In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form).

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2 CHAPTER 8 In Chapter 4, you used a web to organize the connections you found between each of the different representations of lines. These connections enabled you to use any representation (such as a graph, rule, situation, or table) to find any of the other representations. In this chapter, a quadratics web will challenge you to find connections between the different representations of a parabola. Through this endeavor, you will learn how to rewrite quadratic equations by using a process called factoring. You will also discover and use a very important property of zero. In this chapter, you will learn: Quadratics Think about these questions throughout this chapter:? How can I rewrite it? What s the connection? What s special about zero? What information do I need? Is there another method? How to factor a quadratic expression completely. How to find the roots of a quadratic equation, if they exist. How to move from all representations of a parabola (rule, graph, table, and situation) to each of the other representations directly. Section 8.1 In this section, you will develop a method to change a quadratic equation written as a sum into its product form (also called its factored form). Section 8.2 Through a fun application, you will find ways to generate each representation of a parabola from each of the others. You will also develop a method to solve quadratic equations using the Zero Product Property. Yo! Section 8.3 In this section, you will be introduced to another method to solve quadratic equations called the Quadratic Formula. 604 Algebra Connections

3 Chapter 8 Teacher Guide Section Lesson Days Lesson Objectives Materials Homework Introduction to Factoring Quadratics Factoring with Generic Rectangles Factoring with Special Cases Factoring Completely Investigating a Parabola Multiple Representations for Quadratics Algebra tiles 8-6 to 8-11 Algebra tiles 8-16 to 8-21 Overhead algebra tiles 8-27 to 8-32 Overhead algebra tiles (optional) Poster graph paper and markers OR overhead graph paper and pens Lesson Res. Pg. Poster paper and markers Colored pencils or pens 8-37 to to to Zero Product Property None 8-65 to Solving Quadratic Equations by Factoring None 8-76 to Completing the Quadratic Web Graphing calculators Overhead graphing calculator 8-85 to Introduction to the Quadratic Formula More Solving Quadratic Equations Overhead graphing calculator OR Lesson Res. Pg. on a transparency 8-96 to None to Choosing a Strategy Yo-yo (optional) to Chapter Closure Total: Varied Format Options 12 days plus optional time for Chapter Closure Chapter 8: Quadratics 605

4 Overview of the Chapter In Chapter 8, students will broaden their understanding of multiple representations, systems of equations, and solving equations to include quadratic equations. This chapter has many learning objectives: Students will learn how to factor quadratic and simple cubic equations. Students will learn how to solve quadratic equations by factoring and using the Zero Product Property and will recognize that these solutions are the x-intercepts of a corresponding parabola. Students will learn how to use the Quadratic Formula to solve quadratic equations. Students will examine when a parabola has zero, one, or two x-intercepts. Students will apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. Students begin this chapter by learning how to factor quadratics in Section 8.1. They will learn this very similarly to the way they learned how to multiply binomials: they will first factor with tiles and then will develop a shortcut method using a generic rectangle. Section 8.2 challenges students to find connections between the different representations of a parabola so that they can use any representation to get any of the others. Through this endeavor, students will need to find out how to graph a parabola directly from its rule, which requires a new algebraic process. Students then discover the Zero Product Property and learn how it can help them solve quadratic equations. Finally, students confront a quadratic equation that cannot be factored but does in fact have roots. This sets up the need for the Quadratic Formula, which is introduced and derived in Section 8.3. (See more notes about its derivation below.) Necessary Materials This text assumes that students have daily access to a scientific calculator, graph paper, and a ruler or straightedge. Furthermore, please note that most lessons in this chapter require the use of algebra tiles, which were previously used in Chapters 2, 3, 5, and 6. When algebra tiles is listed anywhere as a necessary material, please be prepared with: Algebra tiles, enough for each student or pair of students. Cornerpiece, one per student or pair of students. Overhead algebra tiles and cornerpiece for teacher use. Note: See the teacher notes in Chapter 2 to find out how to purchase algebra tiles or how to create them from resource pages. In addition, Lesson assumes that students have access to graphing technology, such as graphing calculators or computers with graphing software. The graphing technology allows 606 Algebra Connections

5 students to use a process of Guess and Check to find the equation of the parabola that best fits the information given in each complaint. If the technology is not available, students will need to use graphing shortcuts to check their answers to problem See the Technology Notes section in the front of this Teacher Guide for directions on using a TI-83. For all other calculator models, consult the appropriate calculator user s guide or look online at The Derivation of the Quadratic Formula The Quadratic Formula is introduced at this stage to augment the methods students can use to solve quadratic equations. Using the algebraic skills they already have, students will derive the Quadratic Formula in this chapter. However, in Chapter 12, students will learn how to derive the Quadratic Formula another way: by completing the square. Accuracy of Solutions and Using a Scientific Calculator Before teaching this chapter, consider how you want students to display their answers to problems solved using the Quadratic Formula. For example, would you rather have them express the solutions to x 2! 3x! 5 = 0 as x = or as x! 4.2 or x! "1.2? 3± 29 2 Also, if you have your students use a scientific calculator to evaluate these expressions, what degree of accuracy do you expect? This text leaves these decisions to the teacher and only explicitly asks students to use their calculators to evaluate their answers when the application or context requires an interpretable value. You will need to instruct students exactly how you want them to leave answers when no special instructions are given in the text. This curriculum assumes that students have access to scientific calculators at school and at home. It is common, however, for students comfortable with using a calculator to have difficulty evaluating a complicated expression such as those that arise when using the Quadratic Formula.! !!3 Several problems that ask students to evaluate messy expressions, such as, are assigned in homework prior to Section 8.3. This will give you the opportunity to assess your students ability to use their calculators appropriately and to address common errors before Section 8.3. Where Is This Going? Quadratic equations will continue to be an area of focus throughout Chapters 9 through 12. For example, in Chapter 9, students will examine the solutions to quadratic inequalities. In Chapter 10, students will develop a new method to help solve quadratic equations called completing the square. In addition, students will rely on their understanding of factoring to simplify rational expressions in Chapter 10. Furthermore, in Chapter 12, students will derive the Quadratic Formula by completing the square. Chapter 8: Quadratics 607

6 Lesson How can I find the product? Introduction to Factoring Quadratics Lesson Objective: Students will review how to build rectangles with tiles and will learn shortcuts for finding the dimensions of a completed generic rectangle. Students will discover that the products of the terms in each diagonal of a generic rectangle are equal. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 8-1 through 8-5 Ways of Thinking: Materials: Suggested Lesson Activity: Reversing thinking, generalizing, justifying Algebra tiles Note: When algebra tiles is listed anywhere as a necessary material, please be prepared with: Algebra tiles, enough for each student or pair of students Cornerpiece, one per student or pair of students Overhead algebra tiles and cornerpiece for teacher use General Team Roles transparency (optional) Note: The General Team Roles transparency is the Lesson Resource Page ( General Team Roles ). It is listed as an optional resource throughout many lessons in the rest of the text. When you see General Team Roles transparency listed, it is always the Lesson Resource Page that is being referred to. Have a student volunteer read the lesson introduction. Then have teams start problem 8-1 as a warm-up. This problem asks students to review what they learned in Chapter 5. After about five minutes, move teams on to problem 8-2, which asks students to factor expressions using algebra tiles to investigate whether a sum can always be written as a product. This will help students recognize that not all expressions are factorable. Also, the act of building composite rectangles with algebra tiles will strengthen students understanding of the factoring process that will be introduced in Lesson Specifically, students will need to understand the options for splitting the x-tiles between the two corners of the rectangle. To ensure that there is enough time for problems 8-4 and 8-5, stop teams after 20 minutes. Then teams should move on to problem 8-3, which has students focus on finding the dimensions of a completed generic rectangle. As you circulate, emphasize that students should look for special strategies to find the dimensions. After 10 minutes, pull the class together and ask students to share any shortcuts they discovered. Students may notice that each 608 Algebra Connections

7 dimension is the greatest common factor (GCF) of its corresponding row or column, demonstrated in the diagram at right. Note that this is true when the quadratic expression does not have a common factor. Students will examine what happens when a quadratic has a common factor in Lesson As teams finish, have them post solutions on the board for others to check. Leave the solutions on the board so that they can be used during closure. Color-coding work can help some students keep track of the different parts of the quadratic. For example, writing all x 2 -terms in blue, all x-terms in red, and all constants in green (in both the expression and in the generic rectangle) can help students see how the parts in the generic rectangle relate to the expression. See the example at right. Finally, problem 8-4 asks students to recognize that the product of the terms of a generic rectangle s diagonal equals the product of the terms in the other diagonal. This is an important pattern that will later help students develop an algorithm for factoring quadratics without algebra tiles. blue +1 3x 2x 5 6x 2 2x +5 6x 2 +17x + 5 6x 2 red 2x 5 15x 15x green Closure: (10 minutes) Team Roles: Either individually or as a class, have students respond to problem 8-5, which asks them to write a description of the diagonals pattern introduced in problem 8-4. Be sure that all students not only recognize the pattern, but also test the pattern on the other generic rectangles in problem 8-3. A Math Notes box proving that the pattern always works is not offered until Lesson so that students will not have the pattern in front of them while they try to discover it on their own. However, once students have found the pattern, there is no harm in proving this pattern for closure. If you have not changed your teams or roles recently, you may want to start this chapter with a new seating chart. Remind team members of their roles using the General Team Roles transparency from Chapter 7. Emphasize the need for Facilitators to keep teams together in problem 8-2. Some good phrases for Facilitators to use are, Did everyone get? and Are we all ready to move on? Resource Managers will not only need to get the algebra tiles for the team, but they will also be responsible for making sure the tiles are returned in complete sets. Homework: Problems 8-6 through 8-11 Chapter 8: Quadratics 609

8 8.1.1 How can I find the product? Introduction to Factoring Quadratics Student pages for this lesson are In Chapter 5 you learned how to multiply algebraic expressions using algebra tiles and generic rectangles. This section will focus on reversing this process: How can you find a product when given a sum? 8-1. Review what you know about products and sums below. a. Write the area of the rectangle at right as a product and as a sum. Remember that the product represents the area found by multiplying the length by the width, while the sum is the result of adding the areas inside the rectangle. [ (x + 4)(y + x + 2) = xy + x 2 + 6x + 4 y + 8 ] x y x b. Use a generic rectangle to multiply (6x!1)(3x + 2). Write your solution as a sum. [ 18x 2 + 9x! 2 ] 8-2. The process of changing a sum to a product is called factoring. Can every expression be factored? That is, does every sum have a product that can be represented with tiles? Investigate this question by building rectangles with algebra tiles for the following expressions. For each one, write the area as a sum and as a product. If you cannot build a rectangle, be prepared to convince the class that no rectangle exists (and thus the expression cannot be factored). [ a: (2x + 3)(x + 2), b: (2x + 1)(3x + 2), c: no solution, d: (2x + y)(y + 3) ; Conclusion: Not every expression can be factored. ] a. 2x 2 + 7x + 6 b. 6x 2 + 7x + 2 c. x 2 + 4x +1 d. 2xy + 6x + y 2 + 3y 610 Algebra Connections

9 8-3. Work with your team to find the sum and the product for the following generic rectangles. Are there any special strategies you discovered that can help you determine the dimensions of the rectangle? Be sure to share these strategies with your teammates. [ a: (3x +1)(2x + 5) = 6x x + 5, b: (5x! 2)(y + 3) = 5xy +15x! 2y! 6, c: (4x! 3)(3x + 4) = 12x 2 + 7x! 12 ] a. b. c. 2x 5!2y!6!9 x!12 6x 2 15x 5xy 15x 12x 2 16x 8-4. While working on problem 8-3, Casey noticed a pattern with the diagonals of each generic rectangle. However, just before she shared her pattern with the rest of her team, she was called out of class! The drawing on her paper looked like the diagram below. Can you figure out what the two diagonals have in common? [ The product of each diagonal is equal: 6x 2! 5 = 30x 2 and 2x!15x = 30x 2. ] 2x 5 6x 2 15x 8-5. Does Casey s pattern always work? Verify that her pattern works for all of the 2-by-2 generic rectangles in problem 8-3. Then describe Casey s pattern for the diagonals of a 2-by-2 generic rectangle in your Learning Log. Be sure to include an example. Title this entry Diagonals of a Generic Rectangle and include today s date. [ Diagonals: part (a) are both 30x 2, part (b) are both!30xy, part (c) are both!144x 2. Students should state that this pattern always works and describe the pattern in their own words. Typical response: The product of one diagonal always equals the product of the other diagonal. ] Chapter 8: Quadratics 611

10 ETHODS AND MEANINGS MATH NOTES New Vocabulary to Describe Algebraic Expressions Since algebraic expressions come in many different forms, there are special words used to help describe these expressions. For example, if the expression can be written in the form ax 2 + bx + c and if a is not 0, it is called a quadratic expression. Study the examples of quadratic expressions below. Examples of quadratic expressions: x 2! 15x m 2! 25 12! 3k 2 + 5k The way an expression is written can also be named. When an expression is written in product form, it is described as being factored. When factored, each of the expressions being multiplied is called a factor. For example, the factored form of x 2! 15x + 26 is (x!13)(x! 2), so x!13 and x! 2 are each factors of the original expression. Finally, the number of terms in an expression can help you name the expression to others. If the expression has one term, it is called a monomial, while an expression with two terms is called a binomial. If the expression has three terms, it is called a trinomial. Study the examples below. Examples of monomials: 15xy 2 and!2m Examples of binomials: 16m 2! 25 and 7h h Examples of trinomials: 12! 3k 2 + 5k and x 2! 15x Write the area of the rectangle at right as a sum and as a product. [ (2x! 3)(x + 2y! 4) = 2x 2 + 4xy! 11x! 6y + 12 ]!3x!6y 2x 2 4xy 12!8x 612 Algebra Connections

11 8-7. Multiply the expressions below using a generic rectangle. Then verify Casey s pattern (that the product of one diagonal equals the product of the other diagonal). [ a: 12x x! 5, b: 4x 2! 28x + 49 ] a. (4x!1)(3x + 5) b. (2x! 7) Remember that a Diamond Problem is a pattern for which the product of two numbers is placed on top, while the sum of the same two numbers is placed on bottom. (This pattern is demonstrated in the diamond at right.) Copy and complete each Diamond Problem below. [ Find solutions in the diamonds below. ] x xy x+y y a. b. c. d e. f. 6x 2 2x 3x 5x!7x 2 7x x 6x 8-9. For each line below, name the slope and y-intercept. [ a: m = 2, (0,!! 1 2) ; b: m =!3, (0,!!7) ; c: m =! 2 3, (0,!8) ; d: m = 0, (0,!!2) ] a. y =!1+4 x 2 b. 3x + y =!7 c. y =!2 3 x + 8 d. y =! On graph paper, graph y = x 2! 2x! 8. a. Name the y-intercept. What is the connection between the y-intercept and the rule y = x 2! 2x! 8? [ (0, 8); It is the constant in the equation. ] b. Name the x-intercepts. [ ( 2, 0) and (4, 0); Students may notice that the product of the x-intercepts equals the constant term. ] c. Find the lowest point of the graph, the vertex. [ (1, 9); Its x-coordinate is midway between the x-intercepts. ] Calculate the value of each expression below. [ a:!1, b:! 7.24, c:! " 4.24 ] a. 5! 36 b c.!2! 5 Chapter 8: Quadratics 613

12 Lesson Is there a shortcut? Factoring with Generic Rectangles Lesson Objective: Students will develop an algorithm to factor quadratic expressions without algebra tiles. Length of Activity: One day (approximately 50 minutes) Core Problems: Ways of Thinking: Materials: Suggested Lesson Activity: Problems 8-12 through 8-15 (parts (a) and (b)) Making connections, generalizing Algebra tiles Introduce today s lesson and use problem 8-12 as a quick warm-up for students to review what they learned in Lesson Although problem 8-13 is written so that students can investigate factoring in teams, it is recommended that it instead be used as an outline for a whole-class discussion. For part (a), have students build a composite rectangle with tiles in order to factor (as shown in the diagram at right). Then, as a class, figure out how to split the 10 x-terms in the generic rectangle for part (b). At this point, expect students to guess and check how to place the x-terms into the rectangle so that it is factorable. For example, they may first try to split the 10 x-terms by placing 5x into each corner. However, when finding the dimensions, they will quickly learn that there are no dimensions that, when multiplied, will create that generic rectangle. Let students struggle for a bit, as this will help them appreciate the use of the Diamond 8 Problems later. Have overhead 3x algebra tiles ready to demonstrate 2 all of the different combinations for 10x and to verify the solution. At this point we expect that students will be glad to find a shortcut that will help them decide how to split the x-term in the generic rectangle. Part (c) hints that Casey s pattern can help factor another trinomial, 2x 2 + 7x + 6. Specifically, we know that the product of the two missing terms equals the product of the terms on the other diagonal (6 and 2x 2 ), based on the pattern introduced in problem 8-4. Also, the number of x-tiles that are placed in opposite corners of a generic rectangle must add up to the total number of x-tiles in the original expression (which, in this case, is 7x).? 2x 2 6? x x x 614 Algebra Connections

13 Part (d) of problem 8-13 points out that since students know the sum and product of the two missing terms, they can create and solve a Diamond Problem (see diagram at right) to find the missing terms in the generic rectangle. Once the generic rectangle is complete, students can find common factors in rows and columns to find the dimensions of the generic rectangle. To recap, this four-step process will help students factor all factorable quadratic expressions: 1. Place the x 2 -terms and constant terms of the quadratic expression in opposite corners of the generic rectangle. Determine the sum and product of the two remaining corners: The sum is simply the x-term of the quadratic expression, while the product is equal to the product of the x 2 -terms and constant terms. 2. Place this sum and product into a Diamond Problem and solve. 3. Place the solutions from the Diamond Problem into the generic rectangle and find the dimensions of the generic rectangle. 4. Write the answer as a product. product 12x 2 Problem 8-14 offers teams a chance to recap the factoring process. Then students practice using this factoring method in problem Note that part (d) of problem 8-15 is not factorable. 7x sum Closure: (5 minutes) Pull the class together and point out that students have accomplished an important goal: They have developed an algorithm to factor quadratics without algebra tiles. Have a few student volunteers recap the factoring process. Ask questions that help students justify statements, such as, Why do you multiply the x 2 -term with the units term and place that at the top of the Diamond Problem? Homework: Problems 8-16 through Is there a shortcut? Factoring with Generic Rectangles Student pages for this lesson are Since mathematics is often described as the study of patterns, it is not surprising that generic rectangles have many patterns. You saw one important pattern in Lesson (Casey s pattern from problem 8-4). Today you will continue to use patterns while you develop a method to factor trinomial expressions. Chapter 8: Quadratics 615

14 8-12. Examine the generic rectangle shown at right. a. Review what you learned in Lesson by writing the area of the rectangle at right as a sum and as a product. [ (5x! 2)(2x! 7) = 10x 2! 39x + 14 ]!35x 14 10x 2! 4x b. Does this generic rectangle fit Casey s pattern for diagonals? Demonstrate that the product of each diagonal is equal. [!35x "! 4x = 10x 2 "14 = 140x 2 ] FACTORING QUADRATICS To develop a method for factoring without algebra tiles, first study how to factor with algebra tiles, and then look for connections within a generic rectangle. a. Using algebra tiles, factor 2x 2 + 5x + 3 ; that is, use the tiles to build a rectangle, and then write its area as a product. [ (2x + 3)(x +1) ] b. To factor with tiles (like you did in part (a)), you need to determine how the tiles need to be arranged to form a rectangle. Using a generic rectangle to factor requires a different process. Miguel wants to use a generic rectangle to factor 3x x + 8. He knows that 3x 2 and 8 go into the rectangle in the locations shown at right. Finish the rectangle by deciding how to place the ten x-terms. Then write the area as a product. [ One corner should contain 4x, while the other should contain 6x ; (3x + 4)(x + 2). ] 3x 2 8 c. Kelly wants to find a shortcut to factor 2x 2 + 7x + 6. She knows that 2x 2 and 6 go into the rectangle in the locations shown at right. She also remembers Casey s pattern for diagonals. Without actually factoring yet, what do you know about the missing two parts of the generic rectangle? [ Their sum is 7x, and their product is 12x 2. ] d. To complete Kelly s generic rectangle, you need two x-terms that have a sum of 7x and a product of 12x 2. Create and solve a Diamond Problem that represents this situation. [ The product 12x 2 should be placed at the top of the diamond problem, 7x at the bottom, and terms 3x and 4x should be in the middle. ]? 6 2x 2? product sum e. Use your results from the Diamond Problem to complete the generic rectangle for 2x 2 + 7x + 6, and then write the area as a product of factors. [ (2x + 3)(x + 2) ] 616 Algebra Connections

15 8-14. Factoring with a generic rectangle is especially convenient when algebra tiles are not available or when the number of necessary tiles becomes too large to manage. Using a Diamond Problem helps avoid guessing and checking, which can at times be challenging. Use the process from problem 8-13 to factor 6x x The questions below will guide your process. a. When given a trinomial, such as 6x x + 12, what two parts of a generic rectangle can you quickly complete? [ One corner contains 6x 2, and the opposite corner contains 12. ] b. How can you set up a Diamond Problem to help factor a trinomial such as 6x x + 12? What goes on the top? What goes on the bottom? [ The product of the x 2 and units terms (in this case, 72x 2 ) goes on top, while the x-term (17x ) goes on bottom. ] c. Solve the Diamond Problem for 6x x + 12 and complete its generic rectangle. d. Write the area of the rectangle as a product. [ (2x + 3)(3x + 4) ] product sum Use the process you developed in problem 8-13 to factor the following quadratics, if possible. If a quadratic cannot be factored, justify your conclusion. [ a: (x + 3)(x + 6), b: (4x! 3)(x + 5), c: (2x! 3)(2x! 1), d: not factorable because there are no integers that multiply to get!9x 2 (the diagonal of the generic rectangle) and add to get 5x. ] a. x 2 + 9x +18 b. 4x 2 +17x!15 c. 4x 2!8x + 3 d. 3x 2 + 5x!3 MATH NOTES ETHODS AND MEANINGS Diagonals of Generic Rectangles Why does Casey s pattern from problem 8-4 work? That is, why does the product of the d ad bd terms in one diagonal of a 2-by-2 generic rectangle always equal the c ac bc product of the terms in the other diagonal? a b Examine the generic rectangle at right for (a + b)(c + d). Notice Product = abcd Product = abcd that each of the resulting diagonals have a product of abcd. Thus, the product of the terms in the diagonals are equal. Chapter 8: Quadratics 617

16 8-16. Use the process you developed in problem 8-13 to factor the following quadratics, if possible. [ a: (x! 6)(x + 2), b: (2x + 1) 2, c: (x! 5)(2x +1), d: (x + 4)(3x! 2) ] a. x 2! 4x!12 b. 4x 2 + 4x +1 c. 2x 2! 9x! 5 d. 3x 2 +10x! For each rule represented below, state the x- and y-intercepts, if possible. [ a: x-intercepts ( 1, 0) and (3, 0), y-intercept: (0, 3); b: x-intercept (2, 0), no y-intercept; c: x-intercepts ( 3, 0), ( 1, 0), and (1, 0), y-intercept (0, 2); d: x-intercept (8, 0), y-intercept (0, 20) ] a. b. c. d. 5x! 2y = 40 x y Graph y = x 2! 9 on graph paper. a. Name the y-intercept. What is the connection between the y-intercept and the rule y = x 2! 9? [ (0, 9); It is the constant in the equation. ] b. Name the x-intercepts. What is the connection between the x-intercepts and the rule y = x 2! 9? [ (3, 0) and ( 3, 0); Students may notice that the product of the x-intercepts equals the constant term. ] Find the point of intersection for each system. [ a: (6, 9), b: (0, 2) ] a. y = 2x! 3 x + y =15 b. 3x = y! 2 6x = 4! 2y 618 Algebra Connections

17 8-20. Solve each equation below for the given variable, if possible. [ a: x =! 10 23, b: all numbers, c: c = 0 ] a. 4 x 5 = x!2 7 b.!3(2b! 7) =!3b + 21! 3b c. 6! 2(c! 3) = Find the equation of the line that passes through the points ( 800, 200) and ( 400, 300). [ y = 1 4 x ] Lesson How can I factor this? Factoring with Special Cases Lesson Objective: Students will continue to practice their factoring skills while learning about special cases: quadratics with missing terms, quadratics that are not in standard form, and quadratics with more than one possible factored form. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 8-22 through 8-24 Ways of Thinking: Materials: Suggested Lesson Activity: Justifying, applying and extending, generalizing Overhead algebra tiles General Team Roles transparency (optional) Start class with an overview of the lesson objective. Then move teams into problem 8-22, which is a review of previous work and is an opportunity for you to assess where students are in their understanding. As you circulate, look for and correct common errors, and be available to answer teams questions. Be sure that students are completing both a Diamond Problem and a generic rectangle for each quadratic. Ask questions that help you know if students understand the factoring process. For example, when students state that 6x 2 must go at the top of a Diamond Problem, ask them to explain how they know it is 6x 2. If a student has part of a generic rectangle completed, ask him or her to justify one or two of the entries. Note that the quadratic in part (a) of problem 8-22 is a perfect square trinomial, so while debriefing this problem, you may want to point out that the answer can be written two ways: as (x + 3)(x + 3) or as (x + 3) 2. The quadratic in part (c) is not factorable. Problem 8-23 introduces students to quadratics that are not in standard form or that have terms missing. Remind teams that you expect to hear mathematical discussions for each case. Expect some questions from Chapter 8: Quadratics 619

18 teams that are unsure about putting 0 into their Diamond Problem. Note that the expression in part (d) of problem 8-23 (40!100m) is not quadratic; however, students can still use the same process to factor it. In this case, since the m 2 -term is 0, the generic rectangle will have 0 and 40. Since the product of the missing terms is 0 and the sum is 100m, a Diamond Problem can help determine the other entries in the generic rectangle, as shown at right. Help students recognize that this generic rectangle can be abbreviated into a generic rectangle with only two parts, like the one shown at right. 100m Finally, problem 8-24 introduces students to a quadratic expression that has two factored forms, which will later lead to an investigation of what it means to be completely factored in Lesson After teams have finished problem 8-24, debrief as a whole class by having different volunteers demonstrate their solutions, so that students can see that there is more than one factored form. Have overhead algebra tiles ready to demonstrate that the rectangles have different arrangements indeed, as shown below m 0 100m m 40 If time allows, problem 8-25 offers a challenge for teams to use their factoring and multiplication skills. Closure: (8 minutes) Team Roles: Problem 8-26 asks students to explain how to factor a quadratic in their own words. Since this is an important entry, you may want to work together as a class to create a strong description that can be entered into each student s Learning Log. A Math Notes box in Lesson will also summarize this process for later reference. Today s work will depend on good communication between team members. Use the General Team Roles transparency to review the roles at the beginning of the lesson. Facilitators should keep their teams together as students apply their factoring skills. They should also make sure team members agree on an answer before moving to the next problem and should ask, Is another answer possible? Homework: Problems 8-27 through 8-32 Note: Problem 8-32 is a preview of the Zero Product Property. Furthermore, problem 8-28 is important to assign, as it introduces the idea of square root. This idea will be used in homework problems to prepare students to use the Quadratic Formula in Section 8.3. Plan on taking some time to go over problem 8-28 before you begin Lesson Algebra Connections

19 8.1.3 How can I factor this? Factoring with Special Cases Student pages for this lesson are Practice your new method for factoring quadratic expressions without tiles as you consider special types of quadratic expressions Factor each quadratic below, if possible. Use a Diamond Problem and generic rectangle for each one. [ a: (x + 3) 2, b: (2x + 3)(x +1), c: not factorable, d: (3m + 7)(m! 2) ] a. x 2 + 6x + 9 b. 2x 2 + 5x + 3 c. x 2 + 5x! 7 d. 3m 2 + m! SPECIAL CASES Most quadratics are written in the form ax 2 + bx + c. But what if a term is missing? Or what if the terms are in a different order? Consider these questions while you factor the expressions below. Share your ideas with your teammates and be prepared to demonstrate your process for the class. [ a: (3x! 2)(3x + 2), b: 4x(3x! 4), c: (4k! 3)(2k! 1), d: 20(2! 5m) ] a. 9x 2! 4 b. 12x 2! 16x c k 2! 10k d. 40!100m Now turn your attention to the quadratic below. Use a generic rectangle and Diamond Problem to factor this expression. Compare your answer with your teammates answers. Is there more than one possible answer? [ Students should get either (2x! 6)(2x + 1) or (x! 3)(4x + 2). Note that some may get the error (2x! 6)(4x + 2) by looking only for common factors. Be sure to correct this by asking the students to verify their answer by multiplying the factors. ] 4x 2!10x! 6 Chapter 8: Quadratics 621

20 8-25. The multiplication table below has factors along the top row and left column. Their product is where the row and column intersect. With your team, complete the table with all of the factors and products. Multiply x! 2 2x + 1 x + 7 x 2 + 5x! 14 2x x + 7 3x + 1 3x 2! 5x! 2 6x 2 + 5x In your Learning Log, explain how to factor a quadratic expression. Be sure to offer examples to demonstrate your understanding. Include an explanation of how to deal with special cases, such as when a term is missing or when the terms are not in standard order. Title this entry Factoring Quadratics and include today s date. ETHODS AND MEANINGS MATH NOTES Standard Form of a Quadratic A quadratic expression in the form ax 2 + bx + c is said to be in standard form. Notice that the terms are in order from greatest exponent to least. Examples of quadratic expressions in standard form: 3m 2 + m!1, x 2! 9, and 3x 2 + 5x. Notice that in the second example, b = 0, while in the third example, c = The perimeter of a triangle is 51 cm. The longest side is twice the length of the shortest side. The third side is 3 cm longer than the shortest side. How long is each side? Write an equation that represents the problem and then solve it. [ s + 2s + s + 3 = 51 ; 12, 24, and 15 cm ] 622 Algebra Connections

21 8-28. Remember that a square is a rectangle with four equal sides. a. If a square has an area of 81 square units, how long is each side? [ 9 units ] b. Find the length of the side of a square with area 225 square units. [ 15 units ] c. Find the length of the side of a square with area 10 square units. [ 10 units ] d. Find the area of a square with side 11 units. [ 121 square units ] Factor the following quadratics, if possible. [ a: (k! 2)(k! 10), b: (2x + 7)(3x! 2), c: (x! 4) 2, d: (3m + 1)(3m!1) ] a. k 2!12k + 20 b. 6x 2 +17x!14 c. x 2! 8x +16 d. 9m 2! Examine the two equations below. Where do they intersect? [ (2, 5) ] y = 4x! 3 y = 9x! Find the equation of a line perpendicular to the one graphed at right that passes through the point (6, 2). [ y =!x + 8 ] Solve each equation below for x. Check each solution. [ a: 5, b: 6, c: 5 or 6, d:! 1 4, e: 8, f:! 1 4 or 8 ] a. 2x!10 = 0 b. x + 6 = 0 c. (2x!10)(x + 6) = 0 d. 4x +1 = 0 e. x! 8 = 0 f. (4 x +1)(x! 8) = 0 Chapter 8: Quadratics 623

22 Lesson Can it still be factored? Factoring Completely Lesson Objective: Students will complete their focus on factoring by considering expressions that can be factored first with a common factor and then again using the quadratic factoring method. Length of Activity: One day (approximately 50 minutes) Core Problems: Ways of Thinking: Materials: Suggested Lesson Activity: Problems 8-33 through 8-36 (parts (a) and (b)) Applying and extending, making connections Overhead algebra tiles (optional) Problem 8-32, from homework, may have raised some questions for your students because they are not used to equations having more than one possible answer. Be prepared to address any questions regarding these problems at the beginning of the lesson. You may want to ask students who found possible solutions to explain their thinking to the class. Choose a volunteer to read the lesson introduction, which connects today s lesson with what students learned about factors of numbers in previous courses. Then start teams with a factoring warm-up, problem This collection of expressions includes perfect square trinomials (which give students a nice short-hand answer), missing terms, a quadratic that is not in standard form, and a quadratic with more than one possible factored form. After about 15 minutes, ask volunteers to post their solutions on the board or overhead. Look for students who got different answers to part (d) of problem 8-33 and ask them to post their answers. This will enable the class to look at the similarities and differences between both results. Leave this work on the board or save transparencies of the work for a discussion during closure of this lesson. Then have teams move on to problem 8-34, which asks them to analyze why some quadratics have unique factored forms while others have more than one possible factored form. As you circulate, ask questions that help teams focus on the terms of each quadratic, such as, What do you notice about all of the terms of 12t 2! 10t + 2? Is that true for the other quadratics as well? Students should recognize that each term is a multiple of 2. Problem 8-34 introduces students to the idea of a common factor. If teams get frustrated, be ready to pull the class together to review common factors. Problem 8-35 gives students an opportunity to practice factoring completely and gives you the opportunity to work with any teams that might be struggling. 624 Algebra Connections

23 Closure: (10 minutes) During the introduction of the lesson, students were told that while 12 can be rewritten as 3 4, as 2 6, as 1 12, or as 2 2 3, only is considered to be factored completely, since the factors are prime and cannot be factored themselves. For closure, connect this with the two solutions found for part (d) of problem 8-33: (3n + 3)(n + 2) and (n +1)(3n + 6). Neither of these expressions is completely factored, because each contains a binomial that can be factored further. It is also a good idea to point out that each of the answers for part (d) of problem 8-33 can be factored again to find the most factored form. For example, if 3n + 3 from (3n + 3)(n + 2) is factored, the end result is 3(n + 1)(n + 2). Likewise, if 3n + 6 from (n +1)(3n + 6) is factored, the result is the same. Although factoring out a common factor first often makes factoring easier by lowering the coefficients of the expression, it does not need to happen first. If students forget to factor out a common factor first, they can always factor it out later. Homework: Problems 8-37 through Can it still be factored? Factoring Completely Student pages for this lesson are There are many ways to write the number 12 as a product of factors. For example, 12 can be rewritten as 3 4, as 2 6, as 1 12, or as While each of these products is accurate, only is considered to be factored completely, since the factors are prime and cannot be factored themselves. During this lesson you will learn more about what it means for a quadratic expression to be factored completely Review what you have learned by factoring the following expressions, if possible. [ a: (3x! 2) 2, b: (9m + 1)(9m! 1), c: (x! 4)(x! 7), d: Students will probably get one of two possible answers: (3n + 3)(n + 2) or (n + 1)(3n + 6). ] a. 9x 2! 12x + 4 b. 81m 2! 1 c x 2! 11x d. 3n 2 + 9n + 6 Chapter 8: Quadratics 625

24 8-34. Compare your solutions for problem 8-33 with the rest of your class. a. Is there more than one factored form of 3n 2 + 9n + 6? Why or why not? [ Yes, because there are two different arrangements of tiles that build a rectangle. ] b. Why does 3n 2 + 9n + 6 have more than one factored form while the other quadratics in problem 8-33 only have one possible answer? Look for clues in the original expression ( 3n 2 + 9n + 6 ) and in the different factored forms. [ Because there is a common factor of 3 in each of the terms of the original expression and in one of the two binomials in either of the two partially factored forms. ] c. Without factoring, predict which quadratic expressions below may have more than one factored form. Be prepared to defend your choice to the rest of the class. [ (i) and (iii) both have common factors, so they could have more than one factored form. ] i. 12t 2! 10t + 2 ii. 5 p 2! 23p! 10 iii. 10x x! 15 iv. 3k 2 + 7k! FACTORING COMPLETELY In part (c) of problem 8-34, you should have noticed that each term in 12t 2! 10t + 2 is divisible by 2. That is, it has a common factor of 2. a. What is the common factor for 10x x! 15? [ 5 ] b. For an expression to be completely factored, each factor must have all common factors separated out. Sometimes it is easiest to do this first. Since 5 is a common factor of 10x x! 15, you can factor 10x x! 15 using a special generic rectangle, which is shown below. Find the length of this generic rectangle and write its area as a product of its length and width. [ 5(2x 2 + 5x! 3) ] 5 10x x! 15 c. Can the result be factored even more? That is, can either factor from the result from part (b) above also be factored? Factor any possible expressions and write your solution as a product of all three factors. [ Yes; 2x 2 + 5x! 3 can be factored as (2x!1)(x + 3). Therefore, the final factored form of 10x x! 15 is 5(2x! 1)(x + 3). ] 626 Algebra Connections

25 8-36. Factor each of the following expressions as completely as possible. [ a: 5(x + 4)(x!1), b: 3x(x + 3)(x! 5), c: 2(x + 5)(x! 5), d: y(x! 5)(x + 2) ] a. 5x 2 +15x! 20 b. 3x 3! 6x 2! 45x c. 2x 2! 50 d. x 2 y! 3xy!10y ETHODS AND MEANINGS MATH NOTES Review the process of factoring quadratics developed in problem 8-13 and outlined below. This example demonstrates how to factor 3x 2 +10x Place the x 2 - and constant terms of the quadratic expression in opposite corners of the generic rectangle. Determine the sum and product of the two remaining corners: The sum is simply the x-term of the quadratic expression, while the product is equal to the product of the x 2 - and constant terms. 2. Place this sum and product into a Diamond Problem and solve it. Factoring Quadratic Expressions 3x 2 6x 24x 2 10x 8 4x 3. Place the solutions from the Diamond Problem into the generic rectangle and find the dimensions of the generic rectangle. 4. Write your answer as a product: (3x + 4)(x + 2). 2 x 6x 3x 2 3x 8 4x Factor the quadratic expressions below. If the quadratic is not factorable, explain why not. [ a: (2x + 5)(x!1), b: (x! 3)(x + 2), c: (3x +1)(x + 4), d: It is not factorable because no integers have a product of 14 and a sum of 5. ] a. 2x 2 + 3x! 5 b. x 2! x! 6 c. 3x 2 +13x + 4 d. 2x 2 + 5x + 7 Chapter 8: Quadratics 627

26 8-38. A line has intercepts (4, 0) and (0, 3). Find the equation of the line. [ y = 3 4 x! 3 ] As Jhalil and Joman practice for the SAT, their scores on practice tests rise. Jhalil s current score is 850, and it is rising by 10 points per week. On the other hand, Joman s current score is 570 and is growing by 50 points per week. a. When will Joman s score catch up to Jhalil s? [ in 7 weeks ] b. If the SAT test is in 12 weeks, who will score highest? [ Joman will score more with 1170 points, while Jhalil will have 970. ] Mary says that you can find an x-intercept by substituting 0 for x, while Michelle says that you need to substitute 0 for y. a. Who, if anyone, is correct and why? [ Michelle is correct. One way to view this is graphically: The x-intercept always has a y-coordinate of 0 because it lies on the x-axis. ] b. Use the correct approach to find the x-intercept of! 4x + 5y = 16. [ ( 4, 0) ] Find three consecutive numbers whose sum is 138 by writing and solving an equation. [ 45, 46, 47; x + (x + 1) + (x + 2) = 138 ] Match each rule below with its corresponding graph. Can you do this without making any tables? Explain your selections. [ a: 2, b: 3, c: 1 ] a. y =!x 2! 2 b. y = x 2! 2 c. y =!x Algebra Connections

27 Lesson What do I know about a parabola? Investigating a Parabola Lesson Objective: Students will describe a parabola, using its intercepts, vertex, symmetry, and whether it is pointing up or down. Length of Activity: One day (approximately 50 minutes) Core Problems: Problem 8-43 Ways of Thinking: Materials: Suggested Lesson Activity: Generalizing, making connections Poster graph paper and markers OR overhead graph paper and overhead pens Have a student volunteer read the problem statement for problem 8-43, Functions of America. Be sure that all students understand the goals of this investigation. Describe your expectations for the team posters (such as being readable and full of mathematics). Emphasize that completed posters should address all four bullets in the student text. If you use team roles, you may want to discuss the importance of each student fulfilling his or her responsibility in their team. See suggestions in the Team Roles section below. Assign each team a parabola from the list below and get teams started on their investigation. Each parabola will offer new insights during presentations. See the notes below: y = x 2! 2x! 8 This parabola has x-intercepts at x = 4 and 2. Observations can be that the y-intercept is the constant term in the equation and that the product of the roots is the y-intercept. y =!x y = x 2! 4x + 5 This parabola points downward, and its y-intercept is the vertex. Its x-intercepts are at (2, 0) and ( 2, 0), and the y-axis is its line of symmetry. This parabola has no x-intercepts. y = x 2! 2x + 1 This parabola has only one x-intercept: (1, 0). y = x 2! 6x + 5 This parabola has x-intercepts at x = 1 and x = 5. If students graph it using a narrow domain (such as! 4 " x " 4 ), they will not get a complete graph with both intercepts. Chapter 8: Quadratics 629

28 y =!x 2 + 3x + 4 This parabola has x-intercepts at x = 4 and x = 1. The vertex of the parabola occurs at x = 1.5, which will not show up in a table with integer inputs. Therefore, students need to use the idea of symmetry to find the vertex. y =!x 2 + 2x! 1 y = x 2 + 5x + 1 This parabola has only one x-intercept at (1, 0) and is the same as y = x 2! 2x + 1 but flipped over. (Students may connect this with flipping tiles to find the opposite expression in Chapter 2.) This is a parabola with no integer x-intercepts, so the x-intercepts need to be estimated. As you circulate, be prepared to help students with the calculations for their rules, especially teams with rules that include!x 2. Be sure to ask students to justify their conclusions. Here are some questions that can help students support their reasons and use mathematical reasoning: Do you predict that all parabolas behave that way? How do you know? How do you know if that point is the highest (or lowest) point on the parabola? Why does your parabola point downward? What part of the rule causes the parabola to behave that way? Are there any patterns in the table? Do all parabolas have this (these) pattern(s)? Why or why not? If any team is stuck on the meaning of the word symmetry, you may want to point out the Math Notes box at the end of the lesson. This box not only defines the word, but it also offers examples of some lines of symmetry. Remind students to put their findings on their team poster or transparency. Leave enough time for quick presentations. Closure: (15 minutes) Have teams quickly share their findings. If there is not much time remaining to have complete presentations, then encourage one team member (such as the Recorder/Reporter) to make a short statement on one or two important observations that no other team has yet mentioned. As teams share, create a class list of all of the observations from the investigation. Record each statement using the language the student uses, as long as the rest of the class understands the statement. Expect some simplistic statements along with some unique and interesting observations. 630 Algebra Connections

29 The list should contain all of the key elements that we want students to use for the rest of this chapter when describing parabolas: x-intercepts, y-intercepts, symmetry, vertex, and whether the parabola is pointing up or down. The posters of student work will be referred to in Chapter 11 when students do similar investigations of other types of functions, such as y = x and y = x 3. Team Roles: If you use team roles, emphasize that the Recorder/Reporter has the important task of organizing the creation of the poster or presentation or team report. If you are going to require individual reports from each student, then the Recorder/Reporter is responsible to help teammates describe the team s findings. On the other hand, the team will need the Facilitator to make sure that all team members understand the task and agree with the findings. The Resource Manager will be responsible for quickly asking for assistance from the teacher when the entire team is stuck. Finally, the team will need the Task Manager to be very mindful of the time remaining for the investigation. This student needs to remind the other team members respectfully to stay on task and to make sure that the roles are being fulfilled. Homework: Problems 8-45 through What do I know about a parabola? Investigating a Parabola Student pages for this lesson are In previous chapters, you have investigated linear equations. In Section 8.2, you will study parabolas. You will learn all you can about their shape, study different equations used to graph them, and see how they can be used in real-life situations. Chapter 8: Quadratics 631

30 8-43. FUNCTIONS OF AMERICA Congratulations! Your work at the Line Factory was so successful that the small local company grew into a national corporation called Functions of America. Recently your company has had some growing pains, and your new boss has turned to your team for help. See her memo below. To: Your study team From: Ms. Freda Function, CEO Re: New product line MEMO I have heard that while lines are very popular, there is a new craze in Europe to have non-linear designs. I recently visited Paris and Milan and discovered that we are behind the times! Please investigate a new function called a parabola. I d like a full report at the end of today with any information your team can give me about its shape and equation. Spare no detail! I d like to know everything you can tell me about how the rule for a parabola affects its shape. I d also like to know about any special points on a parabola or any patterns that exist in its table. Remember, the company is only as good as its employees! I need you to uncover the secrets that our competitors do not know. Sincerely, Ms. Function, CEO Your Task: Your team will be assigned its own parabola to study. Investigate your team s parabola and be ready to describe everything you can about it by using its graph, rule, and table. Answer the questions below to get your investigation started. You may answer them in any order; however, do not limit yourselves to these questions! Does your parabola have any symmetry? That is, can you fold the graph of your parabola so that each side of the fold exactly matches the other? If so, where would the fold be? Do you think this works for all parabolas? Why or why not? Problem continues on next page 632 Algebra Connections

31 8-43. Problem continued from previous page. Is there a highest or lowest point on the graph of your parabola? If so, where is it? This point is called a vertex. How can you describe the parabola at this point? Are there any special points on your parabola? Which points do you think are important to know? Are there any special points that you expected but do not exist for your parabola? What connection(s) do these points have with the rule of your parabola? How would you describe the shape of your parabola? For example, would you describe your parabola as pointing up or down? Do the sides of the parabola ever go straight up or down (vertically)? Why or why not? Is there anything else special about its shape? [ See the Suggested Lesson Activity for this lesson for notes about each parabola below. ] List of Parabolas: y = x 2! 2x! 8 y =!x y = x 2! 4x + 5 y = x 2! 2x + 1 y = x 2! 6x + 5 y =!x 2 + 3x + 4 y =!x 2 + 2x! 1 y = x 2 + 5x Prepare a poster for the CEO detailing your findings from your parabola investigation. Include any insights you and your teammates found. Explain your conclusions and justify your statements. Remember to include a complete graph of your parabola with all special points carefully labeled. Chapter 8: Quadratics 633

32 MATH NOTES OOKING DEEPER Symmetry When a graph or picture can be folded so that both sides of the fold will perfectly match, it is said to have symmetry. The line where the fold would be is called the line of symmetry. Some shapes have more than one line of symmetry. See the examples below. This shape has one line of symmetry. This shape has two lines of symmetry. This shape has eight lines of symmetry. This graph has two lines of symmetry Calculate the value of each expression below. [ a: 2, b: 3, c: 6.1 ] a b.!1+ 49!2 c.!10! Find the equation of the line that goes through the points ( 15, 70) and (5, 10). [ y =!3x + 25 ] Change 6x! 2y = 10 to slope-intercept ( y = mx + b) form. Then state the slope (m) and the y-intercept (b). [ y = 3x! 5 ; m = 3 and b =!5 ] Copy the figure at right onto your paper. Then draw any lines of symmetry. [ There is only one line of symmetry: horizontal through the middle. ] 634 Algebra Connections

33 8-49. For each rule represented below, state the x- and y-intercepts. [ a: x-intercepts ( 2, 0) and (0, 0), y-intercept (0, 0); b: x-intercepts ( 3, 0) and (5, 0), y-intercept (0, 3); c: x-intercepts ( 1, 0) and (1, 0), y-intercept (0, 1); d: x-intercept (9, 0), y-intercept (0, 6) ] a. b. c. d. 2x + 3y =18 x y Use a generic rectangle to multiply each expression below. [ a: 6x 2 + x! 12, b: 25x 2! 20x + 4 ] a. (3x! 4)(2x + 3) b. (5x! 2) 2 Lesson What s the connection? Multiple Representations for Quadratics Lesson Objective: Students will identify connections between different representations of quadratics: an equation, a table, a context, and a graph. Students will also connect the intercepts and vertex of a parabola to a context: the launch, maximum height, and landing of a water balloon. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 8-51 and 8-52 Ways of Thinking: Materials: Suggested Lesson Activity: Making connections, reversing thinking, applying and extending Lesson Resource Page, one per student Poster paper and markers for the class quadratic web Colored pencils or pens The introduction to today s lesson reminds students of the multiple representations web they created for linear equations in Chapter 4. For Chapter 8: Quadratics 635

34 Closure: (15 minutes) linear equations, the situations they studied were tile patterns and various linear contexts. Today students will begin constructing a similar web to keep track of connections they know for quadratic equations. Place the quadratic web (as shown below with no connections made) on a large poster paper in a visible spot in the classroom. During the next four lessons, add arrows on the poster when the class decides it has found the connection between those representations. After reading the lesson introduction, have students work on problem There are many possible answers, so encourage students to be creative! While debriefing problem 8-51, take the opportunity to point out that not every curve is a parabola or part of a parabola. For instance, to many students, a semi-circle seems like part of a parabola. On the other hand, you can point out the interesting fact that, due to the laws of physics, an object thrown, kicked, or flung into the air on Earth will always rise and fall in the shape of a parabola (assuming no other factors such as wind or spin affect it). This will provide a good lead-in to problem Ask a few student volunteers to read the problem statement for problem You may want to have a ball Jen Maggie 30 Al toss as a demonstration of a Imp 25 parabolic curve in front of the class. 20 Distribute the resource page and 15 make sure students understand the 10 task. Students will be given 5 information about four waterballoon launches, each in a different representation. They will need to Solution to part (a) analyze the information given and create a table and graph on the resource page for each launch. See solution graph above. This is a full day, and problem 8-53 may be used as an extension if time is available. Pull the class together and revisit the web (shown again in problem 8-52). Ask the class to use arrows to connect any representations they can. Discuss any discrepancies between different teams webs. You can refer to the various contestants in the waterballoon contest as concrete examples for students to test whether they are Graph Graph QUADRATIC WEB Table Situation QUADRATIC WEB Table Situation Rule or Equation Rule or Equation 636 Algebra Connections

35 currently able to move around the web in particular ways. At this point, students are able to make the connections shown above. Students should recognize that there are a few connections still missing. Problem 8-53 reviews the tile pattern problems of Chapter 4 in which students took the situation (the pattern) and wrote a rule to represent the number of tiles in the x th figure. However, if time is not available, you can remind students of their work from problem 4-1 by displaying one of the team posters from that lesson. Team Roles: As students work on the resource page, the Recorder/Reporter should make sure that teammates can clearly see each other s work to enhance communication. Homework: Problems 8-54 through What s the connection? Multiple Representations for Quadratics Student pages for this lesson are In Chapter 4 you completed a web for the different representations of linear equations. You discovered special shortcuts to help you move from one representation to another. For example, given a linear equation, you can now draw the corresponding graph as well as determine an equation from a graph. Graph Table Rule or Equation Today you will explore the connections between the different representations for quadratics. As you work, keep in mind the following questions: Situation What representations are you using? What is the connection between the various representations? What do you know about a parabola? Chapter 8: Quadratics 637

36 8-51. WATER-BALLOON CONTEST Every year Newtown High School holds a water-balloon competition during halftime of their homecoming game. Each contestant uses a catapult to launch a water balloon from the ground on the football field. This year you are the judge! You must decide which contestants win the prizes for Longest Distance and Highest Launch. Fortunately, you have a computer that will collect data for each throw. The computer uses x to represent horizontal distance in yards from the goal line and y to represent the height in yards. Jen s data The announcer shouts, Maggie Nanimos, you re up first! She runs down and places her catapult at the 3-yard line. After Maggie s launch, the computer reports that the balloon traveled along the parabola y =!x 2 +17x! 42. Then you hear, Jen Erus, you re next! Jen runs down to the field, places her catapult at the goal line, and releases the balloon. The tracking computer reports the path of the balloon with the graph at right. The third contestant, Imp Ecable, accidentally launches the balloon before you are ready. The balloon launches, you hear a roar from the crowd, turn around, and SPLAT! The balloon soaks you and your computer! You only have time to write down the following partial information about the balloon s path before your computer fizzles: x (yards) y (yards) Finally, the announcer calls for the last contestant, Al Truistic. With your computer broken, you decide to record the balloon s height and distance by hand. Al releases the balloon from the 10-yard line. The balloon reaches a height of 27 yards and lands at the 16-yard line. a. Obtain the Lesson Resource Page from your teacher. For each contestant, create a table and graph using the information provided for each toss. Determine which of these contestants should win the Longest Distance and Highest Throw contests. [ Longest Distance: Maggie, Highest Throw: Jen ] b. Find the x-intercepts of each parabola. What information do the x-intercepts tell you about each balloon toss? [ Jen: (0, 0) and (8, 0), Maggie: (3, 0) and (14, 0), Imp: (2, 0) and (12, 0), Al: (10, 0) and (16, 0); the x-intercepts tell where the balloon was launched and where it landed. ] c. Find the vertex of each parabola. What information does the vertex tell you about each balloon throw? [ Jen: (4, 32), Maggie: (8.5, 30.25), Imp: (7, 25), and Al: (13, 27); it tells where the balloon was at its maximum height. ] 638 Algebra Connections

37 8-52. Today you have explored the four different representations of quadratics: table, graph, equation, and a description of a physical situation involving motion. Draw the representations of the web as shown below in your Learning Log and label it Quadratic Web. a. Draw in arrows showing the connections that you currently know how to make between different representations. Be prepared to justify a connection for the class. [ We expect students to connect rule table, table graph, graph situation, and table situation. In problem 4-1 and in other problems, students found rules for quadratic tile Graph QUADRATIC WEB Table Situation patterns. Also, in Chapter 3, students connected table rule with the Silent Board Game, but they will continue to strengthen this connection. ] b. What connections are still missing? [ table rule, rule graph, and rule situation ] Rule or Equation SITUATION TO RULE Review how to write a rule from a situation by examining the tile pattern below. QUADRATIC WEB Table Graph Rule or Equation Figure 1 Figure 2 Figure 3 Situation a. Write a rule to represent the number of tiles in Figure x. [ One way to write the rule is y = (x + 1)(x + 2) + 2. ] b. Is the rule from part (a) quadratic? Explain how you know. [ Yes; students can multiply it out to find that y = x 2 + 3x + 4. ] c. If you have not done so already, add this pathway to your web from problem Graph y = x 2! 8x + 7 and label its vertex, x-intercepts, and y-intercepts. [ vertex: (4, 9), x-intercepts: (1, 0) and (7, 0), y-intercept: (0, 7) ] Chapter 8: Quadratics 639

38 3 + 0 = What is special about the number zero? Think about this as you answer the questions below. a. Find each sum: + 3 = [ 3 ]! = [ 7 ] = [ 6 ] 0 + (! 2) = [ 2 ] b. What is special about adding zero? Write a sentence that begins, When you add zero to a number, [ it does not change the value of the number. ] c. Julia is thinking of two numbers a and b. When she adds them together, she gets a sum of b. Does that tell you anything about either of Julia s numbers? [ It tells us that a = 0. ] d. Find each product: 3! 0 = [ 0 ] (!7) "0 = [ 0 ] 0!6 = [ 0 ] 0!("2) = [ 0 ] e. What is special about multiplying by zero? Write a sentence that begins, When you multiply a number by zero, [ the result is always 0. ] Based on the tables below, say as much as you can about the x- and y-intercepts of the corresponding graphs. [ a: x-intercepts (2, 0), ( 4, 0), and (3, 0), y-intercept: (0, 18); b: x-intercepts (3, 0) and (8, 0), y-intercept: (0, 3); c: x-intercept (1, 0) and y-intercept (0, 4) ] a. x y b. x y c. x y For the line described by the equation y = 2x + 6 : a. What is the x-intercept? [ ( 3, 0) ] b. What is the slope of any line perpendicular to the given line? [! 1 2 ] Solve the following systems of equations using any method. Check your solution if possible. [ a: no solution, b: (7, 2) ] a. 6x! 2y = 10 3x! y = 2 b. x! 3y = 1 y = 16! 2x 640 Algebra Connections

39 Lesson How are quadratic rules and graphs connected? Zero Product Property Lesson Objective: Students will learn that they can sketch the graph of a quadratic rule quickly, using its intercepts. Students will also learn how to find the x-intercepts of a parabola by factoring the corresponding quadratic equation and applying the Zero Product Property. Length of Activity: One day (approximately 50 minutes) Core Problems: Problem 8-59 through 6-64 Ways of Thinking: Materials: Making connections, justifying None Suggested Lesson Activity: Today students develop a method to sketch the graph of a parabola from its rule using its intercepts. Students discover that although they can easily determine the y-intercept from the equation, they do not yet know how to find the x-intercepts. The class studies the special properties of zero to discover the Zero Product Property, which, combined with factoring, allows students to find the roots of a quadratic equation. There are four main parts to this lesson. Part 1: After reading the lesson introduction, start teams on problem 8-59 as a warm-up or work this problem out as a class. During this problem, students will discover that with the x-intercepts and the y-intercept, they can sketch the graph of a parabola without a table. Be on the lookout for asymmetric parabolas sketched by students who mistakenly assume the vertex must lie on the y-intercept. Use this opportunity to emphasize that the vertex must lie on the line of symmetry, which must be halfway between the intercepts. Then have teams work on problem 8-60, which reminds them that they can quickly find the y-intercept from the rule. Students also create a quadratic equation ( 0 = 2x 2 + 5x! 12 ) by substituting 0 for y. As teams work on problem 8-60, circulate to check for understanding. Part (a) of problem 8-60 may confuse students, so be ready to direct their attention to a graph. Pointing at an x-intercept, ask, What do you know about y here? Once teams have realized that they do not yet have the algebra tools to solve 0 = 2x 2 + 5x! 12 for x, move on to Part 2. Part 2: Note: Problem 8-61 is included in the student text for team use. However, this problem may be best used as the foundation for a wholeclass discussion. One way to start this discussion is to write, WHAT IS SPECIAL ABOUT ZERO? on the board or overhead and challenge students to Chapter 8: Quadratics 641

40 come up with as many ideas as they can. Students who have completed problem 8-55 from the previous lesson s homework may offer ideas about what happens when you add zero to a number and what happens when you multiply a number by zero. Then ask two students to come to the front of the class. Give each student a secret number, making sure that one of the two numbers is zero. Ask the volunteers to share their numbers with each other and to tell the class what the product of their numbers is. Then ask the class as a whole what they can conclude about the secret numbers. This can be repeated with three or more students to ensure that students understand that the Zero Product Property applies even when there are more than two numbers multiplied. Tell students that the property they just discovered is called the Zero Product Property. Have a volunteer describe the Zero Product Property in words, beginning with something like, If the product of two or more numbers is zero, then Be sure students recognize that there needs to be a product to use the Zero Product Property. Part 3: Once students have recorded the Zero Product Property, return to the quadratic they found in part (c) of problem 8-60: 0 = 2x 2 + 5x! 12 Ask, Can we use the Zero Product Property to solve this? Why or why not? Then have teams factor the quadratic so that they have a product equal to zero: 0 = (x + 4)(2x! 3) Ask, If (x + 4) times (2x! 3) equals 0, then what does the Zero Product Property tell us? When a student volunteers that one of the expressions must equal zero, take the opportunity to show students how you want them to record this information. One method is shown below: x + 4 = 0 or 2x! 3 = 0 Ask teams to solve these equations and explain why it makes sense that there are two solutions. Then have teams proceed to parts (c) and (d) of problem Part 4: If time allows, problems 8-63 and 8-64 offer teams a chance to practice their new method of sketching a parabola without a table. Closure: (5 minutes) Have a team volunteer to post their solution to problem 8-63 on the board so that other students can check their work. Then add the arrow on the poster of the quadratic web (rule graph) and ask students what connections they still need to complete the web. Homework: Problems 8-65 through Algebra Connections

41 8.2.3 How are quadratic rules and graphs connected? Zero Product Property Student pages for this lesson are You already know a lot about quadratics and parabolas, and you have made several connections between their different representations on the quadratic web. Today you are going to develop a method to sketch a parabola from its equation without a table. Graph QUADRATIC WEB Table Situation Rule or Equation WHAT DO YOU NEED TO SKETCH A PARABOLA? How many points do you need in order to sketch a parabola? 1? 10? 50? Think about this as you answer the questions below. (Note: A sketch does not need to be exact. The parabola merely needs to be reasonably placed with important points clearly labeled.) a. Can you sketch a parabola if you only know where its y-intercept is? For example, if the y-intercept of a parabola is at (0, 15), can you sketch its graph? Why or why not? [ No; the y-intercept is not enough information. ] b. What about the two x-intercepts of the parabola? If you only know where the x- intercepts are, can you draw the parabola? For example, if the x-intercepts are at ( 3, 0) and (5, 0), can you predict the path of the parabola? [ No; the parabola could vary in width and direction. ] c. Can you sketch a parabola with only its x-intercepts and y-intercept? To test this idea, sketch the graph of a parabola y = x 2! 2x!15 with x-intercepts ( 3, 0) and (5, 0) and y-intercept (0, 15). [ Yes; solution shown at right. ] Chapter 8: Quadratics 643

42 8-60. In problem 8-59, you learned that if you can find the intercepts of a parabola from a rule, then you can sketch its graph without a table. a. What is true about the value of y for all x-intercepts? What is true about the value of x for all y-intercepts? Review your knowledge of intercepts and describe it here. [ y = 0 for all x-intercepts and x = 0 for all y-intercepts. ] b. If x = 0 at the y-intercept, find the y-intercept of the graph of y = 2x 2 + 5x! 12. [ (0, 12) ] c. Since the x-intercept occurs when y = 0, write the equation that you would need to solve to find the x-intercepts of the graph of y = 2x 2 + 5x! 12. [ 0 = 2x 2 + 5x! 12 ] d. The solutions of the equation 2x 2 + 5x! 12 = 0 are called its roots and are the zeros of 2x 2 + 5x! 12. At this point, can you solve 2x 2 + 5x! 12 = 0 for x? Explain why or why not. [ Not yet, because it has an x 2 term. ] ZERO PRODUCT PROPERTY The equation you wrote in part (c) of problem 8-60 is called a quadratic equation. To solve it, you need to examine what you know about zero. Study the special properties of zero below. Nathan, Nancy, and Gaston are playing a game where Nathan and Nancy each think of a number and then give Gaston a clue about their numbers. Using the clue, Gaston must tell them everything that he knows about their numbers. a. Nathan and Nancy s first clue for Gaston is that when you multiply their numbers together, the result is zero. What conclusion can Gaston make? [ At least one of the two numbers must be zero. ] b. Disappointed that Gaston came so close to figuring out their numbers, Nathan and Nancy invite Nadia over to make things harder. Nathan, Nancy, and Nadia all think of secret numbers. This time Gaston is told that when their three secret numbers are multiplied together, the answer is zero. What can Gaston conclude this time? [ At least one of the three numbers must be zero. ] c. Does it matter how many numbers are multiplied? If the product is zero, what do you know about one of the numbers? This property is called the Zero Product Property. With the class, write a description of this property in your Learning Log. Title this entry Zero Product Property and include today s date. [ Typical response: If the product of two or more numbers is zero, then you know that one of the numbers must be zero. ] 644 Algebra Connections

43 8-62. How can you use the Zero Product Property to help you solve the quadratic equation 0 = 2x 2 + 5x! 12 from part (c) of problem 8-60? a. Examine the quadratic equation. Is there a product that equals zero? If not, how can you rewrite the quadratic expression as a product? [ 0 = (2x! 3)(x + 4) ] b. Now that the equation is written as a product of factors equaling zero, you can use the Zero Product Property to solve it. Since you know that one of the factors must be zero, you can set up two smaller equations to help you solve for x. Use one factor at a time and determine what x-value makes it equal to zero. [ 2x! 3 = 0 or x + 4 = 0, so x = 3 2 or x =! 4.] c. What do these solutions represent? What do they tell you? [ The x-intercepts are at ( 3 2,!0) and ( 4, 0). ] d. You now know the roots of the equation 0 = 2x 2 + 5x! 12 (also called the zeros of 2x 2 + 5x! 12 ). Use the roots to find the x-intercepts of the graph of the parabola y = 2x 2 + 5x! 12. Then sketch a graph of the parabola. [ The solution graph is shown at right. ] Use a similar process to sketch the parabola y = x 2 + x! 6 by using its intercepts. [ This parabola should have x-intercepts ( 3, 0) and (2, 0) and y-intercept (0, 6). ] Sketch the parabola y = 2x 2 + 6x + 4 by using its intercepts. [ x-intercepts: ( 1, 0) and ( 2, 0), y-intercept: (0, 4), solution graph shown at right ] MATH NOTES ETHODS AND MEANINGS Zero Product Property When the product of two or more numbers is zero, one of those numbers must be zero. This is known as the Zero Product Property. If the two numbers are a and b, this property can be written as follows: If a and b are two numbers where a!!b = 0, then a = 0 or b = 0. For example, if (2x! 3)(x + 5) = 0, then 2x! 3 = 0 or x + 5 = 0. Solving yields the solutions x = 3 or x =!5. This property helps you solve quadratic 2 equations when the equation can be written as a product of factors. Chapter 8: Quadratics 645

44 8-65. Compare the two equations below. ( x + 2)( x! 1) = 0 and ( x + 2) + ( x! 1) = 0 a. How are the equations different? [ One is a product and the other is a sum. ] b. Solve both equations. [ first: x =!2 or x = 1 ; second: x =! 1 2 ] For each equation below, solve for x. [ a: x = 2 or x =!8, b: x = 3 or x = 1, c: x =!10 or x = 2.5, d: x = 7 ] a. (x! 2)(x +8) = 0 b. (3x! 9)(x!1) = 0 c. (x +10)(2x! 5) = 0 d. (x! 7) 2 = Examine the system of equations below. 5x! 2y = 4 x = 0 a. Before solving this system, Danielle noticed that the point of intersection is also the y-intercept of 5x! 2y = 4. Explain how she knows this. [ The line x = 0 is the y-axis, so this system is actually finding where the line 5x! 2y = 4 crosses the y-axis. ] b. Find the point of intersection of the two rules above. [ (0, 2) ] The x-intercepts of the graph of y = 2x 2! 16x + 30 are (3, 0) and (5, 0). a. What is the x-coordinate of the vertex? How do you know? [ 4; Since the vertex lies on the line of symmetry, it must lie halfway between the x-intercepts. ] b. Use your answer to part (a) above to find the y-coordinate of the vertex. Then write the vertex as a point (x, y). [ (4, 2) ] Factor each quadratic below completely. [ a: 2(x! 2)(x +1), b: 4(x! 3) 2 ] a. 2x 2! 2x! 4 b. 4x 2! 24x Algebra Connections

45 8-70. The symbol represents less than or equal to, while the < symbol represents less than. a. Similarly, translate and >. [ The symbol represents greater than or equal to and the symbol > represents greater than. ] b. How can you write an expression that states that 5 is greater than 3? [ 5 > 3 ] c. Write another expression that states that x is less than or equal to 9. [ x! 9 ] d. Translate the expression!2 < 7 into words. [ 2 is less than 7. ] Lesson What new connection can I make? Solving Quadratic Equations by Factoring Lesson Objective: Students will review using the Zero Product Property to find the x-intercepts of a parabola and will then practice solving quadratic equations by factoring. Students will also learn how to write a quadratic rule from a table. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 7-73 through 7-75 Ways of Thinking: Materials: Suggested Lesson Activity: Reversing thinking None Today s lesson involves two essential parts: (1) Students will practice solving quadratic equations using the Zero Product Property, and (2) Students will discover a method for finding a quadratic equation from a table using the x-intercepts. Part 1: Start class with an overview of the lesson. Give teams 10 minutes to review what they learned in Lesson as they find the y-intercept and x-intercepts of y = x 2 + 3x + 2 in problem Then move teams on to problem 8-72, which addresses the fact that part of the process they used in problem 8-71 involved solving a quadratic equation. Problem 8-72 asks students to concentrate on that task by solving six more quadratic equations. Have teams post solutions so that the other teams can check their work and find errors. Note: Parts (c) and (e) are already in factored form. Do not be surprised if students multiply the factors just to factor the result. Let this come up during the debriefing of problem Chapter 8: Quadratics 647

46 Part 2: Problem 8-73 changes the direction of the lesson. After what was essentially review, this part of the lesson dives into new territory: finding a quadratic equation from a table. This will require students to shift their focus from factoring and solving to building an equation. Ask, We know how to make a table for a rule. How can we make a rule from a table? If students are not sure how to start, ask, What pieces of information in the table can help us? and How can that help us? If students are still stuck, ask, What are the roots of this equation? How can they help us? Students should focus on building an equation from its roots. For example, in part (a) of problem 8-73, once students notice that the x-intercepts are ( 3, 0) and (2, 0), then the quadratic equation a(x + 3)(x! 2) = 0 is needed to get those intercepts. Since there could be a GCF multiplied on the quadratic, students can multiply (x + 3) and (x! 2) to verify that the y-intercept matches the constant in the equation. Or, students could substitute the coordinates of another point from the table into y = (x + 3)(x! 2) to verify that it makes the equation true. Note: This course does not expect students to find a comprehensive method for finding the rule for all quadratics from tables. This would require an analysis of finite differences or would require students to solve systems of three equations with three unknowns. Those methods will be learned in a later course. By having students build equations from the x-intercepts, their conceptual understanding of the Zero Product Property will be strengthened. If there is time, problem 8-74 revisits Imp s water-balloon toss from problem This problem offers a review of using symmetry to find roots and the fact that parabolas pointing downward must have a negative x 2 -term. This is a challenging problem, and it is highly recommended that you solve the problem before class to determine if it is appropriate for your students. Closure: (5 minutes) Team Roles: Finish this lesson by asking students to return to their Learning Log entry from Lesson and to add the connection table rule to their web. Add this connection to the class poster web as well. Problem 8-75 asks students to write a short explanation for how to find a quadratic equation from its table. Problem 8-72 asks teams to post solutions, so Recorder/Reporters need to be prepared to bring solutions forward for class discussion. There are multiple approaches to problem 8-73, so the Facilitators should ask, Did anyone solve it differently? Did everyone get the same equation? Homework: Problems 8-76 through Algebra Connections

47 8.2.4 What new connection can I make? Solving Quadratic Equations by Factoring Student pages for this lesson are In Lesson 8.2.3, you developed a method for finding the x-intercepts of a parabola given by y = ax 2 + bx + c by finding the roots of the corresponding quadratic equation, ax 2 + bx + c = 0, or the zeros of ax 2 + bx + c. Today you will learn how to use that skill to solve a wide variety of quadratic equations. You will also revisit the quadratic web, make a connection between the table and rule of a parabola, and then apply this connection to the water-balloon competition you analyzed in problem Graph QUADRATIC WEB Table Situation Rule or Equation Review what you learned in Lesson by sketching the graph of y = x + 3x + 2 without a table. Specifically, find the x-intercepts and the y-intercept of the parabola and sketch its graph. [ The parabola should have y-intercept (0, 2) and x-intercepts ( 1, 0) and ( 2, 0). ] Part of finding the x-intercepts of a parabola involves creating a quadratic equation of the form ax 2 + bx + c = 0 and finding its roots (which are also the zeros of the expression ax 2 + bx + c ). Practice using the Zero Product Property to solve the quadratic equations below. [ a: x = 2 or x = 4, b: x = 1 or x = 4 3, c: x = 5 or x = 3 2, d: x = 0 or x = 6, e: x = 5 or x = 1.5, f: x = 2 or x = 6 ] a. x 2 + 6x + 8 = 0 b. 0 = 3x 2! 7x + 4 c. (x + 5)(!2x + 3) = 0 d. x 2 + 6x = 0 e. 0 = 3(x! 5)(2x + 3) f. x 2 + 4x! 9 = 3 Chapter 8: Quadratics 649

48 8-73. TABLE TO RULE You know how to make a table for a quadratic rule, but how can you write an equation when given the table? Examine this new connection that requires you to reverse your understanding of the Zero Product Property as you find a rule for each table below. What clues in the tables helped you find the rule? [ a: y = (x + 3)(x! 2) = x 2 + x! 6, b: y = (x + 5)(x! 1) = x 2 + 4x! 5 ; Students can build the equation using the x-intercepts and can then multiply to verify the y-intercept or can substitute a point from the table to confirm their rule. ] a. x y b. x y WATER-BALLOON CONTEST REVISITED Remember Imp s water-balloon toss? Since the water balloon was thrown on the computer, you were given only a table of data, shown again below. Find a rule that represents the height of Imp s balloon as it traveled through the air. [ By symmetry, (12, 0) is also an x-intercept. Thus, the quadratic must be of the form y = a(x! 2)(x! 12). Since the parabola points down, a must be negative. Testing a point shows that y =!(x! 2)(x! 12) =!x x! 24 is correct. ] x (yards) y (yards) Find the quadratic web in your Learning Log entry from Lesson In this entry, add a short explanation for how to find a quadratic equation from its table. Then add an arrow to your web for the connection you made today. 650 Algebra Connections

49 8-76. Jamie was given the problem, Find the result when the factors of 65x x! 133 are multiplied together. Before she could answer, her sister, Lauren, said, I know the answer without factoring or multiplying! What was Lauren s answer and how did she know? [ The result must be the original expression because multiplying and factoring are opposite processes; 65x x! 133. ] Solve the equations below for x. Check your solutions. [ a: x = 3 or x =! 2 3, b: x = 2 or x = 5, c: x =!3 or x = 2, d: x = 1 2 or x =! 1 2 ] a. (6x!18)(3x + 2) = 0 b. x 2! 7x + 10 = 0 c. 2x 2 + 2x! 12 = 0 d. 4x 2! 1 = Sketch each parabola below with the given information. [ See graphs below. ] a. A parabola with x-intercepts (2, 0) and (7, 0) and y-intercept (0, 8). b. A parabola with exactly x-intercept at ( 1, 0) and y-intercept (0, 3). c. The parabola represented by the equation y = (x + 5)(x!1) Review the meanings of the inequality symbols in the box at right. Then decide if the statements below are true or false. [ a: true, b: false, c: true, d: true, e: false, f: false ] a. 5 < 7 b.!2 " 9 c. 0! 0 d.!5 >!10 e. 16! "16 f. 1 > 1 < less than less than or equal to > greater than greater than or equal to Calculate the expressions below with a scientific calculator. [ a: 1, b: 1.6, c: 3 ] a.! b ! 3 c ! 3+1 " Find the equation of the line through the points (6, 8) and (0, 0). [ y =! 4 3 x ] a. What is the slope of the line? [! 4 3 ] b. Is the point (3, 4) on the line? How can you tell? [ Yes; it makes the equation true and lies on the graph of the line. ] Chapter 8: Quadratics 651

50 Lesson What s the connection? Completing the Quadratic Web Lesson Objective: Students will practice moving from a quadratic rule to its graph (and vice versa) for parabolas with one or two x-intercepts. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems 8-82 (one part) and 8-83 Ways of Thinking: Materials: Technology Notes: Reversing thinking, justifying, making connections Graphing calculators (or comparable computer technology), one per student Overhead graphing calculator or computer with projection display Problem 8-83 assumes that students have access to graphing technology, such as a graphing calculator or a computer with graphing software. The graphing technology allows students to use a process of Guess and Check to find the equation of the parabola that best fits the information given in each complaint. If the technology is not available, students will need to use graphing shortcuts to check their answers to problem See the Technology Notes section in the front of this Teacher Guide for directions on using a TI-83. For all other calculator models, consult the appropriate calculator user s guide or look online at Before class begins, be sure to test your equipment (both for teachers and students) to make sure that there are no surprises, such as dead batteries or software that does not run. If you are using TI-83 calculators, you may want to preset your calculators to be in the mode shown at right. Suggested Lesson Activity: Note: Do not pass out the graphing technology until problem Start class today by asking a volunteer to read the lesson introduction. Point out that the arrow from graph to rule is still missing from the quadratic web. Today s goal is to complete the web. Problem 8-82 asks teams to match each graph with its rule. Students need to justify their matches using all of the tools they have for working with quadratics. If time is short, each parabola could be assigned to a single team to present to the class rather than each team doing all six parabolas. Then distribute the graphing technology. Read problem 8-83 together as a class and make sure that students understand the goals of the task. Clearly state what type of product you are expecting. For example, are you expecting students to write a formal letter back to each customer? 652 Algebra Connections

51 Are you expecting teams to design a poster with the information for the client? Then have the teams begin addressing the concern stated in the first letter. After 10 minutes or so, ask one or two teams to share their findings, thus helping any teams that are struggling. Ask teams questions that help them generalize the process. For example, ask, If you have the graph of the parabola, how can you build its equation? then Is that always true if a parabola has only one x-intercept? or How can you verify that your equation is correct? Have teams continue to work on problem 8-83 as you circulate. Ask questions that probe students understanding. Closure: (15 minutes) Add the arrow connecting graph rule to the class quadratics web. If there is time, problem 8-84 offers students a chance to bring closure to problem 8-51, Water-Balloon Contest. This problem asks students to find the rules for the two remaining launches: one from the graph of Jen s launch, and one from the description of Al s launch. Expect students to use Guess and Check to find the stretch factor of the parabola. Homework: Problems 8-85 through 8-90 Note: Problem 8-87 shows students the need for the Quadratic Formula that will be introduced in Lesson The quadratic y = 2x 2 + 2x! 1 is not factorable, yet the parabola clearly has x-intercepts. Emphasize the need for students to spend time tonight on this problem What s the connection? Completing the Quadratic Web Student pages for this lesson are In just three lessons you have almost completed the quadratic web. Revisit the web posted in your classroom. What connections, if any, still need to be made? QUADRATIC WEB Table Today you will focus on how to get a quadratic rule from a graph and a situation. As you work, ask yourself the following questions: Graph Rule or Equation Which representation am I given? Which representation am I looking for? Situation How can I reverse this process? Is there another way? Chapter 8: Quadratics 653

52 8-82. Several parabolas and quadratic rules are shown below. Match each graph with its rule. Justify your choices and share any shortcuts you find with your teammates. (Note: Not every rule will be matched with a parabola.) [ (1) b, (2) e, (3) a, (4) g, (5) d, (6) i ] Graphs: (1) y (2) y (3) y ( 7, 0) (4, 0) x ( 6, 0) ( 4, 0) x ( 3, 0) x (4) (5) (6) y y y (0, 0) x (4, 0) ( 2, 0) (2, 0) x (! 1 2, 0) ( 3 2, 0) x Rules: a. y = (x + 3) 2 b. y = x 2 + 3x! 28 c. y = x 2! 11x + 28 d. y = x 2! 4 e. y = x x + 24 f. y = 2x x + 5 g. y = x 2! 4x h. y = (x! 3) 2 i. y = 4x 2! 4x! Algebra Connections

53 8-83. QUALITY CONTROL, Part One Congratulations! With your promotion, you are now the Quality Assurance Representative of the Function Factory. Your job is to make sure your clients are happy. Whenever a client writes to the company, you must reply with clear directions that will solve his or her problem. Your boss has provided graphing technology and a team of fellow employees to help you fulfill your job description. Your Task: 1. Carefully read the complaints below. Study each situation with your grapher. Work with your team to resolve each situation. 2. Write each customer a friendly response that offers a solution to his or her problem. Remember that the customers are not parabola experts! Do not assume that they know anything about parabolas. [ Solutions: Letter A: The client should order the parabola y = (x! 1)(x + 6); Letter B: Students should recommend the parabola y = (x! 5) 2 ; Letter C: Students should recommend the parabola y =!(x + 3)(x! 2). ] Dear Ms. Quadratic, A I followed all of the directions given in your brochure on how to order a parabola. I tried to order a parabola that passed through the points (1, 0) and ( 6, 0), only to have you send me the wrong one! Please tell me how to order the correct parabola. Your immediate reply is appreciated. Perturbed in Pennsylvania Dear Ms. Quadratic, I am a very dissatisfied customer. I want a parabola that hits the x-axis only once at (5, 0), yet I see NO mention of this type of parabola in your pamphlet. Your company mission statement assures me that my needs will be met no matter what. How should I order my special parabola? Sincerely, Troubled in Texas B Dear Ms. Quadratic, Please help! I have searched through your entire brochure and did not see a parabola that would fit my needs. All I want is a parabola that looks like this: Every time I order an equation to give me this parabola you always send me a different one! I refuse to pay for any parabola but the one shown above. Please tell me how I should find the equation of this parabola or I will take my business elsewhere! Thank you, Agitated in Alaska C Chapter 8: Quadratics 655

54 8-84. EXTRA! EXTRA! A journalist from the school newspaper wants to publish the results from the waterballoon contest. She wants a rule for each toss so that she can program her computer to create a graph for her article. You already have rules for the tosses made by Maggie and Imp from problems 8-51 and Jen s data a. Examine the graph at right that represents the height of Jen s toss. Find the rule for this parabola. [ y =!2x(x! 8) =!2x x ] b. Al released his balloon from the 10-yard line, and it landed at the 16-yard line. If the ball reached a height of 27 yards, what equation represents the path of his toss? [ y =!3(x! 10)(x!16) ] QUALITY CONTROL, Part Two Lots O Dough, a wealthy customer, would like to order a variety of parabolas. However, he is feeling pressed for time and said that he will pay you lots of extra money if you complete his order for him. Of course you agreed! He sent you sketches of each parabola that he would like to receive. Determine a possible equation for each parabola so that you can pass this information on to the Manufacturing Department. [ a: y = x 2 + 2x! 8, b: y = x 2! 6x + 9, c: y = x 2! 7x, d: y =!x 2! 4x + 5 ] a. y b. y ( 4,0) (2, 0) x (3, 0) x c. y d. y (0, 0) x (7, 0) ( 5, 0) (1, 0) x 656 Algebra Connections

55 8-86. Find the slope and y-intercept of the graph of 6y! 3x = 24. [ m = 1, (0, 4) ] Examine the graph of y = 2x 2 + 2x!1 at right. a. Estimate the zeros of 2x 2 + 2x! 1 from the graph. [ 1.4 and 0.3 ] b. What happens if you try to use the Zero Product Property to find the roots of 2x 2 + 2x! 1 = 0? [ The quadratic is not factorable. ] Solve the equations below for x. Check your solutions. [ a: x = 4 or x = 10, b: x = 8 or x = 1.5 ] a. x 2 + 6x! 40 = 0 b. 2x x! 24 = Calculate the expressions below. Then compare your answers from (a) and (b) to those in problem What do you notice? [ a: 4, b: 10, c: 8, d: 1.5; They are the same. ] a.! !(4)(1)(!40) 2 " 1 b.!6! 6 2!(4)(1)(!40) 2 " 1 c.!13! 13 2!(4)(2)(!24) 2 " 2 d.! !(4)(2)(!24) 2 " Use any method to solve the systems of equations below. [ a: (1, 1), b: ( 2, 1 2 ) ] a. 2x! 3y = 5 4x + y = 3 b. m =!3 + 2n 4m + 6n =!5 Chapter 8: Quadratics 657

56 Lesson What if it s not factorable? Introduction to the Quadratic Formula Lesson Objective: Students will learn how to use the Quadratic Formula to solve quadratic equations that are not factorable. Length of Activity: One day (approximately 50 minutes) Core Problems: Ways of Thinking: Materials: Suggested Lesson Activity: Problems 8-91 through 8-94 (parts (a) and (b)) Making connections Overhead graphing calculator/computer OR Lesson Resource Page on a transparency To start, have teams work on problem After most teams have discovered that the quadratic is not factorable, discuss parts (a) and (b) together as a whole class. Expect students to state that the parabola does not cross the x-axis in part (b). Be ready with an overhead graphing calculator to display the graph for part (c). This graph will dispel the notion that a non-factorable quadratic has no roots. Use the tracing feature (or root-finding feature) of the calculator to estimate the x-intercepts of y = x 2! 3x! 7. Note: If you do not have a graphing calculator or computer with comparable software, then use the large graph of y = x 2! 3x! 7 provided on a resource page. Copy this resource page onto a transparency and have it ready to display in class. Problem 8-92 introduces students to the Quadratic Formula. This problem may best be worked out as a whole class so you can demonstrate how you would like students to leave their solutions (in either decimal or exact form). Note: Students may, at this time, be unfamiliar with the radical symbol and the square-root operation. If so, take a few minutes to discuss this now, referring back to problem Then move teams on to problem 8-93, which directs students attention to the derivation of the Quadratic Formula in the Math Notes box at the end of the lesson. Decide before class how much time you want students to spend on this. Later, in Chapter 12, the Quadratic Formula will be derived by completing the square. At this point, it is most important for students to recognize that the formula does not just come out of the blue but instead is a solution of a general quadratic equation. If you do spend time on the derivation given in the Math Notes box, you will need to help 658 Algebra Connections

57 students understand why the ± symbol is necessary once you take the square root of both sides. Problem 8-94 asks students to use the Quadratic Formula to solve several quadratics. As students work on these problems, circulate and look for common errors. Some students will have difficulty placing a, b, and c into the formula, while others will be challenged when evaluating the expression. Once teams are close to finishing problem 8-94, pass out graphing calculators (or comparable graphing technology) so that students can check their solutions graphically. Since part (d) of problem 8-94 has no solution, the parabola y = x 2! 5x + 9 does not intersect the x-axis. Checking the graph will help students validate this conclusion. Closure: (8 minutes) Debrief part (d) of problem 8-94 with the class. Then ask students to summarize how to use the Quadratic Formula in their Learning Logs in problem Homework: Problems 8-96 through What if it s not factorable? Introduction to the Quadratic Formula Student pages for this lesson are In Section 8.2 you developed a method to find the x-intercepts of a parabola by factoring and using the Zero Product Property. Today you will learn a new method to solve quadratic equations. Yo! Use the Zero Product Property to find the roots of x 2! 3x! 7 = 0. a. What happened? [ The quadratic is not factorable. ] b. What does this result tell you about the roots? [ Students may assume that the parabola will have no x-intercepts because the quadratic is not factorable. That is okay at this point. Part (c) will correct this belief. Actually, this result merely tells us that the x-intercepts are not rational. ] c. Your teacher will display the graph of y = x 2! 3x! 7 for the class. Did the graph confirm your answer to part (b)? Estimate the roots using the graph. [ The intercepts are 1.5 and 4.5. ] Chapter 8: Quadratics 659

58 8-92. QUADRATIC FORMULA Since a parabola can have x-intercepts even when its corresponding quadratic equation is not factorable, another way to find the roots of a quadratic equation is needed. a. One way to calculate the roots of a quadratic equation is by using the Quadratic Formula, shown below. This formula uses values a, b, and c from a quadratic equation written in standard form (explained in the next paragraph). x =!b ± b2! 4ac 2a When the quadratic equation is written in standard form (i.e., it looks like ax 2 + bx + c = 0 ), then a is the number of x 2 -terms, b is the number of x-terms, and c is the constant. If x 2! 3x! 7 = 0, then what are a, b, and c? [ a = 1, b =!3, c =!7 ] b. The Quadratic Formula calculates two possible answers by using the ± symbol. This symbol (read as plus or minus ) is shorthand notation that tells you to calculate the formula twice: once with addition and once with subtraction in the numerator. Therefore, every Quadratic Formula problem is really two different problems unless the value of b 2! 4ac is 0. ± x =!b + b2! 4ac 2a or x =!b! b2! 4ac 2a Carefully substitute a, b, and c from x 2! 3x! 7 = 0 into the Quadratic Formula. Evaluate each expression (once using addition and once using subtraction) to solve for x. Do these solutions match those from part (c) of problem 8-91? 3± 37 [ and 1.5; yes ] The Quadratic Formula is only one of the tools you can use to solve quadratic equations. a. What are the other methods that you can use? [ graphing and factoring with the Zero Product Property ] b. You may be thinking, Where did this formula come from? Why does it work? You can find the formula by starting with a generic quadratic ax 2 + bx + c = 0 and using your algebra skills to solve for x. See the Math Notes box for this lesson to learn about one way this formula can be derived. Later, in Chapter 12, you will learn another formal method to derive the Quadratic Formula. 660 Algebra Connections

59 8-94. Use the Quadratic Formula to solve the equations below for x, if possible. Check your solutions. [ a: x = 2 or x =! 1 3, b: x = 7 or x = 2.5, c: x = 0.5 or x = 0.75, d: no solution ] a. 3x 2 + 7x + 2 = 0 b. 2x 2! 9x! 35 = 0 c. 8x x + 3 = 0 d. x 2! 5x + 9 = In your Learning Log, describe how to use the Quadratic Formula. Be sure to include an example. Title this entry Quadratic Formula and include today s date. MATH NOTES OOKING DEEPER Deriving the Quadratic Formula Why is x =!b ± b2!4ac a solution of ax 2a 2 + bx + c = 0? One way to derive this formula is shown below. 1. Begin with the quadratic equation in standard form. 2. Multiply each side by 4a. 3. Add b 2! 4ac to each side in order to get a factorable quadratic on the left. 4. The left side can be factored as (2ax + b) 2, which is demonstrated in the generic rectangle shown at right. 5. Take the square root of each side. Since both the positive and negative values of a number can be squared to give the same result (for example, 4 2 and (! 4) 2 both equal 16), then there are two possible square root values: b 2! 4ac and b 2! 4ac. 6. Now continue to solve for x by subtracting b from both sides and dividing by 2a. Notice that a cannot equal zero or else you will get an error! However, if a = 0, then this equation would not be quadratic and you would not use this formula. 2ax ax 2 + bx + c = 0 4a(ax 2 + bx + c) = 0 4a 2 x 2 + 4abx + 4ac = 0 4a 2 x 2 + 4abx + b 2 = b 2! 4ac b 2abx 4a 2 x 2 2ax b 2 2abx b (2ax + b) 2 = b 2! 4ac 2ax + b = ± b 2! 4ac 2ax =!b ± b 2! 4ac x =!b± b2!4ac 2a 7. Thus, x =!b ± b2!4ac 2a are solutions of the equation ax 2 + bx + c = 0. Chapter 8: Quadratics 661

60 8-96. Solve the following quadratic equations by factoring and using the Zero Product Property. Be sure to check your solutions. [ a: x = 6 or x = 7, b: x = 2 3 or x = 4, c: x = 0 or x = 5, d: x = 3 or x = 5 ] a. x 2!13x + 42 = 0 b. 0 = 3x x! 8 c. 2x 2!10x = 0 d. 4x 2 +8x! 60 = Use the Quadratic Formula to solve x 2!13x + 42 = 0. Did your solution match the solution from part (a) of problem 8-96? [ x = 6 or 7; yes ] Does a quadratic equation always have two solutions? That is, does a parabola always intersect the x-axis twice? [ no ] a. If possible, draw an example of a parabola that only intersects the x-axis once. [ The parabola should be tangent to the x-axis. ] b. What does it mean if the quadratic equation has no solution? Draw a possible parabola that would cause this to happen. [ Answers vary, but the parabola should not cross the x-axis. ] Find the equation of the line through the point ( 2, 8) with slope 1 2. [ y = 1 2 x + 9 ] For each of the following equations, indicate whether its graph would be a line or a parabola. [ line: (a) and (c); parabola: (b) and (d) ] a. 5x + 2y = 7 b. y = 3x 2 c. y = 3 d. 4x 2 + 3x = 7 + y Multiple Choice: Which equations below are equivalent to: [ A and D ] 1 (6x! 14) + 5x = 2! 3x + 8? 2 a. 3x! 7 + 5x =10! 3x b. 3x!14 + 5x = 2! 3x + 8 c. 8x!14 =10! 3x d. 6x!14 +10x = 4! 6x Algebra Connections

61 Review the descriptions for the inequality symbols <,, >, and in problem Then decide if the statements below are true or false. [ a: false, b: true, c: true, d: true, e: true, f: false, g: true, h: false ] a. 11 <!13 b. 5! 2 " 10 c. 13 >!3(2! 6) d. 4! 4 e. 9! "9 f.!2 >!2 g.!16 <!15 h. 0 > 6 Lesson What if the equation is not in standard form? More Solving Quadratic Equations Lesson Objective: Students will learn how to solve quadratic equations that have no solution, have only one solution, or are not in standard form. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems and Ways of Thinking: Materials: Suggested Lesson Activity: Closure: (5 minutes) Applying and extending None To start this lesson, we recommend that you ask students to review what they learned in Lesson Then have students solve problem 8-103, which requires them to use both the Quadratic Formula and the Zero Product Property to solve a quadratic equation. Problem requires teams to solve a quadratic equation that is not in standard form or has one of the terms missing. Note that part (c) intentionally has large coefficients and part (d) has only one solution. While debriefing the solutions to problem 8-104, it may be a good opportunity to point out that the Quadratic Formula helped solve part (c) in a situation where factoring would have been very challenging. Also point out for part (d) that since the b 2! 4ac portion of the formula is equal to 0, then there is only one solution. Finally, ask teams to work on problem 8-105, which requires the use of the Quadratic Formula. In this problem, students apply their skills with quadratic equations to approximate real data. Have students review the information in the Math Notes box for this lesson. This box contains a summary of solving quadratic equations using both the Zero Product Property and the Quadratic Formula. Homework: Problems through Chapter 8: Quadratics 663

62 8.3.2 What if the equation is not in standard form? More Solving Quadratic Equations Yo! Student pages for this lesson are Today you will apply and extend what you know about solving quadratic equations For the quadratic equation 6x x! 10 = 0 : a. Solve it using the Zero Product Property. [ (3x! 2)(2x + 5) = 0, x = 2 3 or! 5 2 ] b. Solve it using the Quadratic Formula. [ a = 6, b = 11, c =!10, x = 2 3 or! 5 2 ] c. Did the solutions from parts (a) and (b) match? If not, why not? [ Yes, although some students may get a decimal answer for one and not the other, and not initially recognize them as equal. ] As the Math Notes box from Lesson demonstrated, the Quadratic Formula can solve any quadratic equation ax 2 + bx + c = 0 if a! 0. But what if the equation is not in standard form? What if terms are missing? Consider these questions as you solve the quadratic equations below. Share your ideas with your teammates and be prepared to demonstrate your process for the class. [ a: x = 5.5 or x =!5.5, b: x = 2 or x =! 1 2, c: x =!3 or x = 14, d: x = 5 6 ] a. 4x 2!121 = 0 b. 2x 2! 2! 3x = 0 c. 15x 2!165x = 630 d. 36x = 60x THE SAINT LOUIS GATEWAY ARCH y The Saint Louis Gateway Arch (pictured at right) has a shape much like a parabola. Suppose the Gateway Arch can be approximated by y = 630! x 2, where both x and y represent distances in feet and the origin is the point on the ground directly below the arch s apex (its highest point). x a. Find the x-intercepts of the Gateway Arch. What does this information tell you? Use a calculator to evaluate your answers. [ 315 and 315 feet; The base of the arch is 315 feet from the center in both directions. ] b. How wide is the arch at its base? [ 630 feet ] c. How tall is the arch? How did you find your solution? [ 630 feet; y-intercept ] d. Draw a quick sketch of the arch on graph paper, labeling the axes with all of the values you know. 664 Algebra Connections

63 ETHODS AND MEANINGS MATH NOTES Solving a Quadratic Equation So far in this course, you have learned two algebraic methods to solve a quadratic equation of the form ax 2 + bx + c = 0. One of these methods, the Zero Product Property, requires the equation to be a product of factors that equal zero. In this case, the quadratic equation must be factored, as shown in Example 1 below. Another strategy uses the Quadratic Formula, as demonstrated in Example 2 below. Notice that each strategy results in the same answer. Example 1: Solve 3x 2 + x! 14 = 0 for x using the Zero Product Property. Solution: First, factor the quadratic so it is written as a product: (3x + 7)(x! 2) = 0. The Zero Product Property states that if the product of two terms is 0, then at least one of the factors must be 0. Thus, 3x + 7 = 0 or x! 2 = 0. Solving these equations for x reveals that x =! 7 3 or that x = 2. Example 2: Solve 3x 2 + x! 14 = 0 for x using the Quadratic Formula. Solution: First, identify a, b, and c. a equals the number of x 2 -terms, b equals the number of x terms, and c equals the constant. For 3x 2 + x! 14 = 0, a = 3, b = 1, and c =!14. Substitute the values of a, b, and c into the Quadratic Formula and evaluate the expression twice: once with addition and once with subtraction. Examine this method below: x =!1+ 12!4(3)(!14) =! =! = 12 6 = 2 2 " 3 or x =!1! 12!4(3)(!14) =!1! =!1!13 6 =!14 6 =! " 3 Chapter 8: Quadratics 665

64 Solve the following quadratic equations by factoring and using the Zero Product Property. Then check your solutions. [ a: x = 5, b: x = 1 3 or x = 6, c: x = 1 or x = 5 3, d: x = ± 3 4 ] a. x 2!10x + 25 = 0 b. 0 = 3x 2 +17x! 6 c. 3x 2! 2x = 5 d. 16x 2! 9 = Use the Quadratic Formula to solve part (b) of problem above. Did your solution match the solution you got by factoring and using the Zero Product Property (in part (b) of problem 8-106)? [ x = 1 3 or x = 6; yes ] Find the equation of each parabola below based on the given information. [ a: y = (x + 3)(x! 1) = x 2 + 2x! 3, b: y = (x! 2)(x + 2) = x 2! 4 ] a. b. x y Solve the following problem using any method. Write your solution as a sentence. [ If x = the width, then x(2x + 5) = 403 ; the width is 13 cm. ] The length of a rectangle is 5 cm longer than twice the length of the width. If the area of the rectangle is 403 square centimeters, what is the width? Which of the points below is a solution to 4x!3y = 10? Note: More than one point may make this equation true. [ Both (b) and (c) are solutions. ] a. (1, 2) b. (4, 2) c. (7, 6) d. (4, 3) 666 Algebra Connections

65 Kristen loves shortcuts. She figured out that she can find x- and y-intercepts for any line without graphing! For example, she knows that the x-intercept for 5x! 3y = 15 is (3, 0) just by examining the rule. a. What is her shortcut? [ While students will find many different patterns, she solved for x when y = 0. ] b. Does this shortcut work for the y-intercept? Try it and then test your result by changing 5x! 3y = 15 into y = mx + b form. [ The y-intercept is (0, 5), and a shortcut is to solve for y when x = 0; y = 3 5 x! 5. ] c. Use this shortcut to find the x- and y-intercepts of 3x! 2y = 24. [ x: (8, 0), y: (0, 12) ] Lesson Which method should I use? Choosing a Strategy Lesson Objective: Students will practice using the Zero Product Property and the Quadratic Formula by deciding which method is best to try first for different types of equations. Students will also solve quadratics that are not in standard form and will solve an application problem using a quadratic equation. REMINDER: The Additional Topics Lesson 1.4 was designed to follow Lesson It focuses on the division of polynomials. Length of Activity: One day (approximately 50 minutes) Core Problems: Problems and Ways of Thinking: Justifying Materials: Yo-yo (optional demonstration tool for problem 8-115) Suggested Lesson Activity: Today s lesson consists primarily of teamwork with class debriefing. Introduce the goal of the lesson: that students improve their quadraticsolving skills by deciding which strategy is most efficient for different types of quadratic equations. If time allows, students will also examine an application that requires the Quadratic Formula. Give teams about 30 minutes to solve the equations in problem It is not necessary for all students to solve all six quadratic equations. If you think that time will run short, select four quadratics for each team, but make sure each quadratic is assigned to at least one team for effective debriefing. As you circulate, ask, Why didn t you try factoring first? What about the equation told you that using the Quadratic Formula might be easier? and Do you think that factoring and solving was easier than using the Quadratic Formula to solve this quadratic? Why or why not? Chapter 8: Quadratics 667

66 Also use this time to catch and correct errors. Ask questions that help students recognize their errors, such as, Is the equation in standard form?, What does the equation need to look like for you to use the Zero Product Property?, or If b is negative, then what is b? Assign each team a problem to put up on the board so that students can compare their strategies and solutions. Problem asks students to summarize how they know which strategy to try first when solving a quadratic equation. It is recommended that you pull the class together and come up with a class list, perhaps organized in a table like the one below, and then ask students to copy the list into their Learning Logs. Zero Product Property Looks factorable Nice numbers Already factored and equal to 0 Quadratic Formula Decimals or fractional coefficients Large numbers Not factorable Then have teams work on problem 8-114, and, if time allows, problem Problem helps students solve a quadratic of the form (ax + b)(cx + d) = k for k! 0, while problem requires students to solve a quadratic equation to solve an application. In this application problem, the height of a yo-yo with respect to Moe s hand is represented by a quadratic equation. Closure: (8 minutes) Team Roles: Process the results from problems and so students understand any mistakes they may have made or misunderstandings they may have had. Note: the Math Notes box for this lesson provides an overview on simplifying radicals. While not central to this lesson, this provides an opportunity for you to introduce this idea to the extent you believe is appropriate for your students. Problem should generate team discussions about quadratic strategies. Students may be tempted to split-up the quadratics among team members to save time. However, this effectively eliminates all meaningful mathematical discussion among team members. Remind Facilitators that they should make sure that the team works together on each problem so that a discussion about which method to use and why can occur. Also, the Task Managers should make sure that all team members are heard and can justify their reasoning to the rest of the team. Homework: Problems through Algebra Connections

67 8.3.3 Which method should I use? Choosing a Strategy Yo! Student pages for this lesson are You now have two algebraic methods to solve quadratic equations: using the Zero Product Property and using the Quadratic Formula. How can you decide which strategy is best to try first? By the end of this lesson, you should have some strategies to help you determine which method to try first when solving a quadratic equation Examine the quadratic equations below with your team. For each equation: [ a: x =!3 or x =!9, b: x =!17.6 or x =!0.36, c: x =! 4 3 or x = 1 2, d: x = 4, e: no solution, f: x = 2.5 or x =!0.9 ] Decide which strategy is best to try first. Solve the equation. If your first strategy does not work, switch to the other strategy. Check your solution(s). Be prepared to share your process with the class. a. x 2 +12x + 27 = 0 b. 0.5x 2 + 9x = 0 c. (3x + 4)(2x!1) = 0 d. x = 8x e. x 2 + 5! 2x = 0 f. 20x 2! 30x = 2x With the class, decide when it is best to solve a quadratic by factoring and when you should go directly to the Quadratic Formula. Copy your observations in your Learning Log. Title this entry Choosing a Strategy to Solve Quadratics and include today s date. Chapter 8: Quadratics 669

68 While solving (x! 5)(x + 2) =!6, Kyle decided that x must equal 5 or 2. Not so fast! exclaimed Stanton, The product does not equal zero. We need to change the equation first. a. What is Stanton talking about? [ The Zero Product Property only works when a product equals zero. ] b. How can the equation be rewritten? Discuss this with your team and use your algebraic tools to rewrite the equation so that it can be solved. [ Students need to multiply the binomials and add six to both sides. The result: x 2! 3x! 4 = 0. ] c. Solve the resulting equation from part (b) for x. Do your solutions match Kyle s? [ x = 4 or x = 1; no ] MOE S YO Moe is playing with a yo-yo. He throws the yo-yo down and then pulls it back up. The motion of the yo-yo is represented by the equation y = 2x 2! 4.8x, where x represents the number of seconds since the yo-yo left Moe s hand, and y represents the vertical height in inches of the yo-yo with respect to Moe s hand. Note that when the yo-yo is in Moe s hand, y = 0, and when the yo-yo is below his hand, y is negative. a. How long is Moe s yo-yo in the air before it comes back to Moe s hand? Write and solve a quadratic equation to find the times that the yo-yo is in Moe s hand. [ x = 0 seconds and x = 2.4 seconds, so it is in the air for 2.4 seconds. ] b. At what time does the yo-yo turn around? Use what you know about parabolas to help you. [ at x = 1.2 seconds ] c. How long is the yo-yo s string? That is, what is y when the yo-yo changes direction? [ 2.88 feet ] d. Draw a sketch of the graph representing the motion of Moe s yo-yo. On the sketch, label the important points: when the yo-yo is in Moe s hand and when it changes direction. [ The sketch should have x-intercepts (0, 0) and (2.4, 0) and vertex (1.2, 2.88). ] 670 Algebra Connections

69 MATH NOTES OOKING DEEPER Simplifying Square Roots Before calculators were universally available, people who wanted to use approximate decimal values for numbers like 45 had a few options: 1. Carry around copies of long square-root tables. 2. Use Guess and Check repeatedly to get desired accuracy. 3. Simplify the square roots. A square root is simplified when there are no more perfect square factors (square numbers such as 4, 25, and 81) under the radical sign. Simplifying square roots was by far the fastest method. People factored the number as the product of integers hoping to find at least one perfect square number. They memorized approximations of the square roots of the integers from one to ten. Then they could figure out the decimal value by multiplying these memorized facts with the roots of the square numbers. Here are some examples of this method. Example 1: Simplify 45. First rewrite 45 in an equivalent factored form so that one of the factors is a perfect square. Simplify the square root of the perfect square. Verify with your calculator that both 3 5 and 45! Examine Example 2 and Example 3 at right. Note that in Example 3, 72 was rewritten as 36 2, rather than as 9 8 or 4 18, because 36 is the largest perfect square factor of 72. However, since Example 1 45 = 9!!5 = 9!! 5 = 3 5 Example 2 Example = 9!! 3 = 36!! 2 = 3 3 = 6 2 4! 18 = 2 9! 2 = 2 9! 2 = 2! 3 2 = 6 2 and 9! 8 = 3 4! 2 = 3 4! 2 = 3! 2 2 = 6 2, you can still get the same answer if you simplify it using different methods. When you take the square root of an integer that is not a perfect square, the result is a decimal that never repeats itself and never ends. This result is called an irrational number. The irrational numbers and the rational numbers together form the numbers we use in this course, which are called real numbers. Generally, since it is now the age of technology, when a decimal approximation of an irrational square root is desired, a calculator is used. However, for an exact answer, the number must be written using the symbol. Chapter 8: Quadratics 671

70 Write and solve an equation (or system of equations) for the situation described below. Define your variable(s) and write your solution as a sentence. [ If n = # nickels and q = # of quarters, then 0.05n q = 1.90, n = 2q + 3, and n = 13, so Daria has 13 nickels. ] Daria has 18 coins that are all nickels and quarters. The number of nickels is 3 more than twice the number of quarters. If she has $1.90 in all, how many nickels does Daria have? Solve the following quadratic equations using any method. [ a: x = ± 0.08, b: x = 2 9 or x = 4, c: no solution, d: x 1.4 or x 17.4 ] a x 2! 64 = 0 b. 9x 2! 8 =!34x c. 2x 2! 4x + 7 = 0 d. 3.2x + 0.2x 2! 5 = Find a rule that represents the number of tiles in Figure x for the tile pattern at right. [ While the expressions may vary, each should be equivalent to y = x 2 + 4x + 3. ] Figure 1 Figure 2 Figure Solve the equations below for x. Check your solutions. [ a: x = 2, b: x = 15, c: x = 2, d: all numbers ] a. 3x 2 + 3x = 6 + 3x 2 b. 5 x = 1 3 c. 5! (2x!3) =!3x + 6 d. 6(x! 3) + 2x = 4(2x +1)! Line L passes through the points ( 44, 42) and ( 31, 94), while line M has the rule y = 6 + 3x. Which line is steeper? Justify your answer. [ Line L has slope 4, while line M has slope 3. Therefore, line L is steeper. ] Multiple Choice: Which line below is perpendicular to the line 2x! 5y = 3? [ D ] a. 2x + 5y = 7 b.!2x + 5y = 4 c. 5x! 2y =!1 d. 5x + 2y = Algebra Connections

71 Chapter 8 Closure What have I learned? Reflection and Synthesis Closure Objective: Length of Activity: Materials: Team Brainstorm Making Connections The objective of the chapter closure depends on which closure activity you choose to use. However, some of the main objectives of each activity are for the students to reflect on what they have learned and for the teacher to be able to assess informally where the students are in their understanding. The time needed for chapter closure varies from class to class and depends on the needs of the students. While some form of closure is strongly recommended for all classes, it can be part of a lesson, a homework assignment, or a full day s lesson. Five closure activities are provided from which you can choose. Note: It is not expected (nor is it recommended) that you assign all five parts. For teachers using closure activity #1 Team Brainstorm: Poster paper and colored markers (optional), one per team For teachers using closure activity #2 Making Connections: Chapter 8 Closure Resource Page: Concept Map Cards (2 pages in all), one of each page per team. If these cards are not cut apart ahead of time, you will also need to provide scissors. Poster paper, glue sticks, and colored markers (optional), one each per team For teachers using closure activity #3 Summarizing My Understanding: Chapter 8 Closure Resource Page: Quadratic Multiple Representations GO (Graphic Organizer), one per student If this is the first time you are doing a Team Brainstorm with your class, be sure to review the Team Brainstorm section of the course notes for closure preceding Chapter 1. Listing the topics from the chapter can help students remember what they have learned and help them bring the big ideas in focus as they begin other closure activities. Students responses can help you become aware of and fill gaps in students thinking from the chapter. Making Connections can help students avoid the feeling that algebra is a collection of unrelated procedures. This activity will ensure that they see how the topics they have been studying relate to one another. To make a meaningful concept map, students will need teacher support. It is strongly recommended that you read the Making Connections section of the course notes for chapter closure preceding Chapter 1 before doing this activity with the class. It is also recommended that Chapter 8: Quadratics 673

72 you make your own concept map before having students make them. Having cards available with the topics and vocabulary from the chapter can help students test out their ideas about how concepts connect before drawing the map. A resource page is provided for this purpose. Or, students could brainstorm the vocabulary first and write them on sticky notes or index cards. Below is an example of what a concept map could look like. The students directions ask them to sketch an example for each idea or term. This is intended merely to be a visual reminder of the meaning. It would be a good idea to show a few examples to the class (an empty table for the term table, a set of axes for the term axes, etc.). Phrase describing connection Phrase describing connection Phrase describing connection Phrase describing connection Phrase describing connection If students have made concept maps for previous chapters, you may want to find a way for students to continue building on this concept map. Producing a bulletin board with a class concept map can be a great way to find consensus on the connections as well as validate the connections that students have developed throughout the course. Summarizing My Understanding This part of closure gives students the opportunity to write as much as they possibly can about a topic. For this chapter, students can use a GO page similar to the one from Chapter 3, in which they illustrate a problem in all the various representations of the quadratic web. Posters of this activity could make great classroom references. For a more in-depth discussion of GO pages and their uses, refer to the course notes for chapter closure preceding Chapter 1. Situation Having students write a chapter on solving systems of equations or asking students to create a lesson on how to write a mathematical sentence from a word problem would also be excellent closure for this chapter. For more information on having students write chapters in a math book, see the notes on chapter closure preceding Chapter 1. Table Multiple Representations Graph Rule 674 Algebra Connections

73 What Have I Learned? How Am I Thinking? This section gives students the opportunity to see if they can work with the current topics at the expected level. It also provides teachers with the opportunity to see the topics with which students are struggling and whether any topics need revisiting or emphasizing in later work. Students can work on this in class or for homework, but it is a good idea to find time for students to ask questions before any formal assessment. This course has been designed to help students recognize five different ways they think while they solve problems and when they try to understand a concept. In this chapter, students are asked to focus on three Ways of Thinking and write about times when they used these Ways of Thinking. You may want to start this activity by reviewing each Way of Thinking. Summaries of each can be found in the closure sections in Chapters 1 through 5. If your students do not have access to these chapters, you may want to copy these sections before class and distribute them during this activity for students to review. Chapter 8 Closure What have I learned? Reflection and Synthesis Student pages for this lesson are The activities below offer you a chance to reflect on what you have learned during this chapter. As you work, look for concepts that you feel very comfortable with, ideas that you would like to learn more about, and topics you need more help with. Look for connections between ideas as well as connections with material you learned previously. TEAM BRAINSTORM With your team, brainstorm a list for each of the following topics. Be as detailed as you can. How long can you make your list? Challenge yourselves. Be prepared to share your team s ideas with the class. Topics: What have you studied in this chapter? What ideas and words were important in what you learned? Remember to be as detailed as you can. Ways of Thinking: What Ways of Thinking did you use in this chapter? When did you use them? Connections: What topics, ideas, and words that you learned before this chapter are connected to the new ideas in this chapter? Again, make your list as long as you can. Chapter 8: Quadratics 675

74 MAKING CONNECTIONS The following is a list of the vocabulary used in this chapter. The words that appear in bold are new to this chapter. Make sure that you are familiar with all of these words and know what they mean. Refer to the glossary or index for any words that you do not yet understand. binomial factor generic rectangle graph monomial parabola product quadratic equation Quadratic Formula root solution standard form for quadratics sum symmetry trinomial vertex x-intercept x y table y-intercept Zero Product Property Make a concept map showing all of the connections you can find among the key words and ideas listed above. To show a connection between two words, draw a line between them and explain the connection, as shown in the example below. A word can be connected to any other word as long as there is a justified connection. For each key word or idea, provide a sketch that illustrates the idea (see the example below). Word A Example: These are connected because Word B Example: Your teacher may provide you with vocabulary cards to help you get started. If you use the cards to plan your concept map, be sure either to re-draw your concept map on your paper or to glue the vocabulary cards to a poster with all of the connections explained for others to see and understand. While you are making your map, your team may think of related words or ideas that are not listed above. Be sure to include these ideas on your concept map. SUMMARIZING MY UNDERSTANDING This section gives you an opportunity to show what you know about certain math topics or ideas. Your teacher will give you directions for exactly how to do this. Your teacher may also provide a GO page to work on. The GO stands for Graphic Organizer, a tool you can use to organize your thoughts and communicate your ideas clearly. 676 Algebra Connections

75 WHAT HAVE I LEARNED? This section will help you evaluate which types of problems you have seen with which you feel comfortable and those with which you need more help. Even if your teacher does not assign this section, it is a good idea to try these problems and find out for yourself what you know and what you need to work on. Solve each problem as completely as you can. The table at the end of the closure section has answers to these problems. It also tells you where you can find additional help and practice on problems like these. CL For the graph of the line at right: a. Find the slope. b. Find the y-intercept. c. Find an equation. d. Find an equation of a line perpendicular to this one that passes through (0, 7). CL Factor and use the Zero Product Property to find the roots of the following quadratic equations. a. 0 = x 2! 7x + 12 b. 0 = 6x 2! 23x + 20 c. 0 = x 2! 9 d. 0 = x x + 36 CL Use the Quadratic Formula to solve these equations. a. 0 = x 2! 7x + 3 b. 3x 2 + 5x +1= 0 Chapter 8: Quadratics 677

76 CL Use the graph at right to answer the questions below. a. One of these lines represents Feng, and one represents Wai. Write an equation for each girl s line. b. The two girls are riding bikes. How fast does each girl ride? Distance (miles) Wei Feng c. When do Feng and Wai meet? At that point, how far are they from school? Time (hrs) CL Factor each expression below completely. a. 3x x + 30 b. 7x 2! 63 CL Find the coordinates of the y-intercept, x-intercepts, and vertex of y = x 2! 2x!15. Show all of the work that you do to find these points. CL Solve for x using the method of your choice. a. 0 = 2x 2! 5x! 33 b. 0 = 3x 2! 4x! 1 CL Quinn started off with twice as much candy as Denali, but then he ate 4 pieces. When Quinn and Denali put their candy together, they now have a total of 50 pieces. How many pieces of candy did Denali start with? CL Check your answers using the table at the end of the closure section. Which problems do you feel confident about? Which problems were hard? Use the table to make a list of topics you need help on and a list of topics you need to practice more. 678 Algebra Connections

77 HOW AM I THINKING? This course focuses on five different Ways of Thinking: reversing thinking, justifying, generalizing, making connections, and applying and extending understanding. These are some of the ways in which you think while trying to make sense of a concept or to solve a problem (even outside of math class). During this chapter, you have probably used each Way of Thinking multiple times without even realizing it! Choose three of these Ways of Thinking that you remember using while working in this chapter. For each Way of Thinking that you choose, show and explain where you used it and how you used it. Describe why thinking in this way helped you solve a particular problem or understand something new. (For instance, explain why you wanted to generalize in this particular case, or why it was useful to see these particular connections.) Be sure to include examples to demonstrate your thinking. Answers and Support for Closure Activity #4 What Have I Learned? Problem Solution Need Help? More Practice CL a. m =! 2 3 b. y-intercept: 4 c. y =! 2 3 x + 4 d. y = 3 2 x + 7 Lessons 7.1.3, 7.1.5, 7.2.2, 7.2.3, and Math Notes boxes Problems 8-9, 8-31, 8-57, 8-86, 8-99, and CL a. x = 4 or x = 3 b. x = 5 2 or x = 4 3 c. x = 3 or x = 3 d. x = 6 Lessons 8.1.4, 8.2.3, and Math Notes boxes Problems 8-66, 8-72, 8-77, 8-88, 8-96, and ± 37 CL a. x = ( x! 6.54 or 0.46) 2!5± 13 b. x = ( x! "0.23or 1.43) 6 Problem 8-92, Lessons and Math Notes boxes Problems 8-94, 8-97, 8-104, and Chapter 8: Quadratics 679

78 Problem Solution Need Help? More Practice CL a. Feng: y = 4x Wai: y = 6x! 12 b. Feng rides at 4 miles per hour; Wai rides at 6 miles per hour. c. Feng and Wai meet after 6 hours. At that point, they are 24 miles from school. Lessons and Math Notes boxes See Lessons and CL a. 3(x + 2)(x + 5) b. 7(x + 3)(x! 3) Problems 8-13, 8-14, and 8-35; Lesson Math Notes box Problems 8-15, 8-16, 8-22, 8-23, 8-24, 8-29, 8-33, 8-36, 8-37, and 8-69 CL y-intercept:!15 x-intercepts: 5 and!3 vertex: (1, 16) Problems 8-43, 8-51, 8-59, 8-60, 8-61, and 8-62 Problems 8-10, 8-54, 8-68, and 8-82 CL a. x = 11 2 or x =!3 4± 28 b. x = = 4±2 7 = 2± ( x! 1.55 or x! " 0.22 ) Lessons 8.1.4, 8.2.3, 8.3.1, Problems and CL Denali has 18 pieces of candy. Lesson Math Notes box Problem Algebra Connections

79 Lesson Resource Page Problem 8-51 Water-Balloon Contest Height in Air (in yards) Distance Along Ground (in yards) Maggie s Toss Jen s Toss Imp s Toss Al s Toss x y x y x y x y Chapter 8: Quadratics 681

80 Lesson Resource Page Problem 8-91 f (x) = x 2! 3x! Algebra Connections

81 Chapter 8 Closure Resource Page: Concept Map Cards Page 1 of 2 binomial factor generic rectangle graph monomial parabola product Quadratic Formula quadratic equation root Chapter 8: Quadratics 683

82 Chapter 8 Closure Resource Page: Concept Map Cards Page 2 of 2 solution standard form for quadratics sum symmetry trinomial vertex x-intercept x y table y-intercept Zero Product Property 684 Algebra Connections

83 Chapter 8 Closure Resource Page: Quadratic Multiple Representations GO Situation Graph Multiple Representations Table Rule Chapter 8: Quadratics 685

84 686 Algebra Connections