€¦ · ALGEBRA THROUGH VISUAL PATTERNS VOLUME 1 Introduction vii LESSON 1 Tile Patterns & Graphing 1 LESSON 2 Positive & Negative Integers 31 LESSON 3 Integer Addition & Subtraction

A Math Learning Center publication adapted and arranged by

EUGENE MAIER and LARRY LINNEN

ALGEBRA THROUGH VISUAL PATTERNS, VOLUME 2

A Math Learning Center Resource

Copyright © 2005, 2004 by The Math Learning Center, PO Box 12929,

Salem, Oregon 97309. Tel. 503 370–8130. All rights reserved.

QP388 P0405

The Math Learning Center is a nonprofiit organization serving the

education community. Our mission is to inspire and enable individuals to

discover and develop their mathematical confidence and ability. We offer

innovative and standards-based professional development, curriculum,

materials, and resources to support learning and teaching. To find out more

visit us at www.mathlearningcenter.org.

The Math Learning Center grants permission to classroom teachers to

reproduce blackline masters in appropriate quantities for their classroom use.

This project was supported, in part, by the National ScienceFoundation. Opinions expressed are those of the authors and notnecessarily those of the Foundation.

Prepared for publication on Macintosh Desktop Publishing system.

Printed in the United States of America.

ISBN 1-886131-60-0

ALGEBRA THROUGH VISUAL PATTERNS

VOLUME 1

Introduction vii

LESSON 1 Tile Patterns & Graphing 1

LESSON 2 Positive & Negative Integers 31

LESSON 3 Integer Addition & Subtraction 47

LESSON 4 Integer Multiplication & Division 57

LESSON 5 Counting Piece Patterns & Graphs 73

LESSON 6 Modeling Algebraic Expressions 91

LESSON 7 Seeing & Solving Equations 113

LESSON 8 Extended Counting Piece Patterns 135

LESSON 9 Squares & Square Roots 163

LESSON 10 Linear & Quadratic Equations 185

LESSON 11 Complete Sequences 217

LESSON 12 Sketching Solutions 251

LESSON 13 Analyzing Graphs 281

LESSON 14 Complex Numbers 315

Appendix 333

VOLUME 2

START-UP FOCUS FOLLOW-UP

ALGEBRA THROUGH VISUAL PATTERNS | 163

THE BIG IDEASquare roots are viewed as the lengths of sides ofsquares. Methods of constructing a square of any givenintegral area, and thus the square root of any positiveinteger, are developed. One of these constructionsleads to the Pythagorean Theorem.

OverviewStudents construct squares ofintegral areas and establish therelationship between squaresand square roots.

Materials� Centimeter grid paper (see

Appendix), 2-3 sheets perstudent, 1 transparency.

� Start-Up Master 9.1,1 transparency.

OverviewStudents dissect squares andreassemble the pieces to formtwo squares and, conversely,dissect two squares andreassemble the pieces to forma single square. In the process,they arrive at the PythagoreanTheorem. They dissect rect-angles and reassemble thepieces to form squares and, inso doing, construct squareroots. Students examine therelationship between products(quotients, sums) of squareroots and square roots ofproducts (quotients, sums).

Materials� Centimeter grid paper (see

Appendix), 2-3 sheets perstudent

� Scissors, 1 pair per student

� Start-Up Master 9.1, 1 copyper student and 1 transpar-ency

� Focus Masters 9.1-9.2,1 copy of each per student.

� Focus Master 9.3, 1 copy perstudent and 1 transparency.

OverviewStudents solve problemsinvolving squares and squareroots, using the PythagoreanTheorem as necessary. Theyrelate the arithmetic mean andthe geometric mean of twopositive numbers to the con-struction of squares.

Materials� Follow-Up 9, 1 copy per

student.

LESSON 9SQUARES & SQUARE ROOTS

164 | ALGEBRA THROUGH VISUAL PATTERNS

TEACHER NOTES


START-UP

Overview Materials

SQUARES AND SQUARE ROOTS LESSON 9

Students construct squares of inte-gral areas and establish the relation-ship between squares and squareroots.

� Centimeter grid paper(see Appendix), 2 to 3sheets per student, 1transparency.

1 Distribute centimeter grid paperto the students. Tell them that1 square represents 1 unit of area.For each of the integers 1 through25 ask them to construct, if possible,a square whose vertices are gridintersection points and whose areais the given integer. For each squarethey draw, ask the students to indi-cate its area and the length of itsside. Discuss.

1 If your students are familiar with the basic properties ofsquare roots, you may wish to omit this lesson.

A student may believe they are finished when they have con-structed all the squares whose sides lie along a gridline. If thishappens, you can simply tell the student there are more. Nor-mally, someone in the class will discover a square that “tilts.”

Of the integers 1 through 25, there are 13 for which a squareexists that satisfies the conditions of Action 1. Start-Up Master9.1 attached at the end of this activity shows a square of eacharea. A square of area 25 can also be obtained by carrying out a3,4 pattern as described below.

One way to obtain a square that fits the conditions is to pick twointersection points as successive vertices. In the instance shownbelow, one can get from point P to point Q by going 3 units inone direction and 1 in the other. Repeating this 3,1 pattern, asshown, results in a square.

Square generated by a 3,1 pattern.


ACTIONS COMMENTS

1

1

1

1

3

3

3

3

P

Q

continued next page


START-UP BLACKLINE MASTER 9.1

1 1

2

2

4 25

89 3

10

16 41713

2025

185

5

10

17

20

8

13

18

ACTIONS COMMENTS


START-UPSQUARES AND SQUARE ROOTS LESSON 9

2 Discuss with the students whythey can be certain that the 13integers mentioned in Comment 1are the only ones in the range 1through 25 for which squares existthat satisfy the conditions of Action 1.

1 continuedThe area of this square can be found by subtracting the area of theshaded regions from the area of the circumscribed square (see thefigure). Note that regions A and C combine to form a rectangle ofarea 3 as do rectangles B and D. Thus, the area of the inscribedsquare is 16 – 6, or 10. Since the area of the square is 10, thelength of its side is 10. An approximation of 10 can be obtainedby measuring the side of the square with a centimeter ruler.

If n is non-negative, the positive square root of n, written n, is thelength of the side of a square of area n.

2 One way to see there are only 13 different areas is to notethat if a square is to have area no greater than 25, then thedistance between successive vertices must be less than or equalto 5. Thus, if P and Q are successive vertices, Q must lie on orwithin a circle of radius 5 whose center is at P. In the sketch, the13 intersections marked with an x are possibilities for Q that leadto 13 differently sized squares. Any other choice for Q leads to asquare the same size as one of these 13.

A

D C

B

x x xx

xx

xx

xx

xx

xP

ACTIONS COMMENTS


START-UPSQUARES AND SQUARE ROOTS LESSON 9

20 or

2 5

5

5

5

5

5

5

3 Point out to the students that 20 = 2 5. Ask the students toexamine the squares they haveconstructed for other relationshipsof this type.

3 The square of area 20 is composed of 4 squares of area 5, asshown in the sketch. Hence, the side of a square of area 20 is twicethe length of the side of a square of area 5. Thus, 20 = 2 5.

The square of area 8 is composed of 4 squares of area 2 and thesquare of area 18 is composed of 9 squares of area 2. Thus, 8 = 2 2 and 18 = 3 2.

4 The areas of the squares are, respectively, 29, 32, 37, and 52.Square b) can be divided into squares of area 8 or squares of area2 to show that 32 = 2 8 = 4 2. Also square d) forms 4 squaresof area 13, giving 52 = 2 13.

Some students may notice that the area of a square is equal tothe sum of the squares of the numbers in the pattern that form it,e.g., 29 = 52 + 22. If so, tell them to keep that observation in mindas the lesson continues.

4 Ask the students to constructsquares, and find their areas andside lengths, using the followingpatterns:

a) 5,2

b) 4,4

c) 6,1

d) 4,6

Ask the students to examine theirsquares for relationships betweensquare roots like those discussed inAction 3.

168 | ALGEBRA THROUGH VISUAL PATTERNS © THE MATH LEARNING CENTER



1 1

2

2

4 25

89 3

10

16 41713

2025

185

5

10

17

20

8

13

18


FOCUS

Overview Materials


Students dissect squares and reas-semble the pieces to form twosquares and, conversely, dissect twosquares and reassemble the piecesto form a single square. In the pro-cess, they arrive at the PythagoreanTheorem. They dissect rectanglesand reassemble the pieces to formsquares and, in so doing, constructsquare roots. Students examine therelationship between products(quotients, sums) of square rootsand square roots of products (quo-tients, sums).

� Centimeter grid paper(see Appendix), 2 or 3sheets per student

� Scissors, 1 pair perstudent

� Start-Up Master 9.1,1 copy per student and1 transparency

1 Give each student a sheet ofcentimeter grid paper and a pair ofscissors. Ask them to construct asquare using a 4, 2 pattern. Havethem divide their square into 3parts as shown below. Then ask thestudents to cut out these 3 partsand reassemble them to form 2adjacent squares, and determine thelength of the sides of these squares.

1 Rotating the right triangles around a vertex, as shown, trans-forms the original square into 2 squares. The lengths of the sidesof the squares are 2 and 4. Notice that these are the lengths ofthe legs of the rotated right triangles.

� Focus Masters 9.1-9.2,1 copy of each perstudent.

� Focus Master 9.3, 1 copyper student and 1 trans-parency.

ACTIONS COMMENTS

ACTIONS COMMENTS


FOCUSSQUARES AND SQUARE ROOTS LESSON 9

2 Ask the students to construct asquare using a 6,3 pattern and thenuse the method of Action 1 totransform it into a 3 x 3 and a 6 x 6square.

3 Distribute a copy of Focus Mas-ter 9.1 to each student. Have thestudents cut out the square andtriangle. Then have them dissect thesquare and reassemble it into 2squares whose sides have length aand b. Discuss.

2 You can ask for volunteers to describe how they proceeded.

3 The students can cut out the triangle and use it to divide thesquare into the following regions, which can be cut out andreassembled into the desired squares.

Notice that the area of the large square is c2 and the areas of the2 smaller squares are a2 and b2. Thus, c2 = a2 + b2. Since c is thehypotenuse of the given right triangle and a and b are the legs ofthe triangle, we have shown that the square of the hypotenuse ofa right triangle is the sum of the square of its legs. This result isknown as the Pythagorean Theorem.

b

b

a a


FOCUS BLACKLINE MASTER 9.1

c2

c a

b

c

ACTIONS COMMENTS



4 a) The distance d is the hypotenuse of a right triangle whoselegs are 3 and 7. (See the figure.) Thus, d2 = 32 + 72 = 58. Henced = 58 ≈ 7.62.

a)

c)

5 This is the converse of the dissection done in Action 1. It canbe accomplished by locating the point P on the long side of thefigure, and then cutting and reassembling as shown.

4 Ask the students to use thePythagorean Theorem to find thefollowing distances on a coordinategraph:

a) The distance d from the origin tothe point (3,7).

b) The distance d between the points(2,10) and (5,4).

c) The distance d between the points(–3,–6) and (7,–2).

5 Distribute a copy of Focus Mas-ter 9.2 to each student. Ask them todissect the two squares shownbelow so they can be reassembledinto a single square. Ask for volun-teers to show how they dissectedthe squares.

d2 = 32 + 62 = 45; d = 45 ≈ 6.71

b)

–2

8

7

6

5

4

3

2

1

–1 1 2 3 40

d

(3,7)

11

10

9

8

7

6

5

4

3

2

1

1 2 3 4 5 6

d

(2,10)

0

6

3

(5,4) d

(7,–2)

10

4

(–3,–6)

–2

1

–1

–2

–3

–4

–5

–6

–7

–3 –1 1 2 3 4–4 5 6 7 8

cut

cut

a

a

b

b

P

d2 = 102 + 42 = 116; d = 116 ≈ 10.77

ACTIONS COMMENTS



6 Place a transparency of Start-UpMaster 9.1 on the overhead. Discusswith the students how squares notappearing on the transparency canbe constructed using the method ofAction 5. Then distribute a copy ofStart-Up Master 9.1 to each student.Ask them to dissect and reassemblesquares from Start-Up Master 9.1 toform squares of area 11 and 21.

7 Tell the students that squares canalso be constructed by dissectingand reassembling rectangles, forexample, a square of area 21 can beconstructed from a 3 x 7 rectangle.To begin an investigation of howsquare can be dissected and reas-sembled into squares, distributecentimeter grid paper to the stu-dents, ask them to dissect a 9 x 16rectangle so that it can be reas-sembled into a square. Ask them todo this by making as few cuts aspossible.

6 Figure 1 shows a square of area 11 constructed from squaresof areas 2 and 9. The students may find it helpful to tape thesquares together in the position shown before dissecting them.

Figure 2 shows a square of area 21 constructed from squares ofarea 5 and 16. A square of area 21 can also be constructed fromsquares of areas 1 and 20 or areas 4 and 17 or areas 8 and 13.

Note that all the squares of area less than 25 that do not appearon Start-Up Master 9.1 can be constructed by this method.

7 Transforming a 9 x 16 rectangle into a square is simpler thantransforming a 3 x 7 rectangle into a square since, in the formercase, the side of the square is integral.

The students will recognize that they need to form a 12 x 12square and they most likely will proceed by cutting the 9 x 16rectangle along grid lines. By doing this, a square can be obtainedby making 3 cuts as shown below.

Figure 1 Figure 2

a

b2

9b

a

9

2

11

5

16b

a 5

16

b

a

21

cut 1

cut 2

cut 3

ACTIONS COMMENTS



However, only 2 cuts are required:

The students are not likely to find this dissection. One way toproceed is to tell the students that you can get a square bymaking only 2 cuts and show them the first cut in silhouette onthe overhead. (See the figure below.) Since the students know thelength of the side of the square is 12, they have a clue to thelocation of this diagonal cut.

cut 1

cut 2

ACTIONS COMMENTS



8 Distribute a copy of Focus Mas-ter 9.3 to each student. Tell thestudents to cut off the bottomportion of the page and set it asidefor use in the next Action. Then askthe students to cut out one of therectangles in the top portion ofFocus Master 9.3 and dissect itusing the “diagonal cut” method ofAction 7 and reassemble the result-ing pieces to form a square. Havethem measure their resulting figureto determine whether or not it is asquare. If not, ask them to dissect asecond rectangle so that the result-ing figure is more “squarelike.”

9 Using a transparency of thebottom half of Focus Master 9.3 onthe overhead, discuss with thestudents how cut lines can be deter-mined so that the resulting piecescan be arranged to form a square.Then ask the students to dissect therectangle on the bottom half ofFocus Master 9.3 and reassemble itto form a square.

8 For the dissection shown below, the reassembled pieces donot form a square.

The sides of the constructed rectangle can be compared bymoving the top piece of the rectangle to the position shownbelow. A more “squarelike” figure would be obtained by lengthen-ing distance AE somewhat.

9 Note that the dissection in Action 8 would have resulted in asquare if, in the last figure in Comment 8, vertices A and C coincided:

A D

CB

E

A

D

C

E

A

E

B

D

C

A B

D

C

E



A D

B C

CUT

ACTIONS COMMENTS



If dotted lines are added to the previous figure to show theoriginal location of the pieces, the figure becomes:

Notice that the sum of the two angles at the top of thefigure is a right angle and segments AB and CD have thesame length, since they were originally opposite sides ofa rectangle. These observations point to the followingprocedure, illustrated below for determining cut lines.

Beginning with rectangle ABCD, the cut lines can bedetermined as follows.

1. Locate point P on an extension of BC at a distance hfrom C. (See Figure 1.)

2. Extend side CD. (See Figure 1.)

3. Place a sheet of paper with a square corner so thatthe corner is on the extension of CD and the edges ofthe paper go through points B and P. (See Figure 2.)

4. Let E be the intersection of the edge of the paperand segment AD. The first cut line is the segment BE. Thedistance d from A to E is the side of the desired square.(See Figure 2.)

5. Using a square corner of a sheet of paper as a guide,starting from a point F on BC at a distance d from C,draw the perpendicular from F to the first cut line. Thisis the second cut line. (See Figure 3.)

AB

D

C

A

B

A D

CB

h

P

h

A

B P

sheet ofpaper

Ed

CB

F

sheet ofpaper

d

first cutline

second cutline

Figure 1

Figure 2

Figure 3

ACTIONS COMMENTS



10 Using the procedure described in Comment 6, a 3 x 7rectangle can be dissected and reassembled as shown below toobtain a square of area 21. Note that the dimension of the squareis the distance between vertex C of the rectangle and squarecorner S. Hence, the distance CS is 21.

11 If the students start with a 1 by 7 rectangle, an additionalcut is required. (Additional cuts are required whenever the lengthof a rectangle is more than 4 times its height.)

10 Distribute centimeter gridpaper to the students. Tell them thateach square is 1 unit of area. Thenask them to dissect a rectangle ofarea 21 and reassemble it to form asquare.

11 Repeat Action 10 to obtain asquare of area 7.

C

d

dd

S

d d d

d

If the rectangle is cut on the dotted lines it can be reassembledto form a square.

ACTIONS COMMENTS



12 If the procedure of Comment 9 is carried out on a3 x 5 rectangle, the result is a square of area 15. The dimension ofthis square is distance CS in the following figure. Hence, segmentCS has length 15.

Segment CS has length 15.

13 The students may arrive at their conclusions by consideringspecific values for S and T. If so, you can ask them if they havereason to believe their conclusions hold for all values of S and T.The equality of the expressions in a) and b) are useful in simplify-ing radicals.

a) A general conclusion can be reached by noting that S T isthe length of the side of a square whose area is ( S T )( S T )and ST is the length of the side of a square whose area is ST.Since ( S T )( S T ) = ( S S )( T T ) = ST, these two squareshave the same area. Hence the lengths of their sides are equal.

b) Since ( S ⁄ T )( S ⁄ T ) = S S ⁄ T T = S⁄T, both S ⁄ T and S⁄T arethe lengths of the sides of squares whose area is S⁄T. Hence thetwo expressions are equal.

12 Ask the students to construct aline segment of length 15.

13 Write the following pairs ofexpressions on the overhead orchalkboard:

a) S T ; ST

b) S⁄ T ; S⁄T

c) S + T; S + T

For each pair of expressions, askthe students to determine, for posi-tive numbers S and T, whether thefirst expression is less than, equal to,or greater than the second expres-sion. Discuss their conclusions.

S

C

ST

S T ST

STS TS T( ) S T( )

S S=( ) T T( )

=ST

continued next page

⁄

ACTIONS COMMENTS



13 continuedc) As shown in the sketch below, S + T is the combined lengthof adjacent squares of areas S and T respectively. If the area Swhere distributed around the square of area T to obtain a squareof area S + T, the length of the side of this square would be lessthen the combined lengths of the sides of the original squares.Hence, S + T > S + T , that is the sum of the square roots oftwo positive numbers is greater than the square root of theirsum.

Some students may be curious about the relationship betweenS – T and S – T. Assuming S is greater than T so that S – T is

positive, from the last statement in c) above, the sum of thesquare roots of S – T and T is greater than the square root oftheir sum, which is S. Thus S – T + T > S and so, subtracting

T from both expressions, S – T > S – T .

As determined in a) and b) above, ST = S T and S⁄T = S ⁄ T .These results can be used to “simplify radicals.” For example, 45 = (9)(5) = 9 5 = 3 5; 7⁄5 = 35⁄25 = 35⁄ 25 = 35⁄5.

S

T

S

T

S T+ S + T

© THE MATH LEARNING CENTER ALGEBRA THROUGH VISUAL PATTERNS | 179



c2

c a

b

c







A D

B C

CUT



1 The area of a rectangles is 250 square inches. Its length is twice its width. Find its

dimensions.

2 A 14 foot ladder is leaning against a wall. The foot of the ladder is 6 feet from the wall.

How far is the top of the ladder above the ground?

3 Find the area of an equilateral triangle whose sides are 10 inches long.

4 a) The edge of a cube is 1 inch long. Find the length of a diagonal of the cube.

(A diagonal of a cube is a line segment that connects two vertices of the cube and goes

through its center.)

b) What is the length of a diagonal if an edge is s inches long?

5 Find the perimeter and area of the quadrilateral shown below. Each grid square is

1 square centimeter.

FOLLOW-UP BLACKLINE MASTER 9

6

5

4

3

2

1

1 2 3 4

d

5

a

b

c

6 a) Non-square rectangle R has sides of length a and b, where a < b.

Square S has the same area as rectangle R.

Square T has the same perimeter as rectangle R.

Find the lengths of the sides of squares S and T.

b) (Challenge) The geometric mean of a and b is the length of the side of square S. The

arithmetic mean of a and b is the length of the side of square T. Determine which of these

means is the greater. Explain how you arrived at your conclusion.



ANSWERS TO FOLLOW-UP 9

1

2w2 = 250 w2 = 125 = 25 x 5 w = 5 5

The dimensions are 5 5 by 10 5.

2

3

4 a) A diagonal d of a cube is the hypotenuse of atriangle whose legs are an edge of the cube and adiagonal of the base. The edge of a cube is 1 and thediagonal of a base is 2. Hence, d2 = 12 + ( 2)2 =1 + 2 = 3. So d = 3.

b) d = 3 s

h2 + 62 = 142

h2 = 142 – 62 = 160h = 160 ≈ 12.6 ft

h2 + 52 = 102

h2 = 102 – 52 = 75 = (25)3h = 5 3;

Area = 5h = 25 3

a2 = 22 + 42 = 20; a = 20b2 = 22+ 32 = 13; b = 13 c = 4d2 = 52 + 22 = 29; d = 29

Perimeter = 20 + 13 +4 + 29 ≈ 17.5 cm

Area = 5 x 6 – 4 – 3 – 4 – 5 = 30 – 16 = 14 sq cm

w

2w

h 14 ft.

6 ft.

h10

5

6

5

4

3

2

1

1 2 3 4

d

5

a

b

c

6

5

4

3

2

1

1 2 3 4

55

4

4 3

5

continued



6 a)

b) Square T is larger than square S, hence the arithmeticmean is greater than the geometric mean. One can seethat square T is larger than square S by comparing areas.Since R and S have the same area, T can be comparedwith R.

If R is superimposed on T as shown in the followingsketch, the area of region A is (a⁄2 + b⁄2)(b⁄2 – a⁄2) whilethat of region B is a(b⁄2 – a⁄2). Since a < b, a⁄2 + a⁄2 < a⁄2 + b⁄2 .Thus, the area of A is greater than the area of B. Hence T,which consists of regions A and C, has a greater areathan R, which consists of regions B and C.

ANSWERS TO FOLLOW-UP 9 (CONT.)

a

b

ab

TS

R

ab aba + b____

2

(a + b)2____

4

(a⁄2 + b⁄2) – a = (b⁄2 – a⁄2)

b – (a⁄2 – b⁄2) = (b⁄2 – a⁄2)a⁄2 + b⁄2

A

C B

T

R

a



THE BIG IDEAAlgebra Pieces are used to develop strategies for solv-ing linear and quadratic equations and systems ofequations. Coordinate graphs of the values of thearrangements establish a relationship between algebraand geometry and illustrate solutions to systems oflinear and quadratic equations.

OverviewStudents relate graphs ofpoints that lie along a linearpath to sequences of countingpiece arrangements and formu-las for the nth arrangement ofsuch sequences.

Materials� Algebra pieces including

frames), 1 set per student.

� Start-Up Masters 10.1 and10.3, 1 copy of each perstudent and 1 transparencyof each.


� Algebra Pieces for theoverhead.

OverviewStudents examine relationshipsbetween Algebra Piece, graphi-cal, and symbolic representa-tions of the nth arrangementsof extended sequences ofcounting piece arrangements.They use Algebra Pieces andgraphs to represent and solvelinear and quadratic equations.

Materials� Algebra Pieces (including

frames), 1 set per student.

� Focus Masters 10.1, 10.2, and10.4, 1 transparency of each.

� Focus Masters 10.3, 10.5, and10.6, 1 copy of each perstudent and 1 transparencyof each.

� Coordinate grid paper (seeAppendix), 2 sheets pergroup and 1 transparency.


� 1⁄4″ grid paper, 2-4 sheetsper student.

OverviewStudents create sequencesthat satisfy specific conditions.They write formulas for, graph,and solve linear and quadraticequations. They complete thesquare to solve quadraticequations.


student.

� Coordinate grid paper (seeAppendix), 8 sheets perstudent.

LESSON 10LINEAR & QUADRATICEQUATIONS


TEACHER NOTES


START-UP

Overview Materials

LINEAR & QUADRATIC EQUATIONS LESSON 10

Students relate graphs of points thatlie along a linear path to sequencesof counting piece arrangements andformulas for the nth arrangement ofsuch sequences.

� Algebra pieces includingframes), 1 set per student.

� Start-Up Masters 10.1and 10.3, 1 copy of eachper student and 1 trans-parency of each.

1 Give Algebra Pieces to eachstudent. Write the following charton the overhead. Have the studentsform the –2nd through 2nd and thenth arrangements of an extendedsequence of counting piece arrange-ments which fits the data on thechart (n) indicates the arrangementnumber and v(n) is the value ofarrangement n). Ask the students towrite a formula for v(n). Discuss.

n … –2 –1 0 1 2 …

v(n) … –3 –1 1 3 5 …

1 Various extended sequences of arrangements are possible.Shown below are two possibilities with corresponding formulas.Be sure the students indicate the arrangement numbers for theirsequences.



ACTIONS COMMENTS

Arrangement No.:

Value, v (n ) :

. . .–2

(–2) + (–1)

–1

(–1) + 0

0

0 + 1

1

1 + 2

2

2 + 3

. . .n

n + (n + 1)

. . .

Arrangement No.:

Value, v (n ) :

. . .–2

2(–2) + 1

–1

2(–1) + 1

0

2(0) + 1

1

2(1) + 1

2

2(2) + 1

. . .n

2(n) + 1

. . .

ACTIONS COMMENTS


START-UPLINEAR & QUADRATIC EQUATIONS LESSON 10

2 Shown below is one extended sequence that fits the data.

For this sequence, v(–3) = –10, v(3) = 8 and v(4) = 11. Coordinatepoints for these 3 cases are circled on the copy of Start-UpMaster 10.1 shown on the left. The ordered pair (–4,–13) lies offthe graph.

It may be instructive here to review the use of terms such ashorizontal axis, vertical axis, and origin. (The origin is the coordi-nate (0,0) which is the point of intersection of the horizontal andvertical axes.)

2 Give each student a copy ofStart-Up Master 10.1. Have thestudents form the –3rd through 3rdand the nth arrangements of an ex-tended sequence of counting piecearrangements which fits the datadisplayed in graphical form on Start-Up Master 10.1. Ask them to deter-mine v(–4), v(–3), v(3), and v(4) forthe sequence and, if possible, addthis information to their graph.



v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

1–2 3 4–1 2–3–4

11

v(n) = __________________________

v (n ) :

. . .–3

–10

–2

–7

–1

–4

0

–1

1

2

. . .2

5

3

8

ACTIONS COMMENTS



3 Ask the students to:

a) record in the space provided onStart-Up Master 10.1 a formula forv(n) for the sequence they con-structed in Action 2;

b) label the coordinates of eachpoint on the graph;

c) record 4 or 5 observations aboutthe graph.

Discuss, encouraging observationsabout relationships between thenumbers in the students’ formulasfor v(n) and their graphs.

3 a) The arrangements shown in Comment 2 suggest the formulav(n) = 3n – 1. The students may have other equivalent formulas.

b) The coordinates are labeled on the copy of Start-Up Master10.1 shown on the left below.

c) Students’ observations will vary. Following are examples ofobservations that students have made about the graph.

The points of the graph lie along the path of a straight line.

The points are equally spaced.

Moving left to r ight, to get from one point to the next, go 1 unit to theright and 3 units up.

The increase in height from point to point is always the same.

There are only points on the graph where n is an integer.

Plotting points for v(n) = 3n – 1 is just like plotting points for v(n) = 3nafter shifting the coordinate axes down 1 unit.

In the formula, v(n) = 3n – 1, 3 is the coefficient of n and is theamount the value, v(n), increases as n increases by 1. The constantterm, –1, is the value of the 0th arrangement. It indicates wherethe graph intersects the vertical axis. When the points on a graphlie along a straight line, the graph is called linear.

Some students may draw a line connecting the points of the graph,implying there are arrangements for non-integral values of n. Thestudents may even suggest ways of constructing such arrange-ments (see Lesson 11); however, for this extended sequence,there are only points on the graph for integral values of n.

The intent throughout this lesson is to promote intuitions aboutrelationships between graphs, formulas, and the sequences ofarrangements the graphs and formulas represent. Terminologysuch as slope and x- or y-intercept are introduced in Lesson 11,after extended sequences of arrangements are augmented sotheir graphs are continuous.



v(n) = __________________________3n – 1

v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

1–2 3 4–1 2–3–4

11

(–3,–10)

(–2,–7)

(–1,–4)

(0, 1)

(1,2)

(2 5)

(3,8)

(4,11)

ACTIONS COMMENTS



4 Place a transparency of Start-UpMaster 10.2 on the overhead, re-vealing the top half only (see nextpage). Tell the students the arrange-ment shown is the nth arrangementof an extended sequence of count-ing piece arrangements. Ask thestudents to form the –3rd to 3rdarrangements of this sequence.

5 Distribute a copy of Start-UpMaster 10.3 to each student. Forthe sequence of Action 4, ask thestudents to record a formula for v(n),construct its graph (see completedactivity below), and record theirobservations about the graph. Intro-duce the concept of function, andthe domain and range of a function.

4 Arrangements numbered –3 through 3 are shown on thebottom half of Start-Up Master 10.2. Recall that a –n-framecontains red tile if n is positive and black tile if n is negative. Itcontains no tile if n is 0.

5 The formula for v(n) can be written in various forms. Onepossibility is v(n) = 4 – n. Another is v(n) = 4 + (–n).

In general, a function is a rule that relates 2 sets by assigning eachelement in the 1st set (called the domain) to exactly one elementin the 2nd set (called the range). Hence, the relationship v(n) = 4+ (–n) is a function that relates the variable n to v(n) so that, forany arrangement number, n, there is exactly one value of thearrangement, v(n). The set of all values for n—in this case, theintegers—is the domain of the function v(n) = 4 + (–n). The set ofall possible values for v(n)—in this case, also the integers—is therange of the function. (In Lesson 11, students explore functionswhose domains and ranges include all real numbers.)



v(n) = __________________________

Observations about the graph:

4 – n

v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

12




v(n) = __________________________

v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

1–2 3 4–1 2–3–4

11




……

–3 –2 –1 0 1 2 3




v(n) = __________________________

Observations about the graph:

v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

12


TEACHER NOTES


FOCUS

Overview Materials


Students examine relationshipsbetween Algebra Piece, graphical,and symbolic representations of thenth arrangements of extendedsequences of counting piece arrange-ments. They use Algebra Pieces andgraphs to represent and solve linearand quadratic equations.

� Algebra Pieces (includ-ing frames), 1 set perstudent.

� Focus Masters 10.1, 10.2,and 10.4, 1 transparencyof each.

� Focus Masters 10.3, 10.5,and 10.6, 1 copy of eachper student and 1 trans-parency of each.

1 Arrange the students in groupsand distribute Algebra Pieces toeach student. Ask the groups to formthe –3rd through 3rd and nth ar-rangements of an extended se-quence of counting piece arrange-ments for which v(n) = n2 + 2n + 1.

1 Shown below is one possible set of arrangements.

One possible nth arrangement is shown below. Notice the use ofthe two frames to represent 2n. These frames are needed becausethey may represent a black n-strip or a red n-strip, depending onthe value of n.

� Coordinate grid paper(see Appendix), 2 sheetsper group and 1 trans-parency.


� 1⁄4″ grid paper, 2-4 sheetsper student.

ACTIONS COMMENTS

–3 –2 –1 0 1 2 3Arrangement number

ACTIONS COMMENTS


FOCUSLINEAR & QUADRATIC EQUATIONS LESSON 10

2 Tell the students there exists asequence of square arrangementswhich fits the criterion of Action 1.Ask the groups to show how theirarrangements from Action 1 can beformed into a sequence of squares,using edge pieces to show the valuesof the edges of the squares. Discuss.

2 Some students may have formed square arrangements inAction 1. If so, you may call the other students’ attention to thesearrangements.

Below is a set of square arrangements with edge pieces.

Other edges are possible. For example, here is another possibilityfor the –3rd arrangement:

The nth arrangement formed in Action 1 can be rearranged toform a square with 2 possibilities for edges, as shown below.Notice that the figures show that (n + 1)2 and (–n – 1)2 areequivalent expressions for v(n).

If edge frames were not discussed earlier, you will need to do sonow. Edge frames are edge pieces whose color, like that of frames,differs for positive and negative n. Edge frames are obtained bycutting frames into thirds lengthwise.

Edge frames

Has value n for n Has value –n for npositive, negative, or 0. positive, negative, or 0.

oo

o

o

–n – 1n + 1

oo

Arrangement number–3 –2 –1 0 1 2 3

ACTIONS COMMENTS



continued next page

The use of edge frames is illustrated below.

Note that the area of a square is always positive, while the valueof a square can be positive, negative, or zero. Similarly, the lengthsof the edges of a square are always identical and positive, whilethe values of the edges may be zero, both positive, both negative,or one positive and one negative.

3 A square has value 400 provided its edges all have value 20 or–20. Hence, the nth arrangement, viewed as a square whose edgehas value n + 1, has value 400 provided n + 1 has value 20 or –20.Since n + 1 is 20 when n is 19 and n + 1 is –20 when n is –21, the19th and –21st arrangements have value 400. Thus, the equation(n + 1)2 = 400 has been solved. The solutions are 19 and –21.Note: (–n – 1)2 = 400 also has solutions 19 and –21.

4 The nth arrangement contains a black n2-mat and 4 –n-frames:

3 Ask the students to determinewhich arrangements in the extendedsequence of Actions 1 and 2 have avalue of 400. Discuss the students’methods. Ask them to identify theequation that has been solved.

4 Place a transparency of FocusMaster 10.1 on the overhead. Askthe students to form the nth ar-rangement of this sequence and towrite an expression for v(n). o

o o

o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

v(n) = n2 – 4n

LINEA

R &

QU

AD

RA

TIC

EQU

AT

ION

S

FO

CU

S B

LA

CK

LIN

E M

AS

TE

R 10

……

–3 –2 –1 0 1 2 3

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

oo

o

o

o

o

oo

o o

oo

n

n

n

–n

–n

–n

–n

n

n –n –n n

2 2 –2 –2

ACTIONS COMMENTS



5 Ask the students to determinefor what n the extended sequencein Action 4 has v(n) = 525. Discusstheir strategies. If it isn’t suggestedby students, introduce the methodof solving the quadratic equationn2 – 4n = 525 by completing the square.

6 Ask the students to form the ntharrangement of an extended sequencefor which v(n) = n2 + 4 and thenhave them do the same for an ex-tended sequence for which v(n) =2n2 + 6n – 3. Have the students usetheir Algebra Pieces to determinefor which n the nth arrangements ofthese two extended sequences havethe same value. Ask the students toidentify the equation that has beensolved and to verify their solutions.

5 Adding 4 black tile to an nth arrangement results in a squarearray whose edges have value n – 2 or –n + 2, as shown below.

A square whose value is 525 + 4, or 529, has an edge whose valueis 23 or –23 (a calculator with a square root key is helpful here).If n – 2 is 23, then n is 25, and if n – 2 is –23, then n is –21. Hence,the 25th and –21st arrangements have value 525. Similarly, if theedge has value –n + 2, then –n + 2 = 23 or –23; hence, again,n = –21 or 25.

Historically, the above method of solving a quadratic equation iscalled completing the square. A quadratic equation is an equationthat can be written in the form ax2 + bx + c, where a, b, and c areconstants and a ≠ 0. The word quadratic is derived from the Latinword, quadratus, meaning square.

6 One way of representing the two nth arrangements is shownbelow:

n2 + 4

2n2 + 6n – 3

These two arrangements have the same value if, after an n2-mathas been removed from each of them, the remaining portions

o o

o o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

v(n) + 4 = (n – 2)2 v(n) = (–n + 2)2

o

o

oo

o o

o o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

525

4

n – 223 or –23

ACTIONS COMMENTS



7 Write quadratic equation a)below on the overhead. Then askthe students to find all solutions ofthe equation. Repeat for one ormore of b)-g). Discuss the students’methods.

a) n2 – 6n = 40

b) 2n2 + 38 = 4n2 – 12

c) (n – 1)(n + 3) = 165

d) 4n2 + 4n = 2600

e) n2 – 5n + 6 = 0

f) n2 + n = 6

g) n2 + 3n – 10 = 0

have the same value, i.e., if n2 + 6n – 3 has value 4 or, equivalently,n2 + 6n has value 7.

The method of completing the square is illustrated at the left.Adding 9 black tile to n2 + 6n produces a square array whoseedge has value n + 3.

Thus, n2 + 6n has value 7 if the square array has value 16, i.e., ifits edge has value 4 or –4. If n + 3 is 4, then n is 1; if n + 3 is –4,then n is –7. Hence, the 1st and –7th arrangements of the 2sequences have the same value.

7 Students’ methods of solving these equations may vary.

a) If 9 black tile are added to a collection for n2 – 6n, the result-ing collection can be formed into a square array with edge n – 3(see below). If n2 – 6n has value 40, the square array has value 49and its edge has value 7 or –7. If n – 3 is 7, then n is 10; if n – 3 is–7, then n is –4. So the equation has two solutions: 10 and –4.Note: the square could also have edge –n + 3 = 7 or –7, in whichcase, the solutions are still 10 and –4.

(n2 – 6n) + 9 = 40 + 9 = 49

b) Sketches for 2n2 + 38 and 4n2 – 12 have the same value if2n2 – 12 is 38. This is the case if n2 is 25, that is, if n = 5 or n = –5.

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

n – 3 =7 or –7

continued next page

n2 –12n2n2 n2n2 38n2

38

ACTIONS COMMENTS



7 continuedc) Shown at the left is a representation of (n – 1)(n + 3). Notethe values of the edges differ by 4 and their product is 165. Since11 and 15 differ by 4 and 11 x 15 = 165, and since –11 and –15differ by 4 and –11 x –15 = 165, the array will have value 165 ifthe edges have values 11 and 15 or –11 and –15. If the edges havevalues 11 and 15 then n is 12; if they have values –11 and –15 thenn is –14. (Finding the pair 11 and 15 is facilitated by noting that oneof the pair should be smaller and one larger than 165 ≈ 13.)

Alternatively, adding 4 black tile to the array shown above andremoving collections whose values are 0 leaves a collection ofpieces that can be arranged in a square array that has value 169and edge n + 1. Hence n + 1 is 13 or –13, in which case n = 12 orn = –14.

d) If 1 black tile is added to a collection for 4n2 + 4n, a squarearray with edge 2n + 1 (or –2n – 1) can be formed, as shown atthe left. If the value of the original collection is 2600, the valueof the square array is 2601. Using a calculator, one finds 2601= 51. Hence 2n + 1 is 51 or –51. Thus n = 25 or n = –26.

Using another approach, dividing a collection for 4n2 + 4n by 4results in the collection n2 + n with value 650. From thiscollection, a rectangle with value 650 and edges n by (n + 1)can be formed. Since 25 x 26 = 650 and –25 x –26 = 650,n = 25 or –26.

e) A collection for n2 – 5n + 6 can be formed into a rectangulararray with edges n – 2 and n – 3. The array has value 0 if an edgehas value 0. This is the case if n = 2 or n = 3.

o o

o o

o o

o o

165

11 or –15

15 or –11

n2

n169

n

1

n + 1

n + 1

51 or –51

2600 1

o o

o o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

ACTIONS COMMENTS



Alternatively, by cutting a –n-frame and 2 black tile in halves andadding 1⁄4 of a black tile to a collection for n2 – 5n + 6, theresulting collection can be formed into a square with edge n – 21⁄2.If the original collection has value 0, the square array has value 1⁄4and its edge has value 1⁄2 or –1⁄2. If n – 21⁄2 = 1⁄2 then n = 3 and ifn – 21⁄2 = –1⁄2, then n = 2.

f) Beginning with a collection for n2 + n, if one cuts the n–framein halves and adds 1⁄4 of a black tile, a square array with edgen + 1⁄2 can be formed. This square array has value 61⁄4 and its edgehas value 21⁄2 or –21⁄2. If n + 1⁄2 = 21⁄2, then n = 2; if n + 1⁄2 = – 21⁄2,then n = –3.

g) A collection for n2 + 3n – 10 contains a n2-mat, 3n-frames and 10 red tile. If this collection is arranged asshown on the left and 2 n-frames and 2 –n-frames areadded, the value of the collection is unchanged and theresulting collection can be formed into an array withedges n – 2 and n + 5. This array has value 0 if one of theedges has value 0, that is if n = 2 or n = –5.

Here are other equations you might have students solve:(n – 4)(n + 2) = 0; n2 + 4n – 5 = 0; n2 + 6n = –8;n2 = 7n – 6; 2n2 – 2n = 112. If you create others, be surethey have integer solutions (in later lessons studentsexplore sequences with nonintegral arrangement numbers).

o o

o o

o o

o o

o o

o o

o o

o o

o o o o

o o o o

o o o o

o o o o

o o

o o

o o

o o

0

or

1__4

1__2 – 1__

2

n2

n/2

–n –

6

n +or

1__4

n__21__

2

1__2

n2

6

n

n2 + n n2 + n + 1⁄4 = (n + 1⁄2)2 or (–n – 1⁄2)2

n + 5

n – 2 n2 + 3n – 10

ACTIONS COMMENTS



8 Arrangements numbered –3 through 3 are shown on thebottom half of Focus Master 10.2.

Here are three possibilities for v(n): v(n) = n2 – 2n – 3;v(n) = (n + 1)(n – 3); and v(n) = (–n – 1)(–n + 3).

Edge pieces help to illustrate the latter two formulas:

The coordinate points associated with these arrangements areshown on the completed graph on the left.

Here are some observations made about the graph:

The points of the graph do not lie along the path of a straight line;they lie along the path of a U-shaped curve.

The graph is symmetric about the vertical line that passes throughn = 1. That is , if the graph is folded along the vertical line that goesthrough n = 1, the points to the right of the fold coincide with those tothe left of the fold.

The point (1,–4) is a turning point where the graph stops falling andstarts to rise (looking from left to right).

The smallest value for v(n) is –4. It occurs when n = 1.

When n is greater than 1, as n increases so does v(n). When n is lessthan 1, as n decreases v(n) increases.

The domain of v(n) = n2 – 2n – 3 is the integers and its range is theset of integers greater than or equal to –4.

8 Place a transparency of FocusMaster 10.2 on the overhead, reveal-ing the top half only. Tell the stu-dents that the arrangement shownis the nth arrangement of an ex-tended sequence of counting piecearrangements. Ask them to form the–3rd through 3rd arrangements(with edges) of this sequence. Thendistribute a copy of Focus Master10.3 to each student and have thestudents write a formula for v(n),construct its graph (see completedgraph below), and record their obser-vations about the graph.

oo

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o

oo o o o

o o o o

o o o o

o o o oo o o o

o o o o





v(n) = ______________________

v (n)

n

30

28

26

24

22

20

18

16

14

12

8

6

4

2

–2

–4

2–4 6 8–2 4–6–8

32

10

1 5 73–3 –1–5–7

(n + 1)(n – 3)

ACTIONS COMMENTS



9 Place a copy of Focus Master 10.4on the overhead, and tell the stu-dents that Arrangements I and II arethe nth arrangements of two differ-ent extended sequences. Ask themto write formulas for v1(n), thevalue of the nth arrangement of thefirst sequence and v2(n), the value ofthe nth arrangement of the secondsequence.

10 Give each student a copy ofFocus Master 10.5 and ask them todo the following:

a) record their formulas for v1(n)and v2(n);

b) graph v1(n) and v2(n) on the coor-dinate grid, indicating the points onthe graph of v1(n) with an x and thoseon the graph of v2(n) with an o (seecompleted graph on the next page);

c) record observations about thegraphs and the relationships be-tween them.

Discuss their observations.

9 Here are possible formulas for the given nth arrangements:v1(n) = 6n – 2v2(n) = n2 + 7n – 8

Other formulas are possible. For example, v2(n) = (n + 8)(n – 1);edge pieces may help the students see this formula:

10 The completed graphs are shown on a copy of Focus Master10.5 on the next page.

On the next page are some observations about the graphs. Ifstudents don’t bring these up, you might prompt discussion byposing questions such as: What points, if any, do the 2 graphs havein common? What do the common points on the graphs tell aboutthe 2 sequences of counting piece arrangements? When is v1(n)greater than v2(n)? How do the shapes of the graphs compare?How could you change the equation of v1 so that it doesn’tintersect v2? Are v1 and v2 functions? How do the domains andranges of v1 and v2 compare?, etc. If you wish, you may use thediscussion as an opportunity to introduce inequality notation.

n + 8

n – 1

continued next page



III

ACTIONS COMMENTS



10 continuedTwo points, (2,10) and (–3,–20), are on both graphs. This tells us thatthe 2nd arrangements of the 2 sequences have the same value andthe –3rd arrangements also have the same value. It also tells us theequation 6n – 2 = n2 + 7n – 8 has 2 solutions, n = 2 and n = –3.

Note: you might have the students form the 2nd arrangement ofeach sequence and verify that they have the same value. Likewisefor the –3rd arrangements.

When n is between –3 and 2, the values of the arrangements inSequence I are greater than the values of the arrangements inSequence II. Using inequality notation: if –3 < n < 2, then v1(n) > v2(n).

When n is less than –3 and when n is greater than 2, the value ofSequence II is greater than the value of Sequence I. Using inequalitynotation, if n < –3 or n > 2, then v2(n) > v1(n).

The graph of v1(n) follows the path of a straight line, and the graph ofv2(n) follows the path of a U-shaped curve.

The graph of Sequence I always rises as values of n increase from leftto right. Looking at the graph of Sequence II from left to right, thegraph falls as n increases, until n = –4; v(–4) = v(–3); then after n = –3,as n increases the graph rises.

Once Sequence II starts to rise, it rises faster than Sequence I.

Sequence II has line symmetry about a vertical line that passes midwaybetween n = –3 and n = –4.

If the U-shaped curve that the graph of Sequence II follows is traced,we think the turning point is (–31⁄2,–201⁄4). Since the curve is symmet-ric , we think the turning point is half way between n = – 3 and n = – 4,or at n = – 31⁄2. We found the y-coordinate of the turning point byfinding (31⁄2)2 + 7(–31⁄2) – 8 = – 201⁄4.

A function such as v1(n) whose graph lies on a straight line iscalled a linear function; and a function such as v2(n) whose graphlies on a parabola (i.e., a U-shaped curve) is called a quadraticfunction.



x: v1(n) = ______________________

Observations:

o: v2(n) = ______________________

Observations:

v (n)

32

28

24

20

16

12

8

4

–4

–8

–12

–16

–20

–24

–28

–6 4 6–4 2–8–10

36

–1 3 51–5 –3–7–9 –2–11n

x

x

x

x

x

x

x

x

x

n 2 + 7n – 86n – 2

ACTIONS COMMENTS



11 You might ask the students for their views on the advantagesand disadvantages of graphical versus non-graphical methods. (InLesson 13 students are introduced to graphing calculators, atwhich time their views may change.)

Students will use a variety of ways to solve this equation. Some ofthese ways are shown below.

a) One can view n2 + 7n – 8 as an n + 8 by n – 1array whose value equals 6n – 2. When 6n – 2 isremoved from this array, the remaining portion musthave value 0. If this portion is rearranged and an n-strip and a –n-strip is added to it, its value is un-changed and the result is an n + 3 by n – 2 arraywhose value is 0. Hence, one of the dimensions ofthe array must have value 0. Thus,n = –3 or n = 2.

b) Alternatively, removing an n-strip and a –n-strip from thearrangement of value 0, and then adding 6 black tile to it, resultsin an n + 1 by n array whose value is 6, that is, an array whosedimensions are consecutive integers and whose product is 6.There are two possibilities for such a pair of integers: 2 and 3, or–3 and –2. Since n is the least of the pair, n = 2 or n = –3.

11 Point out to the students thatin Action 10 they used informationfrom their graphs to find the solu-tions of the equation 6n – 2 = n2 +7n – 8. Then ask the students tosolve this equation by a methodthat doesn’t involve graphing. Dis-cuss the methods the students use.

continued next page

(n + 3)(n – 2) = 0

0

6n – 2

added

(n + 1)n = 6

0

n2 + n – 6 = 0

ACTIONS COMMENTS



11 continuedc) One can also find a solution by completing the square. Bytaking the above n + 1 by n array whose value is 6 and cutting then-strip in half and adding 1⁄4 of a black tile, one can form a squareof dimension n + 1⁄2 whose value is 61⁄4 or 25⁄4. Hence, n + 1⁄2 = 5⁄2or n + 1⁄2 = – 5⁄2 and, as before, one has n = 2 or n = –3.

One might record the above actions using algebraic symbols asfollows:

n2 + 7n – 8 = 6n – 2n2 + 7n – 8 – (6n – 2) = 0

n2 + n – 6 = 0n2 + n = 6

n2 + n + 1⁄4 = 6 + 1⁄4 = 25⁄ 4

(n + 1⁄2)2 = 25⁄4n + 1⁄2 = ± 5⁄2

n = –1⁄2 ± 5⁄2n = 2 or n = –3

d) In the above completing the square procedures, one canavoid cutting pieces by introducing an appropriate multipli-cation in the procedure. Once one has determined that acollection of 1 n2-mat and 1 n-strip has value 6, then acollection of 4 n2-mats and 4 n-strips has value 24, so that acollection of 4 n2-mats and 4 n-strips and 1 black tile hasvalue 25. This collection can be formed into a square ofside 2n + 1. Hence, (2n + 1)2 = 25, whence2n + 1 = 5 or 2n + 1 = –5 and, as before, n = 2 or n = –3.

(n + )2 = 61__2

1__4

4n2 + 4n = 24n2 + n = 6

(2n + 1)2 = 25

ACTIONS COMMENTS



12 The activities on Focus Master 10.6 could be completed ashomework and then discussed in class.

This action is intended to acquaint students with the graphs ofconstant, linear, and quadratic graphs. These are considered ingreater detail in subsequent lessons.

A constant function is a function whose range consists of a singlequantity such as v1(n) in pair 2 where the value for every n is –7.Both v1(n) and v2(n) in pair 1 are examples of linear functions. Ingeneral, v(n) is a linear function if v(n) is of the form an + b wherea is not zero. Both v1(n) and v2(n) in pair 3 and v2(n) in pair 2 areexamples of quadratic functions. In general, v(n) is a quadraticfunction if v(n) is of the form an2 + bn + c where a is not zero.

A table of values for v1(n) and v2(n) is useful for generating andorganizing ordered pairs to plot. Following is a table for Pair 1.

12 Give each student coordinategrid paper (see Appendix) and a copyof Focus Master 10.6. Have themcarry out the instructions. Whenthe students are finished, invitevolunteers to share their questions,observations, and conjectures.

n v1(n) v2(n)

–3 11 –24

–2 8 –20

–1 5 –16

0 2 –12

1 –1 –8

2 –4 –4

3 –7 –0

continued next page



1 Formulas for the values of the nth arrangements of 3 pairs of extended sequences are

given below. For each pair of sequences, do the following:

a) Make a table showing v1(n) and v2(n) for n from –3 to 3. Then graph v1(n) and v2(n) on

the same coordinate axes.

b) Determine when v1(n) = v2(n).

Pair 1

v1(n) = –3n + 2 v2(n) = 4n – 12

Pair 2

v1(n) = –7 v2(n) = –n2 – 2n + 8

Pair 3

v1(n) = n2 + 2 v2(n) = –n2 + 4

2 Record your general observations and conjectures about graphing constant, linear, and

quadratic functions.

ACTIONS COMMENTS



12 continuedIn each of the graphs shown here, points on the graph of v1(n) areindicated by an o and those on the graph of v2(n) by an x.

One can determine when v1(n) = v2(n) for Pair 1 and Pair 3 fromthe graphs. However, not all solutions of v1(n) = v2(n) in Pair 2 areapparent from the portion of the graph shown for that pair. Thestudents may deduce by symmetry that, in addition to being equalwhen n = 3, v1(n) and v2(n)are also equal when n = –5. The stu-dents can also find this solution by extending their graph to theleft or through the use of Algebra Pieces.

1

2

3

4

5

6

7

8

9

10

11

12

1 2 3–1–2–3

x

o

v(n)

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

–12

–13

–14

–15

–16

–17

–18

–19

–20

–21

–22

–23

–24

n

x

x

x

x

x

x

o

o

o

o

o

o

1

2

3

4

5

6

7

8

9

10

1 2 3–1–2–3

x

v(n)

nx x

x

xx

x

o o

o

o o

o o

11

12

–1

–2

–3

–4

–5

–6

Pair 1

Pair 2

Pair 3

Following are some conjecturesthe students might make aboutthe graphs:

The graph of a constant function liesalong a horizontal line, that of alinear function along a non-horizon-tal line, and that of a quadraticfunction along a U-shaped curve.

A graph that is linear rises from leftto right if the coefficient of n ispositive and falls if the coefficient isnegative.

For an equation whose graph followsa linear path, the constant moves thegraph up or down from the horizon-tal axis . The coefficient of the n-term determines the “steepness” ofthe graph.

If the coefficient of the n2-term is negative the U opens down (the Uwould spill water). If the coefficient of the n2-term is positive the Uopens up (the U would hold water).

A line and a U-shaped curve can only intersect in 0, 1, or 2 points; 2different lines can intersect in 0 or 1 point; 2 different U-shapedcurves can intersect in 0, 1, or 2 points .

1

2

3

4

5

6

7

8

9

10

1 2 3–1–2–3

x

v(n)

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

n

x

x x

x

x

xo oo oo oo

11

12




……

–3–2

–10

12

3




…

–3 –2 –1 0 1 2 3

…




v(n) = ______________________

v (n)

n

30

28

26

24

22

20

18

16

14

12

8

6

4

2

–2

–4

2–4 6 8–2 4–6–8

32

10

1 5 73–3 –1–5–7




III




x: v1(n) = ______________________

Observations:

o: v2(n) = ______________________

Observations:

v (n)

32

28

24

20

16

12

8

4

–4

–8

–12

–16

–20

–24

–28

–6 4 6–4 2–8–10

36

–1 3 51–5 –3–7–9 –2–11n




1 Formulas for the values of the nth arrangements of 3 pairs of extended sequences are

given below. For each pair of sequences, do the following:

a) Make a table showing v1(n) and v2(n) for n from –3 to 3. Then graph v1(n) and v2(n) on

the same coordinate axes.

b) Determine when v1(n) = v2(n).

Pair 1

v1(n) = –3n + 2 v2(n) = 4n – 12

Pair 2

v1(n) = –7 v2(n) = –n2 – 2n + 8

Pair 3

v1(n) = n2 + 2 v2(n) = –n2 + 4

2 Record your general observations and conjectures about graphing constant, linear, and

quadratic functions.



1 Use Algebra Pieces or sketches to solve the following equations. In each instance, use

algebra symbols to record the steps in your solution.

a) n2 + 2n = 35

b) (n – 5)(n – 1) = 21

c) 2n2 – 6n = 36

2 For each of a)-d) below, create an extended sequence of counting piece arrangements

whose graph meets the given conditions. Then write a formula for the value of its nth

arrangement and sketch its graph.

a) The graph of this sequence is linear, falls from left to right, and contains the point (0,2).

b) The points (1,5) and (2,8) lie on the graph of this sequence.

c) The graph of this sequence is U-shaped and (0,–3) is its lowest point.

d) The graph of this sequence is U-shaped and contains the points (2,–8) and (–2,–8).

3 Create two extended sequences whose graphs are both linear, but do not lie on a hori-

zontal line, and have the point (7,9), but no other point, in common. Then write formulas

for the values of the nth arrangements of both sequences and sketch their graphs.


ANSWERS TO FOLLOW-UP



ANSWERS TO FOLLOW-UPANSWERS TO FOLLOW-UP 10

1 a)

b)

Rearrange and add 4 black:

c)

Double, rearrange, and add 9 black:

n2 + 2n = 35n2 + 2n + 1 = 36 (n + 1)2 = 36 n + 1 = 6 or –6 n = 5 or –7

(n – 5)(n – 1) = 21

n2 – 6n + 5 = 21

n2 – 6n + 9 = 25

(n – 3)2 = 25

n – 3 = 5 or –5

n = 8 or –2

2n2 – 6n = 36

4n2 – 12n = 72

4n2 – 12n + 9 = 81

(2n – 3)2 = 81

2n – 3 = 9 or –9

2n = 12 or –6

n = 6 or –3

2 a) One possibility: b) One possibility: v(n) = –n + 2 v(n) = 3n + 2

c) One possibility: d) One possibility:v(n) =n2 – 3 v(n) = –2n2

3 One possibility: v1(n) = n + 2v2(n) = 2n – 5

n2 + 2n + 1

36

(n – 1)(n – 5)

21

o

o o

o

o o o

o

o o o o

o

o o o

o

o o o o

o

o o o

(n – 3)2 = 25

25

o o

o

o o

o o

o o

o o

o o o o

o o o o

o o

o o

o

o

o o

o o

o

o o

2n2 – 6n

36

o o

o o

o o

o o

o o o o

o o o o

o o

o o

o o o o

o o o o

o o o o

o o o o

(2n – 3)2

81

o o o

o o o

o o o

o o

o o

o o

o o

o

o o

v(n)

n1

2

3

4

5

6

7

–1

–3

1 2 3 4–1–2–3–4

–2

v(n)

n1

2

3

4

5

6

7

8

9

10

11

1 2 3–1–2–3

–4

–6

–5

–1

–3

–2

–7

1

2

3

4

5

6

–1

–3

–2

1 2 3–1–2–3

v(n)

n

1 2 3–1–2–3

v(n)

n

–4

–6

–5

–1

–3

–2

–7–8–9

v(n)

n1

2

3

4

5

6

7

8

9

10

11

–1

–3

–2

1 2 3 4 5 6 7 8 9

12

13

o: v1(n) = n + 2x: v2(n) = 2n – 5

oo

oo

oo

oo

o

x

x

x

x

x

x

x

x

x



THE BIG IDEAComplete sequences of arrangements, which have anarrangement corresponding to every point of a num-ber line, are introduced. The values of these arrange-ments serve as a vehicle for studying linear and qua-dratic graphs and equations.

OverviewAn extended sequence ofarrangements is augmented toprovide an arrangementcorresponding to every pointon the number line.

Materials� Start-Up Masters 11.1-11.2;

1 copy of each per studentand 1 transparency of each.


OverviewGraphs of complete sequencesof arrangements are con-structed, leading to a discussionof lines and parabolas. Thecoordinates of various pointsof graphs are determined byobservation or by using Alge-bra Pieces or sketches to solvethe appropriate equations.

Materials� Algebra pieces for each

student.

� Focus Masters 11.1-11.2,1 copy of each per studentand 1 transparency of each.

� Focus Masters 11.3-11.4,1 transparency of each.

� Coordinate grid paper (seeAppendix), 10 or 11 sheetsper student and 1 transpar-ency.

� Algebra pieces for theoverhead.

OverviewStudents graph straight linesand write their equations. Theyfind the intercepts, turningpoints, and points of intersec-tions of intersecting parabolas,and solve linear and quadraticequations.


student.

� Coordinate grid paper (seeAppendix).

LESSON 11COMPLETE SEQUENCES


TEACHER NOTES


START-UP

Overview Materials

COMPLETE SEQUENCES LESSON 11

An extended sequence of arrange-ments is augmented to provide anarrangement corresponding toevery point on the number line.

� Start-Up Masters 11.1-11.2; 1 copy of each perstudent and 1 transpar-ency of each.

1 Distribute a copy of Start-UpMaster 11.1 to each student. Showthem the Algebra Piece Arrange-ment of 2 n-frames and 1 red tileshown below. Tell them it is the ntharrangement of an extended se-quence of counting piece arrange-ments. Ask them to sketch, in Sec-tion A of Start-Up Master 11.1, the–3rd through 3rd arrangements ofthe sequence, representing countingpieces by grid squares.

2 Distribute a copy of Start-UpMaster 11.2 to each student. Forthe sequence introduced in Action 1,ask the students to record a for-mula for v(n) in the space providedand then construct a graph of v(n).

1 The students can use red and black pens or markers, if avail-able, to fill in squares to represent red and black tile. In thesketches shown below, the hatched squares represent red tile.

2 The completed graph is shown below. Some students may haveother expressions for v(n) that are equivalent to the one shown.

ACTIONS COMMENTS


–3 –2 –1 0 2 31arrangement number



v(n) = 2n – 1

v (n)

n

10

9

8

7

6

5

4

3

2

1

1

–2

3

–4

5

–6

–7

8

–9

10

–11

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

ACTIONS COMMENTS


START-UPCOMPLETE SEQUENCES LESSON 11

3 Mention to the students thatthere is no point on the graph forn = 11⁄2 since there is no 11⁄2tharrangement. Ask the students toimagine that the sequence has beenaugmented to contain such anarrangement. Ask them to sketch, inSection B of Start-Up Master 11.2,what they think the 11⁄2th arrange-ment looks like. Have them computethe value of that arrangement andadd the corresponding point to theirgraph. Discuss their ideas and rea-soning. Repeat for n = 31⁄2 andn = –23⁄4 in Sections C and D ofStart-Up Master 11.2.

3 Below on the left is a sketch of a 11⁄2th arrangement, based onthe pattern of the arrangements in the original sequence. Its netvalue is 2. Thus, (11⁄2,2) is the point on the graph correspondingto this arrangement. Similarly, (31⁄2,6) is the point associated withthe 31⁄2th arrangement.

The net value of the –23⁄4th arrangement is –61⁄2, as illustratedbelow. Its corresponding point on the graph is (–23⁄4,–61⁄2).

v(31⁄2) = 2(31⁄2) – 1 = 6v(11⁄2) = 2(11⁄2) – 1 = 2

B C

D

ACTIONS COMMENTS



4 Have the students do the follow-ing on Start-Up Master 11.1 and11.2:

a) On Start-Up Master 11.2 choosesome point between two integerson the positive part of the n-axisand suppose it represents the posi-tive number P. Then choose somepoint between two integers on thenegative part of the n-axis andsuppose it represents the negativenumber Q.

b) In Sections E and F of Start-UpMaster 11.1, sketch, respectively, thePth and Qth arrangements anddetermine the values of thesearrangements.

c) Add the points associated withthe Pth and Qth arrangements tothe graph in Start-Up Master 11.2.Discuss the methods the studentsuse to carry out these procedures.

4 The students choice of location for P and Q will vary.

Shown below are sketches of the Pth and Qth arrangements for thechoices of P and Q shown on the graph at the left below. The corre-sponding points (P,2P – 1) and (Q,2Q – 1) are shown on the graph.

v(P) = 2P – 1 v(Q) = 2Q – 1

The location of the points on the graph can be determined bymeasuring. For example, one can mark off on the edge of a pieceof paper a segment whose length is the distance between 0 and P,and adjoin to it a segment whose length is P – 1. The sum of thesetwo lengths will be the distance of the point (P,2P – 1) above then-axis. This is illustrated below. Some of the students may locatethe points by noting that all the points of the graph are collinear(i.e., they all lie on the same line) and locate (P,2P – 1) and(Q,2Q – 1) so that collinearity is maintained.



v(n) = 2n – 1

v (n)

n

10

9

8

7

6

5

4

3

2

1

1

–2

–3

–4

5

–6

–7

–8

9

10

–11

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

PQ

ACTIONS COMMENTS



5 Ask the students to imagine thatthe sequence of arrangementsdiscussed in Actions 1-4 has beenaugmented so there is an arrange-ment corresponding to every pointon the n-axis. Ask them how theycould show this on their graphs.Discuss.

Introduce the terms complete se-quence of arrangements and realnumber.

5 The resulting graph is a straight line, only a portion of whichshows in the graph the students have constructed. The actualgraph extends indefinitely in both directions, as indicated by thearrowheads.

A sequence of arrangements which has an arrangement corre-sponding to every point of a coordinate axis will be referred toas a complete sequence of arrangements. The set of numberscorresponding to the points of a coordinate axis is called the setof real numbers. (Every real number may be represented by adecimal, which may or may not terminate.) Thus a completesequence of arrangements is a sequence which has an arrange-ment for every real number.



v(n) = 2n – 1

v (n)

n

10

9

8

7

6

5

4

3

2

1

1

2

–3

–4

–5

6

–7

–8

–9

10

–11

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

PQ

ACTIONS COMMENTS



6 The students may suggest a variety of ways to represent thexth arrangement. It might be represented by a sketch. For ex-ample, the 1st sketch below consists of 2 unshaded strips, eachlabeled x, and a single red counting piece. It is understood thateach strip is to be filled with a collection of counting pieces and/or parts of pieces whose value equals its label. Thus, if x is posi-tive, the strip is filled with black; if x is negative it is filled withred; and if x is 0, it is empty. The 2nd sketch uses edge pieces torepresent 2x as a rectangle with edge values 2 and x.

Alternatively, Algebra Pieces might be used to form a representa-tion. The frames are to be thought of as x-frames rather thann-frames; that is, each frame represents a strip of counting piecesor partial pieces whose value is x.

Henceforth, the letter x will be used when referring to a genericarrangement in a complete sequence of arrangements. (With thisconvention, the horizontal axis in the graph shown in Action 5would be labeled x, the vertical axis labeled v(x) and the formulawritten as v(x) = 2x – 1.) The letter n will be used when referringto a generic arrangement in a sequence of arrangements corre-sponding to integers, e.g., an extended sequence of arrangements.

The above usage follows the customary, but not universal, practiceof using letters like x, y, and z to represent quantities that cantake on any value, integral or not, (i.e., continuous variables) andusing letters like k, m, and n to represent quantities that haveintegral values (i.e., discrete variables). The choice of a letter torepresent a generic arrangement is arbitrary. For example, onemight refer to the zth arrangement and write v(z) = 2z – 1. Inthis case, if frames were used to represent the zth arrangement,they would be referred to as z-frames or –z-frames and havevalues z or –z, respectively.

xx

2x – 12x – 1 x

2

2x – 1

–n -frame

n -frameo o o o

o o o o

–x -frame

x -frameo o o o

o o o o

When arrangementnumbers are integers:

When arrangementnumbers are realnumbers:

6 Ask the students to suppose ageneric point x on the horizontalaxis in Action 5 is selected and tocreate a representation of the xtharrangement. Discuss their repre-sentations.


TEACHER NOTES




A

B

E

F

C D




v(n) =

v (n)

n

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7


FOCUS

Overview Materials


Graphs of complete sequences ofarrangements are constructed,leading to a discussion of lines andparabolas. The coordinates of vari-ous points of graphs are determinedby observation or by using AlgebraPieces or sketches to solve theappropriate equations.

� Algebra pieces for eachstudent.

� Focus Masters 11.1-11.2,1 copy of each per stu-dent and 1 transparencyof each.

� Focus Masters 11.3-11.4,1 transparency of each.

1 Show the students the followingxth arrangement from a completesequence of arrangements. Ask themto write a formula for v(x). Thengive a sheet of coordinate gridpaper to each student and havethem construct a graph consistingof all points (x,y) such that y = v(x).

1 You may want to clarify that the partial counting piece is 1⁄2 ofa black counting piece, so y = v(x) = 7⁄2 – 3x. Note that, in thegraph shown below, the horizontal axis is labeled x. The verticalaxis is labeled y, it could also be labeled v(x).

Some students may write y = 7⁄2 + 3(–x). You may want to discusswhy 7⁄2 – 3x and 7⁄2 + 3(–x) are equivalent expressions. One wayto see this is to note that a collection of 3 –x-frames is theopposite of a collection of 3 x-frames, that is 3(–x) = –(3x), andadding the opposite of a value is equivalent to subtracting thatvalue. Thus, 7⁄2 + 3(–x) = 7⁄2 + (–(3x)) = 7⁄2 – 3x.

y = 7⁄2 – 3x

� Coordinate grid paper(see Appendix), 10 or 11sheets per student and1 transparency.

� Algebra pieces for theoverhead.

ACTIONS COMMENTS

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

11

10

9

8

7

6

5

4

3

2

1

1

2

–3

–4

–5

–6

7

8

9

–10

–11

x

y

–2–3–4 –1–5–6–7–8 1 2 3 4 5 6 7 8

continued next page

ACTIONS COMMENTS


FOCUSCOMPLETE SEQUENCES LESSON 11

2 Tell the students that the point(x, y) is on the graph in Action 1 andthat y = 8. Have them determinewhat x equals. Ask for volunteers todescribe the methods they use.Discuss.

Repeat if y is: a) –9, b) –14.2, c) 100.

1 continuedNotice that for this graph, the values for x included in the graphare all of the real numbers and the values for y are all of the realnumbers. Hence, the domain and range of y = 7⁄2 – 3x are the realnumbers.

2 The students might see from their graph in Action 1, if drawnaccurately, that if the point (x, y) is on their graph and y = 8, thenx = –1.5. If this happens, you might point out that the equation 8= 7⁄2 – 3x has been solved graphically, that is, a solution has beenobtained by graphing the function y = 7⁄2 – 3x and determiningthe point on this graph for which y = 8.

Some fractions can be avoided by doubling the collection ofpieces representing v(x) as shown in the figure below. In thisfigure, we have 2v(x) = 16, hence the 6 –x-frames have a totalvalue of 9, so 2 of them have value 3. If 2 –x-frames have value 3,then 2 x-frames have value –3 and, as before, x = –3⁄2.

Symbolically, the steps in this method of determining x could berecorded as follows:

7⁄2 + 3(–x) = 87 + 6(–x) = 16

6(–x) = 92(–x) = 3

2x = –3x = –3⁄2

a) When y = –9, it is difficult to determine the exact value of xfrom the graph. In this case, if one doubles the xth arrangement,the 6 –x-frames have a total value of –25, so 6 x-frames have atotal value of 25, that is 6x = 25. Hence, x = 25⁄6 = 41⁄6.

b) Proceeding as in b), 6x = 35.4 and x = 5.9.

c) Again proceeding as in b), 6x = –193 and x = –321⁄6.

o o o o

o o o oo o o o

o o o oo o o o

o o o oo o o o

o o o o

o o o o

o o o oo o o o

o o o o

2v (x) = 7 + 6(–x) = 16

ACTIONS COMMENTS



3 The graph shows that v(x) increases by 4 as x increases by 1.Thus, v(1) = 4 + v(0), v(2) = 8 + v(0), v(3) = 12 + v(0) and so forth.Since v(0) = 2, this suggests that the formula is v(x) = 4x + 2,which can be verified for other points on the graph. If the expres-sion v(x) is represented by y, the formula can be written y = 4x + 2.The students may arrive at the result by other methods.

An Algebra Piece arrangement is shown on the left below. A sketchof the xth arrangement is shown on the right.

v(x) = 4x + 2 v(x) = 4x + 2

4 Focus Master 11.2, shown on the next page, may be useful fordiscussion. The number 4 in the product 4x is called the coefficientof x. It tells how much y values change as x values increase by 1(for example, as x changes from 0 to 1, y changes from 2 to 6, anincrease of 4). This rate of change, that is, the change in y for eachunit increase in x, is called the slope of the line. (If y had de-creased as x increased, the change, and hence the slope, wouldhave been negative.)

3 Place a transparency of FocusMaster 11.1 on the overhead andgive a copy to each group. Tell thestudents this is a graph of y = v(x)for a certain complete sequence ofarrangements. Ask them to write aformula for v(x) in the space pro-vided on the bottom of Focus Mas-ter 11.1 and to sketch or constructan Algebra Piece representation ofthe xth arrangement. Discuss.

4 Discuss the students’ ideas aboutways the numbers, 4 and 2, in theformula for v(x) relate to its graph.Introduce the terms coefficient,slope, intercept, and slope-interceptform of the equation of a line.

xxxx

continued next page



v(n) = __________________________

22

20

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

–22

x

y

–4–6–8 –2–3–5–7 1 2 3 4 5 6 7 8–1

ACTIONS COMMENTS



5 Give each student a sheet ofcoordinate grid paper. Tell them thatthe graph of y = v(x) for a certaincomplete sequence of counting piecearrangements is a straight line whichpasses through the points (–2,10)and (4,–8). Ask them to graph andfind a formula for v(x). Then have themconstruct an Algebra Piece repre-sentation or draw a sketch of thexth arrangement. Discuss.

4 continuedThe constant 2 is the value of y when x is 0 or, to put it anotherway, the y-coordinate of the point where the line crosses the y-axis. This value is called the y-intercept of the line.

A line may also cross the x-axis. If it does, the value of y is zero atthe point of intersection with the x–axis and the value of x at thatpoint is called the x-intercept. For y = 4x + 2, the x-intercept is –1⁄2.

The equation y = 4x + 2 is said to be in slope-intercept form wherethe coefficient, 4, of x is the slope of the line and the constant, 2,is the y-intercept. In general, y = ax + b is the equation of a linewhose slope is a and y-intercept is b.

5 Students will need to locate axes and then scale them. In thegraph shown below, the x-axis is scaled so that each subdivisionrepresents 1 unit and the y-axis is scaled so that each subdivisionrepresents 2 units. The graph will appear differently for otherscales, but will still be a straight line.

–4–6–8

18

16

14

12

10

8

6

4

2

2

–4

6

8

–10

12

–14

–16

18

–2–3–5–7 –1 1 2 3 4 5 6 7 8

v (x)

x



10

8

6

4

2

–2–2 –1 1 2

y = 4x + 2

y - intercept

x - intercept

increase of 4

increase of 4

slope

1

1

ACTIONS COMMENTS



If the points (–2,10) and (4,–8) are located on a graph anda straightedge is used to draw a line through them, onesees that the y-intercept of the line is 4.

The slope of the line is –3 since y values decrease by 3 asx values increase by 1. One way to determine this is tonote the y values decrease 18 units (from 10 to –8) as thex values increase 6 units (from –2 to 4), which is equiva-lent to a y-decrease of 3 units for every 1 unit x-increase.

Since the line has slope –3 and y-intercept 4, y = v(x) =–3x + 4. The students may use other methods to find aformula for v(x).

Shown below are various sketches and Algebra Piecerepresentations of y = –3x + 4, some of which includeedge pieces. In the sketches, a numeral alongside the edgeof a rectangle denotes the value of the edge. Note thatthe value of an edge may differ from its length—the lengthof an edge is always positive or zero, while the value of anedge can be positive, negative, or zero. If the value of anedge is positive, the value of the edge and its length are

continued next page

–4–6–8

18

16

14

10

8

6

4

2

–2

–4

–6

8

–10

–12

–14

–16

–18

–2–3–5–7 –1 1 2 3 4 5 6 7 8

increase of 6

decreaseof 18

(–2, 10)

(4, –8)

v (x)

x

slope = –18⁄6 = –3 the same. If the value of an edge is negative, the value ofthe edge and its length are opposites. For example, if thevalue of an edge is –3, its length is 3.

–x

3

–x–3x–x

–x4

x

–3 –3x 4

o o o o

o o o oo o o o

o o o oo o o o

o o o o

o o o o

o o o oo o o o

o o o o

o o o o

o o o oo o o o

o o o oo o o o

o o o o

oo

o o o o

o o o o

ACTIONS COMMENTS



6 Repeat Action 5 for the points(–2,–8) and (4,7).

6 In the graph shown below, the x-axis is scaled so that eachsubdivision represents 1⁄2 unit and the y-axis is scaled so thateach subdivision represents 1 unit. The graph’s appearance willdiffer for other scales, but will still be a straight line.

–2

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–3 –1 1 2 3

v (x)

x

increaseof 15

increaseof 6

4

– 3

v(x) = 2 x – 3v (x) = x – 3

x5__2

5__2

1__2

x

5__2

Using a straightedge to draw a line connecting the 2 given points,one sees that the y-intercept is –3. Also, y values increase by 15as x values increase by 6. This is equivalent to a y-increase of 21⁄2for each x-increase of 1. Thus, the line has slope 5⁄2. Hence, y =v(x) = (5⁄2)x – 3. The students can verify this formula by showingthat it provides the correct values for v(–2) and v(4), namely –8and 7.

On the left below is an Algebra Piece representation in which anx-frame has been cut in half. On the right is a sketch in which thevalues of edges and regions are shown.

ACTIONS COMMENTS



7 Give each student 2 or 3 sheetsof coordinate grid paper. Ask themto construct or sketch an AlgebraPiece representation of the xtharrangement of a complete sequenceof arrangements for which the graphof v(x) is a straight line that satisfiesthe conditions in a) below. Inviteseveral volunteers to sketch theirxth arrangements at the overheadand then have the students graphv(x) and verify that the conditions ina) are met. Discuss relationshipsamong the Algebra Piece represen-tations and the equations and graphsthat represent them. Repeat for b)-h).

a) y-intercept is 4

b) slope is 3

c) slope is –2 and y-intercept is 3

d) slope is 0 and y-intercept is 7

e) slope is 3⁄4

f) slope is –4⁄6 and y-intercept is –3

7 The collection of all equations that satisfy a) form a family oflines whose y-intercept is 4. Similarly, the collections of equationssatisfying b) and e) form families of lines. This idea is investigatedfurther in Lesson 12. The conditions given in c), d), f), and g) eachproduce a unique line.

a) If the graph of v(x) has y-intercept 4, then v(0) = 4. This will bethe case if the “constant” part of the arrangements has value 4.Shown below are 2 possibilities.

v(x) = x + 4 v(x) = –2x + 4

The graphs of the above equations have y-intercept 4 and slopes1 and –2, respectively.

b) If the graph of v(x) is a straight line whose slope is 3, then thevalues of the arrangements must increase by 3 as x increases by 1.One possibility is that v(0) = 0, v(1) = 3, v(2) = 6, and so forth.This will be the case if v(x) = 3x. Other possibilities can be obtainedby adding a constant to this expression, e.g., v(x) = 3x + 2. Thedifference in the graphs of these 2 expressions is that the y-interceptof the 1st is 0 and that of the 2nd is 2.

c) If the graph of v(x) is a straight line and both the slope andy-intercept are given, there is only one possibility. In this case,v(x) = –2x + 3, or an equivalent expression.

o o o o

o o o o

o o o o

o o o o

v(x) = 3x v(x) = 3x + 2

continued next page

ACTIONS COMMENTS



7 continuedd) The line with 0 slope and y-intercept (0,7) has equation y = 7.The slope of any horizontal line y = b, for b a real number, is 0,since y does not change as x increases.

e) Since the slope of this line is 3⁄4, then the line must “rise” (i.e.,change upwards vertically) 3⁄4 unit for every 1 unit of “run” (i.e.,left to right horizontal change). Or, in other words, it must rise 3units for every run of 4 units. Hence, any line whose equation ofthe form y = (3⁄4)x + b, where b is a real number, has slope 3⁄4.

f) The equation for this line is y = (–4⁄6)x – 3. Some students maypoint out this could also be written as y = (–2⁄3)x – 3.

Note that if y = (–4⁄6)x – 3 then 6y = –4x – 18 which can berewritten as 4x + 6y = –18. The latter equation is commonlyreferred to as the standard form of the equation. In general, anequation ax + by = c, where a, b, and c are integers, is in standardform. Notice the variables are on one side of the equation and theconstant is on the other. It is not as easy to “see” the slope andintercept of the graph of a linear equation in standard form ascompared to an equation in slope-intercept form.

8 a) Shown below is an Algebra Piece representation, with edge pieces.

v1(x) = (2 – x)(3 + x)

8 Give each student a sheet ofcoordinate grid paper.

a) Ask the students to build orsketch an Algebra Piece representa-tion of the xth arrangement of acomplete sequence of arrangementsfor which v1(x) = (2 – x)(3 + x).

b) Ask the students to predict whatthe graph of v1(x) looks like. Discusstheir predictions and then ask themto draw the graph. Ask the studentsfor their observations.

y

x

y = b

(0,0)

(x2,b)(x1,b)

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o

o

ACTIONS COMMENTS



b) Since v1(0) = 6, the y-intercept is 6. The students may observethat the x-intercepts are 2 and –3 since v1(x) = 0 for those valuesof x. If x is in the interval between the x-intercepts, both factorsof v1(x) are positive and v1(x) is positive (i.e., for –3 < x < 2, v1(x)> 0). Outside this interval, one factor of v1(x) is positive and theother is negative, so v1(x) is negative (i.e., for x < –3 and x > 2,v1(x) < 0).

An alternate form for v1(x) is –x2 – x + 6, as can be seen from theAlgebra Piece representation shown above.

In the graph of y = v1(x) shown below, every subdivision of thex-axis represents 1⁄2 unit and every subdivision of the y-axisrepresents 1 unit. The shape of the graph will vary for otherscalings of the axes; however, regardless of the scaling, the graphis a parabola that is symmetric about the vertical line x = –1⁄2 andopens downward.

continued next page

y

x–2

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

3 –1 1 2 3 4–4

y = v1(x) = (2 – x)(3 + x)

Some students may find a few points on the graph and connectthese points with straight line segments. If so, you might havethem find more points on the graph to help show that it isrounded rather than angular.

ACTIONS COMMENTS



9 If y = x and y = (2 – x)(3 + x) are graphed on the same coordi-nate system, it appears that the two graphs intersect when x isabout 1.7 and –3.7.

The exact values of x where the 2 graphs intersect can be foundby determining when v1(x) = v2(x), i.e., when (2 – x)(3 + x) = x.This is the case if the value of the circled portion of the AlgebraPiece representation for v(x), shown at the left, is 0. If the valueof the circled collection is 0, the value of its opposite collection,shown below, is also 0.

Thus the portion of the collection consisting of the x2-mat andthe 2 x-strips has value 6. If one black counting piece is added tothis collection, it has value 7 and can be arranged into a squarewith edge (x + 1). Thus, x + 1 equals 7 or – 7 , so x = –1 + 7or x = –1 – 7 . Since 7 ≈ 2.65, x ≈ 1.65 or x ≈ –3.65.

9 Ask the students to graph v2(x) = xon the same coordinate axes astheir sketch of v1(x) = (2 – x)(3 + x).Have them find where the graphsintersect. Discuss.

y

x–2

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

3 –1 1 2 3 4–4

y = (2 – x)(3 + x)

y = x

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

x + 1

x + 1

ACTIONS COMMENTS



10 The students may find it helpful to construct a table of values.

The graph of v(x) is shown below.

y = v(x) = 1⁄2x2 – x – 8

Since v(0) = –8, the y-intercept is –8.

The x-intercepts occur when v(x) = 0, that is when 1⁄2x2 – x – 8 =0. This will be the case if 1⁄2x2 – x = 8, or, doubling, if x2 – 2x = 16.Completing the square, as shown in the Algebra Piece illustration,one has (x – 1)2 = 17. Hence, x – 1 is 17 or – 17 and x = 1 +

17 or x = 1 – 17 . With the aid of a calculator, one finds 1 +17 ≈ 5.12 and 1 – 17 ≈ –3.12.

The “turning” point occurs on the graph’s line of symmetry, whichis x = 1. Since v(1) = –17⁄2, the coordinates of the turning pointare (1,–17⁄2).

10 Give each student a sheet ofcoordinate grid paper. Tell them thatthe value of the xth arrangement ofa certain complete sequence ofarrangements is v(x) where v(x) =1⁄2x2 – x – 8. Ask the students toconstruct a graph of v(x) and to findthe graph’s x- and y-intercepts andthe coordinates of its “turning”point. Discuss.

continued next page

y

x–2

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–3 –1 1 2 3 4–4–5–6–7 5 6 7 8

o o o o

o o o o

o o o o

o o o o 1

16

x2 – 2x + 1 = (x – 1)2 = 17

ACTIONS COMMENTS



11 Give each student a sheet ofcoordinate grid paper. Show thestudents the following pair of ex-pressions:

v1(x) = –4x – 3v2(x) = 1⁄2x2 + 3x – 27⁄2

Tell the students these are thevalues of the xth arrangements oftwo complete sequences of arrange-ments. Ask the students to do thefollowing:

a) graph v1(x) and v2(x) on the samecoordinate axes,

b) determine the x- and y-interceptsof the graphs,

c) determine the values of x forwhich v1(x) = v2(x).

11 a) A transparency of the completed graph can be made fromFocus Master 11.3.

b) For v1(x): x-intercept is –3⁄4; y-intercept is –3. For v2(x):x-intercepts are –9 and 3; y-intercept is –27⁄2.

c) If –4x – 3 = 1⁄2x2 + 3x – 27⁄2, then, doubling both arrangements,–8x – 6 = x2 + 6x – 27 and, adding 8x + 27 to both arrangements,21 = x2 + 14x. Hence, the square shown below has value 70. Thusx + 7 = 70 or – 70. Hence, x = –7 + 70 or x = –7 – 70. Since

70 ≈ 8.37, x ≈ 1.37 or x ≈ –15.37. The common point on thegraphs of v1(x) and v2(x) which occurs at the latter value of x isbeyond the scope of the graph shown above.

x2

xxx

49

x + 7

xxxx

x x xxxxx

xxxx

(x + 7)2 = x2 + 14x + 49 = 21 + 49 = 70



y

x2

24

22

20

18

16

14

12

10

8

6

4

2

2

4

6

8

10

12

14

16

18

20

3 1 1 2 3 44567 5 6

v2(x ) = –4x – 3

v1(x ) = 1⁄2x2 + 3x – 27⁄2

8910111213

ACTIONS COMMENTS



continued next page

12 Repeat Action 11 for the pair:v1(x) = (x – 6)(x + 2)v2(x) = –x2 + 10x – 16

12 a) A transparency of the completed graph can be made fromFocus Master 11.4.

b) Since a product is 0 if, and only if, one of its factors is 0, v1(x) = 0when x – 6 = 0 or x + 2 = 0, that is, if x = 6 or x = –2. Thus thex-intercepts of the graph of v1(x) are 6 and –2. Its y-intercept isv1(0), which is –12.

The x-intercepts of v2(x) occur when v2(x) = 0, that is when anarrangement for –x2 + 10x – 16 has value 0. Thus –x2 + 10x = 16or, taking opposites, x2 – 10x = –16. As shown in the sketch below,this implies (x – 5)2 = 9, in which case x = 8 or x = 2.

The y-intercept is v2(0), which is –16.

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

–16

x – 5

(x – 5)2 = x2 – 10x + 25 = –16 + 25 = 9



y

x2

18

16

14

12

10

8

6

4

2

2

4

6

8

10

12

14

16

18

3 1 1 2 3 445 5 6

v2(x ) = –x2 + 10x – 16

v1(x ) = (x – 6)(x + 2)

7 9 10

ACTIONS COMMENTS



12 continuedc) The values, v1(x) and v2(x), are equal if the two arrangementsshown below have the same value.

Adding x2 – 10x + 12 to both of these arrangements, one has 2x2

– 14x = –4. or, halving, x2 – 7x = –2. Completing the square, asshown below, gives (x – 7⁄2)2 = –2 + 49⁄4 = 41⁄4.

Thus, x – 7⁄2 = 41⁄2 or – 41⁄2. So x = (7 + 41)⁄2 or x = (7 – 41)⁄2. Withthe help of a calculator, one has x ≈ 6.70 or x ≈ 0.30.

x2

x

v1(x) = (x – 6)(x + 2) = x2 – 4x – 12

x

x

x

–12

x– x– x– x– x– x–

–6

2

–x2 –16x

v2(x) = –x2 + 10x – 16

x x x x x x x x x

x2

x

(x – )2 = (x2 – 7x) +

x–

x

–72

x–x–

–12

x494

x– x– x–

–72

–12

x

72

494

494

414= –2 + =

ACTIONS COMMENTS



Alternatively, instead of halving 2x2 – 14x, one could double it, toobtain 4x2 – 28x = –8. Then, completing the square, as shown onthe left, one has (2x – 7)2 = 41. Whence 2x – 7 = 41 or – 41and, as before, x = (7 + 41)⁄2 or x = (7 – 41)⁄2.

You might ask the students to record the steps in their solutionusing algebraic notation. For example, the steps in the last solu-tion might be recorded as follows:

(x – 6)(x + 2) = –x2 + 10x – 16x2 – 4x – 12 = –x2 + 10x – 16

x2 – 4x – 12 + (x2 – 10x + 12) = –x2 + 10x – 16 + (x2 – 10x + 12)2x2 – 14x = –44x2 – 28x = –8

4x2 – 28x + 49 = –8 + 49(2x – 7)2 = 41

2x – 7 = ± 412x = 7 ± 41x = (7 ± 41)⁄2

x

(2x – 7)2 = (4x2 – 28x) + 49 = –8 + 49 = 41

x

x2

x

x

–7x

–7x

–7x –7x 49

–7

–7

x2

x2x2

13 (Optional) a) Have the studentsfind x if 2x2 + 5x –10 = 0. Ask forvolunteers to show their solutions.Discuss the methods the studentsuse. Repeat for 3x2 + 5x – 10 = 0.

b) (Challenge) Ask the students tofind all values of x for which ax2 +bx + c = 0 where a, b, and c arearbitrary integers.

continued next page

13 a) You can ask the volunteers to use Algebra Pieces orsketches to illustrate their thinking.

If 2x2 + 5x – 10 = 0, then 2x2 + 5x = 10 and x2 + 5⁄2x = 5. Com-pleting the square, as shown below, one has (x + 5⁄4)2 = 5 + 25⁄16 =105⁄16. Hence, x + 5⁄4 = 105 )⁄4 or – 105 )⁄4. Thus x = (–5 + 105 )⁄4 or(–5 – 105 )⁄4.

x2

x

(x + )2 = (x2 + x) +

x

54

2516

x

54

54

2516

= 5 + =

54

x54

52

2516

10516

ACTIONS COMMENTS



13 continuedThe introduction of fractions can be delayed by multiplying theoriginal equation by an amount so that the coefficient of x2 is aperfect square and the number of x-strips can be divided evenlyalong the sides of the resulting square. This means that thenumber of x-strips must be divisible by twice the length of theresulting square. This can be accomplished by multiplying theoriginal equation by 4 times the original coefficient of x2, in thiscase 4 x 2 or 8. Increasing amounts by a factor of 8, the equation2x2 + 5x = 10 becomes 16x2 + 40x = 80. Then completing thesquare as shown in the sketch, one has (4x + 5)2 = 80 + 25 = 105.Thus 4x + 5 = 105 or – 105 so 4x = –5 + 105 or 4x = –5 – 105 ,and, as above, x = (–5 + 105 )⁄4 or (–5 – 105 )⁄4.

If 3x2 + 5x – 10 = 0, then 3x2 + 5x = 10. Multiplying by 4 times 3,or 12, this becomes 36x2 + 60x = 120. Completing the square, asshown, one has (6x + 5)2 = 120 + 25 = 145. Hence, 6x + 5 = 145or – 145 and x = (5 + 145 ) ⁄6 or (5 – 145 ) ⁄6.

b) If ax2 + bx + c = 0 , then ax2 + bx = –c. Multiply by 4a: 4a2x2 +4abx = –4ac. Complete the square: (2ax + b)2 = 4a2x2 + 4abx + b2

= –4ac + b2.Thus: 2ax + b = 2b − 4ac or – 2b − 4ac

2ax = – b + 2b − 4ac or –b – 2b − 4ac

x = (–b + 2b − 4ac )⁄2a or (–b – 2b − 4ac )⁄2a.

This result is known as the quadratic formula. You may want to askthe students to use the formula to solve the equations of part a).

If b2 – 4ac < 0, the solutions are complex numbers. These arediscussed in Lesson 14.

16x2

4x(4x + 5)2 = (16x2 + 40x) + 25 = 80 + 25 = 105

20x

5

5

4x

20x 25

36x2

6x(6x + 5)2 = (36x2 + 60x) + 25 = 120 + 25 = 145

30x

5

5

6x

30x 25

4a2x2

(2ax + b)2 = (4a2x2 + 4abx) + b2

= –4ac + b2

= b2 – 4ac

2ax 2abx

2abx b2b

2ax b




v(n) = __________________________

22

20

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

–22

x

y

–4–6–8 –2–3–5–7 1 2 3 4 5 6 7 8–1




10

8

6

4

2

–2–2 –1 1 2

y = 4x + 2

y - intercept

x - intercept

increase of 4

increase of 4

slope

1

1




y

x–2

24

22

20

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

–3 –1 1 2 3 4–4–5–6–7 5 6

v2(x ) = –4x – 3

v1(x ) = 1⁄2x2 + 3x – 27⁄2

–8–9–10–11–12–13




y

x–2

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

–3 –1 1 2 3 4–4–5 5 6

v2(x ) = –x2 + 10x – 16

v1(x ) = (x – 6)(x + 2)

7 8 9 10



1 Using the given xth arrangement from a complete sequence of arrangements, find the

missing values in each table below. Explain how you arrived at your answers.

a) b)

2 For each of a)-d) below, sketch the graph of a straight line that satisfies the given

conditions. Write an equation for the line.

a) slope of 3 and y-intercept of –2

b) passes through points (–2,–9) and (3,11)

c) passes through (2,1) and x-intercept is 3

d) slope is 0 and passes through (–7,–3)

3 Equations a) and b) below each represent the value of the xth arrangement of a com-

plete sequence of arrangements. Graph a) and b) on the same coordinate system. Find all x-

intercepts, y-intercepts, the coordinates of the turning points, and the values of x for which

v1(x) = v2(x). Explain how you arrived at your answers.

a) v1(x) = (x + 4)(x – 3)

b) v2(x) = (2 – x)(5 + x)

4 Solve the following equations for x. Draw sketches to illustrate your methods, then

record the steps in your solutions using algebraic symbols.

a) –x⁄2 + 8 = 3x + 5

b) (x + 4)(x – 3) = –x + 3

c) x2 – x – 12 = –x2 + x


x v(x)___ 1731⁄2 ______ 20031 901⁄2

x v(x)55 ______ –230611⁄4 –121___ 178

x

x

x

–x

–x




1 a) v(x) = 3x – 21/2, so:if v(x) = 17 then 3x = 17 + 21/2 = 18 + 11/2, andx = 61/2;if x = –31/2, then v(x) = 3x – 21/2 = –101/2 – 221/2 = –13;if v(x) = 200, then 3x = 200 + 21/2 = 201 + 11/2 andx = 671/2.

b) v(x) = –2x + 11/2, so:if x = 55, then v(x) = –110 + 11/2 = –1081/2;if v(x) = –230, then –2x = –230 –11/2, so x = 115 +3/4 = 1153/4;if v(x) = 178, then –2x = 178 – 11/2, so x = –89 + 3/4

= –881/4.

2

3 a) x-intercepts: 3 and –4; y-intercept: –12; turningpoint: (–1/2, –121/4)

b) x-intercepts: 2 and –5; y-intercept: –10; turningpoint: (–3/2, 121/4)

y

x

10

9

8

7

6

5

4

3

2

1

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

2–4 6 8–2 4–6–8

11

1 5 73–3 –1–5–7

a) y = 3x – 2

b) y = 4x – 1

c) y = –x + 3d) y = – 3

y

x

20

18

16

14

12

10

8

6

4

2

–2

–4

–6

–8

–10

–12

–14

–16

–18

–20

–22

2–4 6 8–2 4–6–8 1 5 73–3 –1–5–7

v2(x ) = (2 – x)(5 + x)

22

v1(x ) = (x + 4)(x – 3)



3 continuedIf v1(x) = v2(x), adding x2 + 3x + 12 to both arrange-ments, one has 2x2 + 4x = 22, or, halving, x2 + 2x = 11.See the sketches below. Completing the square gives(x + 1)2 = 12. Hence, x + 1 = 12 or – 12. So x =–1 + 12 or x = –1 – 12. Since 12 ≈ 3.46, x ≈ 2.46or –4.46.

Adding x2 + 3x + 12 to both arrangements:

Halving each arrangement:

Completing the square:

2x2 + 4x 22

x2 + 2x 11

(x + 1)2 = x2 + 2x + 1 = 11 + 1 = 12

6

–x + 16

In symbols:–x⁄2 + 8 = 3x + 5–x + 16 = 6x + 10

–x + 16 + (x – 10) = 6x + 10 + (x – 10)6 = 7xx = 6⁄7

∴ x = 6⁄7

v1(x) = (x + 4)(x – 3)= x2 + x – 12

v2(x) = (2 – x)(5 + x)= –x2 – 3x + 10

x2

x

x

x

x–12

x– x– x–

–3

4 xx

–x2

x

–x

10

x– x– x–

5

2 xx

x– x–

x2

xxxx

22

x2

x2

xx

1 1

x2

x

x

1

x

x

1

1

12


8–x x

xx

25

16 xxx

10

xxx

–x

6 xxx

xx

xx

–x⁄2 + 8 3x + 5

7x

6x + 10

4 a)

Doubling both arrangements:

Adding x – 10 to both:

continued



4 b)

Add x + 12 to both arrangements:

Complete the square:

In symbols:(x + 4)(x – 3) = –x + 3

x2 + x – 12 = –x + 3x2 + x – 12 + (x + 12) = –x + 3 + (x + 12)

x2 + 2x = 15x2 + 2x + 1 = 15 + 1

(x + 1)2 = 16x + 1 = ±4

x = – 1 ± 4x = 3, – 5

(x + 1)2 = x2 + 2x + 1 = 15 + 1 = 16x + 1 = 4 or –4x = 3 or –5


x2

xx

15

(x + 4)(x – 3) = x2 + x – 12 –x + 3

x2 + 2x

x2

x

x

x

x–12

x– x– x–

–3

4 xx

3–x

4 c)

x2 – x – 12 –x2 + x

Add x2 – x + 12 to each arrangement:

Double both arrangements:

Complete the square:

(2x – 1)2 = 4x2 – 4x + 1 = 24 + 1 = 252x – 1 = 5 or –5x = 3 or –2

In symbols:x2 – x – 12 = –x2 + x

x2 – x – 12 + (x2 – x + 12) = –x2 + x + (x2 – x + 12)2x2 – 2x = 12

4x2 – 4x = 244x2 – 4x + 1 = 25

(2x – 1)2 = 252x – 1 = ±5

2x = 6, – 4x = 3, – 2

x2

–12–x –x2 x

x2

12–x –xx2

x2 24–x –xx2

x2 –x –xx2

x2 –x

–x

x2

x2 –x

–x

x2

2x –1

2x

1–1

25

x2

x

x

1

x

x

1

1

16



THE BIG IDEASketches that show the essential features of a math-ematical situation are a valuable aid in problem solving.Many word problems found in beginning algebracourses are readily solved with the help of an appro-priate sketch or diagram.

OverviewStudents draw sketches thatillustrate geometric andnumerical situations. They usetheir sketches to help themanswer questions about thesesituations.

Materials� None, other than paper and

pencil.

OverviewStudents use sketches anddiagrams to model mathemati-cal situations. They reasonfrom their sketches to solveproblems and find solutions ofequations.

Materials� Focus Master 12.1, 1 trans-

parency.

� Focus Master 12.2, 1 copyper student.

� Focus Master 12.3 (2 pages),1 copy per student.

OverviewStudents use sketches to solveproblems. They record theirthought processes using alge-braic symbols and equations.


student.

LESSON 12SKETCHING SOLUTIONS


TEACHER NOTES


START-UP

Overview Materials

SKETCHING SOLUTIONS LESSON 12

Students draw sketches that illus-trate geometric and numericalsituations. They use their sketchesto help them answer questionsabout these situations.

� None, other than paperand pencil.

1 Ask the students to draw a sketchof a rectangle.

2 Ask the students to draw andlabel a sketch of a rectangle whoselength is 6 inches longer than itswidth. Ask for volunteers to showtheir sketches. Discuss with thestudents whether or not the sketchesadequately convey the informationgiven, and whether the words andsymbols used are absolutely essential.

3 Tell the students the perimeterof the rectangle in Action 2 is 56inches. Ask them to determine thedimensions of the rectangle, Discussthe methods the students use.

1 Students often have difficulty in drawing sketches that disclosethe essential features of a situation. As a first step in developingtheir sketching skills, the students are asked to sketch a familiarfigure. If your students have had experience using sketches anddiagrams to solve problems visually, you may want to skip theStart-Up and proceed to the Focus.

Most students will draw a rectangle that is wider than it is tall. Itis likely their sketches will contain no words or symbols. Gener-ally, it is unnecessary to label a sketch of a rectangle for thestudents to recognize what has been drawn.

2 Below are some possible sketches.

3 The students will arrive at the dimensions in various ways.Some may observe that if the length is reduced by 6 inches, therectangle becomes a square whose perimeter is 44 inches. Hence,the length of its side is 11 inches. Thus the dimensions of theoriginal rectangle are 11 and 11 + 6, or 17, inches.

Others may note that the sum of the width and length is half of56, or 28, and their difference is 6. Hence they may search fortwo numbers that differ by 6 and add to 28.

Still others may see that the perimeter consists of 2 segments oflength 6 and 4 segments of unknown but equal length. The sum ofthe lengths of these latter segments is 56 – 12 or 44. Hence eachsegment is 11, and the dimensions of the rectangle are 11 and11 + 6, or 17.

ACTIONS COMMENTS

6

6

6

6

d

d

d d

ACTIONS COMMENTS


START-UPSKETCHING SOLUTIONS LESSON 12

4 The last solution in Comment 3 might be recorded as follows.Note, that the algebraic equations become a symbolic way ofrecording one’s thinking. They are the record of a chain of thoughtrather than a series of manipulations carried out according toprescribed rules.

5 Having the students draw sketches of a situation before aproblem is posed focuses their attention on creating a sketch thatportrays the essential features of the situation.

Below are some possible sketches. Notice that, in the last sketchshown, the essential information is carried in the symbols and notthe sketch—that is, if the symbolic phrase “3w + 4” is erased, thedistinguishing feature of the rectangle is lost.

6 The students will use various methods to arrive at the dimen-sions. One way is to note that the perimeter of 48 inches consistsof 2 segments of length 4 and 8 other segments of equal length.Hence, the lengths of the 8 segments total 40 inches, so each is 5.Thus, the dimensions of the rectangle are 5 inches and 3 x 5 + 4= 19 inches.

4 Discuss with the students howalgebraic symbols and equations canbe used to record their thinking infinding the dimensions of the rect-angle in Action 3.

5 Ask the students to draw a sketch,using as few words and symbols aspossible, that portrays a rectangle ofunknown dimensions whose length is4 units longer than 3 times its width.Ask for volunteers to replicate theirdrawings on the overhead. Discusswhether the drawings adequatelyconvey the information given aboutthe rectangle and whether the wordsand symbols used are essential.

Person’s Thinking

The perimeter of 56 consists of 2segments of length 6 and 4 othersegments of the same length thatwe’ll call d.

So, the 4 segments of length d havea total length of 56 – 12 or 44.

Thus, the length of each segment is44 ÷ 4, or 11.

Hence, the dimensions of therectangle are 11 and 11 + 6.

Algebraic Symbols

56 = 2(6) + 4d = 12 + 4d

4d = 56 – 12 = 44

d = 44 ÷ 4 = 11

width = d = 11length = d + 6 = 17

x

4 4x x x

w

3w + 4

6 Tell the students that the perim-eter of the rectangle they drew inAction 2 is 48 inches. Then ask themto determine the dimensions of therectangle. Ask for volunteers todescribe their thinking.

ACTIONS COMMENTS



7 Repeat Action 5 for a rectanglewhose length is 5 inches less thantwice its width. Then ask the stu-dents to determine the dimensionsof the rectangle if its perimeter is32 inches. Have several studentsdescribe their thinking in determin-ing the dimensions of the rectangle.

8 Ask the students to sketch asquare. Then have them sketch anequilateral triangle whose sides are2 units longer than the sides of thesquare. Ask the students to reasonfrom their diagrams to determinethe length of the side of the squareif the square and the triangle haveequal perimeters. Ask for volunteersto show their sketches and describetheir thinking.

9 Ask the students to draw diagramsor sketches which represent anumber and that number increasedby 6. Show the various ways in whichstudents have done this. Then askthe students to use one of thesketches to determine what thenumbers are if their sum is 40.

7 To emphasize the mathematical relationships in the rectangle,it is helpful to discuss the students’ sketches before telling themthe perimeter of the rectangle. Here is one sketch:

The extended rectangle shown above has a perimeter 10 incheslonger than the original rectangle, so its perimeter is 42 inches.These 42 inches are composed of 6 equal lengths. So each ofthese lengths is 7 inches. The width of the original rectangle isone of these lengths, or 7 inches; the length of the rectangle is 5inches less than 2 of these lengths, or 9 inches.

8 The perimeter of the square, in the following drawing, contains4 segments of length s; that of the equilateral triangle contains 3segments of length s and 3 of length 2. Thus, the 3 segments oflength 2 must sum to s. So s is 6.

9 Since numbers have no particular shape, the students mustinvent a way of portraying number. They might do this in a varietyof ways, e.g., as a length or as an area or as a “blob.”

Looking at the sketch on the left above, the smaller number isrepresented by a line segment and the larger number by two linesegments. The sum of the lengths of these segments is 40. Thesmall segment has length 6. Hence, the sum of lengths of theother 2 segments is 34. Since these 2 segments are congruent, thelength of each is 34 ÷ 2, or 17. Hence the 2 numbers are 17 and17 + 6, or 23.

ww

w

5

s

s

s

ss

s

s

2

2

2

1 11 11 16

ACTIONS COMMENTS



10 If the sum of the 2 numbers is 36, in the sketch shownbelow, the sum of the lengths of the 5 congruent segments (1segment in the smaller number and 4 in the larger number) is 36 – 1,or 35. Hence, the length of each is 35 ÷ 5, or 7. Thus the numbersare 7 and 4(7) + 1, or 29.

10 Ask the students to drawsketches that represent 2 numberssuch that 4 times the smaller num-ber is 1 less than the larger. Thenask them to reason from theirsketches to determine the numbersif their sum is 36. Ask for volunteersto show their sketches and explaintheir reasoning. larger number

1

smaller number


FOCUS

Overview Materials


Students use sketches and diagramsto model mathematical situations.They reason from their sketches tosolve problems and find solutions ofequations.

� Focus Master 12.1, 1transparency.

� Focus Master 12.2, 1copy per student.

1 Place a transparency of FocusMaster 12.1 on the overhead, re-vealing only Situation a). Have thestudents draw a sketch or diagramthat represents the given informa-tion. Ask for volunteers to showtheir sketches.

2 Ask the students to reason fromtheir drawing to determine howmany people are in each group. Askfor several volunteers to describetheir thinking.

1 Shown below are 2 possibilities.

2 Looking at the first drawing above, the number of people inthe extended rectangle is 48. This extended rectangle consists of4 equal regions. Hence, each region contains 12 people. One ofthese regions represents the second group; hence, the secondgroup contains 12 people. The first group is 5 less than thenumber in 3 regions, or 31.

You may want to discuss with the students how their thinkingmight be represented in a sequence of algebraic statements. Forexample, in the above argument, if one lets x represent thenumber of people in a region, the above solution might be re-corded symbolically as follows:

x = number in second group,4x = 43 + 5 = 48,x = 12 = number in second group,number in first group = 3x – 5 = 36 – 5 = 31.

� Focus Master 12.3, 1copy per student.

ACTIONS COMMENTS

51st

2nd43

2ndgroup

43

1st group 5



Situations

a) The people at a meeting are separated into 2 groups.

The 1st group has 5 less people than 3 times the number in the 2nd group.

There are 43 people at the meeting.

ACTIONS COMMENTS


FOCUSSKETCHING SOLUTIONS LESSON 12

3 Reveal Situation b) on FocusMaster 12.1. Ask the students todraw a sketch or diagram thatrepresents the given situation andthen use their drawing to answer anappropriate question about thesituation. Ask for volunteers toshare their sketches and solutions.

4 Repeat Action 3 for the remain-ing situations on Focus Master 12.1.

3 To facilitate sharing, you might have overhead pens and half-sheets of blank transparencies available so students can preparetheir drawings prior to presenting them to the class.

Below is one possibility, in which a sketch is used to answer thequestion, “What are the three numbers?”

Each box represents 112 ÷ 7. or 16. Therefore, the numbers are32, 16, and 64.

Following is an algebraic representation of the visual reasoningused above:

If x = 2nd number,2x = 1st number,

2(2x) = 4x = 3rd number

So, 2x + x + 4x = 1127x = 112x = 112⁄7 = 16.

Therefore, the 3 numbers are 16, 32, and 64.

4 From time to time as students present their solutions, youmay wish to discuss how a student’s thinking might be repre-sented in a series of algebraic statements. However, the emphasisof this Action is on sketching situations and reasoning from them.

Following are sample questions and diagrams from which answersto these questions can be deduced.

2nd number

1st number

3rd numberSKETCHING SOLUTIONS LESSON 12


Situations




b)There are 3 numbers.

The 1st number is twice the 2nd number.

The 3rd is twice the 1st.

The sum of the 3 numbers is 112.

c) The sum of 2 numbers is 40.

Their difference is 14.

d)The sides of square A are 2 inches longer than the sides of square B.

The area of square A is 48 square inches greater than the area of square B.

e) Melody has $2.75 in dimes and quarters.

She has 14 coins altogether.

f) Three particular integers are consecutive.

The product of the 1st and 2nd integers is 40 less than the square of the 3rd integer.

g) Karen is 4 times as old as Lucille.

In 6 years, Karen will be 3 times as old as Lucille.

ACTIONS COMMENTS



continued next page

c) What are the numbers?

Solution 2

d) What is the length of the small square?

In each of the following, s is the side of the smaller square, B.

e) How many dimes and how many quarters does Melodyhave?

Solution 1

40 = sumlarge small

smallsmall14

40 – 14 = 261326__

2= 13difference

40

14

larger

smaller

The area, 40, of the shaded region is the sum ofthe 2 numbers; the area, 14, of the unshaded re-gion is the difference. The combined area, 54, ofthe shaded and unshaded regions is twice thelarger number. Hence, the larger number is 54 ÷ 2,or 27. The smaller number is 27 – 14, or 13.

The area of the unshadedborder is 48. Hence, the areaof each of the two 2 x s rect-angles is (48 – 4) ÷ 2 or 22.Thus, s is 11.

The area of each of thefour 1 x s rectangles is(48 – 4) ÷ 4, or 11. Thus,s is 11.

The area of the unshadedborder is 48. Hence, the areaof each of the four 1 x (s + 1)rectangles is 48 ÷ 4, or 12,and s is 11.

The value of each shaded bar is 5 x 14, or 70¢.Hence, the value of each unshaded bar is(275 – 140) ÷ 3, or 45¢. So, there are 9 quartersand 5 dimes.

The smaller number is 26 ÷ 2, or 13.The larger number is 40 – 13, or 27.

Solution 1 Solution 2 Solution 3

1 1

1 1

1

1 1

1s

s + 1

4

B

2

s

2

s

s B

1 1

Bs + 1s + 1

s + 1 1

1

no. of quarters no. of dimes14 coins

5¢5¢5¢5¢5¢

$2.75

$1.40

ACTIONS COMMENTS



5 Write the following on the over-head:

One pump can fill a tank in 6 hours.Another pump can fill it in 4 hours.If both pumps are used, how longwill it take to fill the tank?

Ask the students to create sketchesfrom which they can deduce theanswer to the question. After thestudents have worked for 10 to 15

4 continuedf) What are the 3 integers?

g) How old is Lucille?

5 The intent here is to engage the students in thinking about theproblem and then reflecting on other students’ work. Somestudents may not reach a solution before you distribute the FocusMaster; you might give them the option not to examine it untilthey are ready.

In Solution 1, a line segment representing the tank is divided into12 sections, which is a multiple of both 4 and 6. Pump A will fill 2of these sections in an hour while pump B will fill 3 of them.Together they will fill 5 of them in an hour, so it will require 22/5

hour to fill all 12 sections.

The area of the shaded rectangle is the product ofthe first 2 integers. The area of the unshaded regionis the difference between that product and the squareof the 3rd integer which is given to be 40. So, eachof the 3 unshaded rectangles has area (40 – 4) ÷ 3,or 12. Thus, the 3 numbers are 12, 13, and 14.

Comparison of Karen’s age in 6 years with 3times Lucille’s age in 6 years:

These have the same value if each box represents two6’s, or 12. So, Karen is now 48 and Lucille is 12.

1 11 1

11

1stinteger

3rdinteger

2nd integer

1st integer

2nd integer

3rd integer

agesnow

Karen

ages in6 years

Lucille

Karen

Lucille 6

6

Karen

Lucille (3 times)

6

6 6 6



Situations


















ACTIONS COMMENTS



minutes, distribute a copy of FocusMaster 12.2 to each student, point-ing out that these are sketchesvarious students have made inarriving at an answer to the ques-tion. Discuss with the students theirideas and questions about the think-ing behind the solutions, and howtheir sketches compare with thoseon Focus Master 12.2.

6 Give each student a copy ofFocus Master 12.3 (see followingpage). Select, or have the studentsselect, situations from the FocusMaster. Ask the students to posemathematical questions about theselected situations and then drawdiagrams or sketches from whichthey can determine the answers tothe questions. Discuss.

In Solution 2, it is determined that pumps A and B together canfill 5 tanks in 12 hours and hence, can fill 1 tank in 12/5 hours.

Note that in Solution 1, the length of the interval representing 1hour varies from one part of the sketch to the next, while inSolution 2, the length of an interval representing 1 hour remainsthe same.

The thinking in Solution 3 is similar to that in Solution 1 exceptthat the tank is represented by a rectangle rather than a linesegment.

6 You might ask for volunteers to state a question and show,without comment, the diagrams or sketches, appropriately la-beled, that they used to arrive at an answer. Then ask the otherstudents to suggest how the answer was deduced from thesketches.

Following is a sample question and solution for each of thesituations on the Focus master.

continued next page



One pump can fill a tank in 6 hours. Another pump can fill it in 4 hours. If both pumps

are used, how long will it take to fill the tank?

Pump A fills 4 subdivisions in 1 hour.Pump B fills 6 subdivisions in 1 hour.Together, they fill 10 subdivisions in1 hour:

Solution 1

Solution 2

Solution 3

Pump A

1 hour

Pump B

1 hour

Together

1 hour 1 hour of 1 hr.25

Pump A

Pump B

Time to fill 1 tank:

4 hours

6 hours

6 hours 6 hours

4 hours 4 hours 4 hours

Pump A

Pump B

Tanks filled in12 hours:

Pumps A and B fill 5 tanks in 12 hours.

Pump A fills tankin 1 hour.

hour

Pump B fills tankin 1 hour.

1 hour

1 hour

410

14

16

ACTIONS COMMENTS



6 continueda) How long does it take to empty the tank using only the2nd drain?

b) How many books does each girl have now?

c) How many people are in the room?

Working together, bothdrains empty 7 subdivi-sions in 1 hour. The 1stdrain empties 3, so the2nd drain empties 4 sub-divisions in 1 hour.

It takes the 2nddrain 51⁄4 hours toempty the tank.

Each of the boxes belowcontains the same numberof people; 3 of the boxescontain women and 2 con-tain men:

Doubling the men gives 4boxes of men. Adding 6 to thewomen (each X is a woman),gives 6 more than 3 boxes ofwomen:

If the number of men and women are equal, thelast box of men must contain 6 men. Thus, allboxes contain 6 people and, to begin with thereare 18 women and 12 men in the room.

The difference in length of the top and bottomarrows is the number of books Lisa has. Hence,she has 33 – 20, or 13, books. So, Maria has20 – 13, or 7, books.

1st drain empties tank in 1 hour.1__7

Both drains empty tank in 1 hour.1__3

hr.

1 hr.

1 hr.

1 hr.

1 hr.1 hr.1__4

2nd drain

3Lisa

3033

20

Lisa Maria

MM

WWWWMM

M

M

WW

X X X X X X



More Situations

a A tank has 2 drains of different sizes.

If both drains are used, it takes 3 hours to empty the tank.

If only the first drain is used, it takes 7 hours to empty the tank.

On Tuesday only the 2nd drain is used to empty the tank.

b) Yesterday Maria and Lisa together had 20 library books.

Today Maria and Lisa visited the library; Lisa checked out new books and now has

double the number of books that she had yesterday; Maria returned 3 of her books.

Now Maria and Lisa together have 30 books.

c) Of the people in a room, 3⁄5 are women.

If the number of men is doubled and the number of women increased by 6, there are an

equal number of men and women in the room.

d) On Moe’s walk home from school, after 1 mile he stopped for a drink of water.

Next, Moe walked 1⁄2 the remaining distance and stopped to rest at the park bench.

When Moe reached the park bench, he still needed to walk 1 mile more than 1⁄3 the

total distance from school to his home.

e) A gallon of paint contains 20% red paint and 80% blue paint.

Red paint is added until the mixture contains 50% red paint.

f) Standard quality coffee sells for $18.00 per kg.

Prime quality coffee sells for $24.00 per kg.

Every Saturday morning Moonman’s Coffee Shop grinds a 40kg batch of a standard/prime

blend to sell for $22.50/kg.

continued on back


FOCUS BLACKLINE MASTER 12.3 (CONT.)

g) A collection of nickels, dimes, and quarters has 3 fewer nickels than dimes and 3 more

quarters than dimes.

The collection is worth $4.20.

h) For a school play, Kyle sold 6 adult tickets and 15 student tickets.

Kyle collected $48 for his ticket sales.

Matt sold 8 adult tickets and 7 student tickets for the same school play.

Matt collected $38 for his ticket sales.

i) A nurse had 1200 ml of an 85% sugar solution (i.e., the container is 85% sugar and the

rest is water).

She added enough of a 40% sugar solution to create a 60% solution.

j) On Wednesday, a man drove from Gillette to Spearfish in 1 hour and 30 minutes.

On Thursday, driving 8 miles per hour faster, the man made the return trip in 1 hour and

20 minutes.

k) A student averaged 78 points on 3 history tests.

Her score on the 1st test was 86 points.

Her average for the 1st 2 tests was 3 points more than her score on the 3rd test.

l) Traveling by train and then by bus, a 1200 mile trip took Wally 17 hours.

The train averaged 75 mph and the bus averaged 60 mph.

ACTIONS COMMENTS



d) How far is it from school to Moe’s home?

e) How much pure red paint does Jill add?

Solution 1

The added part is 3⁄5 gallon.

Sketch I: Sketch II:

The areas of the rectangles below in Sketch I represent theamount of red paint in 1 gallon of the mixture and x gallons ofadded red paint. If the resulting mixture is to be 50% red paint,the 2 rectangles should be “leveled off” at 50. This will be thecase if, in Sketch II, area A = area B. Since area A is 30 andarea B is 50x, the areas are equal if 30 = 50x, that is, if x = 3⁄5.Hence, 3⁄5 gallon of red paint must be added.

parkbench home

total distance

school

A B

1__3

water

1 1

1__3

1__3

total distancetotal distance

B B

1 11

gal

20% =

red red

1__5 gal1__

5 gal1__5 gal1__

5 gal1__5 gal1__

5 gal1__5 gal1__

5

original gallon added part

gallons1

20%

0%x

100%

red paintin mixture

addedred

paint

gallons1

A

B

50%

x

bluepaint

blue

red

red

red

red

0%

100%

20%

0%

100%

0%

100%

50%

50%

30%

The 3 segments of length 1 comprise the otherthird of the distance. Hence, the distance fromschool to home is 9 miles.

Segments A and B are equal. Thus, replacing A with B.

Distance from school to home:

Solution 2

continued next page

ACTIONS COMMENTS



6 continuedf) How much prime coffee and how much standard coffeeare needed to produce 40 kg of blend?

g) How many of each coin are there?

Sketch I: Sketch II:

The heights of the rectangles represent the numberof coins and their bases the values, so the sum ofthe areas of the rectangles is the total value of thecollection. The value of the unshaded portion is$1.80. Hence, the value of the shaded rectangle is$4.20 – $1.80, or $2.40. Since the value of its base is40¢, its height is 2.40 ÷ .40 = 6. Thus, there are 6nickels, 9 dimes, and 12 quarters.

24

18

040 kg

value ofpremium

value ofstandard

10

A

B22.5

18

0

4.5

40

22.524

1.5

30

priceper kg

The areas of the rectangles in Sketch I represent the values ofthe coffees in the blend. If the blend is to sell for $22.50, the 2rectangles should “level off” at 22.5. This will be the case if, inSketch II, area A = area B. Since the height of B is 1⁄3 the heightof A, for the areas to be equal, the base of B must be 3 timesthe base of A. So, since the base of A plus the base of B is 40,the base of A is 10 and the base of B is 30. Hence, there shouldbe 10 kg of standard coffee and 30 kg of premium coffee.

5¢10¢ 25¢

75¢

75¢30¢

40¢values of coins

3

3numbersof coins

?

ACTIONS COMMENTS



I. Kyle’s sales: II. Matt’s sales:

III. Kyle’s sales increasedby a factor of 1⁄3:

IV. Removing II from III:

Sketch I:

Sketch II:

h) What is the cost of student and adult tickets?

i) How much of the 40% solution did the nurse add?

In the sketches A is the cost of an adult ticketand S is the cost of a student ticket. Vertical di-mensions represent the number of tickets sold.

Increasing Kyle’s sales by a factor of 1⁄3—so Kyleand Matt have the same number of adult sales—and removing Matt’s sales from the result, asshown in sketch IV, shows that 13 student tick-ets cost $26, so each cost $2. Thus, in sketch I,the 15 student tickets cost $30, so the 6 adulttickets cost $18, and each ticket costs $3.

Alternatively, one could quadruple Kyle’s salesand triple Matt’s sales. Then Kyle would have 24A sales and 60 S sales for a total of $192, whileMatt would have 24 A sales and 21 S sales for atotal of $114. So the $78 difference in sales is theresult of 39 S tickets. Hence, each S ticket is $2.

The areas of the rectangles in Sketch I at the leftrepresent the amount of sugar in the solutions. Ifthe resulting solution is to be 60% sugar, the 2rectangles should “level off” at 60. This will be thecase if, in Sketch II, area A = area B. Since the areaof B is 1200 x 25, or 30,000, and the height of A is20, for the areas to be equal, the width of A mustbe 30,000 ÷ 20, or 1500. Hence, 1500 ml of the 40%solution should be added.

48

A S

38

64

6 8 7

20

8

26 13

A S

A S A S

number oftickets sold

15

? 12000%

40%

85%

milliliters

sugar in40% solution

sugar in85% solution

?

A

B

12000%

25%60%

85%

40%

60%20%

milliliters

continued next page

ACTIONS COMMENTS



6 continuedj) How far is it from Gillette to Spearfish?

k) What were the student’s 3 test scores?

l) How far did the train travel?

The areas of the above rectangles represent distances traveled.Since the distances are the same, the areas are equal. Thus, ifone rectangle is superimposed on the other as shown above,the areas of rectangles A and B are equal. So, the distance be-tween Gillette and Spearfish is 64 x (11⁄2) = 96 miles.

Average score is 78: Moving 1 point from lastscore to each of 1st 2 scores,so average of 1st 2 scores is 3greater than 3rd score:

Moving 7 points from 2nd scoreto 1st score, makes 1st score 86:

The 2nd and 3rd scores are 72and 76, respectively.

The distance traveled is repre-sented by the area of the regionin the 1st sketch to the left. Thisregion can be divided into the 2rectangles shown in the 2ndsketch. The area of the lowerrectangle is 1020 miles. Hence,that of the upper is 180 miles, soits length is 180 ÷ 15, or 12hours. Thus, 12 hours of the tripwere by train, and the distancetraveled by train was 75 x 12, or900, miles.

distance fromGillette to Spearfish

Trip:distance fromSpearfish to

Gillette

Return Trip:1 – 1 =

1 hours

8

x

Area B = Area A

x = 64 mph

AB

= ,

1__3

1__3

1__2

1__6

32__3

x__6

32⁄3

x mph(x + 8) mph

1 hours1__2

x

1 1__2

1 1__3

x⁄6

78 78 78 79 79 76

86

72 7678 78 78 79 79 76

86

72 76

time(train)

1200 miles

speedmph

time(train)

17 hours

time(bus)

1020 miles

15

17

1200 – 1020 =180 miles

606075

ACTIONS COMMENTS



7 Here is one sketch of the rectangle:

One way to determine the dimensions of the rectangle is to find2 numbers which differ by 8 and whose product is 1428. If therectangle were a square, its dimension would be 1428 which isabout 38. Since its not a square, one dimension should be some-what larger than this and one dimension somewhat smaller. If oneguesses the dimensions are 34 and 42, a check will verify that thisis correct. Making an educated guess and then checking to see ifit is correct would be more difficult if the dimensions were notintegers.

Another way to proceed is by "completing the square," as shownin the sketches on the left. If the strip of width 8 in the abovesketch is split in two and half of it is moved to an adjacent side,as shown in Figure 1, the result is a square with a 4 by 4 cornermissing. Adding this corner produces a square of area 1428 + 16,or 1444, and edge x + 4, as shown in Figure 2. Hence, x + 4 is 1444, or 38, and x is 34. So the dimensions of the originalrectangle are 34 and 34 + 8, or 42.

The equation x(x + 8) = 1428 has been solved.

8 All of these equations can be solved by completing the square.

In the sketches that follow, differences are treated as sums, e.g.,x – 2 is thought of as x + (–2) and is portrayed by a line segmentof value x augmented by a segment of value –2.

a) x2 – 4x + 6 = 5

One can complete the square as shown in the following sequenceof sketches. Notice that, since x2 – 4x + 6 = 5, then x2 – 4x = –1and x2 – 4x + 4 = 3.

7 Ask the students to sketch arectangle whose length is 8 unitsgreater than its width. Then tellthem the area of the rectangle is1428 and ask them to find its di-mensions. Discuss the equationsthat have been solved.

8 Ask the students to draw sketchesto solve the following equations:

a) x2 – 4x + 6 = 5

b) x2 + 9x = 400

c) x(3x – 4) = 4

d) 2x(3 – x) = 3

x

x 8

1428

4

x4

1444

x + 4

16

x + 4

Figure 1 Figure 2

x

x 2 –4x x 2

–2x

4–2x–2

x 2 – 4x + 4 3

–2x – 2 x – 2x

xx – 2 x – 2

–1

continued next page

ACTIONS COMMENTS



8 a) continuedIf a square region has value 3, its edges have value 3 or – 3 .Hence, x – 2 = 3 or x – 2 = – 3 , and x = 2 + 3 or x = 2 – 3 .

b) x2 + 9x = 400

Completing the square gives the sequence of sketches shown below.

With the help of a calculator, one finds 420.25 = 20.5. Thus,

x + 4.5 = ± 20.5 and x = 16 or x = –25.

Fractions can be avoided by doubling dimensions as shown in thesketches below.

Since 1681 = 41, 2x + 9 = ±41 and the result follows.

c) x(3x – 4) = 4

In the following sequence, the second rectangular region isobtained from the first by increasing its height, and therefore itsarea, by a factor of 3.

= ± 420.25

x 2 9x x 2 x9__2

x9__2

9__2

81__4 81__

4

= 400 + 81__4

x + x + 4.5

x 2 + 9x +420.25

x

x 9__2

9__2

400

4x 2

18x

18x

81 4x 2 + 36x + 81

= 4(x 2 + 9x) + 81

= 4(400) + 81

2x + 9

1681

2x + 992x

9

2x

x 4

–2

–2 4

1612 12

3x –4 –43x 3x

3x

3x – 2

3x 3x – 2 = ±43x = 6 or 3x = –2x = 2 or x = –2⁄3

ACTIONS COMMENTS



d) 2x(3 – x) = 3

In the following sequence, the second rectangular region isobtained from the first by changing the value of its base, andtherefore its area, by a factor of –2.

Many sequences of sketches shown in the solutions above, andelsewhere in this lesson, contain more figures than may be in thesketches the students draw. In a number of instances, severalfigures shown in a sequence of sketches could be combined into asingle figure, especially if an oral presentation is being madeconcurrently, or if solutions are being developed for private useand not for the benefit of a reader.

2x

–x 3

2x

2x –6

2x

2x –3

9

2x – 3

3

–3

–63 –6

2x – 3 = ± 3 2x = 3 ± 3

x = 1⁄2(3 ± 3 )


TEACHER NOTES




Situations





















One pump can fill a tank in 6 hours. Another pump can fill it in 4 hours. If both pumps

are used, how long will it take to fill the tank?

Pump A fills 4 subdivisions in 1 hour.Pump B fills 6 subdivisions in 1 hour.Together, they fill 10 subdivisions in1 hour:

Solution 1

Solution 2

Solution 3

Pump A

1 hour

Pump B

1 hour

Together

1 hour 1 hour of 1 hr.2__5

Pump A

Pump B

Time to fill 1 tank:

4 hours

6 hours

6 hours 6 hours

4 hours 4 hours 4 hours

Pump A

Pump B

Tanks filled in12 hours:

Pumps A and B fill 5 tanks in 12 hours.

Pump A fills tankin 1 hour.

hour

Pump B fills tankin 1 hour.

1 hour

1 hour

4__10

1__4

1__6




More Situations

a) A tank has 2 drains of different sizes.

If both drains are used, it takes 3 hours to empty the tank.

If only the first drain is used, it takes 7 hours to empty the tank.

On Tuesday only the 2nd drain is used to empty the tank.

b) Yesterday Maria and Lisa together had 20 library books.

Today Maria and Lisa visited the library; Lisa checked out new books and now has

double the number of books that she had yesterday; Maria returned 3 of her books.

Now Maria and Lisa together have 30 books.

c) Of the people in a room, 3⁄5 are women.

If the number of men is doubled and the number of women increased by 6, there are

an equal number of men and women in the room.

d) On Moe’s walk home from school, after 1 mile he stopped for a drink of water.

Next, Moe walked 1⁄2 the remaining distance and stopped to rest at the park bench.

When Moe reached the park bench, he still needed to walk 1 mile more than 1⁄3 the

total distance from school to his home.

e) A gallon of paint contains 20% red paint and 80% blue paint.

Red paint is added until the mixture contains 50% red paint.

f) Standard quality coffee sells for $18.00 per kg.

Prime quality coffee sells for $24.00 per kg.

Every Saturday morning Moonman’s Coffee Shop grinds a 40kg batch of a standard/

prime blend to sell for $22.50/kg.

continued on back



FOCUS BLACKLINE MASTER 12.3 (CONT.)

g) A collection of nickels, dimes, and quarters has 3 fewer nickels than dimes and 3 more

quarters than dimes.

The collection is worth $4.20.

h) For a school play, Kyle sold 6 adult tickets and 15 student tickets.

Kyle collected $48 for his ticket sales.

Matt sold 8 adult tickets and 7 student tickets for the same school play.

Matt collected $38 for his ticket sales.

i) A nurse had 1200 ml of an 85% sugar solution (i.e., the container is 85% sugar and the

rest is water).

She added enough of a 40% sugar solution to create a 60% solution.

j) On Wednesday, a man drove from Gillette to Spearfish in 1 hour and 30 minutes.

On Thursday, driving 8 miles per hour faster, the man made the return trip in 1 hour and

20 minutes.

k) A student averaged 78 points on 3 history tests.

Her score on the 1st test was 86 points.

Her average for the 1st 2 tests was 3 points more than her score on the 3rd test.

l) Traveling by train and then by bus, a 1200 mile trip took Wally 17 hours.

The train averaged 75 mph and the bus averaged 60 mph.



1 i) For each of the following problems, use sketches to solve the problem. Label the

sketches and add brief comments as necessary to communicate your thought processes.

ii) For problems a) and b), translate the steps in your thought process into a sequence of

statements using algebraic symbols and equations.

a) The difference between two numbers is 6 and the sum of their squares is 1476. What are

the numbers?

b) The sum of 2 numbers is 32 and the sum of their squares is 520. What are the numbers?

c) The perimeter of a certain rectangle is 92 inches and its area is 493 square inches. What

are its dimensions?

d) Two cars start from points 400 miles apart and travel toward each other. They meet after

4 hours. Find the speed of each car if one travels 20 miles per hour faster than the other.

e) At Henry High School, 1 less than 1⁄5 of the students are seniors, 3 less than 1⁄4 are

juniors, 7⁄20 are freshmen, and the remaining 28 students are sophomores. How many stu-

dents attend Henry High?

f) If 40 cc of a 40% acid solution, 70 cc of a 50% acid solution, and 50 cc of pure acid are

combined, what % acid solution results?

g) How many cubic centimeters of pure sulfuric acid must be added to 100 cc of a 40%

solution to obtain a 60% solution?

h) A 40 foot by 60 foot garden is bordered by a sidewalk of uniform width. The area of the

sidewalk is 864 square feet. What is its width?





1 a) Let the two numbers be x and x + 6.

Sketches:

The value of the following region is 1476:

The value of half of the region is 738:

Rearrange:

The two numbers are 24 and 30 or –24 and –30.

x x 2 x 2 6x

x x 6

6x6 36

x x 2 6x

x 6

18

x 2

x + 3

9

x + 3

9

Algebraic Statements:

x2 + (x + 6)2 = 1476

2x2 + 6x + 6x + 36 = 1476

x2 + 6x + 18 = 738

(x + 3)2 + 9 = 738(x + 3)2 = 729x + 3 = 27 or –27x = 24 or x = –30If x = 24, x + 6 = 30.If x = –30, x + 6 = –24.

continued



b) Let the smaller number be x and the larger number be x + 2d.

Sketches:

The value of the following region is 520:

The value of half the region is 260:

c) Let x be the width of the rectangle and x + 2d itslength. Then x + (x + 2d) is half the perimeter, or 46.

The following region has value 493:

Rearranged it forms a 23 x 23 square with a d x dcorner missing;

So d2 = 36. Hence d = 6, so 2x = 46 – 12 = 34. Thusx = 17 and the dimensions of the rectangle are 17and 29.

x x 2d

x x 2 x 2 2dx

2dx2d 4d2

32

x 2

x + d

d2

x + d

d2

16

Since 162 = 256, d2 = 4 and, since d is positive d= 2, so x = 16 – 2, or 14 and x + 2d = 18. Thenumbers are 14 and 18.

Algebraic Statements:

x2 + (x + 2d)2 = 520,x2 + x2 + 4dx + 4d2 = 520 and2x + 2d = 32,

x2 + 2dx + 2d2 = 260 andx + d = 16,

(x + d)2 + d2 = 260,162 + d2 = 260,256 + d2 = 260,d2 = 4.Since d is positive, d = 2.Thus x = 16 – d = 14 andx + 2d = 18.

x

x + 2d

493

46

x 2

x + d

xd

x + d

23

23 xd

d2 = (23)2 – 493 = 529 – 493 = 36


continued



d) In the following sketch, the base of a rectanglerepresents time, the height represents rate of travel, sothe area represents distance traveled.

In the following, r is the rate of the slower car and r + 20is the rate of the faster car. The total area of the regionis 400.

r = 320 ÷ 8 = 40.

The rates are 40 mph and 60 mph.

e)

The above rectangle is divided into 20 equal parts. The28 sophomores constitute 4 more than 4 of theseparts, so each part represents (28 – 4) ÷ 4, or 6, students.

Hence, there are 6 x 20, or 120, students. (42 are fresh-man, 28 are sophomores, 27 are juniors, and 23 areseniors.)

f) In the following, the base of a rectangle representsthe amount of a solution and the height the percentof acid in the solution. Thus, the area represents theamount of acid in the solution.

If the heights of the above rectangles are “leveledoff,” the height of the resulting rectangle will be(1600 + 3500 + 5000) ÷ 160, which is 63.125 Hence,the solution is slightly more than 63% acid.

g) x is the amount of pure acid to be added.

If the two rectangles are “leveled off ” to a height of60%, the areas of the two shaded rectangles areequal. The area of the rectangle on the left is 20(100),or 2000. Hence the base x of the rectangle on theright is 2000 ÷ 40, or 50. Thus 50 cc of pure acidmust be added.

4

r 320

4

r

2080

FastCarSlow

Car

1

Freshmen

JuniorsSeniors

Sophomores

3

40 70

160050% 3500

50

100%5000

160

40%

100

40%

20%

3500

x

100%2000

40%2000


continued



A 40 x 60 rectangulargarden with an 864 squarefoot border of uniformwidth, w:

Rearranging theborder:

Moving 10 feet from the end of rectangles aand b to the end of rectangles c and d, and thencompleting the square:

2w + 50 = 3364 = 58,so, w = 4 (i.e., the widthof the border is 4 feet).

h)


864

6040

864

864

2500

3364

e f

g h

c d

a

b

a b

e f

g h

c

dw

w

w

w w

w

60

40

50

2w

502w 2w + 50


TEACHER NOTES


THE BIG IDEAAnalyzing the graphs and Algebra Piece representationsof families of linear and quadratic equations promptsintuitions and conjectures about the general charac-teristics of linear and quadratic functions. Exploring avariety of equation solving options—“by-hand” graphs,graphing calculators, Algebra Pieces, algebra symbols,and mental methods—provides students a powerful“tool kit” of problem-solving strategies.

OverviewStudents explore the graphingcalculator and discuss theirsuccesses and challenges withthe graphing calculator.

Materials� Start-Up Master 13.1, 1 copy

per student.

� Graphing calculators, 1 perstudent.

� Graphing calculator for theoverhead (recommended).

OverviewStudents use graphing calcula-tors to graph, solve, and evalu-ate linear and quadratic equa-tions and inequalities. Specialfunctions of graphing calcula-tors provide information aboutgraphs of everyday situations.Students compare the advan-tages and disadvantages of thegraphing calculator as a toolfor solving and graphingequations to by-hand graphing,Algebra Piece, and symbolicmethods, and mental strategies.

Materials� Focus Master 13.1, 1 trans-

parency.

� Focus Masters 13.2-13.5,1 copy of each per studentand 1 transparency of each.

� Algebra Pieces for eachstudent.


� Graphing calculators,1 per student.

� Graphing calculator for theoverhead (recommended).

OverviewStudents use graphs to repre-sent and solve equations andsystems of equations. Theywrite math expressions thatrepresent graphs of equationsand inequalities. They usegraphs to solve problemsregarding ice cream sales.


student.

� Coordinate grid paper (seeAppendix), 6 sheets perstudent.

LESSON 13ANALYZING GRAPHS



TEACHER NOTES


START-UP

Overview Materials

ANALYZING GRAPHS LESSON 13

Students explore the graphingcalculator and discuss their suc-cesses and challenges with thegraphing calculator.

� Start-Up Master 13.1,1 copy per student.

� Graphing calculators, 1per student.

1 Distribute graphing calculatorsand manuals. Hand out copies ofStart-Up Master 12.1 (see followingpages) and suggest that they mightuse this sheet for ideas on thisStart-Up activity. Then ask studentsto privately explore with theircalculators and find at least twothings that they can do with agraphing calculator that they didn’tknow about previously. Allow 5-10minutes for this exploration.

Ask students to share with anotherstudent one thing they discovered.

Ask for volunteers to present theirdicoveries at the overhead.

1 With their graphing calculators students should be encouragedto find out how to perform specific functions that interest them.Providing manuals for reference is recommended.

Students should be encouraged to demonstrate their findings onthe overhead graphing calculator.

� Graphing calculator forthe overhead (recom-mended).

ACTIONS COMMENTS


TEACHER NOTES



Individually use your graphing calculator to determine which graphing calculator functionsare clear to you. Use the calculator manual as needed. Some suggestions for investigationfollow.

1 This list below contains functions that you will need to know. Investigate those thatinterest you.

ON/OFFCLEAR the screenshow blank coordinate axes in the calculator viewing screenmove the cursor around a blank coordinate axeschange the viewing WINDOW sizeFORMAT the axesdetermine the “standard” WINDOW size on my calculator (on many it is –10 ≤ x ≤ 10and –10 ≤ y ≤ 10)enter an equation y =GRAPH an equation y =TRACE a graph (What shows on the screen when you do this?)ZOOM in on a graphZOOM in again—and againZOOM out on a graphZOOM back to the standard windowTRACE the graph of a function to determine the approximate value of the functionat x = 0, x = 19.75, and x = –37.5TRACE the graph of a function to determine the value of x when y = 75, when y = –75GRAPH 2 equations on the same coordinate axes.TRACE to approximate the intersection of 2 graphsZOOM and TRACE to improve your approximationDRAW a horizontal line on coordinate axes and slide the line up and downDRAW a vertical line on coordinate axes and slide the line left and right

___ view a table of coordinates of 2 equations listed simultaneouslyuse a table to find when 0 = 5x + 1clear MEMoryreset defaults in MEMory


continued on back



START-UP BLACKLINE MASTER 13.1 (CONT.)

view a TABLE of x- and y-coordinates of an equation___ view a table of coordinates of 2 equations listed simultaneously

use a table to find when 0 = 5x + 1clear MEMoryreset defaults in MEMorysolve equations using the “solver” functionuse the “maximum” and minimum” functions to find the turning point of a parabolause the “intersect” function to find the intersection of 2 graphsuse the “zero” function to find the x-intercepts of a graphuse the “value” function to find v(x) for specific values of xset the graphing style to shade the region above a graph; the region below a graph

2 Determine how to perform two calculator functions that are new to you and thatinterest you. Share how these functions work with a classmate.

3 List some functions that are not clearly understood.


FOCUS

Overview Materials


Students use graphing calculators tograph, solve, and evaluate linear andquadratic equations and inequalities.Special functions of graphing calcu-lators provide information aboutgraphs of everyday situations. Stu-dents compare the advantages anddisadvantages of the graphing calcu-lator as a tool for solving and graph-ing equations to by-hand graphing,Algebra Piece, and symbolic methods,and mental strategies.

� Focus Master 13.1, 1transparency.

� Focus Masters 13.2-13.5,1 copy of each perstudent and 1 transpar-ency of each.

� Algebra Pieces for eachstudent.

1 Distribute Algebra Pieces to eachstudent. Write the formula v(x) =–3x + 5 on the overhead and tellthe students that this formula rep-resents the xth arrangement of acomplete sequence of counting piecearrangements. Ask the students toform the –3rd, –2nd, –1st, 0th, 1st,2nd, 3rd, and xth arrangements ofthis sequence. Discuss.

2 Ask the students to leave thearrangements formed in Action 1 ontheir desks/tables and to imaginethe graph of y = –3x + 5 in enoughdetail that they can “see” importantfeatures of the graph. Ask for volun-teers to sketch and explain theirideas at the overhead. Use this as acontext for recalling the terms slope,x-intercept, and y-intercept, andhow to determine the value of each.

If it hasn’t come up previously, pointout to students that an x-interceptof a graph is also referred to as azero of the equation, since it is apoint where the value of y is zero.

1 An Algebra Piece arrangement for the given sequence is shownbelow.

2 In the above sequence, each time the arrangement numberincreases by 1, the value of the arrangement decreases by 3.Hence, the graph of y = –3x + 5 is a line that falls from left toright at the rate of 3 vertical units for every 1 horizontal unit, i.e.,its slope is –3. The line passes through the y-axis at the point(0,5), the y-intercept.

Some students may predict the x-intercept as “a point on the x-axisbetween x = 1 and x = 2, and closer to 2.” Others may mentallysolve the equation –3x + 5 = 0 to determine the x-intercept isx = 5⁄3. And, some may use Algebra Piece representations ofy = –3x + 5 or use algebra symbols to solve for x when y = 0.

Still others may note that, since the slope is –3, the line drops 3units vertically for every 1 unit of horizontal change (to theright), or down 1 unit for every 1⁄3 unit to the right. Hence, sincethe y–intercept is at (0,5), one can drop down 3 units and moveto the right 1 unit to locate the point (1,2) and from there movedown 2 units and over 2⁄3 unit to locate the x-intercept.


� Graphing calculators,1 per student.

� Graphing calculator forthe overhead (recom-mended).

ACTIONS COMMENTS

–2nd –1sto o o o

o o o o

o o o o

o o o

o o o

o o o

0 1st 2nd

……

ACTIONS COMMENTS


FOCUSANALYZING GRAPHS LESSON 13

3 If it didn’t come up in Actions 1or 2, ask the students to determinehow the slope, y-intercept, andx-intercept of the graph relate tothe arrangements formed in Action 1.Discuss.

4 Distribute graphing calculators (ifstudents don’t have them). Ask thestudents to graph v(x) = –3x + 5 ontheir calculators, and to determinevarious methods of using a graphingcalculator to determine a) below.Discuss their ideas regarding theadvantages and disadvantages of thegraphing calculator methods whencompared to: hand graphing, mentalstrategies, and algebraic procedures(either with Algebra Pieces or withsymbols representing the pieces).Repeat for b) and c).

a) the x-intercept

b) the y-intercept

c) the point where x = 49

3 The slope of a line is the ratio of the difference in the valuesof 2 arrangements to the difference in the corresponding arrange-ment numbers. The value of the y-intercept is the value of the 0tharrangement. The x-intercept is the number of the arrangementwhose value is 0.

4 If the calculators were used by other classes, students may needto clear or turn off graphs that were stored in the calculator.

Throughout this lesson students have opportunities to use thecalculator functions that were explored on Start-Up 12.1, andthey are introduced to other functions as needed or appropriatefor use in the activity. The names of functions and menus that arereferenced in this lesson may vary among calculator brands, andsome brands may not have some of the functions. Hence, you mayneed to adapt some actions according to the calculators used byyour students.

a) The students should notice that the TRACE function can give avery close “approximation” for the x-intercept, but not necessar-ily an exact value. A series of traces and zooms for y = –3x + 5 isshown below at the left. Each ZOOM obtains a closer approxima-tion of the x-intercept.

Many students may suggest that mentally calculating the x-intercept(by mentally determining when 0 = –3x + 5) is simple and there-fore using the calculator is not needed to compute the x-inter-

X = 1.91 Y = –.74 X = 1.65 Y = –.05

X = 1.67 Y = –.02 X = 1.666 Y = –.0033

Using ZOOM and TRACE to locate the x-intercept:

1st trace: Trace after 1st zoom:

Trace after 2nd zoom: Trace after 3rd zoom:

cept. And others may note that symbolic proceduresare quick and exact for this equation. Two importantpurposes of this lesson are for students to: 1) developa “tool kit” of options for graphing and solvingequations and 2) develop a sense for the appropri-ate uses of the available options.

On many calculators the “zero,” “intersect,” and“solver” functions are all appropriate for locatingthe x-intercept. Students may also use the tablefunction to locate the x-intercept.

b) Students may feel that mentally evaluating theequation at x = 0, by substituting 0 for x to gety = (–3)(0) + 5 = 5 is the most “reasonable”approach for finding the y-intercept of this equation.They might also use ZOOM and TRACE to locate they-intercept.

ACTIONS COMMENTS



continued next page

c) Students may graph y = –3x + 5 in a standard window andattempt tracing the graph to determine the y value for x = 49.However, this is not possible if x = 49 is outside the viewingwindow. (Note: in graphing calculators there is a standard defaultviewing window, such as [–10,10] for y, and [–10,10] for x.) Hence,use the WINDOW function to resize the window so that x = 49 isincluded and so the corresponding y-value also appears (seeexample at the left). This requires making a mental estimate ofthe y-value when x = 49. For example, one could note thaty = –3(49) + 5 should be a little more than –3(50), and therefore,set the minimum y at –150.

Some students may “see” v(x) using Algebra Pieces:

5 As students discuss the advantages and disadvantages ofvarious techniques, you might encourage them to make connec-tions among the representations they use for these techniques.For example, a particular counting piece arrangement corre-sponds to a point on the graph; a point on the graph can bedescribed by a pair of values, x and y; and the relationship be-tween the values x and y can be described by a general formula(in this case, y = –3x + 5). Understanding connections amongthese mathematical representations empowers students asalgebraic thinkers.

The intent here is to have students continue exploring varioustechniques for solving and evaluating equations while developingcomfort with the techniques and a sense about their appropriateuses. Following are some methods that students may suggest. Ifcomputers are available, you might have the students explore theuse of one or more computer graphing utilities.

a) One way to solve –3x + 5 = –75 is to use Algebra Pieces, asshown below:

5 Ask the students to determinevarious methods of identifying thepoints on y = –3x + 5 where thefollowing are true. Discuss.

a) y = –75

b) y = 102⁄3

c) x = –28.75

d) x = 10

e) x = 257

Use WINDOW to set minimum and maximums:x min 0 x max 50y min –150 y max 20

The y-value at x = 49.X= 49 Y= –142

y = –3x + 5

o o o o

o o o o

o o o o

o o o o

o o o o

o o o o

3(–49) + 5If 49–49

–49

–49

then

–3x –3x +5

o o o o

o o o o

o o o o

o o o o

o o o o

o o o oo o o o

o o o o

o o o o

o o o o

o o o o

o o o o

== –75 so, –80 and thus, = 80/3 = 262/3

ACTIONS COMMENTS



1

5 continuedInstead of Algebra Pieces, one can use symbols representingAlgebra Piece actions, as shown here:

–3x + 5 = –75–3x + 5 – 5 = –75 – 5

–3x = –80–3x⁄3 = –80⁄3

–x = –80⁄3x = 80⁄3

Another possibility is to adjust the window, then graph bothy = –3x + 5 and y = –75 simultaneously on a graphing calculator,and finally zoom and trace to approximate the x-value wherethese graphs intersect. The line y = –75 can be graphed by usingthe Y = function and the GRAPH function on the calculator, or byusing the “horizontal” function from the DRAW menu to form ahorizontal line and then slide the horizontal line vertically until itintersects y = –3x + 5 at y = –75. In the graphs shown at the leftthe horizontal axis was set for –10 ≤ x ≤ 50, and the vertical axiswas set for –150 ≤ y ≤ 20.

Students could also adjust the window so y = – 75 is included,then graph the lines y = –3x + 5 and y = –75 simultaneously onthe calculator, and finally select the function “intersect” todetermine the point of intersection.

Yet another method of solving –3x + 5 = –75 is to use whatever“solver” function is available.

Still another method is to use the TABLE function from the graph-ing calculator to view a table of values for y = –3x + 5. Scroll tothe entry closest to y = –75. Increments in x may need adjust-ment in order to locate an x-value that produces a y-value closerto y = –75.

b) x = –18⁄9

c) If = –28.75,then –3x + 5 = –3(–28.75) + 5

= 86.25 + 5= 91.25

X= 26.38 Y= –75.16

X= 26.66 Y= –74.99

y = –3x + 5

y = –75

y = –3x + 5

y = –75

ACTIONS COMMENTS



One could also set the window of the graphing calculator toinclude x = –28.75 and y = v(–28.75), and then trace and zoom toapproximate v(–28.75) or use a “value” function. To determine awindow that will include v(–28.75), one could set the upperbound for y at v(–30), which is easy to compute mentally. Anotherapproach is to locate the y-value that corresponds to x = –28.75in a table for y = –3x + 5.

d)-e) Methods similar to those used for c) are appropriate for d)and e).

6 a) You may need to remind the students to imagine and predictthe graphs at this point, not to draw or use their calculators.Students may mention differences in the slope, both its steepnessand rise/fall. They may predict that I and IV are mirror images ofeach other across the y-axis, as are lines II and III; and many maypoint out they all have y-intercept 5.

b) Here are graphs of the 4 equations. You might suggest thatstudents be sure to determine the equation associated with eachgraph on their calculators.

c) Some students may describe these as a “family of lines with acommon y-intercept.” Hence, other family members could beequations of any lines whose y-intercept is 5. Other students maysuggest that the lines must have slopes of +1, –1, +3, or –3. Stillothers may say that, for every line in the family, if the line hasslope m, then another family member must have slope –m, andothers may suggest any lines whose slopes are integers andwhose y-intercepts are 5 belong in the family.

6 Place a transparency of FocusMaster 13.1 on the overhead, reveal-ing Part a) only. Discuss the stu-dents’ ideas. Then reveal and discussParts b)-d).

y

x

(0, 5)

–5

–5

5

5

y = x + 5

y = 3x + 5 y = –3x + 5

y = –x + 5

continued next page



I y = –3x + 5

II y = –x + 5

III y = x + 5

IV y = 3x + 5

a) Imagine the graph of each of equations I-IV. What similarities and differences do you

predict about the graphs?

b) Now graph the 4 equations simultaneously on your graphing calculators. Do the results

agree with your predictions? What else do you notice?

c) Equations I-IV are a “family” of equations. What characteristic(s) do you think make

these equations a family? What are two other equations that could belong to this family?

d) What are similarities and differences among Algebra Piece representations of the xth

arrangements of the sequences represented by equations I-IV?

ACTIONS COMMENTS



7 Give each student a copy ofFocus Master 13.2 and repeatAction 6 for several of the equationfamilies listed. Encourage conjec-tures and generalizations aboutrelationships between the graph of aline or parabola and the constant,coefficients, and variables in theequation for the line or parabola.Encourage discussion about theinformation revealed by differentforms of an equation (e.g., factoredor expanded forms of a quadratic).

6 continuedd) For example, the xth arrangements of the sequences repre-sented by these equations each contain only x-frames and 5 unitsor –x-frames and 5 units.

7 Students could explore these in groups or individually ashomework. And you might create other families for students toexamine, based on mathematical ideas or relationships you feelstudents need to discuss further or based on prior conjecturesthat students have made.

The intent here is for students to continue their search forinsights about relationships among equations, their graphs, andtheir Algebra Piece representations.

Conjectures that surface may shift discussion in a number ofdirections; the direction to pursue can be based on students’interest and needs.

1) Changes in the y-intercept generate a family of parallel lines,each with slope –3 in this case.

2) Notice that graphs of equations I and II are identical to theparabola y = x2 after translating it 6 units down (I) or up (II) they-axis. Similarly, graphs of III and IV are translations of the parabola,y = –x2, 6 units down (III) or up (IV) the y-axis. Graphs of II and IIIare reflections of each other across the x-axis, as are graphs of Iand IV. Graphs of I and III are reflections of each other across theline y = –6, while graphs of II and IV are reflections of each otheracross the line y = 6.

3) Some may refer to this as a family of quadratic functionswhose graphs are parabolas with vertices are on the y-axis. This isalso true for the quadratics in 2) above.

y

x–5

–5

5

5

y = –3x + 5y = –3x + 2

y = –3x – 2y = –3x – 5



For each equation family below, record the following on separate paper:

a) your predictions about the graphs of the 4 equations,

b) your observations about calculator graphs of the equations,

c) the characteristic(s) that you think make the equations a family,

d) two additional equations that would fit in the family,

e) similarities and differences among Algebra Piece representations of the 4 equations.

1 I y = –3x + 5

II y = –3x – 5

III y = –3x + 2

IV y = –3x – 2

2 I y = x2 – 6

II y = x2 + 6

III y = –x2 – 6

IV y = –(x2 – 6)

3 I y = 4x2

II y = (1⁄4)x2

III y = –3x2 – 6

IV y = ( 3⁄4)x2

4 I y = x(x – 3)

II y = x2 – 2x

III y = x2 + 2x

IV y = x(x + 3)

5 I y = (x – 3)(x – 4)

II y = x2 – 7x + 12

III y = (x + 2)(x + 3)

IV y = (x + 1)(x – 2)

6 I y = (x – 2)(x – 5)

II y = 2(x – 2)(x – 5)

III y = –(x – 2)(x – 5)

IV y = –2(x2 – 7x + 10)

7 I 28x + 8y = 0

II 7x + 2y = 6

III 14x + 4y = 4

IV 21x + 6y = –12

8 I y = –5x + 2⁄3

II 3y = –15x + 2

III 0 = –5x – y + 2⁄3

IV –2 = –15x – 3y

ACTIONS COMMENTS



4) These are all quadratic functions and, when written in standardform, the coefficients of the x2 and x terms are integers and theconstant is 0. When expressed in factored form, all have x as onefactor.

Students may be interested in pursuing the effects of addingvarious x-terms to the equation y = x2. One interestingconjecture that may come up is that the vertices of all theparabolas of the form y = x2 + bx, where b is a real number,lie on the parabola y = –x2 (see left); and vertices of allparabolas of the form y = –x2 + bx lie on the parabolay = x2.

5) Some students may call this a family of parabolas that can bewritten in the form y = x2 + bx + c, where b and c are real num-bers not equal to zero. Some may notice that the x-intercepts ofa parabola are easy to identify when the equation is in factoredform (assuming it factors). For example, if the equation is of theform y = (x – r)(x – s), where r and s are real numbers, thex-intercepts are x = r and x = s.

Notice that when the two factors in equation I are multiplied, theproduct is equation II. Hence, equations I and II have identicalgraphs, with x-intercepts 3 and 4.

6) Students may describe this as a family of parabolas whosevertices lie on the vertical line midway between x = 2 and x = 5,i.e., parabolas whose vertices lie on the vertical line x = 3.5. Foreach family member, its reflection across the x-axis is also in thefamily. Students may make conjectures about equations for lineswhose graphs are reflections of each other across the x-axis.Encourage this. To prompt thinking you might pose questions,such as, “How can an equation be altered to create a graph that isa reflection across the y-axis? across the line y = x?” Note:replacing x with –x in an equation creates a reflection across they-axis, and exchanging x and y in an equation creates a reflectionacross the line y = x; however, students may not reach theseconclusions.

7) These equations are all written in standard linear form (ax + by= c). Rewriting each equation in slope-intercept form shows thatall 4 lines have the same slope but different y-intercepts. Hence,this is a family of parallel lines with slope –7⁄2. Notice, the stan-dard form of a linear equation does not give away as manyexplicit “clues” about its graph as does the slope-intercept form.

8) The graphs of these equations are all identical; hence theequations are all equivalent. Students may add other equationsequivalent to these, or they may add a set of equivalent equationsthat represent another line.

y = x 2

y = x 2 + 2x

y = –x 2y = x 2 + –3x

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8 Solving this system of equations means to find the value of x forwhich 2x + 1 = x2 – 2, i.e., to solve the equations simultaneously. Interms of the graph, solving the system means finding the points ofintersection of the two graphs. In terms of sequences of countingpiece arrangements, solving the system is equivalent to finding thevalues of x for which the 2 different xth arrangements of thesequences represented by the equations have the same value.

a) The diagrams below show a trace and zoom to locate x = 2.97and x = 3.03 as approximations for one solution. Additionalzooms improve the approximation, suggesting the graphs inter-sect at x = 3.

Trace:

Zoom and then trace again:

A series of zooms and traces of the other intersection pointsuggests x = –1 as a solution. One can verify that x = 3 and –1 aresolutions of the system by testing those points in equations for y1and y2. Since v1(3) = 2(3) + 1 = 7 = 32 – 2 = v2(3), and v1(–1) =2(–1) + 1 = –1 = (–1)2 – 2 = v2(–1), the points (3,7) and (–1,–1)are intersection points of these graphs.

8 Write the equations y1 = 2x + 1and y2 = x2 – 2 on the overhead orboard. Point out that these arereferred to as a system of 2 equa-tions in 2 variables. Ask the groupsto solve this system of equations,using each of the methods listedbelow. Discuss.

a) the TRACE and ZOOM functions onthe graphing calculator

b) Algebra Pieces (or sketches ofthe pieces)

c) algebraic symbols

d) the calculator “solver” function

e) the calculator “intersect” func-tion

f) the calculator TABLE function

y2 = x 2 – 2

y1 = 2x + 1

y2 = x 2 – 2

y1 = 2x + 1

X= 2.97 Y= 6.87

X= 3.03 Y= 7.06

ACTIONS COMMENTS



b) Illustrated below is an Algebra Piece solution.

c) One can also imagine the Algebra Piece actions and recordsymbolic procedures that represent those actions. For example:

d) One can use the “solver” function to determine when thedifference (2x + 1) – (x2 – 2) = 0.

e) If students’ calculators do not have this function, the studentsmay have other methods to suggest.

f) This requires generating side-by-side tables for the 2 equations,and then scrolling to locate the values of x for which the y-valuesfrom the 2 tables are equal.

Adding –2x + 2 to bothcollections produces these2 collections.

“Completing the squares” by adding1 unit to the upper right hand cornerof each collection produces this dia-gram. Since the squares are equal invalue, their edges must be equal.Hence, x – 1 = 2 or x – 1 = –2, so x = 3or x = –1. Therefore, x = 3 and x = –1are the x-coordinates of the intersec-tion points of the 2 graphs.

x2 – 2 = 2x + 1 (form the 2 equal collections)(x2 – 2) + (–2x + 2) = (2x + 1) + (–2x + 2) (add –2x + 2 to both collections)

x2 – 2x = 3 (simplify)x2 – 2x + 1 = 3 + 1 (add 1 to both collections)

(x – 1)2 = 22 or (–2)2 (form squares of each collection)x – 1 = 2 or –2 (take the square root of the value ofso, x = 3 or –1 each square)

=

o o o

o o o

o o o

o o o?

=

o o o

o o o

o o o

o o o

=

x2 – 2 = 2x + 1

x2 – 2x = 3

x2 – 2x + 1 or (x – 1)2 = 4

ACTIONS COMMENTS



9 Write the following system ofequations on the overhead and askthe students to determine thesolution(s), if any, to each system,using the approach of their choice.Ask students to verify each solutionusing a second method, and so thatone of their methods utilizes thegraphing calculator and one doesnot. Discuss as needed.

y = 4 + 2x; y = x + 3

Repeat Action 9 for other systemsof equations selected from thefollowing list.

a) y = 7 – x2; y = –7 + x2

b) y = 3x – 2; y = 3x + 1

c) y = 2x + 7; y = 4x2 – 3x + 2

d) 2x + 3y = 4; x – y = 7

e) y – 2x = 5; x + 3y = 6

f) y = x2 + 7; y = 0

g) y = –2x2 – 1; y = 0

9 Allow plenty of time for students to explore and discuss theirapproaches and results with their classmates before discussing asa class. Students may use a hand graph, a calculator graph, AlgebraPiece procedures, symbolic procedures, mental strategies, thesolver or intersect functions on the calculator, tables generatedby hand or by a calculator, or combinations of these; or, they mayinvent other strategies.

In the following example, zooming and tracing yields the estimate(–1.01,1.97). This suggests that the lines intersect at, or very near,x = –1. Since, y = 4 + 2x and y = x + 3 both equal 2 when x = –1,the point (–1,2) is the exact point of intersection of the 2 graphs.

An algebraic solution based on what students imagine as AlgebraPiece actions might look like the following:

4 + 2x = x + 3 (form the 2 equal collections)4 + x = 3 (remove x from both collections)1 + x = 0 (remove 3 units from both collections)

x = –1 (add –1 to both collections)

a) Here is a solution using algebra symbols to represent AlgebraPieces:

7 – x2 = –7 + x2

14 = 2x2

7 = x2

7 = x or – 7 = x

Since 7 – (± 7 )2 = 0 , and –7 + (± 7 )2 = 0, the graphs intersectat the points (– 7 ,0) and ( 7 ,0). Note that these are exactpoints of intersection; calculator methods give decimal approxi-mations for the x-coordinates.

b) There is no value of x for which these 2 expressions are equal.Some students may reason that it is not possible to add 1 to anumber and produce the same result as subtracting 2 from thenumber. Or, students may reason that since these 2 graphs aredifferent straight lines with the same slope, they are parallel andhence, never intersect. Zooming out on the calculator graph (seediagram on following page) can help verify this; however, it isimportant to note that, when 2 lines are not parallel, it is possibleto miss an intersection point by not zooming out far enough.

y = 4 + 2x

y = x + 3

X= -1.01 Y= 1.97

ACTIONS COMMENTS



Here is a representa-tion using sketchesof Algebra Pieces:

Adding –3x to each collection above leaves:

But it is not possible that –2 = 1, so there are no solutions tothe system.

c) As shown in the calculator display below, one approximation ofan intersection point is (–.638,5.72). The other intersection pointis outside the window. Changing the window ranges for x and yenables one to approximate the other intersection point.

Using Algebra Pieces (see following page) to complete the squarefor this quadratic equation illustrates that these graphs do notintersect at a point whose coordinates are whole or rationalnumbers. Rather, the coordinates of the points of intersection areirrational numbers.

X= –.638 Y= 5.72

y = 2x + 7

y = 4x 2 – 3x + 2

=

– 2 = 1

y = 3x – 2

y = 3x + 1

3x – 2

=

= 3x + 1

continued next page

ACTIONS COMMENTS



Adding –2x to both collections in theabove diagram produces the followingcollections:

9 continued

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o

o

o o

1—4– x

7––16

7––16

5–4

2x x

1––169––

16

9––16

5–4

=of aunitleftover

(2x – )2 + 7

So, (2x – )2 = 6 .

=

–5—4

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

o o o

74x 2 – 5x + 2 =

=

= 2x + 74x2 – 3x + 2

o o o

o o o

o o o

o o o

o o o

o o o

Cutting and rearranging thepieces in the above collectionproduce the following (notethat the pieces on the upperand right edges of the squareare not edge pieces; rather,they are quartered –x-frames):

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Subtracting 7⁄16 from both collections above produces(2x – 5⁄4)2 = 69⁄16 = 105⁄16, and so 2x – 5⁄4 = ± 105

16 . Thus,x = 5

4 ± 10516

2

and the graphs cross at approximately x ≈ 1.906

and x ≈ –.656. When x = –.656, 2x + 7 = 4x2 – 3x + 2 ≈ 5.69. When x≈ 1.906, 2x + 7 = 4x2 – 3x + 2 ≈ 10.8. Thus, the 2 points ofintersection for these graphs occur at approximately (–.656,5.69)and (1.906,10.8). These are rational approximations to irrationalcoordinates.

This example illustrates the convenience of the graphing calculatorfor quickly finding approximate solutions to equations. Recall thatthe TRACE function obtained the approximation x = –.638, which isclose to the algebraic approximation of x = –.656. Repeatedzooms and traces would improve the calculator approximation.

d) In order to graph these equations on the calculator, one mustrewrite them in slope intercept form as y = (–2⁄3)x + 4⁄3 andy = x – 7. Then one can graph and trace to find the approximateintersections, or use the “intersect” function.

Or, one could use the “solver” function to determine when 0 =[(–2⁄3)x + 4⁄3] – (x – 7).

One way to solve this system of equations symbolically is to solve(–2⁄3)x + 4⁄3 = x – 7 for x.

Another symbolic method is to solve for y in one equation,substitute that value in the other equation, and then solve for x.For example, solving the equation x – y = 7 for y, one gets y = x – 7.Then substituting x – 7 for y in the equation 2x + 3y = 4 producesthe new equation 2x + 3(x – 7) = 4. Hence, 2x + 3x – 21 = 4, so5x = 25, and thus, x = 5. This is an example of the method calledsolving by substitution.

e) Students will need to rewrite these equations in slope inter-cept form before graphing them on the calculator, using the“solver” function on the calculator, or solving them using AlgebraPieces or symbols. The method of substitution described in d)could also be used here.

f)-g) There are no solutions in the set of real numbers to eitherof these systems since neither the parabola y = x2 + 7 nor theparabola y = –2x2 – 1 intersects the x-axis (i.e., the line y = 0).Note that if x2 + 7 = 0, then x2 = –7. But there is no real numberwhose square is negative. Solutions for equations like these arediscussed in the next lesson, Complex Numbers.

ACTIONS COMMENTS



10 Give each student a copy ofFocus Master 13.3 and ask the themto carry out the instructions.

Discuss their results. As needed,clarify graphing calculator proce-dures and graphing conventionssuch as the use of dotted lines andopen/closed circles.

10 It may be helpful to have the students complete a) and thendiscuss before continuing with the others. Encourage students towrite mathematical statements that use as few words as possible,but so that someone reading their statements could recreate thegraph exactly.

a) One statement describing this graph is y = 2 for x ≥ 2. This is afunction whose domain is the real numbers ≥ 2, and 2 is the onlynumber in the range. Note: Students may also point out that thisgraph is a ray whose end point is (2,2).

Note that the filled in dot at the point (2,2) implies that the point(2,2) is part of the graph. If the dot were not filled in then thedomain would be x > 2 and the point (2,2) would not be consid-ered on the graph. Students may not be familiar with this nota-tion, but will probably speculate what it means and you can clarifyas needed.

b) An equation for this function is .The domain is all real numbers and the range is the reals ≥ 1.

c) The domain and range of this function, y = x, are the realnumbers.

d) You may need to tell students that all points in the shadedregion are part of the graph to be described. This graph includesthe set of all ordered pairs, (x,y) for which x and y are real numbersand y ≥ –10. The fact that the line y = –10 is a solid line impliesthe points on the line y = –10 are included in the graph. Refer toyour calculator's manual to determine how to shade a region.

e) The dotted line implies the points on the line are not includedin the graph, but all of the shaded region is part of the graph.Hence, this is a graph of all the ordered pairs (x,y) for which xand y are real numbers and y > 15.

f) This is a graph of the function y = –20 for x < 40. The domain isall real numbers x < 40, and the range contains only the number –20.

g) This is a graph of all points on or below the parabola y = –x2 + 4.It can also be described as a graph of the inequality y ≤ –x2 + 4.The y-values are all real numbers less than or equal to 4.

h) This is a graph of the function y = 2x2 for x ≥ 0. The domainand range are the real numbers ≥ 0.

i) This is a graph of the inequality y > x for all real numbers x.

y = { x + 1, x ≥ 0–x + 1, x < 0



y

x1

1

y

x1

1

y

x1

1

y

x5

5

y

x15

15

y

x20

10

y

x1

1

y

x2

1

y

x5

5

a) b) c)

d) e) f)

g) h) i)

1 Write a mathematical statement to describe each graph.

2 Recreate each graph on a graphing calculator.

ACTIONS COMMENTS



11 Give each student a copy ofFocus Master 13.4 and have themcarry out the instructions.

12 Pick one or more of the stu-dents’ questions to explore. Discusstheir results.

11 Students may wonder: Is this a fair head start for the son?Who would win a 100 meter race? When would the man catch upwith the child? What length race would be most fair? etc. Whatare equations and graphs that could represent this situation?

12 If students have difficulty deciding what to explore, youmight suggest something, basing your choice on its mathematicalpotential. Have 1-cm grid paper available as needed.

As an example, the following discussion explores the question“Who would be favored to win a 100 meter race?”

There are several approaches that students could use to investi-gate the above question, such as to make a chart or table ofvalues, and/or to graph the times and distances traveled by bothrunners. A graph is illustrated on the following page. A table canbe produced by hand or by using the table function on thegraphing calculator. Notice that since Marcus runs 20 meters in 5seconds, then he runs 4 meters in 1 second. Similarly, sinceFranko runs 20 meters in 3 seconds, he runs 20⁄3 meters in 1 second.

Marcus Frankotime in distance time in distanceseconds in meters seconds in meters

0 30 0 01 34 1 62⁄32 38 2 131⁄33 42 3 20

11 74 11 731⁄312 78 12 8013 82 13 862⁄314 86 14 931⁄315 90 15 10016 9417 9818 102

… …… …

continued next page



Franko and his son, Marcus, plan to race one another on a track.

Marcus can run 20 meters in 5 seconds.

Franko can run 20 meters in 3 seconds.

They have agreed that Marcus will start 30 meters ahead of Franko.

Pose several questions about this situation.

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12 continued

The table and graph show that Franko catches up with Marcusbetween 11 and 12 seconds. Also, at 12 seconds, Franko is at the80 meter mark, and Marcus is at the 78 meter mark, so Franko isfavored to win a 100 meter race.

160

140

120

100

80

60

40

20

Dis

tan

ce in

Met

ers

Time in Seconds–Franko–Marcus

5 10 15

ACTIONS COMMENTS



13 If it hasn’t already been sug-gested, ask the students to write anequation for y1, the distance ofMarcus from the starting line xseconds after the start of the race,and an equation for y2 the distanceof Franko from the starting line xseconds after the start of the race.Ask for volunteers to share theirequations and have the class deter-mine whether the equations accu-rately represent the distances. Thenask the students to use calculatorgraphs of the equations to determinea fair length for the race (or toverify their results from Action 13).

13 Students may verify their equations by testing variousnumbers of seconds to see if the meters traveled match those intheir tables from Action 13. For example, their equation for y1should yield 50 meters at 5 seconds (the initial 30 plus 20 more),70 meters at 10 seconds, and so forth.

One possible pair of equations is y1 = 4x + 30 and y2 = (20⁄3)x.

In order to find the race length that makes the 30 meter headstart fair, students must determine where Franko catches Marcus(i.e., where the graphs cross, or where the distances run are equal).To do this, they can graph y1 = 4x + 30 and y2 = (20⁄3)x and usetrace to determine the point of intersection (see diagram below).

Many students will probably suggest a “fair” race is about 75 meters.An extension question could be, “How much of a head start doesMarcus need for 100 meters to be the length of a fair race?”

X= 11.7 Y= 76.9

y1 = 4x + 30

y2 = ( )x

X= 11.25 Y= 75

20__3

1st trace:x min 0 x max 20y min 0 y max 150

Trace after several zooms:

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14 These situations could be explored in class, and/or as ahomework activity. Either way, it is helpful to encourage studentsto discuss and compare ideas with classmates.

Situation 1. Examples of questions students may pose include:How much does it cost to drive each vehicle 10 miles? 20 miles?etc. Which company has the better deal? Is there a number ofmiles for which the cost will be the same for both companies? Asan example, the following discussion addresses the question:Which company has the better deal?

Students could make a table, write and solve a system of equa-tions describing each company’s cost, or graph the equations byhand or on a graphing calculator and look for the points wherethe graphs intersect.

If y1 is the cost at We Hardly Try, y2 is the cost at Rent-A-Wreck,and x is the number of miles driven, equations for the total rentalcost from each company could be represented as follows:

y1 = $10 + .10xy2 = .15x

A symbolic solution to this system could look like the following:10 + .10x = .15x

10 + .10x – .10x = .15x – .10x10 = .05x

10 (1⁄.05) = .05x (1⁄.05)200 = x

When x = 200, y1 = 30 = y2. Hence, at 200 miles the cost ofdriving a car from either company is the same. For less than 200miles, Rent-A-Wreck is a better deal, while We Hardly Try is abetter deal for more than 200 miles. Traces on a graphing calcula-tor illustrate this:

14 Give each student a copy ofFocus Master 13.5 and ask them tocomplete Situation 1. Discuss theirresults. Then repeat for Situations 2and 3.

1st trace:x min 0 x max 300 (miles)y min 0 y max 50 (cost) Trace after several zooms:

X= 201 Y= 30.1

RAW

WHT

X= 200 Y= 30RAW

WHT


For each of the 3 Situations shown below, please do the following:

a) Make a diagram or sketch that illustrates the important mathematical relationships in

the situation.

b) Write several mathematical questions that a person might investigate about the situation.

c) Investigate one or more of your mathematical questions.

d) After you complete your investigation of each question, write a summary that includes a

statement of the question, an explanation of your solution process, your answer to the

question, and verification that your answer works.

Situation 1

The Rent-A-Wreck and the We Hardly Try car rental companies charge the

following prices:

We Hardly Try charges an initial fee of $10 and then charges $.10 per mile. Rent-A-Wreck

does not charge an initial fee, but charges $.15 per mile.

Situation 2

The Saucey Pizza Company charges $7 for a pizza. The ingredients and labor for each pizza

cost $2.50. The overhead costs (lights, water, heat, rent, etc.) are $100 per day.

Situation 3

Michael, the golf pro at U-Drive-It Golf Range, claims that when he hits the ball from the

lower level tee, the height h of the ball after t seconds is: h = 80t – 16t2.

Michael also claims that when he hits the ball from the elevated tee, the ball reaches the

following height in t seconds: h = 20 + 80t – 16t2.


ACTIONS COMMENTS



Situation 2. Examples of questions that may come up include: IfSaucey’s sells 100 pizzas, how much money will they make or losethat day? How many pizzas do they need to sell in a day to makea profit? Should they change their pricing? Following is one way ofanswering the question: How many pizzas do they need to sell ina day to make a profit?

Since a pizza is sold for $7 and the ingredients and labor cost$2.50, the profit on each pizza is $7 – $2.50 = $4.50. If Saucey’ssells x pizzas in a day, then the amount of daily profit, y, is y = $4.50x.The minimum number of pizzas that must be sold to have a totalincome greater than the daily overhead cost of $100 can be foundby tracing a graph of y = 4.5x to determine the value of x when yexceeds $100, as shown at the left. Since y exceeds 100 betweenx = 22 and x = 23, Saucey must sell at least 23 pizzas per day orlose money.

Another approach to answering this question is to simultaneouslygraph the equation representing the daily cost, y1 = 100 + 2.50x(where x is the number of pizzas sold), and the equation for dailyincome, y2 = 7x. The intersection of the graphs is the point atwhich Saucey’s breaks even (see graph at the left). This alsoindicates Saucey must sell 23 pizzas in order for daily income toexceed daily cost.

Situation 3. Here are some questions students may pose: At whattime does the golf ball reach its highest point? How long does ittake before the golf ball hits the ground? How high is the elevatedtee? Where did those formulas come from?

X= 22.3 Y= 100.53

profit: y = 4.5x

X= 22.8 Y= 157

income: y2 = 7x

cost: y1 = 100 + 2.5x

x min 0 x max 50 (pizzas)y min 0 y max 200 (dollars)

x min 0 x max 50 (pizzas)y min 0 y max 200 (dollars)

lower level tee elevated teeseconds height in feet height in feet

x y1 = 80x – 16x2 y2 = 80x – 16x 2 + 20

0 0 200.5 36 561.0 64 841.5 84 1042.0 96 1162.5 100 1203.0 96 1163.5 84 1044.0 64 844.5 36 565.0 0 205.5 –44 –24

continued next page

An example of a table generated by the calculator is shownat the left, where y1 describes the height of the ball when hitfrom the lower tee, and y2 describes the height of the ballfrom the elevated tee. The table suggests the ball reaches ahigh point after 2.5 seconds, and then starts back down again,hitting the ground after 5 seconds (a table with smallerincrements can be used to see if the high point is slightlymore or less than 2.5). Notice the table lists negative heights,but since the ball stops descending at ground level, those donot make sense and are irrelevant.

Graphing and tracing both equations on a graphing calculator,one can find the high point of the golf ball and the time whenthe ball hits the ground (see diagram on the next page).

ACTIONS COMMENTS



14 continuedNotice, in the diagram below, the 2 graphs are the same shape,the graph of y2 is a vertical translation 20 units above the graphof y1, and it takes longer for the ball hit from the elevated tee tohit the ground.

The coefficients of the variables have physical significance. Forexample, the “80” means the golf ball rises at a rate of 80 feet/secwhen it is first hit. On the other hand, the “–16” is the effect ofthe pull of the earth’s gravity on the golf ball (the force of gravityis a downward force of 16 feet per second squared (i.e., per x2,where x is the number of seconds). Gravity slows the height gainof the ball over time and eventually the ball starts to descend.The 20 in the elevated tee equation is the height of the ball attime 0, so the elevated tee is 20 feet above the ground level.

y2

x = 2.55y1 = 89.96y2 = 119.96

y1

x = 5y1 = 0

x = 5.24y2 = –.12

y1 = 80x – 16x2

y2 = 80x – 16x2 + 20

A ball hit from the elevated tee stays 20feet higher, and hits the ground about aquarter second later. Note: the graphs ofthe equations are not the paths of theballs, but rather give the heights of theball for various times.




I y = –3x + 5

II y = –x + 5

III y = x + 5

IV y = 3x + 5

a) Imagine the graph of each of equations I-IV. What similarities and differences do you

predict about the graphs?

b) Now graph the 4 equations simultaneously on your graphing calculators. Do the results

agree with your predictions? What else do you notice?

c) Equations I-IV are a “family” of equations. What characteristic(s) do you think make

these equations a family? What are two other equations that could belong to this family?

d) What are similarities and differences among Algebra Piece representations of the xth

arrangements of the sequences represented by equations I-IV?




For each equation family below, record the following on separate paper:

a) your predictions about the graphs of the 4 equations,

b) your observations about calculator graphs of the equations,

c) the characteristic(s) that you think make the equations a family,

d) two additional equations that would fit in the family,

e) similarities and differences among Algebra Piece representations of the 4 equations.

1 I y = –3x + 5

II y = –3x – 5

III y = –3x + 2

IV y = –3x – 2

2 I y = x2 – 6

II y = x2 + 6

III y = –x2 – 6

IV y = –(x2 – 6)

3 I y = 4x2

II y = (1⁄4)x2

III y = –3x2 – 6

IV y = (–3⁄4)x2

4 I y = x(x – 3)

II y = x2 – 2x

III y = x2 + 2x

IV y = x(x + 3)

5 I y = (x – 3)(x – 4)

II y = x2 – 7x + 12

III y = (x + 2)(x + 3)

IV y = (x + 1)(x – 2)

6 I y = (x – 2)(x – 5)

II y = 2(x – 2)(x – 5)

III y = –(x – 2)(x – 5)

IV y = –2(x2 – 7x + 10)

7 I 28x + 8y = 0

II 7x + 2y = 6

III 14x + 4y = 4

IV 21x + 6y = –12

8 I y = –5x + 2⁄3

II 3y = –15x + 2

III 0 = –5x – y + 2⁄3

IV –2 = –15x – 3y




y

x1

1

y

x1

1

y

x1

1

y

x5

5

y

x15

15

y

x20

10

y

x1

1

y

x2

1

y

x5

5

a) b) c)

d) e) f)

g) h) i)

1 Write a mathematical statement to describe each graph.

2 Recreate each graph on a graphing calculator.




Franko and his son, Marcus, plan to race one another on a track.

Marcus can run 20 meters in 5 seconds.

Franko can run 20 meters in 3 seconds.

They have agreed that Marcus will start 30 meters ahead of Franko.

Pose several questions about this situation.



For each of the 3 Situations shown below, please do the following:

a) Make a diagram or sketch that illustrates the important mathematical relationships in

the situation.

b) Write several mathematical questions that a person might investigate about the situation.

c) Investigate one or more of your mathematical questions.

d) After you complete your investigation of each question, write a summary that includes a

statement of the question, an explanation of your solution process, your answer to the

question, and verification that your answer works.

Situation 1

The Rent-A-Wreck and the We Hardly Try car rental companies charge the following prices:

We Hardly Try charges an initial fee of $10 and then charges $.10 per mile. Rent-A-Wreck

does not charge an initial fee, but charges $.15 per mile.

Situation 2

The Saucey Pizza Company charges $7 for a pizza. The ingredients and labor for each pizza

cost $2.50. The overhead costs (lights, water, heat, rent, etc.) are $100 per day.

Situation 3

Michael, the golf pro at U-Drive-It Golf Range, claims that when he hits the ball from the

lower level tee, the height h of the ball after t seconds is: h = 80t – 16t2.

Michael also claims that when he hits the ball from the elevated tee, the ball reaches the

following height in t seconds: h = 20 + 80t – 16t2.




1 For each of the following families of 3 equations, graph the equations on 1 coordi-

nate axis, and list the characteristics that make the equations a family. Then create and

graph 2 or more additional equations that have those characteristics. Label each graph

with its equation.

a) y = 4x – 1 b) y = 3x – 2 c) y = x2 + 3

y = 4x + 2 y = x⁄5 – 2 y = –2x2 + 3

y = 4x – 5 y = –6x – 2 y = x2⁄4 + 3

d) y = 1⁄x + 5 e) y = 4(x – 2)(x + 5)

y = 1⁄x – 4 y = –3(x – 2)(x + 5)

y = 1⁄x + 3 y = (x – 2)(x + 5)

2 For each of the following systems of equations, use your calculator to find a solution.

a) 6x + 3y = 5 b) y = x2 – 8x + 18 c) y = 2⁄x + 3

2y – 3x = 12 2x + y = 7 y = x2 + 4

d) y = x2⁄2 – x⁄2 + 3 e) y = x2 – 3x – 2 f) y = –x2 – x – 2

y = –x2 – 3x + 5 x2 + 3y – 18 = 0 y = (x + 1)(x – 2)

3 Verify your solutions to Problems 2a) and 2b) by solving each using another method.

Show your thinking and reasoning.

4 Discuss the advantages and disadvantages of using the graphing calculator to solve

equations. Give examples to illustrate your ideas.





1 For each of the following families of 3 equations,graph the equations on 1 coordinate axis, and list thecharacteristics that make the equations a family.

a) y = 4x – 1y = 4x + 2y = 4x – 5

These equations produce lines that are parallel withslopes of 4 and cross the x–axis at 1, –2, and 5 re-spectively.

b) y = 3x – 2y = x⁄5 – 2y = –6x – 2

These equations produce three intersecting lines thatintersect at (0,–2) with slopes of 3, 1⁄5, and –6 respec-tively.

c) y = x2 + 3y = –2x2 + 3y = x2⁄4 + 3

These equations produce three parabolas, all withvertices at (0,3) and a vertical axis of x = 0. Theequation y = –2x2 + 3 produces a parabola that opensdownward with a maximum vertex at (0,3). The othertwo equations produce parabolas that open upwardeach with a minimum vertex at (0,3).

d) y = 1⁄x + 5y = 1⁄x – 4y = 1⁄x + 3

These equations produce graphs of inverse variations,each with a vertical asymptote of the y–axis. Horizontalasymptotes are y = 5, y = –4, and y = 3, respectively.

e) y = 4(x – 2)(x + 5)y = –3(x – 2)(x + 5)y = (x – 2)(x + 5)

These equations produce parabolas that each intersectthe x–axis at (–5,0) and (2,0). The equationy = –3(x – 2)(x + 5) produces a parabola that opensdownward. The screen shots above illustrate theminimum and maximum vertices for each equation.

continued



2 For each of the following systems of equations useyour calculator to find the solutions:

a) 6x + 3y = 52y – 3x = 12

Graph the equations and choose INTERSECT from theCALC (or appropriate) menu. By repeatedly pressingENTER (or its equivalent) you will see the followingscreens, with the last being the intersection of thetwo equations. Only the INTERSECTION screen is shownfor the remaining problems.

b) y = x2 – 8x + 182x + y = 7

There is no intersectionof these two equations,or no solution.

c) y = 2⁄x + 3y = x2 + 4

d) y = x2⁄2 – x⁄2 +3

y = –x2 – 3x + 5

There are two points of intersection, or two solutions,as indicated by the screen shots below:

e) y = x2 – 3x – 2x2 + 3y –18

There are two points of intersection, or two solutions,as indicated by the screen shots below:

f) y = –x2 –x –2y = (x + 1)(x – 2)

The graphs of these twoequations touch at theindicated point.

3 Verify your solutions to Problems 2a) and 2b) bysolving each using anothermethod. Show yourthinking and reasoning.

Answers will vary, butstudents might solvesymbolically or they mightuse the table function intheir calculator, as shownfor problem 2a:

4 Discuss the advantages and disadvantages of usingthe graphing calculator to solve equations. Giveexamples to illustrate your ideas.

Answers will vary.


© THE MATH LEARNING CENTER



THE BIG IDEAComplex numbers are introduced to provide squareroots for negative real numbers. Green and yellowcomplex number pieces, along with red and blackcounting pieces, are used to carry out computationsinvolving complex numbers.

OverviewGreen and yellow complexnumber pieces are introducedto provide square roots fornegative numbers.

Materials� Black and red counting pieces,

20 per student.

� Black and red edge pieces,12 per student.

� Green and yellow complexnumber pieces, 20 perstudent.

� Green and yellow edgepieces, 12 per student.

� Black and red counting andedge pieces for the over-head.

� Green and yellow complexnumber and edge pieces forthe overhead.

OverviewStudents use tile pieces toperform computations withcomplex numbers. They solveequations involving complexnumbers.

Materials� Black and red counting pieces,

20 per student.

� Black and red edge pieces,12 per student.

� Green and yellow complexnumber pieces, 20 perstudent.

� Green and yellow edge pieces,12 per student.

� Black and red counting andedge pieces for the overhead.

� Green and yellow complexnumber and edge pieces forthe overhead.

OverviewStudents perform computationsand solve equations involvingcomplex numbers. They associ-ate complex numbers withpoints on a coordinate grid.


student.

� Counting and complexnumber pieces for use athome.

� Grid paper (see Appendix),1 sheet.

LESSON 14COMPLEX NUMBERS


TEACHER NOTES


START-UP

Overview Materials

COMPLEX NUMBERS LESSON 14

Green and yellow complex numberpieces are introduced to providesquare roots for negative numbers.

� Black and red countingpieces, 20 per student.

� Black and red edgepieces, 12 per student.

� Green and yellow com-plex number pieces, 20per student.

1 Distribute black and red count-ing and edge pieces to the students.Ask them to construct a squarearray, with edge pieces, whose valueis 9. Discuss.

2 Ask the students to construct asquare array, with edge pieces,whose value is –9. Discuss.

1 There are two possibilities; an array with black edge pieces oran array with red edge pieces:

If a square array has two identical, adjacent edges, the value ofthat edge is called a square root of the value of the array. Thus 9has two square roots: 3 and –3. If x is positive, the symbol x isused to designate the positive square root of x (cf. Start-Up 9).

2 A red array has one black edge and one red edge:

Since a square red array does not have two identical edges, thereis no positive or negative number that is a square root of –9. Ingeneral if x is a negative number, there is no positive or negativenumber that is a square root of x.


� Black and red countingand edge pieces for theoverhead.

� Green and yellow com-plex number and edgepieces for the overhead.

ACTIONS COMMENTS

ACTIONS COMMENTS


START-UPCOMPLEX NUMBERS LESSON 14

3 In these materials, colored tile and edge pieces are repre-sented as follows:

black red green yellow

In an array, if a row and column both have green edges, the piecewhich lies at their intersection will be red.

4 Green and yellow are opposites. Thus, the net value of thefollowing collection is 3 yellow.

The net value of the following collection is 2 black plus 3 green.

In order to distinguish between net values in the black/red systemand net values in the green/yellow system, the letter i will beused to indicate net values in the green/yellow system. Thus, acollection of 4 black tile has net value 4 while a collection of 4green tile has net value 4i. A collection of 4 red tile has net value–4 while a collection of 4 yellow tile has net value –4i.

3 Distribute green and yellowcomplex counting and edge piecesto the students. Tell the studentsthe purpose of these pieces is toprovide square roots for negativenumbers; in particular, if both edgesof an array are green, the array willbe red. Illustrate.

4 Discuss the net value of collec-tions of green/yellow and black/redpieces. Introduce notation andterminology for these collections.

ACTIONS COMMENTS


START-UPCOMPLEX NUMBERS LESSON 14

Shown below are two other collections and their net values.

Numbers of the form a + bi are called complex numbers. If a and bare integers, a + bi is called a complex or Gaussian integer. Animaginary number is a complex number for which b ≠ 0. A realnumber is a complex number for which b = 0, a pure imaginarynumber is an imaginary number for which a = 0.

3 – 3i –4


TEACHER NOTES


FOCUS

Overview Materials


Students use tile pieces to performcomputations with complex num-bers. They solve equations involvingcomplex numbers.

� Black and red countingpieces, 20 per student.

� Black and red edgepieces, 12 per student.

� Green and yellow com-plex number pieces, 20per student.

1 Ask students to use countingpieces to find the sum and differ-ence of 3 + 2i and 4 – 5i. Discuss.

1 The sum (3 + 2i) + (4 – 5i) can be found by combining acollection whose value is 3 + 2i with a collection whose value is4 – 5 i and then finding the value of the combined collection.

7 – 3i

The difference (3 + 2i) – (4 – 5i) can be found by combining acollection whose value is 3 + 2i with a collection whose value isthe opposite of 4 – 5i and then finding the value of the combinedcollection.

Alternatively, the difference can be found by forming a collectionwith net value 3 + 2i from which a collection with value 4 – 5ican be removed:


� Black and red countingand edge pieces for theoverhead.

� Green and yellow com-plex number and edgepieces for the overhead.

ACTIONS COMMENTS

3 + 2i

4 – 5i

3 + 2i

(3 + 2i) – (4 – 5i) = –1 + 7i

ACTIONS COMMENTS


FOCUSCOMPLEX NUMBERS LESSON 14

2 Ask the students to form acounting piece array for which oneedge has value 1 + 2i and the otheredge has value 2 + 3i. Then ask thestudents to form an array to findthe product (1 + 2i)(2 + 3i). Discuss.

2 You may want to remind the students that a piece in the arraywill be red if both of its edges are green. Also, that in the com-plex numbers, 1 maintains its role as a multiplicative identity. Thismeans that a piece which has a black edge, i. e., an edge whosevalue is 1, will have the color of its other edge; in particular apiece with black and green edges will be green.

An array with the given edges has four sections as shown.

A tile in section I will be black. A tile in section III will be redsince both of its edges are green. A tile in either section II or IVwill be green since one edge is black and the other green.

The completed array appears below. It’s value is –4 + 7i.

IV III

I II

2i

1

2 3i

(1 + 2i)(2 + 3i) = –4 + 7i

ACTIONS COMMENTS



3 Since tile in sections I, II, and IV of the array shown belowhave a black edge, they will have the color of the other edge.Hence tile in section I are black, those in II are yellow and thosein IV are green.

The tile in Section III have one green and one yellow edge. Thestudents may offer various reasons why these tile are black. Oneargument might be that a tile with two green edges is red, andhence changing one of these edges to its opposite will change thevalue of the tile to its opposite.

Another way to see this is to form an array that has one edgeconsisting of a single green edge piece and the other edge con-sisting of 1 green and 1 yellow edge piece. Since this edge hasvalue 0, the value of the array is 0. Thus, since the tile with bothedges green is red, the other tile, which has 1 green and 1 yellowedge, must be black.

The completed array is shown below.

3 Ask the students to form an arrayfor the product (1 + 2i)(2 – 3i).Discuss.

IV III

I II

2i

1

2 –3i

G

G

Y

BR

(1 + 2i)(2 – 3i) = 8 + i

ACTIONS COMMENTS



4 Since a tile 1 with 1 black edge has the color of the otheredge, the first row and column of the table are determined. Also,a tile with 2 red edges is black, one with 2 green edges is red,and, as determined in the last action, a tile with 1 green and 1yellow edge is black. Thus the following is known:

The remaining entries can be determined by methods similar tothose described in Comment 3. For example, the color of a tilewith 1 red and 1 green edge can be determinedby forming an array in which one edge is greenand the other edge is black and red. The com-pleted table is shown below.

Replacing colors by values in the above table produces themultiplication table shown to the left.

4 Ask the students to complete thefollowing table showing the color ofa tile with the given edges.

B

R

G

Y

B R G Yedge

1edge2

B

R

G

Y

B

B

R

G

Y

R

R

B

G

G

R

B

Y

Y

B

edge1edge

2

x

1

–1

i

–i

1

1

–1

i

–i

–1

–1

1

–i

i

i

i

–i

–1

1

–i

–i

i

1

–1

B

R

G

Y

B

B

R

G

Y

R

R

B

Y

G

G

G

Y

R

B

Y

Y

G

B

R

edge1edge

2

B

G

R

YG

ACTIONS COMMENTS



5 Ask the students to use numberpieces to find the following:

a) (3 – 2i) + (4 + 3i)

b) (4 + 6i) – (–6 + 4i)

c) (4 – 5i) – (6 – i)

d) (2 – 3i)(4 – 2i)

e) (3 + i)(3 – 2i)

f) (–1 + 3i)(1+ 3i)

g) 5i ÷ (1 + 2i)

h) (16 + 2i) ÷ (2 – 3i)

5 a) 7 + i

b) 10 + 2i

c) –2 – 4i

d)

(2 – 3i)(4 – 2i) = 2 – 16i

e)

(3 + i)(3 – 2i) = 11 – 3i

f)

(–1 + 3i)(1+ 3i) = –10

continued next page

ACTIONS COMMENTS



5 continuedg) If an array whose value is 5i is constructed so one edge hasvalue 1 + 2i, the value of the other edge is the desired quotient.Since the real part of 5i is 0, the array must have an equal numberof black and red pieces.

The array shown here has value 5i. Its left edge has value 1 + 2i.The value of the other edge is 2 + i. Hence, 5i ÷ (1 + 2i) = 2 + i.

h) The students will devise various strategies for constructing anarray whose value is 16 + 2i and has an edge whose value is 2 – 3i.

One way to proceed is to lay out an edge of 2 black and 3 yellowand then consider which of black or red and which of green oryellow must be in the other edge. Since the resulting array mustcontain at least 16 black tile, colors should be chosen that produceboth black and green tile, with more of the former. As shown onthe left, a selection of black leads to a column of 2 black and3 yellow and a selection of green leads to a column of 2 greenand 3 black. Two of the former and 4 of the latter will produce anarray whose net value is 16 + 2i, as shown below.

2 – 3i

1 + 2i

5i

2 – 3i

16 + 2i

(16 + 2i) ÷ (2 – 3i) = 2 + 4i

5i ÷ (1 + 2i) = 2 + i

ACTIONS COMMENTS



6 Ask the students to find allsolutions to the following equa-tions:

a) x2 = –16

b) x2 + 6x + 34 = 0

c) x2 – 4x + 20 = 10

7 (Optional.) Ask the students touse counting pieces to find a squareroot of 2i. Then ask them how theymight use counting pieces, alongwith a scissors, to find a squareroot of i. Discuss.

6 a) A 4 x 4 red square will either have two green edges or twoyellow edges. Hence, x = 4i or x = –4i.

b) Since x2 + 6x + 9 = (x + 3)2 (see the figure), x2 + 6x + 34 =(x + 3)2 + 25. Hence, if x2 + 6x + 34 = 0, then (x + 3)2 = –25.Therefore x + 3 is the edge of a 5 x 5 red square. Thus x + 3 = 5ior x + 3 = –5i. So x = –3 + 5i or x = –3 – 5i.

c) If x2 – 4x + 20 = 10 then (x – 2)2 = x2 – 4x + 4 = –6. Thus x + 2is the edge of a 6 x 6 red square. Hence x – 2 = i 6 or x – 2 =–i 6. So x = 2 + i 6 or x = 2 – i 6.

7 The following array, with two adjacent edges of the same value,shows that 1 + i is a square root of 2i.

By cutting tile in half, one obtains a square whose value is i:

If the tile are cut in half again, one can rearrange the abovesquare into a square with two adjacent edges of equal value:

The length of each edge piece is half the diagonal of a unit square,or 2⁄2 (see the sketch). Thus, 2⁄2 + 2⁄2i is a square root of i. Sinceeach edge could consist of a red and a yellow edge piece, ratherthan a black and a green, – 2⁄2 – 2⁄2i is also a square root of i.

3

x 2

x

x

3

3x 9

3x

continued next page

ACTIONS COMMENTS



7 continuedNote that no new colors—only scissors—are necessary to obtainsquare roots for i, that is, no new colors are needed to solve theequation x2 = i. This is an illustration of the Fundamental Theoremof Algebra: Every polynomial equation with complex numbers ascoefficients has solutions which are complex numbers.

1

2

2

The square shown here has area 2.Its edges are diagonals of unit squares.



1 Compute:

a) (3 – 5i) – (4 – 2i)

b) (2 – 3i)(4 + i) + (1 – 2i)(5 – 3i)

c) (22 – 7 i) ÷ (2 – 3i)

2 Find all solutions of the following equations:

a) (x + 1)2 = –36

b) x2 + 10x + 100 = 0

c) x2 – 3x + 9 = 0

3 The complex numbers can be associated with points in a coordinate plane by letting

a + bi correspond to the point with coordinates (a, b). For example 3 + 5i corresponds to

the point (3, 5).

Suppose a parallelogram has vertices at the origin O and at the points P and Q which corre-

spond, respectively to 5 – 3i and 6 + 4i. If OP and OQ are two sides of the parallelogram,

find the coordinates of the fourth vertex, S, and the complex number corresponding to it.

How is the complex number corresponding to S related to the complex numbers corre-

sponding to points P and Q?


ANSWERS TO FOLLOW-UP




1 a) A collection for 3 – 5i consists of 3 black and 5 yellow tile; the opposite of a collection for 4 – 2i consistsof 4 red and 2 green tile. Combining these collections and eliminating pairs of opposite tile yields a collectionof 1 red and 3 yellow tile. Hence, (3 – 5i) – (4 – 2i) = –1 –3i.

b)

(2 – 3i)(4 + i) = 11 – 10i; (1 – 2i) (5 – 3i) = –1 – 13i

(2 – 3i)(4 + i) + (1 – 2i) (5 – 3i) = 10 – 23i

c)

(22 – 7i) ÷ (2 – 3i) = 5 + 4i

2 a) x = –1 ± 6i

b)

c)

Completing the square:x2 + 10x + 100 = 0x2 + 10x + 25 = –75(x + 5)2 = –75 = 25(–3)

x + 5 = ±5 3 ix = –5 ± 5 3 i

Multiplying by 4 and then completing the square:x2 – 3x + 9 = 0

4x2 – 12x + 36 = 0(2x – 3)2 = –45 = 9(–5)

2x – 3 = ±3 5 i2x = 3 ± 5 ix = (3 ± 5 i)⁄2

x 2

–3x

–3x

–3x –3x 9

x 2

x 2 x 2

x

–3

x

x x –3

5

x2

x

x

5

5x 25

5x



3 Coordinates of S are (11, 1) which is associated with the complex number 11 + i. This is the sum of thenumbers corresponding to P and Q.


5

4

3

2

1

1 2 3 4 5

Q(6,4)

0–1

–2

–3

–4

6 7 8 9 10 11 12

P(5,–3)

S(11,1)


TEACHER NOTES

APPENDIX


1⁄4-Inch Grid

APPENDIX


1-Centimeter Grid

APPENDIX


Coordinate Grids

APPENDIX


Red and Black Counting Pieces (print back-to-back with page 337)

APPENDIX


Back for Red and Black Counting Pieces

APPENDIX


Blank Counting Pieces

APPENDIX

Algebra Pieces (print back-to-back with page 340)

APPENDIX

Back for Algebra Pieces

APPENDIX

n-Frames (print back-to-back with page 342)

APPENDIX


Back for n-Frames

Documents

€¦ · ALGEBRA THROUGH VISUAL PATTERNS VOLUME 1 Introduction vii LESSON 1 Tile Patterns & Graphing 1 LESSON 2 Positive & Negative Integers 31 LESSON 3 Integer Addition & Subtraction