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NYS COMMON CORE MATHEMATICS CURRICULUM 7•3 Lesson 8

Lesson 8: Using If-Then Moves in Solving Equations

Student Outcomes � Students understand and use the addition, subtraction, multiplication, division, and substitution properties of

equality to solve word problems leading to equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟 where 𝑝𝑝, 𝑞𝑞,and 𝑟𝑟 are specific rational numbers.

� Students understand that any equation with rational coefficients can be written as an equation withexpressions that involve only integer coefficients by multiplying both sides by the least common multiple of allthe rational number terms.

Lesson Notes The intent of this lesson is for students to make the transition from an arithmetic approach of solving a word problem to an algebraic approach of solving the same problem. Recall from Module 2 that the process for solving linear equations is to isolate the variable by making 0s and 1s. In this module, the emphasis will be for students to rewrite an equation using if-then moves into a form where the solution is easily recognizable. The main issue is that, in later grades, equations are rarely solved by “isolating the variable” (e.g., How do you isolate the variable for 3𝑝𝑝2 − 8 = −2𝑝𝑝?). Instead, students learn how to rewrite equations into different forms where the solutions are easy to recognize.

� Examples of Grade 7 forms: The equation23𝑝𝑝 + 27 = 31 is put into the form 𝑝𝑝 = 6, where it is easy to

recognize that the solution is 6.� Example of an Algebra I form: The equation 3𝑝𝑝2 − 8 = −2𝑝𝑝 is put into factored form (3𝑝𝑝 − 4)(𝑝𝑝 + 2) = 0,

where it is easy to recognize that the solutions are �43 ,−2�.

� Example of a Algebra II and Precalculus form: The equation sin3 𝑝𝑝 + sin 𝑝𝑝 cos2 𝑝𝑝 = cos 𝑝𝑝 sin2 𝑝𝑝 + cos3 𝑝𝑝 issimplified to tan 𝑝𝑝 = 1, where it is easy to recognize that the solutions are �𝜋𝜋4 + 𝑘𝑘𝜋𝜋 � 𝑘𝑘 integer�.

Regardless of the type of equation students are studying, the if-then moves play an essential role in rewriting equations into different useful forms for solving, graphing, etc.

The FAQs on solving equations below are designed to help teachers understand the structure of the next set of lessons. Before reading the FAQ, it may be helpful to review the properties of operations and the properties of equality listed in Table 3 and Table 4 of the Common Core Learning Standards (CCLS).

What are the “if-then moves”? Recall the following if-then moves from Lesson 21 of Module 2:

1. Addition property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 + 𝑐𝑐 = 𝑏𝑏 + 𝑐𝑐.2. Subtraction property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 − 𝑐𝑐 = 𝑏𝑏 − 𝑐𝑐.3. Multiplication property of equality: If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 × 𝑐𝑐 = 𝑏𝑏 × 𝑐𝑐.4. Division property of equality: If 𝑎𝑎 = 𝑏𝑏 and 𝑐𝑐 ≠ 0, then 𝑎𝑎 ÷ 𝑐𝑐 = 𝑏𝑏 ÷ 𝑐𝑐.

All eight properties of equality listed in Table 4 of the CCLS are if-then statements used in solving equations, but these four properties are separated out and collectively called the if-then moves.

Lesson 8: Using If-Then Moves in Solving Equations Date: 9/11/14

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What points should I try to communicate to my students about solving equations? The goal is to make three important points about solving equations.

� The technique for solving equations of the form 𝑝𝑝𝑝𝑝 + 𝑞𝑞 = 𝑟𝑟 and 𝑝𝑝(𝑝𝑝 + 𝑞𝑞) = 𝑟𝑟 is to rewrite them into the form𝑝𝑝 = “a number,”

using the properties of operations (Lessons 1–6) and the properties of equality (i.e., the if-then moves) to make 0s and 1s. This technique is sometimes called “isolating the variable,” but that name really only applies to linear equations. You might mention that students will learn other techniques for other types of equations in later grades.

� The properties of operations are used to modify one side of an equation at a time by changing the expressionon a side into another equivalent expression.

� The if-then moves are used to modify both sides of an equation simultaneously in a controlled way. The twoexpressions in the new equation are different than the two expressions in the old equation, but the solutionsare the same.

How do if-then statements show up when solving equations? We will continue to use the normal convention of writing a sequence of equations underneath each other, linked together by if-then moves and/or properties of operations. For example, the sequence of equations and reasons for solving 3𝑝𝑝 = 3 is as follows:

13

(3𝑝𝑝) =13

(3) If-then move: multiply both sides by 13.

�13 ⋅ 3� 𝑝𝑝 = 1

3 (3) Associative property

1 ⋅ 𝑝𝑝 = 1 Multiplicative inverse𝑝𝑝 = 1. Multiplicative identity

This is a welcomed abbreviation for the if-then statements used in Lesson 21 of Module 2:

If 3𝑝𝑝 = 3, then 13

(3𝑝𝑝) =13

(3) by the if-then move of multiplying both sides by 13

.

If 13

(3𝑝𝑝) =13

(3), then �13 ⋅ 3� 𝑝𝑝 = 1

3 (3) by the associative property.

If �13 ⋅ 3� 𝑝𝑝 = 1

3 (3), then 1 ⋅ 𝑝𝑝 = 1 by the multiplicative inverse property.

If 1 ⋅ 𝑝𝑝 = 1, then 𝑝𝑝 = 1 by the multiplicative identity property.

The abbreviated form is visually much easier for students to understand provided that you explain to your students that each pair of equations is part of an if-then statement.

In the unabbreviated if-then statements above, it looks like the properties of operations are also if-then statements. Are they? No. The properties of operations are not if-then statements themselves; most of them (associative, commutative, distributive, etc.) are statements about equivalent expressions. However, they are often used with combinations of the transitive and substitution properties of equality, which are if-then statements. For example, the transitive property states in this situation that if two expressions are equivalent, and if one of the expressions is substituted for the other in a true equation, then the resulting equation is also true (if 𝑎𝑎 = 𝑏𝑏 and 𝑏𝑏 = 𝑐𝑐, then 𝑎𝑎 = 𝑐𝑐).







Thus, the sentence above, “If 13

(3𝑝𝑝) =13

(3), then �13 ⋅ 3� 𝑝𝑝 = 1

3 (3) by the associative property,” can be expanded as

follows:

1. In solving the equation 13

(3𝑝𝑝) =13

(3), we assume 𝑝𝑝 is a number that makes this equation true.

2. �13 ⋅ 3� 𝑝𝑝 = 1

3 (3𝑝𝑝) is true by the associative property.

3. Therefore, we can replace the expression 13

(3𝑝𝑝) in the equation 13

(3𝑝𝑝) =13

(3) with the equivalent expression

�13 ⋅ 3� 𝑝𝑝 by the transitive property of equality.

(You might check that this fits the form of the transitive property described in the CCLS: If 𝑎𝑎 = 𝑏𝑏 and 𝑏𝑏 = 𝑐𝑐, then 𝑎𝑎 = 𝑐𝑐.)

Teachers do not necessarily need to drill down to this level of detail when solving equations with students. Carefully monitor students for understanding and drill down to this level as needed.

Should I show every step in solving an equation? Yes and no: Please use your best judgment given the needs of your students. We generally do not write every step on the board when solving an equation. Otherwise, we would need to include discussions like the one above about the transitive property, which can throw off the lesson pace and detract from understanding. Here are general guidelines to follow when solving an equation with a class, which should work well with how these lessons are designed:

1. It is almost always better to initially include more steps than less. A good rule of thumb is to double the number ofsteps you would personally need to solve an equation. As adults, we do a lot more calculating in our heads than werealize. Doubling the number of steps slows down the pace of the lesson, which can be enormously beneficial toyour students.

2. As students catch on, look for ways to shorten the number of steps (for example, using any order/any grouping tocollect all like terms at once rather than showing each associative/commutative property). Regardless, it is stillimportant to verbally describe or ask for the properties being used in each step.

3. Write the reason (on the board) if it is one of the main concepts being learned in a lesson. For example, the nextfew lessons focus on if-then moves. Writing the if-then moves on the board calls them out to your students andhelps them focus on the main concept. As students become comfortable using the language of if-then moves,further reduce what you write on the board but verbally describe (or ask students to describe) the properties beingused in each step.

We end with a quote from the High School, Algebra Progressions that encapsulates this entire FAQ:

“In the process of learning to solve equations, students learn certain ‘if-then’ moves, for example, ‘if 𝑝𝑝 = 𝑦𝑦, then 𝑝𝑝 + 2 = 𝑦𝑦 + 2.’ The danger in learning algebra is that students emerge with nothing but the moves, which may make it difficult to detect incorrect or made-up moves later on. Thus, the first requirement in the standards in this domain is that students understand that solving equations is a process of reasoning. This does not necessarily mean that they always write out the full text; part of the advantage of algebraic notation is its compactness. Once students know what the code stands for, they can start writing in code.”







Classwork Opening Exercise (5 minutes)

Opening Exercise

Recall and summarize the if-then moves.

If a number is added or subtracted to both sides of a true equation, then the resulting equation is also true:

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂 + 𝒄𝒄 = 𝒃𝒃 + 𝒄𝒄.

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂 − 𝒄𝒄 = 𝒃𝒃 − 𝒄𝒄.

If a number is multiplied or divided to each side of a true equation, then the resulting equation is also true:

If 𝒂𝒂 = 𝒃𝒃, then 𝒂𝒂𝒄𝒄 = 𝒃𝒃𝒄𝒄.

If 𝒂𝒂 = 𝒃𝒃 and 𝒄𝒄 ≠ 𝟎𝟎, then 𝒂𝒂 ÷ 𝒄𝒄 = 𝒃𝒃 ÷ 𝒄𝒄.

Write 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖 in as many true equations as you can using the if-then moves. Identify which if-then move you used.

Answers will vary, but some examples are as follows:

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟑𝟑 + 𝟓𝟓+ 𝟒𝟒 = 𝟖𝟖+ 𝟒𝟒. Add 𝟒𝟒 to both sides.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟑𝟑 + 𝟓𝟓 − 𝟒𝟒 = 𝟖𝟖 − 𝟒𝟒. Subtract 𝟒𝟒 from both sides.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then 𝟒𝟒(𝟑𝟑 + 𝟓𝟓) = 𝟒𝟒(𝟖𝟖). Multiply both sides by 𝟒𝟒.

If 𝟑𝟑 + 𝟓𝟓 = 𝟖𝟖, then (𝟑𝟑 + 𝟓𝟓) ÷ 𝟒𝟒 = 𝟖𝟖 ÷ 𝟒𝟒. Divide both sides by 𝟒𝟒.

Example 1 (10 minutes)

Example 1

Julia, Keller, and Israel are volunteer firefighters. On Saturday, the volunteer fire department held its annual coin drop fundraiser at a streetlight. After one hour, Keller had collected $𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 more than Julia, and Israel had collected $𝟏𝟏𝟓𝟓 less than Keller. The three firefighters collected $𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 in total. How much did each person collect?

Find the solution using a tape diagram.

Julia

Keller

Israel

𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 + 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 = 𝟐𝟐𝟎𝟎

𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 + 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 − 𝟐𝟐𝟎𝟎 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓

𝟑𝟑 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓÷ 𝟑𝟑 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓

𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓

What were the operations we used to get our answer?

First, we added 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 and 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 to get 𝟐𝟐𝟎𝟎. Next, we subtracted 𝟐𝟐𝟎𝟎 from 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓. Finally, we divided 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 by 𝟑𝟑 to get 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓.

𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎

𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎 𝟏𝟏𝟓𝟓







The amount of money Julia collected is 𝒋𝒋 dollars. Write an expression to represent the amount of money Keller collected in dollars.

𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎

Using the expressions for Julia and Keller, write an expression to represent the amount of money Israel collected in dollars.

𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 − 𝟏𝟏𝟓𝟓

or

𝒋𝒋 + 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎

Using the expressions written above, write an equation in terms of 𝒋𝒋 that can be used to find the amount each person collected.

𝒋𝒋 + (𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎) + (𝒋𝒋+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎) = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓

Solve the equation written above to determine the amount of money each person collected, and describe any if-then moves used.

𝒋𝒋 + (𝒋𝒋 + 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎) + (𝒋𝒋+ 𝟒𝟒𝟐𝟐.𝟓𝟓𝟎𝟎) = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓

𝟑𝟑𝒋𝒋 + 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 Any order, any grouping

(𝟑𝟑𝒋𝒋+ 𝟐𝟐𝟎𝟎) − 𝟐𝟐𝟎𝟎 = 𝟏𝟏𝟒𝟒𝟓𝟓.𝟗𝟗𝟓𝟓 − 𝟐𝟐𝟎𝟎 If-then move: Subtract 𝟐𝟐𝟎𝟎 from both sides (to make a 𝟎𝟎).

𝟑𝟑𝒋𝒋 + 𝟎𝟎 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 Any grouping, additive inverse

𝟑𝟑𝒋𝒋 = 𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓 Additive identity

�𝟏𝟏𝟑𝟑� (𝟑𝟑𝒋𝒋) = (𝟓𝟓𝟓𝟓.𝟗𝟗𝟓𝟓) �𝟏𝟏𝟑𝟑� If-then move: Multiply both sides by 𝟏𝟏𝟑𝟑

(to make a 𝟏𝟏).

�𝟏𝟏𝟑𝟑 ⋅ 𝟑𝟑� 𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Associative property

𝟏𝟏 ⋅ 𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Multiplicative inverse

𝒋𝒋 = 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓 Multiplicative identity

If Julia collected $𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓, then Keller collected 𝟏𝟏𝟖𝟖.𝟔𝟔𝟓𝟓+ 𝟒𝟒𝟒𝟒.𝟓𝟓𝟎𝟎 = $𝟔𝟔𝟏𝟏.𝟏𝟏𝟓𝟓, and Israel collected 𝟔𝟔𝟏𝟏.𝟏𝟏𝟓𝟓 − 𝟏𝟏𝟓𝟓 = $𝟒𝟒𝟔𝟔.𝟏𝟏𝟓𝟓.

Discussion (5 minutes)

Have students present the models they created based upon the given relationships and then have the class compare different correct models and/or discuss why the incorrect models were incorrect. Some possible questions from the different models are as follows:

� How does the tape diagram translate into the initial equation?à Each unknown unit represents how much Julia collected: 𝑗𝑗 dollars.

� The initial step to solve the equation algebraically is to collect all like terms on the left hand side of theequation using the any order, any grouping property.

The goal is to rewrite the equation into the form 𝑝𝑝 = “a number” by making zeros and ones.

� How can we make a zero or one?à Zeros are made with addition, and ones are made through multiplication and division.

Scaffolding: Teachers may need to review the process of solving an equation algebraically from Module 2, Lessons 17, 22, and 23.







à We can make a 0 by subtracting 70 from both sides, or we can make a 1 by multiplying both sides by 13

.

(Both are correct if-then moves, but point out to students that making a 1 will result in extra calculations.)

à Let us subtract 70 from both sides. The if-then move of subtracting 70 from both sides will change bothexpressions of the equation (left and right sides) to new nonequivalent expressions, but the newexpression will have the same solution as the old one did.

� In subtracting 70 from both sides, what do 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐 represent in the if-then move, “If 𝑎𝑎 = 𝑏𝑏, then 𝑎𝑎 − 𝑐𝑐 =𝑏𝑏 − 𝑐𝑐”?à In this specific example, 𝑎𝑎 represents the left side of the equation, 3𝑗𝑗 + 70, 𝑏𝑏 represents the right side of

the equation, 125.95, and 𝑐𝑐 is 70.� Continue to simplify the new equation using the properties of operations until reaching the equation

3𝑗𝑗 = 55.95. Can we make a zero or a one?

à Yes, we can make a one by multiplying both sides by 13

. Since we are assuming that 𝑗𝑗 is a number that

makes the equation 3𝑗𝑗 = 55.95 true, we can apply the if-then move of multiplying both sides by 13

. The

resulting equation will also be true. � How is the arithmetic approach (the tape diagram with arithmetic) similar to the algebraic approach (solving

an equation)?à The operations performed in solving the equation algebraically are the same operations done

arithmetically.� How can the equation 3𝑗𝑗 + 70 = 125.95 be written so that the equation will contain only integers? What

would the new equation be?à You can multiply each term by 100. The equivalent equation would be 300𝑗𝑗 + 7000 = 12595.

Show students that solving this problem also leads to 𝑗𝑗 = 18.65.

� What if, instead, we used the amount Keller collected: 𝑘𝑘 dollars. Would that be okay? How would the moneycollected by the other people then be defined?à Yes, that would be okay. Since Keller has $42.50 more than Julia, then Julia would have $42.50 less than

Keller. Julia’s money would be 𝑘𝑘 − 42.50. Since Israel’s money is $15.00 less than Keller, his money is𝑘𝑘 − 15.

� The expressions defining each person’s amount differ depending on who we choose to represent the othertwo people. Complete the chart to show how the statements vary when 𝑝𝑝 changes.

In terms of Julia Israel Keller Julia’s amount (𝑗𝑗) 𝑗𝑗 𝑗𝑗 + 27.50 𝑗𝑗 + 42 Israel’s amount (𝑖𝑖) 𝑖𝑖 − 27.50 𝑖𝑖 𝑖𝑖 + 15 Keller’s amount (𝑘𝑘) 𝑘𝑘 − 42.50 𝑘𝑘 − 15 𝑘𝑘

à If time, set up and solve the equation in terms of 𝑘𝑘. Show students that the equation and solution aredifferent than the equation based upon Julia’s amount, but that the solution, $61.15, matches howmuch Keller collected.

MP.2






NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 1 ALGEBRA I

Lesson 1: Graphs of Piecewise Linear Functions

Student Outcomes � Students define appropriate quantities from a situation (a “graphing story”), choose and interpret the scale

and the origin for the graph, and graph the piecewise linear function described in the video. They understandthe relationship between physical measurements and their representation on a graph.

Classwork Exploratory Challenge (20 minutes)

Show the first 1: 08 minutes of the video below, telling the class that our goal will simply be to describe in words the motion of the man. (Note: Be sure to stop the video at 1:08 because after that the answers to the graphing questions are given.)

Elevation vs. Time #2 [http://www.mrmeyer.com/graphingstories1/graphingstories2.mov. This is the second video under “Download Options” at the site http://blog.mrmeyer.com/?p=213 called “Elevation vs. Time #2.”]

After viewing the video, have students share out loud their ideas on describing the motion. Some might speak in terms of speed, distance traveled over time, or change of elevation. All approaches are valid. Help students begin to shape their ideas with precise language.

Direct the class to focus on the change of elevation of the man over time and begin to put into words specific details linking elevation with time.

� How high do you think he was at the top of the stairs? How did you estimate that elevation?

� Were there intervals of time when his elevation wasn’t changing? Was he still moving?� Did his elevation ever increase? When?

Help students discern statements relevant to the chosen variable of elevation.

If students do not naturally do so, suggest representing this information on a graph. As per the discussion that follows, display a set of axes on the board with vertical axis labeled in units relevant to the elevation.

Ask these types of questions:

� How should we label the vertical axis? What unit of measurementshould we choose (feet or meters)?

� How should we label the horizontal axis? What unit of measurementshould we choose?

� Should we measure the man’s elevation to his feet or to his head onthe graph?

MP.1

Lesson 1: Graphs of Piecewise Linear Functions Date: 5/25/14





http://www.mrmeyer.com/graphingstories1/graphingstories2.mov

http://blog.mrmeyer.com/?p=213


� The man starts at the top of the stairs. Where would that be located on the graph?� Show me with your hand what the general shape of the graph should look like.

Give time for students to draw the graph of the story (alone or in pairs). Lead a discussion through the issues of formalizing the diagram: The labels and units of the axes, a title for the graph, the meaning of a point plotted on the graph, a method for finding points to plot on the graph, and so on.

Note: The graph shown at the end of the video is incorrect! The man starts at “30 feet above the ground,” which is clearly false. You might ask students, “Can you find the error made in the video?”


Present the following graph and question.

Example 1

Here is an elevation-versus-time graph of a person’s motion. Can we describe what the person might have been doing?

Have students discuss this question in pairs or in small groups. It will take some imagination to create a context that matches the shape of the graph, and there will likely be debate.

Additional questions to ask:

� What is happening in the story when the graph is increasing, decreasing, constant over time?à Answers will vary depending on the story: a person is “walking up a hill,” etc.

� What does it mean for one part of the graph to be steeper than another?à The person is climbing or descending faster than in the other part.

� How does the slope of each line segment relate to the context of the person’s elevation?à The slope gives the average change in elevation per minute.

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� Is it reasonable that a person moving up and down a vertical ladder could have produced this elevation versustime graph?à It is unlikely because the speed is too slow: 2.5 feet per minute. If the same graph had units in seconds

then it would be reasonable.

� Is it possible for someone walking on a hill to produce this elevation versus time graph and return to herstarting point at the 10-minute mark? If it is, describe what the hill might look like.à Yes, the hill could have a long path with a gentle slope that would zigzag back up to the top and then a

shorter, slightly steeper path back down to the beginning position.

� What was the average rate of change of the person’s elevation between time 0 minutes and time 4 minutes?

à 1

ft/min, or 2.5 ft/min.

These types of questions help students understand that the graph represents only elevation, not speed or horizontal distance from the starting point. This is an important observation.

Closing (5 minutes)

Ask the following:

� How would you describe the graph of Example 1 to a friend?� What type of equation(s) would be required to create this graph?

Introduce the following definition to your students and discuss briefly. (We will return to this definition later in the year.)

Piecewise-Defined Linear Function: Given non-overlapping intervals on the real number line, a (real) piecewise linear function is a function from the union of the intervals on the real number line that is defined by (possibly different) linear functions on each interval.

Point out that all graphs we studied today are graphs of piecewise linear functions. Remind students (see Standard 8.F.A.3) that the graphs of linear functions are straight lines, and show how each segment in one of the graphs studiedtoday is part of a straight line as in:

or






NYS COMMON CORE MATHEMATICS CURRICULUM M3 Lesson 2 GEOMETRY

Lesson 2: Properties of Area

Student Outcomes � Students understand properties of area:

1. Students understand that the area of a set in the plane is a number greater than or equal to zero thatmeasures the size of the set and not the shape.

2. The area of a rectangle is given by the formula length × width. The area of a triangle is given by theformula 1

2× base × height. A polygonal region is the union of finitely many non-overlapping triangular

regions; its area is the sum of the areas of the triangles. 3. Congruent regions have the same area.4. The area of the union of two regions is the sum of the areas minus the area of the intersection.5. The area of the difference of two regions, where one is contained in the other, is the difference of the

areas.

Lesson Notes In this lesson, we make precise what we mean about area and the properties of area. We already know the area formulas for rectangles and triangles; this will be our starting point. In fact, the basic definition of area and most of the area properties listed in the student outcomes above were first explored by students in third grade (3.MD.C.3 3.MD.C.6, 3.MD.C.7). Since their introduction, students have had continuous exposure to these properties in a varietyof situations involving triangles, circles, etc. (4.MD.A.2, 5.NF.B.4b, 6.G.A.1, 6.G.A.4). It is the goal of this lesson to statethe properties learned in earlier grades explicitly. In that sense, this lesson is a summative experience for studentsrather than an introductory experience. Furthermore, the review is preparatory to the exploration of volume, which willcome later. The examination will allow us to show the parallels between area and volume more explicitly, which helpsset the stage for understanding why volume formulas for cylinders, pyramids, and cones work and, consequently, theapplication of each of those formulas (G-GMD.A.1, G-GMD.A.3).

If these facts seem brand new to students, please refer to the following lessons:

• Property 1, Grade 3, Module 4, Lessons 1–4

• Property 2, Grade 3, Module 4, Topic B focuses on rectangles. Triangles are not studied in Grade 3, Module 4,but there is practice decomposing regions into rectangles and adding up areas of smaller rectangles in Lesson13. Introduction of length × width happens in Grade 4, Module 3, Lesson 1.

• Property 3, though not addressed explicitly, is observed in Grade 3, Module 4, Lesson 5, Problem 1(a) and 1(c).

• Property 4, Grade 3, Module 4, Lesson 13

• Property 5, Grade 3, Module 4, Lesson 13, Problem 3

This lesson does cover some new material. We introduce some notation for set operations for regions in the plane: specific notation related to a union of two regions, ; an intersection of two regions, ; and a subset of a region . Students begin by exploring the properties of area, which are then solidified in a whole-class discussion.

Lesson 2: Properties of Area Date: 10/6/14






This treatment of area and area properties in this lesson is usually referred to as Jordan measure by mathematicians. It is part of the underlying theory of area needed by students for learning integral calculus. While it is not necessary to bring up this term with your students, we encourage you to read up on it by searching for it on the Internet (the brave of heart might find Terence Tao’s online book on measure theory fascinating, even if it is just to ponder the questions he poses in the opening to Chapter 1).

If more time is needed for students to relate their previous experience working with area with these explicit properties, consider splitting the lesson over two days. This means a readjustment of pacing so as not to address all five properties in one day.

Classwork Exploratory Challenge/Exercises 1–4 (15 minutes)

The exercises below relate to the properties of area that students know; these exercises facilitate the conversation around the formal language of the properties in the following Discussion. The exercises are meant to be quick; divide the class into groups so that each group works on a separate problem. Then have each group present their work.

Exploratory Challenge/Exercises 1–4

1. Two congruent triangles are shown below.

a. Calculate the area of each triangle.

𝟏𝟏𝟒𝟒

(𝟏𝟏𝟒𝟒.𝟔𝟔)(𝟖𝟖.𝟒𝟒) = 𝟓𝟓𝟒𝟒.𝟗𝟗𝟒𝟒

b. Circle the transformations that, if applied to the first triangle, would always result in a new triangle with thesame area:

Translation Rotation Dilation Reflection







c. Explain your answer to part (b).

Two congruent figures have equal area. If two figures are congruent, it means that there exists a transformation that maps one figure onto the other completely. For this reason, it makes sense that theywould have equal area because the figures would cover the exact same region.

2. a. Calculate the area of the shaded figure below.

𝟒𝟒�𝟏𝟏𝟒𝟒�

(𝟑𝟑)(𝟑𝟑) = 𝟗𝟗

𝟐𝟐(𝟑𝟑) = 𝟒𝟒𝟏𝟏

The area of the figure is 𝟗𝟗+ 𝟒𝟒𝟏𝟏 = 𝟑𝟑𝟎𝟎.

b. Explain how you determined the area of the figure.

First, I realized that the two shapes at the ends of the figure were triangles with a base of 𝟑𝟑 and a height of 𝟑𝟑 and the shape in the middle was a rectangle with dimensions 𝟑𝟑 × 𝟐𝟐. To find the area of the shaded figure, I found the sum of all three shapes.

3. Two triangles and are shown below. The two triangles overlap forming .

a. The base of figure is comprised of segments of the following lengths: = 𝟒𝟒, = 𝟑𝟑, and = 𝟒𝟒.Calculate the area of the figure .

The area of : 𝟏𝟏𝟒𝟒 (𝟒𝟒)(𝟐𝟐) = 𝟏𝟏𝟒𝟒

The area of : 𝟏𝟏𝟒𝟒 (𝟒𝟒)(𝟓𝟓) = 𝟓𝟓

The area of : 𝟏𝟏𝟒𝟒 (𝟎𝟎.𝟗𝟗)(𝟑𝟑) = 𝟏𝟏.𝟑𝟑𝟓𝟓

The area of figure is 𝟏𝟏𝟒𝟒 + 𝟓𝟓 − 𝟏𝟏.𝟑𝟑𝟓𝟓 = 𝟏𝟏𝟐𝟐.𝟔𝟔𝟓𝟓.

MP.7







b. Explain how you determined the area of the figure.

Since the area of is counted twice, it needs to be subtracted from the sum of the overlapping triangles.

4. A rectangle with dimensions 𝟒𝟒𝟏𝟏.𝟔𝟔× 𝟏𝟏𝟒𝟒 has a right triangle with a base 𝟗𝟗.𝟔𝟔 and a height of 𝟐𝟐.𝟒𝟒 cut out of therectangle.

a. Find the area of the shaded region.

The area of the rectangle: (𝟏𝟏𝟒𝟒)(𝟒𝟒𝟏𝟏.𝟔𝟔) = 𝟒𝟒𝟓𝟓𝟗𝟗.𝟒𝟒

The area of the triangle: 𝟏𝟏𝟒𝟒 (𝟐𝟐.𝟒𝟒)(𝟗𝟗.𝟔𝟔) = 𝟑𝟑𝟒𝟒.𝟓𝟓𝟔𝟔

The area of the shaded region is 𝟒𝟒𝟓𝟓𝟗𝟗.𝟒𝟒 − 𝟑𝟑𝟒𝟒.𝟓𝟓𝟔𝟔 = 𝟒𝟒𝟒𝟒𝟒𝟒.𝟔𝟔𝟒𝟒.

b. Explain how you determined the area of the shaded region.

I subtracted the area of the triangle from the area of the rectangle to determine the shaded region.


The Discussion formalizes the properties of area that students have been studying since Grade 3, debriefs the exercises in the Exploratory Challenge, and introduces set notation appropriate for discussing area. Have the properties displayed in a central location as you refer to each one; and have students complete a foldable organizer on an 8.5 × 11 in. sheet of paper like the example here, where they record each property, and new set notation and definitions, as it is discussed.

� (Property 1) We describe area of a set in the plane as a number, greaterthan or equal to zero, that measures the size of the set and not the shape.How would you describe area in your own words?

Prompt students to think back to Lesson 1, where this question was asked and reviewed as part of the Discussion.

à Area is a way of quantifying the size of a region without any reference to the shape of the region.� (First half of Property 2) The area of a rectangle is given by the formula length × width. The area of a triangle

is given by the formula 12

× base × height. How can we use Exercise 1 to support this?

à The two congruent right triangles can be fitted together to form a rectangle with dimensions12.6 × 8.4. These dimensions are the length and width, or base and height, of the rectangle.

à Since the two right triangles are congruent, each must have half the area of the entire rectangle or anarea described by the formula 1

2× base × height.

MP.7







Figure 1 Figure 2

� (Property 3) Notice that the congruent triangles in the Exercise 1 each have the same area as the other.� (Second half of Property 2) We describe a polygonal region as the union of finitely many non-overlapping

triangular regions. The area of the polygonal region is the sum of the areas of the triangles. Let’s see what thisproperty has to do with Exercise 2.

� Explain how you calculated the area for the figure in Exercise 2.

Select students to share their strategies for calculating the area of the figure. Sample responses are noted in the exercise. Most students will likely find the sum of the area of the rectangle and the area of the two triangles.

� Would your answer be any different if we divided the rectangle into two congruent triangles?

Show the figure below. Provide time for students to check via calculation or discuss in pairs.

à No, the area of the figure is the same whether we consider the middle portion of the figure as arectangle or two congruent triangles.

� Triangular regions overlap if there is a point that lies in the interior of each. One way to find the area of such aregion is to split it into non-overlapping triangular regions and add the areas of the resulting triangular regionsas shown below. Figure 2 is the same region as Figure 1 but is one (of many) possible decomposition into non-overlapping triangles.

� This mode of determining area can be done for any polygonal region.

Provide time for students to informally verify this fact by drawing quadrilaterals and pentagons (ones that are not regular) and showing that each is the union of triangles.

� (Property 4) The area of the union of two regions is the sum of the areas minus the area of the intersection.Exercise 3 can be used to break down this property, particularly the terms union and intersection. How wouldyou describe these terms?

Take several responses from students; some may explain their calculation by identifying the overlapping region of the two triangles as the intersection and the union as the region defined by the boundary of the two shapes. Then use the points below to explicitly demonstrate each union and intersection and what they have to do with determining the area of a region.







� If and are regions in the plane, then denotes the union of the two regions; that is, all points that lie in or in , including points that lie in both and .

� Notice that the area of the union of the two regions is not the sum of the areas of the two regions, whichwould count the overlapping portion of the figure twice.

� If and are regions in the plane, then denotes the intersection of thetwo regions; that is, all points that lie in both and .

� (Property 4) Use this notation to show that the area of the union of two regions is the sum of the areas minusthe area of the intersection.à Area( ) = Area( ) + Area( ) − Area( )

Discuss Property 4 in terms of two regions that coincide at a vertex or an edge. This elicits the idea that the area of a segment must be 0, since there is nothing to subtract when regions overlap at a vertex or an edge.

Here is an optional proof:

Two squares 1 and 2 meet along a common edge , or 1 2 = .

Since 1 2 is a 2𝑠𝑠 × 𝑠𝑠 rectangle, its area is 2𝑠𝑠2. The area of each of the squares is 𝑠𝑠2.

Since

Area( 1 2) = Area( 1) + Area( 2) − Area( 1 2)

we get

2𝑠𝑠2 = 𝑠𝑠2 + 𝑠𝑠2 − Area( ).

Solving for Area( ) shows that Area( ) = 0.

Scaffolding: � As notation is introduced,

have students make achart that includes thename, symbol, and simpledrawing that representseach term.

� Consider posting a chart inthe classroom for studentsto reference throughoutthe module.







It is further worth mentioning that when we discuss the area of a triangle or rectangle (or mention the area of any polygon as such), we really mean the area of the triangular region because the area of the triangle itself is zero since it is the area of three line segments, each being zero.

� (Property 5) The area of the difference of two regions where one is contained in the other is the difference ofthe areas. In which exercise was one area contained in the other?à Exercise 4

� If is contained in , or in other words is a subset of , denoted as , it means that all of the points in are also points in .

� What does Property 5 have to do with Exercise 4?à In Exercise 4, all the points of the triangle are also points in the rectangle. That is why the area of the

shaded region is the difference of the areas.� (Property 5) Use set notation to state Property 5: The area of the difference of two regions where one is

contained in the other is the difference of the areas.à When , then Area( − ) = Area( ) − Area( ) is the difference of the areas.







Closing (5 minutes)

Have students review the properties in partners. Ask them to paraphrase each one and to review the relevant set notation and possibly create a simple drawing that helps illustrate each property. Select students to share their paraphrased versions of the properties with the whole class.

1. Students understand that the area of a set in the plane is a number, greater than or equal to zero, thatmeasures the size of the set and not the shape.

2. The area of a rectangle is given by the formula length × width. The area of a triangle is given by theformula 1

2× base × height. A polygonal region is the union of finitely many non-overlapping triangular

regions and has area the sum of the areas of the triangles. 3. Congruent regions have the same area.4. The area of the union of two regions is the sum of the areas minus the area of the intersection.5. The area of the difference of two regions where one is contained in the other is the difference of the areas.

Exit Ticket (5 minutes)

Lesson Summary

SET (description): A set is a well-defined collection of objects. These objects are called elements or members of the set.

SUBSET: A set is a subset of a set if every element of is also an element of . The notation indicates that the set is a subset of set .

UNION: The union of and is the set of all objects that are either elements of or of , or of both. The union is denoted .

INTERSECTION: The intersection of and is the set of all objects that are elements of and also elements of . The intersection is denoted .






NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 1 ALGEBRA II

Lesson 1: Successive Differences in Polynomials Date: 6/5/14


Lesson 1: Successive Differences in Polynomials

Student Outcomes � Students write explicit polynomial expressions for sequences by investigating successive differences of those

sequences.

Lesson Notes This first lesson of the year tells students that this course is about thinking and reasoning with mathematics. It reintroduces the study of polynomials in a surprising new way involving sequences. This gives you a chance to evaluate how much your students recall from Grade 9. The lesson starts with discussions of expressions, polynomials, sequences, and equations. In this lesson, students continue the theme that began in Grade 6 of evaluating and building expressions. Explore ways to test your students’ recall of the vocabulary terms listed at the end of this lesson.

Throughout this lesson, listen carefully to your students’ discussions. Their reactions will inform you how to best approach the rest of the module. The homework set to this lesson should also give you insight into how much they remember from previous grades and how well they can read instructions. In particular, if they have trouble with evaluating or simplifying expressions or solving equations, then you might want to revisit Lessons 6–9 in Grade 9, Module 1, and Lesson 2 in Grade 9, Module 4. If they are having trouble solving equations, use Lessons 10–12, 15–16, and 19 in Grade 9, Module 1 to give them extra practice.

Finally, the use of the term constant may need a bit of extra discussion. It is used throughout this PK–12 curriculum in two ways: either as a constant number (e.g., the 𝑎 in 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is a number chosen once-and-for-all at the beginning of a problem) or as a constant rate (e.g., a copier that reproduces at a constant rate of 40 copies/minute). We offer both uses in this lesson.

Classwork Opening Exercise (7 minutes)

This exercise provides an opportunity to think about and generalize the main concept of today’s lesson: that the second differences of a quadratic polynomial are constant. This generalizes to the 𝑛th differences of a degree 𝑛 polynomial. Your goal is to help students investigate, discuss, and generalize the second and higher differences in this exercise.

Present the exercise to your students and ask them (in groups of two) to study the table and explain to their partner how to calculate each line in the table. If they get stuck, help them find entry points into this question, possibly by drawing segments connecting the successive differences on their papers (e.g., connect 5.76 and 11.56 to 5.8 and ask, “How are these three numbers related?”). This initial problem of the school year is designed to encourage students to persevere and look for and express regularity in repeated reasoning.

Teachers may also use the Opening Exercise to informally assess students’ pattern-finding abilities and fluency with rational numbers.

MP.1 &

MP.8

Scaffolding:

Before presenting the problem below, you might start by displaying the first two rows of the table on the board and asking students to investigate the relationship between them, including making a conjecture about the nature of the relationship.







Opening Exercise

John noticed patterns in the arrangement of numbers in the table below.

Number 𝟐. 𝟒 𝟑. 𝟒 𝟒. 𝟒 𝟓. 𝟒 𝟔. 𝟒

Square 𝟓. 𝟕𝟔 𝟏𝟏. 𝟓𝟔 𝟏𝟗. 𝟑𝟔 𝟐𝟗. 𝟏𝟔 𝟒𝟎. 𝟗𝟔

First Differences 𝟓. 𝟖 𝟕. 𝟖 𝟗. 𝟖 𝟏𝟏. 𝟖

Second Differences 𝟐 𝟐 𝟐

Assuming that the pattern would continue, he used it to find the value of 𝟕. 𝟒𝟐. Explain how he used the pattern to find 𝟕. 𝟒𝟐, and then use the pattern to find 𝟖. 𝟒𝟐.

To find 𝟕. 𝟒𝟐, John assumed the next term in the first differences would have to be 𝟏𝟑. 𝟖 since 𝟏𝟑. 𝟖 is 𝟐 more than 𝟏𝟏. 𝟖. Therefore, the next term in the square numbers would have to be 𝟒𝟎. 𝟗𝟔 + 𝟏𝟑. 𝟖, which is 𝟓𝟒. 𝟕𝟔. Checking with a calculator, we also find 𝟕. 𝟒𝟐 = 𝟓𝟒. 𝟕𝟔.

To find 𝟖. 𝟒𝟐, we follow the same process: The next term in the first differences would have to be 𝟏𝟓. 𝟖, so the next term in the square numbers would be 𝟓𝟒. 𝟕𝟔 + 𝟏𝟓. 𝟖, which is 𝟕𝟎. 𝟓𝟔. Check: 𝟖. 𝟒𝟐 = 𝟕𝟎. 𝟓𝟔.

How would you label each row of numbers in the table?

Number, Square, First Differences, Second Differences

Discuss with students the relationship between each row and the row above it and how to label the rows based upon that relationship. Feel free to have this discussion before or after they find 7.42 and 8.42. They are likely to come up with labels such as “subtract” or “difference” for the third and fourth row. However, guide them to call the third and fourth rows “First Differences” and “Second Differences,” respectively.


The pattern illustrated in the Opening Exercise is a particular case of a general phenomenon about polynomials. In Algebra I, Module 3, we saw how to recognize linear functions and exponential functions by recognizing similar growth patterns; that is, linear functions grow by a “constant difference over successive intervals of equal length,” and exponential functions grow by a “constant factor over successive intervals of equal length.” In this lesson, we generalize the linear growth pattern to polynomials of second degree (quadratic expressions) and third degree (cubic expressions).

Discussion

Let the sequence {𝒂𝟎, 𝒂𝟏, 𝒂𝟐, 𝒂𝟑, … } be generated by evaluating a polynomial expression at the values 𝟎, 𝟏, 𝟐, 𝟑,…. The numbers found by evaluating 𝒂𝟏 − 𝒂𝟎, 𝒂𝟐 − 𝒂𝟏, 𝒂𝟑 − 𝒂𝟐,… form a new sequence which we will call the first differences of the polynomial. The differences between successive terms of the first differences sequence are called the second differences and so on.

It is a good idea to use an actual sequence of numbers such as the square numbers {1, 4, 9, 16, … } to help explain the meaning of the terms “first differences” and “second differences.”

MP.8

Scaffolding:

For students working below grade level, consider using positive integers {1, 2, 3, … } and corresponding squares {1, 4, 9, . . . } instead of using {2.4, 3.4, 4.4, … }.








Although you may be tempted to work through Example 1 using numbers instead of 𝑎 and 𝑏, using symbols 𝑎 and 𝑏 actually makes the structure of the first differences sequence obvious, whereas numbers could hide that structure. Also, working with constant coefficients gives the generalization all at once.

Note: Consider using Example 1 to informally assess students’ fluency with algebraic manipulations.

Example 1

What is the sequence of first differences for the linear polynomial given by 𝒂𝒙 + 𝒃, where 𝒂 and 𝒃 are constant coefficients?

The terms of the first differences sequence are found by subtracting consecutive terms in the sequence generated by the polynomial expression 𝒂𝒙 + 𝒃, namely, {𝒃, 𝒂 + 𝒃, 𝟐𝒂 + 𝒃, 𝟑𝒂 +𝒃, 𝟒𝒂 + 𝒃, … }.

1st term: (𝒂 + 𝒃) − 𝒃 = 𝒂,

2nd term: (𝟐𝒂 + 𝒃) − (𝒂 + 𝒃) = 𝒂,

3rd term: (𝟑𝒂 + 𝒃) − (𝟐𝒂 + 𝒃) = 𝒂,

4th term: (𝟒𝒂 + 𝒃) − (𝟑𝒂 + 𝒃) = 𝒂.

The first differences sequence is {𝒂, 𝒂, 𝒂, 𝒂, … }. For first degree polynomial expressions, the first differences are constant and equal to 𝒂.

What is the sequence of second differences for 𝒂𝒙 + 𝒃?

Since 𝒂 − 𝒂 = 𝟎, the second differences are all 𝟎. Thus the sequence of second differences is {𝟎, 𝟎, 𝟎, 𝟎, … }.

� How is this calculation similar to the arithmetic sequences you studied in Algebra I, Module 3?à The constant derived from the first differences of a linear polynomial is the same constant addend used

to define the arithmetic sequence generated by the polynomial. That is, the 𝑎 in 𝐴(𝑛) = 𝑎𝑛 + 𝑏 for

𝑛 ≥ 0. Written recursively this is 𝐴(0) = 𝑏 and 𝐴(𝑛 + 1) = 𝐴(𝑛) + 𝑎 for 𝑛 ≥ 0.

For Examples 2 and 3, let students work in groups of two to fill in the blanks of the tables (3 min. max for each table). Walk around the room, checking student work for understanding. Afterward, discuss the paragraphs below each table as a whole class.

Scaffolding:

Try starting the example by first asking students to generate sequences of first differences for 2𝑥 + 3, 3𝑥 − 1, and 4𝑥 + 2. For example, the sequence generated by 2𝑥 + 3 is {3, 5, 7, 9, … } and its sequence of first differences is {2, 2, 2, 2 … }.

You can then use these three sequences as a source of examples from which to make and verify conjectures.








Example 2

Find the first, second, and third differences of the polynomial 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 by filling in the blanks in the following table.

𝒙 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 First Differences Second Differences Third Differences 𝟎 𝒄

𝒂 + 𝒃 𝟏 𝒂 + 𝒃 + 𝒄 𝟐𝒂

𝟑𝒂 + 𝒃 𝟎 𝟐 𝟒𝒂 + 𝟐𝒃 + 𝒄 𝟐𝒂

𝟓𝒂 + 𝒃 𝟎 𝟑 𝟗𝒂 + 𝟑𝒃 + 𝒄 𝟐𝒂

𝟕𝒂 + 𝒃 𝟎 𝟒 𝟏𝟔𝒂 + 𝟒𝒃 + 𝒄 𝟐𝒂

𝟗𝒂 + 𝒃 𝟓 𝟐𝟓𝒂 + 𝟓𝒃 + 𝒄

The table shows that the second differences of the polynomial 𝑎𝑥2 + 𝑏𝑥 + 𝑐 all have the constant value 2𝑎. The second differences hold for any sequence of values of 𝑥 where the values in the sequence differ by 1, as the Opening Exercise shows. For example, if we studied the second differences for 𝑥-values 𝜋, 𝜋 + 1, 𝜋 + 2, 𝜋 + 3,…. , we would find that the second differences would also be 2𝑎. In your homework, you will show that this fact is indeed true by finding the second differences for the values 𝑛 + 0, 𝑛 + 1, 𝑛 + 2, 𝑛 + 3, 𝑛 + 4.

Ask students to describe what they notice in the sequences of first, second, and third differences. Have them make a conjecture about the third and fourth differences of a sequence generated by a third degree polynomial.

Students are likely to say that the third differences have the constant value 3𝑎 (which is incorrect). Have them work through the next example to help them discover what the third differences really are. This is a good example of why we need to follow up conjecture based on observation with proof.


Example 3

Find the second, third, and fourth differences of the polynomial 𝒂𝒙𝟑 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅 by filling in the blanks in the following table.

𝒙 𝒂𝒙𝟑 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅 First Differences Second Differences Third Differences Fourth Differences 𝟎 𝒅

𝒂 + 𝒃 + 𝒄 𝟏 𝒂 + 𝒃 + 𝒄 + 𝒅 𝟔𝒂 + 𝟐𝒃

𝟕𝒂 + 𝟑𝒃 + 𝒄 𝟔𝒂 𝟐 𝟖𝒂 + 𝟒𝒃 + 𝟐𝒄 + 𝒅 𝟏𝟐𝒂 + 𝟐𝒃 𝟎

𝟏𝟗𝒂 + 𝟓𝒃 + 𝒄 𝟔𝒂 𝟑 𝟐𝟕𝒂 + 𝟗𝒃 + 𝟑𝒄 + 𝒅 𝟏𝟖𝒂 + 𝟐𝒃 𝟎

𝟑𝟕𝒂 + 𝟕𝒃 + 𝒄 𝟔𝒂 𝟒 𝟔𝟒𝒂 + 𝟏𝟔𝒃 + 𝟒𝒄 + 𝒅 𝟐𝟒𝒂 + 𝟐𝒃

𝟔𝟏𝒂 + 𝟗𝒃 + 𝒄 𝟓 𝟏𝟐𝟓𝒂 + 𝟐𝟓𝒃 + 𝟓𝒄 + 𝒅

MP.7 &

MP.8







The third differences of 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑 all have the constant value 6𝑎. Also, if a different sequence of values for 𝑥 that differed by 1 was used instead, the third differences would still have the value 6𝑎.

� Ask students to make a conjecture about the fourth differences of a sequence generated by degree 4polynomial. Students who were paying attention to their (likely wrong) conjecture of the third differencesbefore doing this example may guess that the fourth differences are constant and equal to (1 ⋅ 2 ⋅ 3 ⋅ 4)𝑎,which is 24𝑎. This pattern continues: the 𝑛th differences of any sequence generated by an 𝑛th degreepolynomial with leading coefficient 𝑎 will be constant and have the value 𝑎 ∙ (𝑛!).

� Ask students to make a conjecture about the (𝑛 + 1)st differences of a degree 𝑛 polynomial, for example, the5th differences of a fourth degree polynomial.

Students are now ready to tackle the main goal of this lesson—using differences to recognize polynomial relationships and build polynomial expressions.


When collecting bivariate data on an event or experiment, the data does not announce, “I satisfy a quadratic relationship,” or “I satisfy an exponential relationship.” We need ways to recognize these relationships in order to model them with functions. In Algebra I, Module 3, students studied the conditions upon which they could conclude that the data satisfied a linear or exponential relationship. Either the first differences were constant or “first factors” were constant. By checking that the second or third differences of the data are constant, students now have a way to recognize a quadratic or cubic relationship and can write an equation to describe that relationship (A-CED.A.3, F-BF.A.1a).

Give students an opportunity to attempt this problem in groups of two. Walk around the room helping them find the leading coefficient.

Example 4

What type of relationship does the set of ordered pairs (𝒙, 𝒚) satisfy? How do you know? Fill in the blanks in the table below to help you decide. (The first differences have already been computed for you.)

𝒙 𝒚 First Differences Second Differences Third Differences 𝟎 𝟐

−𝟏 𝟏 𝟏 𝟔

𝟓 𝟔 𝟐 𝟔 𝟏𝟐

𝟏𝟕 𝟔 𝟑 𝟐𝟑 𝟏𝟖

𝟑𝟓 𝟔 𝟒 𝟓𝟖 𝟐𝟒

𝟓𝟗 𝟓 𝟏𝟏𝟕

Since the third differences are constant, the pairs could represent a cubic relationship between 𝒙 and 𝒚.







Find the equation of the form 𝒚 = 𝒂𝒙𝟑 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅 that all ordered pairs (𝒙, 𝒚) above satisfy. Give evidence that your equation is correct.

Since third differences of a cubic polynomial are equal to 𝟔𝒂, using the table above, we get 𝟔𝒂 = 𝟔, so that 𝒂 = 𝟏. Also, since (𝟎, 𝟐) satisfies the equation, we see that 𝒅 = 𝟐. Thus, we need only find 𝒃 and 𝒄. Substituting (𝟏, 𝟏) and (𝟐, 𝟔) into the equation, we get

𝟏 = 𝟏 + 𝒃 + 𝒄 + 𝟐 𝟔 = 𝟖 + 𝟒𝒃 + 𝟐𝒄 + 𝟐.

Subtracting two times the first equation from the second, we get 𝟒 = 𝟔 + 𝟐𝒃 − 𝟐, so that 𝒃 = 𝟎. Substituting 𝟎 in for 𝒃 in the first equation gives 𝒄 = −𝟐. Thus, the equation is 𝒚 = 𝒙𝟑 − 𝟐𝒙 + 𝟐.

� After finding the equation, have students check that the pairs (3, 23) and (4, 58) satisfy the equation.

Help students to persevere in finding the coefficients (MP.1). They will most likely try to plug three ordered pairs into the equation, which gives a 3 × 3 system of linear equations in 𝑎, 𝑏, and 𝑐 after they find that 𝑑 = 2. Using the fact that the third differences of a cubic polynomial are 6𝑎 will greatly simplify the problem. (It implies 𝑎 = 1 immediately, which reduces the system to the easy 2 × 2 system above.) Walk around the room as they work, and ask questions that lead them to realize that they can use the third differences fact if they get too stuck. Alternatively, find a student who used the fact, and then have the class discuss and understand his or her approach.

Closing (7 minutes)

� What are some of the key ideas that we learned today?à Sequences whose second differences are constant satisfy a quadratic relationship.

à Sequences whose third differences are constant satisfy a cubic relationship.

The following terms were introduced and taught in Module 1 of Algebra I. The terms are listed here for completeness and reference.

Relevant Vocabulary

Numerical Symbol: A numerical symbol is a symbol that represents a specific number. Examples: 𝟏, 𝟐, 𝟑, 𝟒, 𝝅, −𝟑. 𝟐.

Variable Symbol: A variable symbol is a symbol that is a placeholder for a number from a specified set of numbers. The set of numbers is called the domain of the variable. Examples: 𝒙, 𝒚, 𝒛.

Algebraic Expression: An algebraic expression is either

1. a numerical symbol or a variable symbol or

2. the result of placing previously generated algebraic expressions into the two blanks of one of the fouroperators ((__)+(__), (__)−(__), (__)×(__), (__)÷(__)) or into the base blank of an exponentiation with an exponent that is a rational number.

Following the definition above, (((𝒙) × (𝒙)) × (𝒙)) + ((𝟑) × (𝒙)) is an algebraic expression, but it is generally written more simply as 𝒙𝟑 + 𝟑𝒙.

Numerical Expression: A numerical expression is an algebraic expression that contains only numerical symbols (no

variable symbols) that evaluates to a single number. Example: The numerical expression (𝟑⋅𝟐)𝟐

𝟏𝟐 evaluates to 𝟑.

Monomial: A monomial is an algebraic expression generated using only the multiplication operator (__×__). The expressions 𝒙𝟑 and 𝟑𝒙 are both monomials.

Binomial: A binomial is the sum of two monomials. The expression 𝒙𝟑 + 𝟑𝒙 is a binomial.







Polynomial Expression: A polynomial expression is a monomial or sum of two or more monomials.

Sequence: A sequence can be thought of as an ordered list of elements. The elements of the list are called the terms of the sequence.

Arithmetic Sequence: A sequence is called arithmetic if there is a real number 𝒅 such that each term in the sequence is the sum of the previous term and 𝒅.

Exit Ticket (5 minutes)




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