 96 □8

□ 512 □4

□ √64

□ √64 3

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Review Problems 5. Let � = (7 − 2�) and 𝑘 = 3�. Write each expression in the form 𝑎 � 𝑏�, where 𝑎 and 𝑏 are real numbers. a. 𝑘 = �

b. 𝑘2

c. �2

d. 𝑘 ⋅ �

6. For each polynomial function, rewrite the polynomial in standard form. Then state its degree and constant term. a. �(�) = (3� + 1)(� + 2)(� − 3)

b. �(�) = −2(3� + 1)(� + 2)(� − 3)

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Unit 5, Lesson 2: Positive and Negative Rational Exponents

Warm-Up: Squares and Cubes Find a solution to each equation. 1. 𝑥2 = 25 2. 𝑧2 = 7 3. 𝑦3 = 8 4. 𝑤3 = 19

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Guided Activity: Revisiting Rational Exponents 1. The graph of 𝑦 = 2𝑥 is shown.

a. Complete the table as much as you can without using a calculator.

𝒙

𝟐𝒙 (using exponents) 𝟐𝒙 (decimal approximation)

1 3

2 3

4 3

5 3

b. Use this graph of 𝑦 = 2𝑥 to estimate the value of the other powers in the table, and write your estimates in the table.

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2. Two numbers raised to rational exponents are shown. 1

23 and 23

a. Complete the table by rewriting each number as a radical and evaluating the numerical expression.

Expression

Radical

Expression Raised to the 3rd Power 1 3

�23 � = ______

2 3

�23 � = ______

4 3

�23 � = ______

b. Write a conjecture about what happens when an expression with a rational exponent is raised to the 3rd power.

c. Complete the table by rewriting each expression using either a rational exponent or a radical.

Rational Exponent

Radical 3

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3. The graph of 𝑦 = 2𝑥 is shown using a different window.

x -

a. Complete the table as much as you can without using a calculator.

𝒙

𝟐𝒙 (using exponents) 𝟐𝒙 (decimal approximation)

5 3

−2

−

2−2

2 3

−

4 3

−

2 3

−1 −1

−

2 3

−

2 3

−

1 3

−

2 3

0 0

b. Use your graph of 𝑦 = 2𝑥 to estimate the value of the other powers in the table, and write your estimates in the table.

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4. Two numbers raised to rational exponents are shown. 1

2−3 and 2−3

a. Complete the table by rewriting each number as a radical and evaluating the numerical expression.

Expression

Radical

Expression Raised to the 3rd Power 1

2−3

�2−3 � = ______

2−3

�2−3 � = ______

2−3

b. Write a conjecture about what happens when an expression with a rational exponent is raised to the 3rd power.

c. Complete the table by rewriting each expression using either a rational exponent or a radical.

Rational Exponent

Radical

1 27

2−4 7

1 28

2− 5

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Collaborative Activity: Any Fraction Can Be an Exponent 1. For each set of 3 numbers, cross out the expression that is not equal to the other two expressions. 4

a. 85 , −

b. 8 5 , c.

85, 5

1 1 − 5 4 4, 8 8

43, 42 ,43

3 3 1 −42 , 4−2 3, 4

2. For each expression, write an equivalent expression using radicals. 3

a. 172 −

b. 31 2 3. For each expression, write an equivalent expression using only exponents. a. � 3 �

b. � 3 56 �

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Collaborative Activity: Make These Exponents Less Complicated Match expressions into groups according to whether they are equal. Be prepared to explain your reasoning.

� 3�

(32 )4

(32)2

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� 3 � ·� 3 �

(32 )2

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Lesson Summary 1

Using exponent rules, it can be said that 34 is 5

3 because (34 )4 = 3. But what about 34 ? 4

Using exponent rules, 34 = (35)4 , which means that 34 = 35. Since 35 = 243, it can be 4 5 said that 34 = 243. Alternatively, the fraction 5 can also be expressed as 1 ⋅5. Using 5 1 4 4 4 exponent rules, 34 = (34 )5 = ( 3 )5.

Additional examples of exponents that are fractions and their equivalents are shown.

𝒙

1 3

𝟓𝒙 (using exponents)

2 3

53 3

𝟓𝒙 (equivalent expression)

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4 3

53 3

5 3

or 3

53 3

or 3

625

or 3

3125

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When a number has a negative exponent, it means that the base can be rewritten as the reciprocal with a positive exponent. For example, 7− 5 =

1 6 1 −5 6 and 7 = 75 . 75

The table shows more examples of exponents that are fractions as well as their radical equivalents.

𝒙

−1

−3

𝟓𝒙 (using exponents)

5−1

5−3

𝟓𝒙 (equivalent expression)

1 5

1 3 � 25�

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1 3

5−3

2 3

� 52 �

−3

53 3

1 � 25� 3

or 3

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Practice Problems 1. Write each expression in the form 𝑎𝑏, without using any radicals. a.

1 12 2

2. Write 32−5 without using exponents or radicals. 3. Match the equivalent expressions. 1. 1

A. 83

2. 1

B. 8−3

C. 8−1

3. 1

4. 1

D. 162 E. 16−2

5. 2

F. 160

6. 4

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Review Problems 4. Complete the table. Use powers of 27 in the top row and radicals or rational numbers in the bottom row.

271

272

27−2 1

1 3

5. What are the solutions to the equation (𝑥 − 1) (𝑥 + 2) = −2? 6. Use exponent rules to explain why ( 5)3 =

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Unit 5, Lesson 3: Squares and Square Roots

Warm-Up: Math Talk: Four Squares Find the solutions of each equation mentally. 1. 𝑥2 = 4 2. 𝑥2 = 2 3. 𝑥2 = 0 4. 𝑥2 = − 1

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Exploration Activity: Finding Square Roots Clare was adding √4 and √9 , and, at first, she wrote √4 + √9 = 2 + 3. But then she remembered that 2 and − 2 both square to make 4, and that 3 and − 3 both square to make 9. She wrote down all the possible combinations: 2+3=5

2 + (−3) = −1 (−2) + 3 = 1

(−2) + (−3) = −5

Then she wondered, “Which of these are the same as √4 + √9 ? All of them? Or only some? Or just one?” How would you answer Clare’s question? Give reasons that support your answer.

Guided Activity: One Solution or Two? 1. The graph of 𝑏 = √𝑎 is shown.

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a. Complete the table with the exact values and label the corresponding points on the graph with the exact values. 𝒂 √𝒂

b. Label the point on the graph that shows the solution to √𝑎 = 4. c. Label the point on the graph that shows the solution to √𝑎 = 5.

d. Label the point on the graph that shows the solution to √𝑎 = √5 . 2. The graph of 𝑡 = 𝑠2 is shown.

a. Label the point(s) on the graph that show(s) the solution(s) to 𝑠2 = 25. b. Label the point(s) on the graph that show(s) the solution(s) 𝑠2 = 5.

c. Label the point(s) on the graph that show(s) the solution(s) to 𝑠2 = 5.

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Lesson Summary The symbol √11 represents the principal square root, or positive square root, of 11. To represent the negative square root, write −√11. 2

• The equation 𝑥2 = 11 has 2 solutions because √11 = 11 and �−√11� = 11.

• The equation √𝑥 = 11 only has 1 solution, 121.

• The equation √𝑥 = √11 only has 1 solution, 11.

• The equation √𝑥 = − 11 doesn’t have any solutions because the left side is positive and the right side is negative. It is not possible for a positive number to equal a negative number.

Practice Problems 1. Select all the solutions to the equation 𝑥2 = 7.

□ −49 □ −14

□ − √7 □ √7 □ 14 □ 49

2. Find the solution(s) to each equation, if there are any. a. 𝑥2 = 9

b. √𝑥 = 3

c. √𝑥 = − 3 520 | Unit 5

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3. a. If 𝑐 is a positive number, how many solutions does 𝑥2 = 𝑐 have? Explain.

b. If 𝑐 is a positive number, how many solutions does √𝑥 = 𝑐 have? Explain.

Review Problems 1

4. Suppose that a friend missed class and never learned what 37 3 means.

1 3

a. Use exponent rules your friend would already know to calculate �37 � .

b. Explain why this means that 37 3 is the cube root of 37.

5. Write each expression without using exponents. 2

a. 5 3

−

b. 4 2

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Unit 5, Lesson 4: Key Features of Square Root Functions

Warm-Up: Equation with Exponents 1. Create a true statement by using any digit from 1 to 9 to fill in each box. For an extra challenge, do not repeat any of the digits.

⦁

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(x )

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Exploration Activity: Comparing Quadratic and Square Root Functions 1. Use a calculator to complete the tables of values for the functions 𝑓(𝑥) = 𝑥2 and 𝑔(𝑥) = √𝑥.

𝒙

−4

𝒇(𝒙) = 𝒙𝟐

𝒇(𝒙)

𝒙

𝒈 (𝒙) = √𝒙

−4

−2

𝒈(𝒙)

−2

−1

2. What do you notice about the output values for both functions?

3. Why are all of the output values for 𝑓(𝑥) positive?

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4. Why do the negative input values for 𝑔(𝑥) result in an error on a calculator?

5. Compare the domain and range for 𝑓(𝑥) and 𝑔(𝑥).

6. Graph 𝑓(𝑥) = 𝑥2 and 𝑔(𝑥) = √𝑥 on the same coordinate plane. 15

x -10

-5

7. On the coordinate plane, graph ℎ(𝑥) = −√𝑥.

8. Describe the relationship between 𝑓(𝑥) = 𝑥2 and 𝑔(𝑥) = √𝑥.

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9. With your partner, discuss whether square root functions such as 𝑔(𝑥) = √𝑥 have a relative maximum or relative minimum. Summarize your discussion.

Guided Activity: Graphing and Identifying Key Features of Square Root Functions 1. The graph of 𝑓(𝑥) = √𝑥 – 2 − 3 is shown.

a. What are the domain and range of 𝑓 (𝑥)?

(18, 1) (11, 0) 0

x 15

(2, −3) b. How does the point (2, − 3) relate to the equation 𝑓(𝑥) = √𝑥 – 2 − 3 and to the domain and range?

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c. Use the graph to identify the key features of 𝑓(𝑥). Key Features of 𝒇(𝒙) = √𝒙 − 𝟐 − 𝟑 Increasing interval(s)

𝑥-intercept(s) Left end behavior

As 𝑥 →_____, 𝑦 →_____.

Right end behavior

As 𝑥 →_____, 𝑦 →_____.

Positive interval(s) Negative interval(s)

2. Consider the function 𝑔(𝑥) = 2√𝑥 + 4 −1.

a. Use the equation and the table of values to graph 𝑔(𝑥). 𝒙

𝒈(𝒙)

−4

−1

−3

b. With your partner, discuss why the specific input values in the table were chosen. Summarize your discussion.

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c. Determine the point where the relative maximum or relative minimum for 𝑔(𝑥) occurs.

d. Complete the table for the key features of 𝑔(𝑥). Not all key features may be applicable for the function.

𝑥-intercept(s)

Key Features of 𝒈(𝒙) = 𝟐√𝒙 + 𝟒 − 𝟏

𝑦-intercept

Increasing interval(s)

Decreasing interval(s)

Domain

Positive interval(s)

Range

Negative Interval(s)

Lesson Summary The graph of a square root function models the relationship 𝑦 = √𝑥. However, since the value of the radicand, the number inside the square root, is restricted to values greater than or equal to 0, the domain of the square root function is also restricted. The key features of a square root function include either a relative maximum or relative minimum, which depends on the equation and the transformations applied to the parent function of 𝑦 = √𝑥. A relative maximum is a point on the graph of a function that is higher than any of the points around it.

A relative minimum is a point on the graph of a function that is lower than any of the points around it.

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• The value of a relative maximum is the 𝑦-value of the point where the function changes direction from increasing to decreasing or the 𝑦-value of an endpoint where the function is always decreasing. • The value of a relative minimum is the 𝑦-value of the point where the function changes direction from decreasing to increasing or the 𝑦-value of an endpoint where the function is always increasing. Notice that a square root function does not have any turning points, so it will always be either increasing or decreasing. The increasing and decreasing intervals will include the relative maximum or minimum. For example, the function 𝑓(𝑥) = √𝑥 – 2 − 1 has a relative minimum at (2, −1) with the increasing interval {𝑥|𝑥 ≥ 2} or [2, ∞).

Practice Problems 1. Consider the function 𝑟(𝑥) = −√𝑥 – 3 + 3. Create a table of values for 𝑟(𝑥) and graph the function. 𝒙

𝒓(𝒙)

3 2 1 x 0

-1 -2 -3 -4

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2. Complete the table for the key features of 𝑟(𝑥) = −√𝑥 – 3 + 3. Not all key features may be applicable for the function.

𝑥-intercept(s)

Key Features of 𝒓(𝒙) = −√𝒙 – 𝟑 + 𝟑 Increasing interval(s)

𝑦-intercept

Decreasing interval(s)

Domain

Positive interval(s)

Range

Negative interval(s)

Relative maximum or relative minimum

Left end behavior Right end behavior

As 𝑥 →_____, 𝑦 →_____. As 𝑥 →_____, 𝑦 →_____.

Review Problems 3. Evaluate each expression. a. (∛–27)2

1 2 2 3

b. �125 �

3 c. �64 � 5

–1

164

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4. Complete the table of key features for the function 𝑓(𝑥) = − 𝑥(𝑥 + 1)(𝑥 − 2)3. Key Features of 𝒇(𝒙) = − 𝒙(𝒙 + 𝟏)(𝒙 – 𝟐)𝟑 Zero(s)

Positive interval(s)

𝑦-intercept

Negative interval(s)

Domain

Left end behavior

As 𝑥 →_____, 𝑦 →_____.

Range

Right end behavior

As 𝑥 →_____, 𝑦 →_____.

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Unit 5, Lesson 5: Cubes and Cube Roots

Warm-Up: Put Your Arm(s) Up How are these graphs the same? How are they different?

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Exploration Activity: Finding Cube Roots with a Graph This activity requires the use of an applet, so please make your way over to the digital platform to find the link. Use the applet to zoom in and out on the graph. How many solutions are there to each of the following equations? Estimate the solution(s) from the graph of 𝑦 = 𝑥3. Check your estimate by substituting it back into the equation. 1. 𝑥3 = 8

2. 𝑥3 = 2

3. 𝑥3 = 0

4. 𝑥3 = −8

5. 𝑥3 = −2

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Guided Activity: Cube Root Equations This activity requires the use of an applet, so please make your way over to the digital platform to find the link. Use the applet to zoom in and out on the graph.

1. Use the graph of 𝑦 = ∛𝑥 to estimate the solution(s) to ∛𝑥 = −4 2. Use the meaning of cube roots to find an exact solution to the equation ∛𝑥 = −4. How close was your estimate?

3. Find the solution of the equation ∛𝑥 = 3.5 using the meaning of cube roots. Use the graph to check that your solution is reasonable.

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Collaborative Activity: Solve These Equations with Cube Roots in Them Here are a lot of equations. • ∛𝑡 + 4 = 3

• ∛𝑝 + 4 − 2 = 1

• ∛3 − 𝑤 − 4 = 0

• ∛2𝑛 + 3 = − 5

• −10 = − ∛𝑎 • ∛𝑧 + 9 = 0

• ∛𝑟3 − 19 = 2

• 5 − ∛𝑘 + 1 = − 1

• 6 − ∛𝑏 = 0

• 4 + ∛−𝑚 + 4 = 6 • −∛𝑐 = 5

• ∛𝑠 − 7 + 3 = 0

1. Without solving, identify 3 equations that you think would be the least difficult to solve and 3 equations that you think would be the most difficult to solve. Be prepared to explain your reasoning.

2. Choose 4 equations and solve them. At least one should be from your “least difficult” list and at least one should be from your “most difficult” list.

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Lesson Summary Every number has exactly 1 cube root. This can be verified using the graph of 𝑦 = 𝑥3 shown.

Suppose 𝑦 = −4. The graph of 𝑦 = 𝑥3 intersects the graph of 𝑦 = −4 in only 1 place. The equation 𝑥3 = −4 will have the solution 𝑥 = −∛4.

This is true for any number 𝑎, where 𝑦 = 𝑎 will cross the graph in exactly 1 place, and 𝑥3 = 𝑎 will have 1 solution, ∛𝑎. In an equation like ∛𝑡 + 6 = 0, the cube root can be isolated, and then each side can be raised to the power of 3 to solve for 𝑡, as shown. ∛𝑡 + 6 = 0

∛𝑡 = −6

𝑡 = (−6)3

𝑡 = −216

Although cubing each side of an equation won’t create an equation with solutions that are different from the original equation, it is still always a good idea to check solutions in the original equation because little mistakes can creep in along the way. ∛−216 + 6 = − 6 + 6 = 0

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Practice Problems 1. Select all equations for which − 3 is a solution. □ 𝑥2 = 9

□ 𝑥2 = − 9 □ 𝑥3 = 27

□ 𝑥3 = − 27 □ −𝑥2 = 9

□ (−𝑥)2 = 9

2. The graph of 𝑦 = ∛𝑥 is shown.

a. Use the graph of 𝑦 = ∛𝑥 to estimate the solution(s) to the following equations. i.

∛𝑥 = 2

ii. ∛𝑥 = −4.5 iii. ∛𝑥 = 3.75 b. Use the meaning of cube roots to find exact solutions to all three equations.

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3. Which are the solutions to the equation 𝑥3 = −125? A. 5

B. − 5

C. both 5 and − 5

D. The equation has no solutions.

Review Problems 4. Which are the solutions to the equation √𝑥 = − 8? A. 64 only

B. −64 only

C. 64 and −64

D. This equation has no solutions.

5. Rewrite the rational function 𝑔(𝑥) = constants.

𝑥 − 9 𝑟 𝑥 in the form 𝑔(𝑥) = 𝑐 + 𝑥 , where 𝑐 and 𝑟 are

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Unit 5, Lesson 6: Key Features of Cube Root Functions

Warm-Up: Key Features of a Function The graph of the function 𝑔(𝑥) = (𝑥 – 2)3 + 1 is shown. y

Complete the table of key features for 𝑔(𝑥).

𝑥-intercept(s)

Key Features of 𝒈(𝒙) = (𝒙 − 𝟐)𝟑 + 𝟏 Increasing interval(s)

Decreasing interval(s)

𝑦-intercept Domain

Positive interval(s)

Range

Negative interval(s)

Left end behavior

As 𝑥 →____, 𝑦 →____.

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Right end behavior

As 𝑥 →____, 𝑦 →____. Unit 5 | 541

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Exploration Activity: Comparing Cubic and Cube Root Functions 1. Functions 𝑤(𝑥) = 𝑥3 and 𝑘(𝑥) = ∛𝑥 are graphed on the coordinate plane and are represented in the table of values. 𝒙

𝒘(𝒙) 𝒌(𝒙)

−8

−512 −2

−3

−2

−1

−1.44

−1.26

−1

1.26

−27

−8

−1

512 2

14 12 10

w(x)

8 6 4

k(x)

2 -8 -7

-6

-5 -4 -3 -2 -1 0 -2

x 1

-4 -6 -8 -10 -12 -14 -16

a. With your partner, discuss your observations of the two functions. Summarize your discussion.

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b. Identify the following key features for 𝑤(𝑥) = 𝑥3 and 𝑘(𝑥) = ∛𝑥.

Interval where the function is increasing

𝒘(𝒙) = 𝒙𝟑

𝒌(𝒙) = ∛𝒙

Interval where the function is decreasing Interval where the function is positive Interval where the function is negative c. Discuss with your partner the relationship between the key features of 𝑤(𝑥) = 𝑥3 and 𝑘(𝑥) = ∛𝑥. Summarize your discussion.

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Guided Activity: Graphing and Identifying Key Features of Cube Root Functions 1. Consider the function 𝑝(𝑥) = 2∛𝑥 + 1 − 3.

a. Complete the table of values for 𝑝(𝑥). Then, graph the function on the coordinate grid.

2 1

𝒙

−2

𝒑(𝒙)

-3 -2 -1 0 -1

x 1

-2 -3

−1

-4 -5

-6 -7

2.375

-8

b. Complete the table for the key features of 𝑝(𝑥). Not all key features may be applicable for the function.

𝑥-intercept(s)

Key Features of 𝒑(𝒙) = 𝟐∛𝒙 + 𝟏 − 𝟑 Increasing interval(s)

Decreasing interval(s)

𝑦-intercept Domain

Positive interval(s)

Range

Negative interval(s)

Left end behavior

As 𝑥 →____, 𝑦 →____.

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Right end behavior

As 𝑥 →____, 𝑦 →____.

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2. A table of values for the cube root function 𝑑 (𝑥) is shown. 𝒙

𝒅(𝒙)

−30 3

−11 4

−4 5

−3 6

−2 7

a. Explain whether there are any domain or range restrictions for the function 𝑑(𝑥).

b. Over which interval is the function increasing?

c. Over which interval is the function decreasing?

d. What is the end behavior for 𝑑(𝑥)? e. Plot the points from the table of values of 𝑑(𝑥) on the coordinate grid. Then, sketch the graph of 𝑑 (𝑥).

f. Use the graph to verify your answers for parts A–D. If necessary, make revisions to your answers.

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Lesson Summary The cube root function 𝑓(𝑥) = ∛𝑥 is considered a parent function. Some key features are the same for all cube root functions, such as the domain and range are (−∞, ∞). Other key features, such as zeros and increasing/decreasing intervals, will be different based on the transformations applied to the parent function. For example, the graph of 𝑔(𝑥) = 3∛𝑥 + 4 + 3 is shown. Some of the key features of (𝑥) are listed. • 𝑥-intercept (−5, 0)

• 𝑦-intercept (0, 3 ∛4 + 3)

• Increasing interval (−∞, ∞) • Positive interval (−5, ∞)

• Negative interval (−∞, −5)

Practice Problems 1. Which of the functions is decreasing? A. 𝑦 = ∛𝑥 – 4 B. 𝑦 = ∛4𝑥

C. 𝑦 = 4 − ∛𝑥

D. 𝑦 = ∛𝑥 − 4

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2. Function 𝑘(𝑥) is given. Graph the function, and determine its key features. 𝒌(𝒙) = 𝟏 ∛𝒙 – 𝟏 + 𝟐 𝟐

𝑦-intercept Domain

Range

Increasing interval(s)

Decreasing interval(s)

Positive interval(s) x

-5

Negative interval(s) Left end behavior Right end behavior

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As 𝑥 →_____, 𝑦 →_____. As 𝑥 →_____, 𝑦 →_____.

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Review Problem 3. Draw a rough sketch of the graph of 𝑔(𝑥) = −𝑥2(𝑥 + 2).

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Unit 5, Lesson 7: Graphing Square Root and Cube Root Functions Given a Written Description

Warm-Up: Which One Doesn’t Belong? 1. Several numbers are shown. Circle the number that is not like the other numbers.

256

343

216

1,024

243

2,601

2. Explain your choice using mathematics.

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Guided Activity: Graphing Square Root and Cube Root Functions from a Written Description 1. Function 𝑘(𝑥) passes through the points (4, 8) and (7, 10) and has the following key features. Domain Range 𝒙-intercept

{𝑥|𝑥 ≥ 3}

None

𝒚-intercept

Increasing Interval

{𝑦|𝑦 ≥ 6}

Decreasing Interval

None

{𝑥|𝑥 ≥ 3} None

a. Is 𝑘(𝑥) a square root or a cube root function? Explain your thinking.

b. Graph 𝑘(𝑥).

11 10 9 8 7 6 5 4 3 2 1 -2 -1 0 -1

x 1

10 11 12

-2

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2. Function 𝑚(𝑥) has the following key features. Domain Range 𝒙-intercept

(−∞, ∞)

(−∞, ∞) (−9, 0)

𝒚-intercept

(0, − 3)

Decreasing Interval

(−∞, ∞)

Increasing Interval

None

a. Is 𝑚(𝑥) a quadratic, square root, or cube root function? Include the key features to justify your answer.

b. The function 𝑚(𝑥) passes through points (−1, −2) and (−2, −1). Graph 𝑚(𝑥).

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Collaborative Activity: Using Key Features to Sketch the Graph of a Function 1. With your partner, circle 2 functions from each column. Column A 𝒏(𝒙)

Domain Range 𝑦-intercept

Decreasing interval Relative maximum

𝒘(𝒙)

Domain

Column B

{𝑥|𝑥 ≥ −1}

𝒂(𝒙)

Domain

{𝑦|𝑦 ≤ −2}

Range

(−∞, ∞)

(0, −3)

{𝑥|𝑥 ≥ −1}

𝑥-intercept

(−1, −2)

𝑦-intercept

(0, 0)

Increasing interval Passes through point 𝒕(𝒙)

Domain

(−∞, ∞) (0, 0)

(−∞, ∞)

(−7, −1)

Range

[0, ∞)

Range

𝑥-intercept

[−3, ∞)

(−∞, ∞)

(1, 0)

(0, −3)

𝑥-intercept

{𝑥|𝑥 ≥ 0}

𝑦-intercept

(8, 0)

Decreasing interval

𝑦-intercept

Increasing interval

𝒑(𝒙)

Domain Range

{𝑥|𝑥 ≥ 4}

𝑥-intercept

Decreasing interval Passes through point

Passes through point 𝒓(𝒙)

Domain

(−∞, ∞) (0, 4)

(−∞, ∞) (−1, 6)

{𝑦|𝑦 ≤ 0}

Range

(−∞, ∞)

(4, 0)

{𝑥|𝑥 ≥ 4}

𝑥-intercept

(8, −4)

𝑦-intercept

(1, 0)

Decreasing interval

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Passes through point

(−∞, ∞) (0, 1)

(−∞, ∞) (9, −2)

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2. Graph the 2 functions you selected from Column A on the same coordinate grid. Be sure to label the graph of each function.

3. Graph the 2 functions you selected from Column B on the same coordinate grid. Be sure to label the graph of each function.

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Lesson Summary Radical functions such as 𝑦 = √𝑥 or 𝑦 = ∛𝑥 can also be referred to as root functions. The key features of a function can be used to graph the function.

• Use the zero(s), or 𝑥-intercept(s), to plot the point(s) where the function crosses the 𝑥-axis. For example, 𝑦 = √𝑥 + 1 has an 𝑥-intercept at (−1, 0). Note that some square root functions may not have an 𝑥-intercept, such as 𝑦 = √𝑥 + 3.

• Use the 𝑦-intercept to plot the point where the function crosses the 𝑦-axis. For example, 𝑦 = √𝑥 + 1 has a 𝑦-intercept at (0, 1). Note that some square root functions may not have a 𝑦-intercept, such as 𝑦 = √𝑥 – 2. • Use the given increasing or decreasing intervals to graph the behavior of the function. Note that radical functions will always either increase or decrease.

• The domain and range of cube root functions will be (−∞, ∞), but the domain and range of square root functions will be restricted based on the relative minimum or maximum value of the function.

Practice Problems 1. Sketch the graph of a radical function with following key features. • Domain {𝑥|𝑥 ≥ − 1} • Range {𝑦 ≤ 3}

• Passes through (0, 2) and (3, 1)

• End behavior of as 𝑥 → − 1, 𝑦 → 3, and as 𝑥 → ∞, 𝑦 → −∞

5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1

x 1

-2 -3 -4 -5 -6

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2. Sketch the graph of the function using the key features provided. Key Features of 𝒉(𝒙)

Domain Range

𝑥-intercept

𝑦-intercept End behavior Decreasing interval Increasing interval Passes through point

(−∞, ∞) (−∞, ∞) (1, 0)

(0, −1)

As 𝑥 → ∞, 𝑓(𝑥) → ∞.

5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1

x 1

-2 -3 -4 -5 -6

As 𝑥 → −∞, 𝑓(𝑥) → −∞. None

−∞ < 𝑥 < ∞ (−1, −2)

Review Problem 3. The graph of a polynomial function 𝑓 is shown.

a. Is the degree of the polynomial odd or even? Explain how you know.

b. What is the constant term of the polynomial?

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Unit 5, Lesson 8: Solving Real-World Problems Using Square Root and Cube Root Functions

Warm-Up: Key Features of Square Root Function 1. The graph of 𝑓(𝑥) = 0.5√𝑥 + 4 is shown.

Identify the key features of the graph. Key Features 𝑥-intercept(s)

𝑦-intercept Domain Range Increasing interval(s)

Decreasing interval(s) Positive interval(s) Negative interval(s) Left end behavior Right end behavior

As 𝑥 →_____, 𝑦 →_____. As 𝑥 →_____, 𝑦 →_____.

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Guided Activity: Graphing Real-World Radical Functions 1. The velocity of an object, 𝑣, in meters per second (m/s), can be determined using the object’s mass, 𝑚, in kilograms (kg), using the equation �2𝐾𝐸 = 𝑣, where 𝐾𝐸 𝑚

is the object’s kinetic energy, in joules (J). If an object has a mass of 3 kg, then the equation 𝑦 = �2𝑥 models this relationship. 3

a. Graph the equation 𝑦 = �2𝑥 . 9

8 7 6 5 4 3 2 1 x

-1

b. What is the kinetic energy when the velocity of the object is 4 m/s?

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c. What is the kinetic energy when the velocity of the object is 6 m/s?

d. Complete the table of key features for the equation 𝑦 = �2𝑥 . 3

Key Features 𝑥-intercept(s)

Increasing interval(s)

𝑦-intercept

Decreasing interval(s)

Domain

Positive interval(s)

Range

Negative interval(s)

Relative minimum or maximum

Left end behavior Right end behavior

As 𝑥 →_____, 𝑦 →_____. As 𝑥 →_____, 𝑦 →_____.

e. Interpret the interval for which the graph is increasing within the context.

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2. The distance, 𝑑, in miles (mi.) from which a person can see the horizon is modeled by the function 𝑑 = �3ℎ , where ℎ is the person’s height in ft. above sea level. 2

a. Complete the table of values for 𝑑 = �3ℎ . 𝒉

𝒅

50 3

b. Graph the equation 𝑑 = �3ℎ . 2

c. Complete the table. Key Features Increasing interval(s)

𝑥-intercept(s) 𝑦-intercept

Decreasing interval(s)

Domain

Positive interval(s)

Range

Negative interval(s)

Relative minimum or maximum

Left end behavior Right end behavior

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As 𝑥 →_____, 𝑦 →_____.

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d. Interpret the point �32, 4� within the context. 3

e. Show the point on the graph from which a person can see 12 mi. to the horizon. What is the ordered pair?

f. Use the graph to estimate the height above sea level that a person is standing if they can see 5 mi. to the horizon.

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Collaborative Activity: Intrepreting Key Features 3 6𝑉 1. The diameter, 𝑑, in inches (in.), of a sphere is given by the formula 𝑑 = � , where 𝑉

𝜋

represents the volume, in cubic inches (cu. in.). A sculptor is preparing clay models for a sculpture that will have spheres of various sizes. The sculptor uses the formula to plan the clay models she will use. 3 6𝑉 . a. Graph the equation 𝑑 = �

𝜋

Diameter (inches)

b. The sculptor knows she needs 1 pound of clay for 75 cu. in. What is the diameter of a sphere that will use 1 pound (lb.) of clay?

6 4 2 x

-2 -4

Volume (cubic inches)

c. The sculptor is considering 3 different spheres whose diameters are 1 in., 2 in., and 3 in. What is the volume of each sphere?

Diameter

Volume

1 in. 2 in. 3 in.

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d. The sculptor would like to determine how many spheres she is limited to making with 1 lb. of clay. She wants the diameter of each sphere to increase by 1 in. How many spheres can she make, and what are their diameters?

350

300 250 200

Speed (m/s)

2. Most tsunamis are caused by earthquakes on converging tectonic plate boundaries. Once a tsunami forms, its speed depends on the depth of the ocean. At sea, a tsunami will not normally be noticed as it passes beneath a boat. In deep water, the top of the wave rarely reaches more than 3 feet (ft.) higher than the ocean swell. A tsunami only becomes hazardous when it approaches land. Similar to all waves, as a tsunami enters shallow water near the coast, it slows and the height increases.

150 100 50

The function 𝑆(𝑑) = √9.8𝑑 estimates the speed, in m/s, of a tsunami based on the depth of water, in meters (m).

-50

5,000

10,000

Depth of water (m)

The graph of 𝑆(𝑑) is shown.

A table of values for 𝑆(𝑑) is also shown. 𝒅 𝑺

1,000

98.995

10,000

313.05

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20,000

442.719

25,000

494.975

30,000

542.218

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a. Interpret the point (25000, 494.975) within the context.

b. Discuss with your partner how the table of values shows that a tsunami slows down as it approaches the coast. Summarize your discussion.

c. Complete the table for the left end behavior of the function. Left End Behavior

Interpretation

As 𝑑 → _________, 𝑆 → _________.

As the depth of the water _____________________, the speed of the wave _____________________.

The deepest part of the ocean is called the Challenger Deep, and it is located several hundred kilometers (km) southwest of the US territorial island of Guam in the western Pacific Ocean. It is at the southern end of the Mariana Trench. Challenger Deep is approximately 36,200 ft. deep. d. There are approximately 3.281 ft. in each meter. Determine the depth of Challenger Deep in m. Round to the nearest thousandth.

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e. Use the information about Challenger Deep to determine a domain and a range that makes sense for the context. Write your answer using interval notation. Domain

Range

Lesson Summary Radical, or root, functions are used to model a variety of real-world phenomena, such as the velocity of a natural disaster, the speed of an athlete, the pressure in a water hose, or the radius of a spherical object. In each of these scenarios, it is important to think through which values of the domain and range will make sense in the context of the problem. 3 3 𝑉, where 𝑉 is the For example, the radius of a sphere, in ft., is modeled by 𝑟 = �

4𝜋

volume of the sphere, in cubic feet (cu. ft.). Although the domain and range for the function is (−∞, ∞), negative values for 𝑟 and 𝑉 do not make sense in this context. It also doesn’t make sense for the values to have an upper limit of infinity because it would be impossible to create a sphere with a radius of ∞. In such cases, it is important to interpret the points on the graph within a real-world context rather than only a mathematical one.

Practice Problems 1. In 1996, an hour-long television show only had 12.5 minutes (min.) of commercials. The function 𝐶(𝑥) = 0.7√𝑥 + 12.5 models the amount of time, 𝐶, in min., of commercials for 𝑥, which represents each year since 1996. a. Complete the table of values. 𝒙

𝑪

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b. Graph the function.

c. Complete the table for the left end behavior. Left End Behavior Interpretation

As 𝑥 → _________, 𝐶 → _________.

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2. The estimated speed 𝑠, in miles per hour (mph), of a car at the end of a race is 3 𝑠 = 234 �

𝑝 , where 𝑝 is the horsepower of the car. 2,744

a. Complete the table of values. Round each value to the nearest tenth. 𝒑 𝒔

b. Graph the function. s

p 0

Review Problem 3. Complete the table. Use powers of 16 in the top row. Use radicals or rational numbers in the second row. 3

1 16

16−4

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16−4

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Unit 5, Lesson 9: Inequivalent Equations

Warm-Up: 2 and 2 What do you notice? What do you wonder? • 𝑥2 = 4

• 𝑥2 = − 4

• (𝑥 − 2)(𝑥 + 2) = 0

• 𝑥 = √4

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Exploration Activity: Careful When You Take the Square Root Tyler was solving the equation 𝑥2 − 1 = 3.

He said, “I can add 1 to each side of the equation and it doesn’t change the equation. I get 𝑥2 = 4.” 1. Priya said, “It does change the equation. It just doesn’t change the solutions!” Then she showed these two graphs.

a. How can you see the solution to the equation 𝑥2 − 1 = 3 in Figure A?

b. How can you see the solution to the equation 𝑥2 = 4 in Figure B?

c. Use the graphs to explain why the equations have the same solutions.

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2. Tyler said, “Now I can take the square root of each side to get the solution to 𝑥2 = 4. The square root of 𝑥2 is 𝑥. The square root of 4 is 2.” He wrote: 𝑥2 = 4

√𝑥2 = √4 𝑥 = 2

Priya said, “But the graphs show that there are two solutions!” What went wrong?

Collaborative Activity: Another Way to Solve Han was solving the equation

𝑥 + 3 2 = 4.

He said, “I know that half of 𝑥 + 3 is 4. So 𝑥 + 3 must be 8, since half of 8 is 4. This means that 𝑥 is 5.” 1. Use Han’s reasoning to solve this equation: (𝑥 + 3)2 = 4.

2. What advice would you give to someone who was going to solve an equation like (𝑥 + 3)2 = 4?

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Exploration Activity: What Happens When You Square Each Side? Mai was solving the equation √𝑥 – 1 = 3.

She said, “I can square each side of the equation to get another equation with the same solutions.” Then she wrote: √𝑥 – 1 = 3

(√𝑥 – 1)2 = 32

𝑥 – 1 = 9

𝑥 = 10

1. Check to see if her solution makes the original equation true.

2. Andre said, “I tried your technique to solve √𝑥 – 1 = − 3 but it didn’t work.” Why doesn’t it work? Explain or show your reasoning.

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Guided Activity: Solve These Equations with Square Roots in Them Find the solution(s) to each of these equations, or explain why there is no solution. 1. √𝑡 +1 = 3

2. −10 = −√𝑎

3. √3 – 𝑤 − 4 = 0

4. √𝑧 + 9 = 0

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Lesson Summary Every positive number has 2 square roots. This idea is evident in the graph of 𝑦 = 𝑥2 shown.

If 𝑦 is a positive number, such as 4, then the line 𝑦 = 4 crosses the graph 𝑦 = 𝑥2 in 2 places, so the equation 𝑥2 = 4 will have 2 solutions, √4 and −√4.

This is true for any positive number 𝑎, where 𝑦 = 𝑎 will cross the graph in 2 places, and 𝑥2 = 𝑎 will have 2 solutions, 𝑥 = √𝑎 and 𝑥 = −√𝑎. When working with a square root in an equation like √𝑡 − 6 = 0, first isolate the square root, and then square each side, as shown.

√𝑡 − 6 √𝑡 𝑡 𝑡

= = = =

0 6 62 36

Be aware that sometimes squaring each side of an equation gives results that aren’t solutions to the original equation. For example, a solution method for the equation √𝑡 + 6 = 0 is shown. √𝑡 + 6 √𝑡 𝑡 𝑡

= = = =

0 −6 (−6)2 36

Note that 36 is not a solution to the original equation because √36 + 6 doesn’t equal 0. In fact, √𝑡 + 6 = 0 has no solutions because it’s impossible for the sum of 2 positive numbers to be 0. Remember that a transformed equation sometimes has solutions that the original equation doesn’t have. Always check your solutions by substituting them into the original equation to verify them.

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Practice Problems 1. Noah solved the equation 5𝑥2 = 45. Here are his steps: 5𝑥2 = 45 𝑥2 = 9

𝑥 = 3

Do you agree with Noah? Explain your reasoning.

2. Find the solution(s) to each equation, or explain why there is no solution. a. √𝑥 + 4 + 7 = 5

b. √47 − 𝑥 − 2 = 4

c. 1 √20 + 𝑥 = 5 2

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3. Which is a solution to the equation √5 − 𝑥 + 13 = 4? A. 86 B. 81 C. 9

D. The equation has no solution.

Review Problems 4. Select all expressions that are equal to □ − 5

□ □

√2

1 . (√2 )5

1 √25

1 √32

□ −�√2� 5

□ −22 −

□ 2 2

5. Which are the solutions to the equation 𝑥2 = 36? A. 6 only

B. −6 only

C. 6 and −6

D. This equation has no solutions.

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Unit 5, Lesson 10: Solving Radical Equations – Part 1

Warm-Up: Math Talk: Radical Equations Solve these equations mentally.

∛𝑥 = 1 √7 = √𝑥 − 1 √100 = 2𝑥 √𝑥 + 1 = −5

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Guided Activity: Getting to the Root of the Problem Find the solution(s) to each of these equations, or explain why there is no solution. 1. √𝑎 − 5 = 5 2. ∛𝑎 − 5 = 5 3. ∛𝑏 = − 2 4. √𝑐 + 2 = 0 5. ∛3 − 𝑑 + 4 = 0 6. √7 = √𝑥 − 1 7. √36 = 3𝑦 8. 22𝑧 = ∛11

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Collaborative Activity: Write Your Own Equation 1. Write an equation that includes a radical symbol with: a. one solution

b. no solutions

c. two solutions

2. Switch with a partner and solve their equations.

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Lesson Summary Whenever an equation has a radical symbol that contains a variable, it can be solved by isolating the radical and then raising each side of the equation to a power that would result in the variable raised to an exponent of 1. For example, the solution of the equation − 4 = ∛5𝑝 + 1 is shown. −4 = ∛5𝑝 + 1

(−4)3 = �∛5𝑝 + 1 �3 −64 = 5𝑝 + 1 −65 = 5𝑝 −13 = 𝑝

Sometimes, this solution method results in an equation with solutions that do not make the original equation true. Therefore, it is good practice to verify that the solutions to the new equation are also solutions to the original equation. In this example, the solution to the transformed equation, 𝑝 = −13, also satisfies the original equation because ∛5(−13) + 1 = −4. Another way to solve radical equations is to reason about what the answer is instead of raising each side to a power. For example, when solving √1 − 𝑥 + 5 = 11, the equation can first be rearranged to get √1 − 𝑥 = 6. Then, reason, “If the positive square root of 1 − 𝑥 is 6, then 1 − 𝑥 must be 36 because the positive square root of 36 is 6. Therefore, 𝑥 must be −35 since 1 − (−35) = 36.” Checking this result using substitution shows that 𝑥 = −35 is a solution to the original equation because √1 − (−35) + 5 = 11.

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Practice Problems 1. Find the solution(s) to each of these equations, or explain why there is no solution. a. √𝑥 + 5 + 7 = 10 b. √𝑥 − 2 + 3 = −2 2. For each equation, decide how many solutions it has and explain how you know. a. (𝑥 − 4)2 = 25 b. √𝑥 − 4 = 5 c. 𝑥3 − 7 = −20 d. 6 ⋅ ∛𝑥 = 0 3. Jada was solving the equation √6 − 𝑥 = −16. She was about to square each side, but then she realized she could give an answer without doing any algebra. What did she realize?

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Review Problems 1. Complete the table. Use powers of 16 in the top row and radicals or rational numbers in the bottom row. 161

16 3

1 4

16−1 1 16

2. Which are the solutions to the equation 𝑥3 = 35? A. ∛35

B. −∛35

C. Both ∛35 and −∛35

D. The equation has no solutions.

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Unit 5, Lesson 11: Solving Radical Equations – Part 2

Warm-Up: Radical Equation Blanks Substitute the digits 0–9, no more than once each, to make both the equations true.

�⎕⎕ = ⎕�⎕ �⎕⎕ = ⎕

Guided Activity: Solving Radical Equations Equations with radicals and rational exponents can be solved by rewriting them as equations that do not have radicals or rational exponents. 1. For the following equations, describe the steps taken to solve each equation. Then, check each solution to determine if the solution is valid. Equation and Solution

√𝑥 = 3 2

Description

Check

(√𝑥 1 = (3) 2 𝑥 = 9

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Equation and Solution

Description

Check

√𝑥 + 7 = 5 √𝑥 = −2 2

(√𝑥) 1 = (−2) 1 𝑥 = 4

ℎ 3 + 11 = 9 1

ℎ 3 = −2

3 1 1

�ℎ 3 � = (−2)3 1 ℎ = −8

∛ℎ − 5 = 3 ∛ℎ = 8 3

(∛ℎ) 1 = (8) 1 ℎ = 512

2. For the equations with square roots, how was 𝑥 isolated?

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3. For the equations with cube roots, how was ℎ isolated?

4. One of the equations has a solution listed that is not valid. Discuss with your partner why the solution is not valid. Summarize your discussion.

5. Kim and Nia were solving the square root equation 18 + √𝑥 = 2√𝑥 + 20.

Kim said the solution to the equation is 𝑥 = 1, but Nia said that there is no solution. a. Who do you agree with?

b. Explain why you agree with this student. Show any work that may support your explanation.

c. Write a note to the student you disagreed with including advice that can be used when solving other square root equations.

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Collaborative Activity: Solving Radical Equations 1. Solve each equation. Then, check each solution, and identify any extraneous solutions. a. 3∛𝑥 + 2 = 12

b. 3√(6 − 𝑥) + 4 = −8

c. √4𝑥 − 7 = √3𝑥 + 9 2. Work with your partner to solve 349 = √(13 − 4𝑥)3 + 6.

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Guided Activity: Solving Radical Equations with an Index Greater than 3 1. Consider the radical equation ∜3𝑥 − 5 = 20.

a. Discuss with your partner how to solve for 𝑥. b. Rewrite ∜3𝑥 − 5 using a fractional exponent.

c. Solve ∜3𝑥 − 5 = 20. Explain your process as you solve for 𝑥.

d. Is the solution extraneous? Justify your answer mathematically.

2. Solve ∜9𝑥 + 9 − 2 = 1.

3. Solve √4𝑥 + 8√4𝑥 = 18.

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Lesson Summary The process of isolating a radical and then raising both sides of the resulting equation to a power may lead to an answer that is actually not one of the solutions of the original equation. This extra answer is called an extraneous solution. Extraneous solutions are solutions that when substituted back into the original equation are not valid. Extraneous solutions might occur when solving equations with an even radical index. Since even index radical functions have restricted domains, as you learned previously in this unit, solutions that are mathematically correct but not viable in the original equations may occur. For example, the equation √𝑥 = −3 has no real solution because no square root of a real number would be equal to −3. When solving the equation algebraically, (√𝑥)2 = (−3)2 results in 𝑥 = 9, which is a mathematically true statement. However, substituting the 𝑥- value into the original equation √9 ≠ −3. Therefore, 𝑥 = 9 is an extraneous solution.

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Practice Problems 1. Consider the equation shown. √4 − 𝑥 = 𝑥 − 2

a. Rewrite the equation so that it does not have radicals.

b. Find the solution(s) to the rewritten equation.

c. Are both solutions of the rewritten equation also solutions to the original equation? Justify your answer.

2. Solve the equation −√𝑥 − 2 = 4. Then, check for extraneous solutions.

3. Solve the equation −√1 − 𝑥 + 𝑥 + 5 = 0. Then, check for extraneous solutions.

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Review Problems 4. Identify the degree, leading coefficient, and constant value of each of the following polynomials. a. 𝑓(𝑥) = 2𝑥5 − 8𝑥2 − 𝑥 − 6 b. ℎ(𝑥) = 𝑥3 − 7𝑥2 − 𝑥 + 2 c. 𝑔(𝑥) = 5𝑥2 − 4𝑥3 + 2𝑥 + 5.4

5. Here are two expressions whose sum is a new expression, 𝐴. (2𝑥2 + 5) + (6𝑥⎕ − 7) = 𝐴

Select all the values that we can put in the box so that 𝐴 is a polynomial.

□ −2 □ −1

□ −0.5 □0

□ 0.5 □1 □2

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