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 Science
Foundation. Opinions expressed are those of the authors and not
necessarily 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
VOLUME 2
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
SQUARES & SQUARE ROOTS
LESSON 9
THE BIG IDEA
Square roots are viewed as the lengths of sides of
squares. Methods of constructing a square of any given
integral area, and thus the square root of any positive
integer, are developed. One of these constructions
leads to the Pythagorean Theorem.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
Students construct squares of
integral areas and establish the
relationship between squares
and square roots.
Students dissect squares and
reassemble the pieces to form
two squares and, conversely,
dissect two squares and
reassemble the pieces to form
a single square. In the process,
they arrive at the Pythagorean
Theorem. They dissect rectangles and reassemble the
pieces to form squares and, in
so doing, construct square
roots. Students examine the
relationship between products
(quotients, sums) of square
roots and square roots of
products (quotients, sums).
Students solve problems
involving squares and square
roots, using the Pythagorean
Theorem as necessary. They
relate the arithmetic mean and
the geometric mean of two
positive numbers to the construction of squares.
Materials
Centimeter grid paper (see
Appendix), 2-3 sheets per
student, 1 transparency.
Start-Up Master 9.1,
1 transparency.
Materials
Follow-Up 9, 1 copy per
student.
Materials
Centimeter grid paper (see
Appendix), 2-3 sheets per
student
Scissors, 1 pair per student
Start-Up Master 9.1, 1 copy
per student and 1 transparency
Focus Masters 9.1-9.2,
1 copy of each per student.
Focus Master 9.3, 1 copy per
student and 1 transparency.
ALGEBRA THROUGH VISUAL PATTERNS | 163
TEACHER NOTES
164 | ALGEBRA THROUGH VISUAL PATTERNS
SQUARES AND SQUARE ROOTS
LESSON 9
START-UP
Overview
Students construct squares of integral areas and establish the relationship between squares and square
roots.
Materials
Centimeter grid paper
(see Appendix), 2 to 3
sheets per student, 1
transparency.
Start-Up Master 9.1,
1 transparency.
COMMENTS
ACTIONS
1 Distribute centimeter grid paper
1
If your students are familiar with the basic properties of
square roots, you may wish to omit this lesson.
to the students. Tell them that
1 square represents 1 unit of area.
For each of the integers 1 through
25 ask them to construct, if possible,
a square whose vertices are grid
intersection points and whose area
is the given integer. For each square
they draw, ask the students to indicate its area and the length of its
side. Discuss.
SQUARES AND SQUARE ROOTS
A student may believe they are finished when they have constructed all the squares whose sides lie along a gridline. If this
happens, you can simply tell the student there are more. Normally, someone in the class will discover a square that “tilts.”
Of the integers 1 through 25, there are 13 for which a square
exists that satisfies the conditions of Action 1. Start-Up Master
9.1 attached at the end of this activity shows a square of each
area. A square of area 25 can also be obtained by carrying out a
3,4 pattern as described below.
LESSON 9
START-UP BLACKLINE MASTER 9.1
One way to obtain a square that fits the conditions is to pick two
intersection points as successive vertices. In the instance shown
below, one can get from point P to point Q by going 3 units in
one direction and 1 in the other. Repeating this 3,1 pattern, as
shown, results in a square.
5
2
5
2
1
4
2
1
8
10
10
8
9
3
3
1
1
13
3
17
17
13
16
4
3
Q
1
18
20
18
1
20
25
P
3
5
Square generated by a 3,1 pattern.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 165
SQUARES AND SQUARE ROOTS
LESSON 9
START-UP
COMMENTS
ACTIONS
1 continued
A
B
D
C
2 Discuss with the students why
they can be certain that the 13
integers mentioned in Comment 1
are the only ones in the range 1
through 25 for which squares exist
that satisfy the conditions of Action 1.
The area of this square can be found by subtracting the area of the
shaded regions from the area of the circumscribed square (see the
figure). Note that regions A and C combine to form a rectangle of
area 3 as do rectangles B and D. Thus, the area of the inscribed
square is 16 – 6, or 10. Since the area of the square is 10, the
length of its side is 10. An approximation of 10 can be obtained
by measuring the side of the square with a centimeter ruler.
If n is non-negative, the positive square root of n, written
length of the side of a square of area n.
2
One way to see there are only 13 different areas is to note
that if a square is to have area no greater than 25, then the
distance between successive vertices must be less than or equal
to 5. Thus, if P and Q are successive vertices, Q must lie on or
within a circle of radius 5 whose center is at P. In the sketch, the
13 intersections marked with an x are possibilities for Q that lead
to 13 differently sized squares. Any other choice for Q leads to a
square the same size as one of these 13.
x
x x x
x x x x
x x x x x
P
166 | ALGEBRA THROUGH VISUAL PATTERNS
n, is the
SQUARES AND SQUARE ROOTS
LESSON 9
START-UP
ACTIONS
3 Point out to the students that
20 = 2 5. Ask the students to
examine the squares they have
constructed for other relationships
of this type.
COMMENTS
3
The square of area 20 is composed of 4 squares of area 5, as
shown in the sketch. Hence, the side of a square of area 20 is twice
the length of the side of a square of area 5. Thus, 20 = 2 5.
20 or
5
2 5
5
5
5
5
5
The square of area 8 is composed of 4 squares of area 2 and the
square of area 18 is composed of 9 squares of area 2. Thus,
8 = 2 2 and 18 = 3 2.
4 Ask the students to construct
squares, and find their areas and
side lengths, using the following
patterns:
a) 5,2
b) 4,4
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 area
2 to show that 32 = 2 8 = 4 2. Also square d) forms 4 squares
of area 13, giving 52 = 2 13.
Some students may notice that the area of a square is equal to
the sum of the squares of the numbers in the pattern that form it,
e.g., 29 = 5 2 + 2 2 . If so, tell them to keep that observation in mind
as the lesson continues.
c) 6,1
d) 4,6
Ask the students to examine their
squares for relationships between
square roots like those discussed in
Action 3.
ALGEBRA THROUGH VISUAL PATTERNS | 167
SQUARES AND SQUARE ROOTS
LESSON 9
START-UP BLACKLINE MASTER 9.1
5
2
5
2
1
4
2
1
8
10
10
8
9
3
13
17
17
13
16
4
18
20
18
20
25
168 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
5
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
Overview
Students dissect squares and reassemble the pieces to form two
squares and, conversely, dissect two
squares and reassemble the pieces
to form a single square. In the process, they arrive at the Pythagorean
Theorem. They dissect rectangles
and reassemble the pieces to form
squares and, in so doing, construct
square roots. Students examine the
relationship between products
(quotients, sums) of square roots
and square roots of products (quotients, sums).
ACTIONS
1 Give each student a sheet of
centimeter grid paper and a pair of
scissors. Ask them to construct a
square using a 4, 2 pattern. Have
them divide their square into 3
parts as shown below. Then ask the
students to cut out these 3 parts
and reassemble them to form 2
adjacent squares, and determine the
length of the sides of these squares.
Materials
Centimeter grid paper
(see Appendix), 2 or 3
sheets per student
Scissors, 1 pair per
student
Start-Up Master 9.1,
1 copy per student and
1 transparency
Focus Masters 9.1-9.2,
1 copy of each per
student.
Focus Master 9.3, 1 copy
per student and 1 transparency.
COMMENTS
1 Rotating the right triangles around a vertex, as shown, transforms the original square into 2 squares. The lengths of the sides
of the squares are 2 and 4. Notice that these are the lengths of
the legs of the rotated right triangles.
ALGEBRA THROUGH VISUAL PATTERNS | 169
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
2 Ask the students to construct a
2
You can ask for volunteers to describe how they proceeded.
square using a 6,3 pattern and then
use the method of Action 1 to
transform it into a 3 x 3 and a 6 x 6
square.
3 Distribute a copy of Focus Master 9.1 to each student. Have the
students cut out the square and
triangle. Then have them dissect the
square and reassemble it into 2
squares whose sides have length a
and b. Discuss.
SQUARES AND SQUARE ROOTS
3
The students can cut out the triangle and use it to divide the
square into the following regions, which can be cut out and
reassembled into the desired squares.
a
a
LESSON 9
FOCUS BLACKLINE MASTER 9.1
b
b
c2
c
c
b
170 | ALGEBRA THROUGH VISUAL PATTERNS
a
Notice that the area of the large square is c 2 and the areas of the
2 smaller squares are a 2 and b 2 . Thus, c 2 = a 2 + b 2 . Since c is the
hypotenuse of the given right triangle and a and b are the legs of
the triangle, we have shown that the square of the hypotenuse of
a right triangle is the sum of the square of its legs. This result is
known as the Pythagorean Theorem.
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
4 Ask the students to use the
4
a) The distance d is the hypotenuse of a right triangle whose
legs are 3 and 7. (See the figure.) Thus, d 2 = 3 2 + 7 2 = 58. Hence
d = 58 ≈ 7.62.
Pythagorean Theorem to find the
following distances on a coordinate
graph:
8
a)
a) The distance d from the origin to
the point (3,7).
(3,7)
7
6
5
b) The distance d between the points
(2,10) and (5,4).
d
4
3
2
c) The distance d between the points
(–3,–6) and (7,–2).
1
–2
–1
1
0
2
3
4
b)
11
(2,10)
10
9
c)
8
6
7
1
–4
d
–3
–2
–1
1
6
–1
5
–2
–3
(5,4)
4
3
–4
3
2
3
4
5
6
7
8
(7,–2)
d
4
–5
2
(–3,–6)
1
–6
–7
0
1
2
3
4
d 2 = 3 2 + 6 2 = 45; d =
5 Distribute a copy of Focus Mas-
10
6
45 ≈ 6.71
d 2 = 10 2 + 4 2 = 116; d =
116 ≈ 10.77
5
This is the converse of the dissection done in Action 1. It can
be accomplished by locating the point P on the long side of the
figure, and then cutting and reassembling as shown.
a
cut
ter 9.2 to each student. Ask them to
dissect the two squares shown
below so they can be reassembled
into a single square. Ask for volunteers to show how they dissected
the squares.
5
b
cut
b
P
a
ALGEBRA THROUGH VISUAL PATTERNS | 171
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
6 Place a transparency of Start-Up
6
Master 9.1 on the overhead. Discuss
with the students how squares not
appearing on the transparency can
be constructed using the method of
Action 5. Then distribute a copy of
Start-Up Master 9.1 to each student.
Ask them to dissect and reassemble
squares from Start-Up Master 9.1 to
form squares of area 11 and 21.
Figure 2 shows a square of area 21 constructed from squares of
area 5 and 16. A square of area 21 can also be constructed from
squares of areas 1 and 20 or areas 4 and 17 or areas 8 and 13.
Figure 1 shows a square of area 11 constructed from squares
of areas 2 and 9. The students may find it helpful to tape the
squares together in the position shown before dissecting them.
Note that all the squares of area less than 25 that do not appear
on Start-Up Master 9.1 can be constructed by this method.
Figure 1
Figure 2
a
a
b
9
5
5
b
2
2
b
21
b
11
9
16
16
a
a
7 Tell the students that squares can
also be constructed by dissecting
and reassembling rectangles, for
example, a square of area 21 can be
constructed from a 3 x 7 rectangle.
To begin an investigation of how
square can be dissected and reassembled into squares, distribute
centimeter grid paper to the students, ask them to dissect a 9 x 16
rectangle so that it can be reassembled into a square. Ask them to
do this by making as few cuts as
possible.
7
Transforming a 9 x 16 rectangle into a square is simpler than
transforming a 3 x 7 rectangle into a square since, in the former
case, the side of the square is integral.
The students will recognize that they need to form a 12 x 12
square and they most likely will proceed by cutting the 9 x 16
rectangle along grid lines. By doing this, a square can be obtained
by making 3 cuts as shown below.
cut 2
cut 1
172 | ALGEBRA THROUGH VISUAL PATTERNS
cut 3
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
ACTIONS
COMMENTS
However, only 2 cuts are required:
cut 1
cut 2
The students are not likely to find this dissection. One way to
proceed is to tell the students that you can get a square by
making only 2 cuts and show them the first cut in silhouette on
the overhead. (See the figure below.) Since the students know the
length of the side of the square is 12, they have a clue to the
location of this diagonal cut.
ALGEBRA THROUGH VISUAL PATTERNS | 173
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
8 Distribute a copy of Focus Master 9.3 to each student. Tell the
students to cut off the bottom
portion of the page and set it aside
for use in the next Action. Then ask
the students to cut out one of the
rectangles in the top portion of
Focus Master 9.3 and dissect it
using the “diagonal cut” method of
Action 7 and reassemble the resulting pieces to form a square. Have
them measure their resulting figure
to determine whether or not it is a
square. If not, ask them to dissect a
second rectangle so that the resulting figure is more “squarelike.”
SQUARES AND SQUARE ROOTS
8
For the dissection shown below, the reassembled pieces do
not form a square.
E
A
E
A
D
D
B
C
C
The sides of the constructed rectangle can be compared by
moving the top piece of the rectangle to the position shown
below. A more “squarelike” figure would be obtained by lengthening distance AE somewhat.
LESSON 9
E
FOCUS BLACKLINE MASTER 9.3
D
A
B
C
CUT
A
D
B
C
9 Using a transparency of the
bottom half of Focus Master 9.3 on
the overhead, discuss with the
students how cut lines can be determined so that the resulting pieces
can be arranged to form a square.
Then ask the students to dissect the
rectangle on the bottom half of
Focus Master 9.3 and reassemble it
to form a square.
174 | ALGEBRA THROUGH VISUAL PATTERNS
9
Note that the dissection in Action 8 would have resulted in a
square if, in the last figure in Comment 8, vertices A and C coincided:
E
D
C
A
B
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
If dotted lines are added to the previous figure to show the
original location of the pieces, the figure becomes:
A
B
C A
B
Notice that the sum of the two angles at the top of the
figure is a right angle and segments AB and CD have the
same length, since they were originally opposite sides of
a rectangle. These observations point to the following
procedure, illustrated below for determining cut lines.
Figure 1
A
D
D
Beginning with rectangle ABCD, the cut lines can be
determined as follows.
h
P
C
B
h
1. Locate point P on an extension of BC at a distance h
from C. (See Figure 1.)
2. Extend side CD. (See Figure 1.)
3. Place a sheet of paper with a square corner so that
the corner is on the extension of CD and the edges of
the paper go through points B and P. (See Figure 2.)
Figure 2
E
d
A
first cut
line
sheet of
paper
B
P
B
second cut
line
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. This
is the second cut line. (See Figure 3.)
sheet of
paper
Figure 3
4. Let E be the intersection of the edge of the paper
and segment AD. The first cut line is the segment BE. The
distance d from A to E is the side of the desired square.
(See Figure 2.)
C
F
d
ALGEBRA THROUGH VISUAL PATTERNS | 175
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
COMMENTS
ACTIONS
10 Distribute centimeter grid
10
Using the procedure described in Comment 6, a 3 x 7
rectangle can be dissected and reassembled as shown below to
obtain a square of area 21. Note that the dimension of the square
is the distance between vertex C of the rectangle and square
corner S. Hence, the distance CS is 21.
paper to the students. Tell them that
each square is 1 unit of area. Then
ask them to dissect a rectangle of
area 21 and reassemble it to form a
square.
S
d
d
d
C
11 Repeat Action 10 to obtain a
square of area 7.
11
If the students start with a 1 by 7 rectangle, an additional
cut is required. (Additional cuts are required whenever the length
of a rectangle is more than 4 times its height.)
d
d
d
d
If the rectangle is cut on the dotted lines it can be reassembled
to form a square.
176 | ALGEBRA THROUGH VISUAL PATTERNS
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
ACTIONS
12 Ask the students to construct a
line segment of length 15.
COMMENTS
12 If the procedure of Comment 9 is carried out on a
3 x 5 rectangle, the result is a square of area 15. The dimension of
this square is distance CS in the following figure. Hence, segment
CS has length 15.
S
C
Segment CS has length
13 Write the following pairs of
expressions on the overhead or
chalkboard:
a)
S T ; ST
b)
S⁄ T
;
S⁄ T
⁄
c) S + T; S + T
For each pair of expressions, ask
the students to determine, for positive numbers S and T, whether the
first expression is less than, equal to,
or greater than the second expression. Discuss their conclusions.
15.
13 The students may arrive at their conclusions by considering
specific values for S and T. If so, you can ask them if they have
reason to believe their conclusions hold for all values of S and T.
The equality of the expressions in a) and b) are useful in simplifying radicals.
a) A general conclusion can be reached by noting that S T is
the 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 squares
have the same area. Hence the lengths of their sides are equal.
S
T
( S T )( S T )
=( S S ) ( T T )
=ST
S
ST
ST
T
ST
b) Since ( S ⁄ T )( S ⁄ T ) = S S ⁄ T T = S ⁄ T, both S ⁄
the lengths of the sides of squares whose area is
two expressions are equal.
T
and S ⁄ T are
Hence the
S⁄ T.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 177
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS
ACTIONS
COMMENTS
13
continued
c) As shown in the sketch below, S + T is the combined length
of adjacent squares of areas S and T respectively. If the area S
where distributed around the square of area T to obtain a square
of area S + T, the length of the side of this square would be less
then the combined lengths of the sides of the original squares.
Hence, S + T > S + T , that is the sum of the square roots of
two positive numbers is greater than the square root of their
sum.
S
T
T
T
S+T
S
S
+
Some students may be curious about the relationship between
S – 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 the
square roots of S – T and T is greater than the square root of
their 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 .
178 | ALGEBRA THROUGH VISUAL PATTERNS
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS BLACKLINE MASTER 9.1
c2
c
c
a
b
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 179
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS BLACKLINE MASTER 9.2
180 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
SQUARES AND SQUARE ROOTS
LESSON 9
FOCUS BLACKLINE MASTER 9.3
CUT
A
D
B
C
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 181
SQUARES AND SQUARE ROOTS
LESSON 9
FOLLOW-UP BLACKLINE MASTER 9
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.
6
b
5
4
3
c
a
2
d
1
1
2
3
4
5
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.
182 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
SQUARES AND SQUARE ROOTS
LESSON 9
ANSWERS TO FOLLOW-UP 9
2w
1
5
6
b
5
4
w
3
c
a
2
d
1
2w 2 = 250
w 2 = 125 = 25 x 5
w=5 5
1
The dimensions are 5 5 by 10 5.
3
4
5
4
5
Perimeter = 20 +
4 + 29 ≈ 17.5 cm
13 +
6
5
2
2
a 2 = 2 2 + 4 2 = 20; a = 20
b 2 = 2 2 + 3 2 = 13; b = 13
c=4
d 2 = 5 2 + 2 2 = 29; d = 29
3
4
4
3
14 ft.
h
4
2
1
h2
62
14 2
+
=
h 2 = 14 2 – 6 2 = 160
h = 160 ≈ 12.6 ft
6 ft.
5
1
2
3
Area = 5 x 6 – 4 – 3 – 4 – 5
= 30 – 16 = 14 sq cm
3
10
h
5
h 2 + 5 2 = 10 2
h 2 = 10 2 – 5 2 = 75 = (25)3
h = 5 3;
Area = 5h = 25 3
4 a) A diagonal d of a cube is the hypotenuse of a
triangle whose legs are an edge of the cube and a
diagonal of the base. The edge of a cube is 1 and the
diagonal of a base is 2. Hence, d 2 = 1 2 + ( 2) 2 =
1 + 2 = 3. So d = 3.
b) d =
3s
continued
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 183
SQUARES AND SQUARE ROOTS
LESSON 9
ANSWERS TO FOLLOW-UP 9 (CONT.)
6
T
a)
S
R
ab
a
a____
+b
2
ab
+ b )2
(a____
4
ab
b
b) Square T is larger than square S, hence the arithmetic
mean is greater than the geometric mean. One can see
that square T is larger than square S by comparing areas.
Since R and S have the same area, T can be compared
with R.
If R is superimposed on T as shown in the following
sketch, the area of region A is ( a ⁄ 2 + b ⁄ 2 )( b ⁄ 2 – a ⁄ 2 ) while
that 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 area
than R, which consists of regions B and C.
T
( a⁄ 2 + b⁄ 2) – a = ( b⁄ 2 – a⁄ 2)
A
R
C
a⁄2
+ b⁄ 2
184 | ALGEBRA THROUGH VISUAL PATTERNS
B
a
b – ( a⁄ 2 – b⁄ 2) = ( b⁄ 2 – a⁄ 2)
© THE MATH LEARNING CENTER
LINEAR & QUADRATIC
EQUATIONS
LESSON 10
THE BIG IDEA
Algebra Pieces are used to develop strategies for solving linear and quadratic equations and systems of
equations. Coordinate graphs of the values of the
arrangements establish a relationship between algebra
and geometry and illustrate solutions to systems of
linear and quadratic equations.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
Students relate graphs of
points that lie along a linear
path to sequences of counting
piece arrangements and formulas for the nth arrangement of
such sequences.
Students examine relationships
between Algebra Piece, graphical, and symbolic representations of the nth arrangements
of extended sequences of
counting piece arrangements.
They use Algebra Pieces and
graphs to represent and solve
linear and quadratic equations.
Students create sequences
that satisfy specific conditions.
They write formulas for, graph,
and solve linear and quadratic
equations. They complete the
square to solve quadratic
equations.
Materials
Algebra pieces including
frames), 1 set per student.
Materials
Follow-Up 10, 1 copy per
student.
Coordinate grid paper (see
Appendix), 8 sheets per
student.
Materials
Start-Up Masters 10.1 and
10.3, 1 copy of each per
student and 1 transparency
of each.
Start-Up Master 10.2,
1 transparency.
Algebra Pieces (including
frames), 1 set per student.
Focus Masters 10.1, 10.2, and
10.4, 1 transparency of each.
Focus Masters 10.3, 10.5, and
10.6, 1 copy of each per
student and 1 transparency
of each.
Coordinate grid paper (see
Appendix), 2 sheets per
group and 1 transparency.
Algebra Pieces for the
overhead.
1⁄ 4″
Algebra Pieces for the
overhead.
grid paper, 2-4 sheets
per student.
ALGEBRA THROUGH VISUAL PATTERNS | 185
TEACHER NOTES
186 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP
Overview
Students relate graphs of points that
lie along a linear path to sequences
of counting piece arrangements and
formulas for the nth arrangement of
such sequences.
Start-Up Masters 10.1
and 10.3, 1 copy of each
per student and 1 transparency of each.
Start-Up Master 10.2,
1 transparency.
Algebra Pieces for the
overhead.
COMMENTS
ACTIONS
1 Give Algebra Pieces to each
student. Write the following chart
on the overhead. Have the students
form the –2nd through 2nd and the
nth arrangements of an extended
sequence of counting piece arrangements which fits the data on the
chart (n) indicates the arrangement
number and v(n) is the value of
arrangement n). Ask the students to
write a formula for v(n). Discuss.
n
Materials
Algebra pieces including
frames), 1 set per student.
…
–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 their
sequences.
...
Arrangement No.:
Value, v (n ) :
...
–2
–1
0
1
2
n
(–2) + (–1)
(–1) + 0
0+1
1+2
2+3
n + (n + 1)
...
Arrangement No.:
Value, v (n ) :
...
...
...
–2
–1
0
1
2
n
2(–2) + 1
2(–1) + 1
2(0) + 1
2(1) + 1
2(2) + 1
2(n ) + 1
ALGEBRA THROUGH VISUAL PATTERNS | 187
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP
COMMENTS
ACTIONS
2 Give each student a copy of
2
Start-Up Master 10.1. Have the
students form the –3rd through 3rd
and the nth arrangements of an extended sequence of counting piece
arrangements which fits the data
displayed in graphical form on StartUp Master 10.1. Ask them to determine v(–4), v(–3), v(3), and v(4) for
the sequence and, if possible, add
this information to their graph.
Shown below is one extended sequence that fits the data.
For this sequence, v(–3) = –10, v(3) = 8 and v(4) = 11. Coordinate
points for these 3 cases are circled on the copy of Start-Up
Master 10.1 shown on the left. The ordered pair (–4,–13) lies off
the graph.
It may be instructive here to review the use of terms such as
horizontal axis, vertical axis, and origin. (The origin is the coordinate (0,0) which is the point of intersection of the horizontal and
vertical axes.)
...
...
v (n ) :
LINEAR & QUADRATIC EQUATIONS
v (n )
11
10
9
8
7
6
5
4
3
2
1
n
–3
–2
–1
–1
–2
–1
0
1
2
3
–10
–7
–4
–1
2
5
8
LESSON 10
START-UP BLACKLINE MASTER 10.1
–4
–3
1
2
3
4
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
v(n) = __________________________
188 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP
COMMENTS
ACTIONS
3 Ask the students to:
3
a) record in the space provided on
Start-Up Master 10.1 a formula for
v(n) for the sequence they constructed in Action 2;
b) The coordinates are labeled on the copy of Start-Up Master
10.1 shown on the left below.
a) The arrangements shown in Comment 2 suggest the formula
v(n) = 3n – 1. The students may have other equivalent formulas.
c) Students’ observations will vary. Following are examples of
observations that students have made about the graph.
b) label the coordinates of each
point on the graph;
The points of the graph lie along the path of a straight line.
c) record 4 or 5 observations about
the graph.
Discuss, encouraging observations
about relationships between the
numbers in the students’ formulas
for v(n) and their graphs.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
The points are equally spaced.
Moving left to right, to get from one point to the next, go 1 unit to the
right 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.
START-UP BLACKLINE MASTER 10.1
Plotting points for v(n) = 3n – 1 is just like plotting points for v(n) = 3n
after shifting the coordinate axes down 1 unit.
v (n )
(4,11)
11
10
9
In the formula, v(n) = 3n – 1, 3 is the coefficient of n and is the
amount the value, v(n), increases as n increases by 1. The constant
term, –1, is the value of the 0th arrangement. It indicates where
the graph intersects the vertical axis. When the points on a graph
lie along a straight line, the graph is called linear.
(3,8)
8
7
6
(2 5)
5
4
3
(1,2)
2
1
n
–4
–3
–2
–1
–1
–2
(–1,–4)
–3
–4
–5
(–2,–7)
–6
1
(0, 1)
2
3
4
Some students may draw a line connecting the points of the graph,
implying there are arrangements for non-integral values of n. The
students may even suggest ways of constructing such arrangements (see Lesson 11); however, for this extended sequence,
there are only points on the graph for integral values of n.
–7
–8
(–3,–10)
–9
–10
–11
3n – 1
v(n) = __________________________
The intent throughout this lesson is to promote intuitions about
relationships between graphs, formulas, and the sequences of
arrangements the graphs and formulas represent. Terminology
such as slope and x- or y-intercept are introduced in Lesson 11,
after extended sequences of arrangements are augmented so
their graphs are continuous.
ALGEBRA THROUGH VISUAL PATTERNS | 189
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP
COMMENTS
ACTIONS
4 Place a transparency of Start-Up
Master 10.2 on the overhead, revealing the top half only (see next
page). Tell the students the arrangement shown is the nth arrangement
of an extended sequence of counting piece arrangements. Ask the
students to form the –3rd to 3rd
arrangements of this sequence.
5 Distribute a copy of Start-Up
5
Master 10.3 to each student. For
the sequence of Action 4, ask the
students to record a formula for v(n),
construct its graph (see completed
activity below), and record their
observations about the graph. Introduce the concept of function, and
the domain and range of a function.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP BLACKLINE MASTER 10.3
v (n )
12
11
10
9
8
7
6
5
4
3
2
1
n
–8 –7 –6 –5 –4 –3 –2 –1
1
2
3
4
5
6
7
8
–1
–2
–3
–4
4–n
v(n) = __________________________
Observations about the graph:
190 | ALGEBRA THROUGH VISUAL PATTERNS
4 Arrangements numbered –3 through 3 are shown on the
bottom half of Start-Up Master 10.2. Recall that a –n-frame
contains red tile if n is positive and black tile if n is negative. It
contains no tile if n is 0.
The formula for v(n) can be written in various forms. One
possibility is v(n) = 4 – n. Another is v(n) = 4 + (–n).
In general, a function is a rule that relates 2 sets by assigning each
element in the 1st set (called the domain) to exactly one element
in 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, for
any arrangement number, n, there is exactly one value of the
arrangement, v(n). The set of all values for n—in this case, the
integers—is the domain of the function v(n) = 4 + (–n). The set of
all possible values for v(n)—in this case, also the integers—is the
range of the function. (In Lesson 11, students explore functions
whose domains and ranges include all real numbers.)
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP BLACKLINE MASTER 10.1
v (n )
11
10
9
8
7
6
5
4
3
2
1
n
–4
–3
–2
–1
–1
1
2
3
4
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
v(n) = __________________________
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 191
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP BLACKLINE MASTER 10.2
…
…
–3
–2
–1
192 | ALGEBRA THROUGH VISUAL PATTERNS
0
1
2
3
© THE MATH LEARNING CENTER
LINEAR & QUADRATIC EQUATIONS
LESSON 10
START-UP BLACKLINE MASTER 10.3
v (n )
12
11
10
9
8
7
6
5
4
3
2
1
n
–8 –7 –6 –5 –4 –3 –2 –1
1
2
3
4
5
6
7
8
–1
–2
–3
–4
v(n) = __________________________
Observations about the graph:
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 193
TEACHER NOTES
194 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
Overview
Students examine relationships
between Algebra Piece, graphical,
and symbolic representations of the
nth arrangements of extended
sequences of counting piece arrangements. They use Algebra Pieces and
graphs to represent and solve linear
and quadratic equations.
Materials
Algebra Pieces (including frames), 1 set per
student.
Focus Masters 10.1, 10.2,
and 10.4, 1 transparency
of each.
Focus Masters 10.3, 10.5,
and 10.6, 1 copy of each
per student and 1 transparency of each.
Coordinate grid paper
(see Appendix), 2 sheets
per group and 1 transparency.
Algebra Pieces for the
overhead.
1 ⁄ 4″
grid paper, 2-4 sheets
per student.
COMMENTS
ACTIONS
1 Arrange the students in groups
1
Shown below is one possible set of arrangements.
and distribute Algebra Pieces to
each student. Ask the groups to form
the –3rd through 3rd and nth arrangements of an extended sequence of counting piece arrangements for which v(n) = n 2 + 2n + 1.
–3
Arrangement number
–2
–1
0
1
2
3
One possible nth arrangement is shown below. Notice the use of
the two frames to represent 2n. These frames are needed because
they may represent a black n-strip or a red n-strip, depending on
the value of n.
ALGEBRA THROUGH VISUAL PATTERNS | 195
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
2 Tell the students there exists a
sequence of square arrangements
which fits the criterion of Action 1.
Ask the groups to show how their
arrangements from Action 1 can be
formed into a sequence of squares,
using edge pieces to show the values
of the edges of the squares. Discuss.
–3
Arrangement number
–2
2 Some students may have formed square arrangements in
Action 1. If so, you may call the other students’ attention to these
arrangements.
Below is a set of square arrangements with edge pieces.
–1
0
1
2
3
Other edges are possible. For example, here is another possibility
for the –3rd arrangement:
The nth arrangement formed in Action 1 can be rearranged to
form a square with 2 possibilities for edges, as shown below.
Notice that the figures show that (n + 1) 2 and (–n – 1) 2 are
equivalent expressions for v(n).
o
n+1
–n – 1
o
o
o
If edge frames were not discussed earlier, you will need to do so
now. Edge frames are edge pieces whose color, like that of frames,
differs for positive and negative n. Edge frames are obtained by
cutting frames into thirds lengthwise.
Has value n for n
positive, negative, or 0.
196 | ALGEBRA THROUGH VISUAL PATTERNS
o
o
Edge frames
Has value – n for n
positive, negative, or 0.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
The use of edge frames is illustrated below.
o
o
–n
–n
n
n
–n
oooo oooo
–2
o
–2
oooo oooo
–n
o
oooo oooo
–n
o
oooo oooo
2
o
2
o
o
n
o
n
o
o
o
n
–n
n
Note that the area of a square is always positive, while the value
of a square can be positive, negative, or zero. Similarly, the lengths
of the edges of a square are always identical and positive, while
the values of the edges may be zero, both positive, both negative,
or one positive and one negative.
3 Ask the students to determine
3
which arrangements in the extended
sequence of Actions 1 and 2 have a
value of 400. Discuss the students’
methods. Ask them to identify the
equation that has been solved.
A square has value 400 provided its edges all have value 20 or
–20. Hence, the nth arrangement, viewed as a square whose edge
has 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, the
19th 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 Place a transparency of Focus
4
oooo oooo
Master 10.1 on the overhead. Ask
the students to form the nth arrangement of this sequence and to
write an expression for v(n).
oooo oooo
–1
0
1
2
3
oooo oooo
–2
FOCUS BLACKLINE MASTER 10
…
LINEAR & QUADRATIC EQUATIONS
…
–3
oooo oooo
The nth arrangement contains a black n 2 -mat and 4 –n-frames:
v (n ) = n 2 – 4n
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 197
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
5 Ask the students to determine
5
525
n–2
23 or –23
6 Ask the students to form the nth
arrangement of an extended sequence
for which v(n) = n 2 + 4 and then
have them do the same for an extended sequence for which v(n) =
2n 2 + 6n – 3. Have the students use
their Algebra Pieces to determine
for which n the nth arrangements of
these two extended sequences have
the same value. Ask the students to
identify the equation that has been
solved and to verify their solutions.
o
oooo oooo
o
o
oooo oooo
oooo oooo
o
v(n) + 4 = (n – 2)2
4
oooo oooo
oooo oooo
oooo oooo
oooo oooo
oooo oooo
Adding 4 black tile to an nth arrangement results in a square
array whose edges have value n – 2 or –n + 2, as shown below.
for what n the extended sequence
in Action 4 has v(n) = 525. Discuss
their strategies. If it isn’t suggested
by students, introduce the method
of solving the quadratic equation
n2 – 4n = 525 by completing the square.
v(n) = (–n + 2)2
A square whose value is 525 + 4, or 529, has an edge whose value
is 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 the
edge 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 is
called completing the square. A quadratic equation is an equation
that can be written in the form ax 2 + bx + c, where a, b, and c are
constants and a ≠ 0. The word quadratic is derived from the Latin
word, quadratus, meaning square.
6 One way of representing the two nth arrangements is shown
below:
n2 + 4
2n2 + 6n – 3
These two arrangements have the same value if, after an n 2 -mat
has been removed from each of them, the remaining portions
198 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
have the same value, i.e., if n 2 + 6n – 3 has value 4 or, equivalently,
n 2 + 6n has value 7.
The method of completing the square is illustrated at the left.
Adding 9 black tile to n 2 + 6n produces a square array whose
edge has value n + 3.
Thus, n 2 + 6n has value 7 if the square array has value 16, i.e., if
its 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 2
sequences have the same value.
7 Write quadratic equation a)
7
below on the overhead. Then ask
the students to find all solutions of
the equation. Repeat for one or
more of b)-g). Discuss the students’
methods.
a) n 2 – 6n = 40
Students’ methods of solving these equations may vary.
a) If 9 black tile are added to a collection for n 2 – 6n, the resulting collection can be formed into a square array with edge n – 3
(see below). If n 2 – 6n has value 40, the square array has value 49
and 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 which
case, the solutions are still 10 and –4.
oooo oooo oooo
oooo oooo oooo
b) 2n 2 + 38 = 4n 2 – 12
c) (n – 1)(n + 3) = 165
d) 4n 2 + 4n = 2600
oooo oooo oooo
n–3=
7 or –7
e) n 2 – 5n + 6 = 0
f) n 2 + n = 6
g) n 2 + 3n – 10 = 0
oooo oooo oooo
( n 2 – 6 n ) + 9 = 40 + 9 = 49
b) Sketches for 2n 2 + 38 and 4n 2 – 12 have the same value if
2n 2 – 12 is 38. This is the case if n 2 is 25, that is, if n = 5 or n = –5.
38
n2
n2
38
n2
n2
n2
n2
–12
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 199
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
165
oooo
oooo
7 continued
c) Shown at the left is a representation of (n – 1)(n + 3). Note
the values of the edges differ by 4 and their product is 165. Since
11 and 15 differ by 4 and 11 x 15 = 165, and since –11 and –15
differ by 4 and –11 x –15 = 165, the array will have value 165 if
the edges have values 11 and 15 or –11 and –15. If the edges have
values 11 and 15 then n is 12; if they have values –11 and –15 then
n is –14. (Finding the pair 11 and 15 is facilitated by noting that one
of the pair should be smaller and one larger than 165 ≈ 13.)
11 or –15
15 or –11
n+1
169
n
1
n2
n
Alternatively, adding 4 black tile to the array shown above and
removing collections whose values are 0 leaves a collection of
pieces that can be arranged in a square array that has value 169
and edge n + 1. Hence n + 1 is 13 or –13, in which case n = 12 or
n = –14.
n+1
2600
1
d) If 1 black tile is added to a collection for 4n 2 + 4n, a square
array with edge 2n + 1 (or –2n – 1) can be formed, as shown at
the left. If the value of the original collection is 2600, the value
of 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 4n 2 + 4n by 4
results in the collection n 2 + n with value 650. From this
collection, 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.
oooo oooo
oooo oooo
51 or –51
oooo oooo oooo
oooo oooo oooo
200 | ALGEBRA THROUGH VISUAL PATTERNS
e) A collection for n 2 – 5n + 6 can be formed into a rectangular
array with edges n – 2 and n – 3. The array has value 0 if an edge
has value 0. This is the case if n = 2 or n = 3.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
1
__
4
or –
oooo oooo
oooo oooo
oo
1
__
2
oo
oooo oooo oo
oooo oooo oo
0
1
__
2
6
6
1
__
4
n/2
n2
n
n2 + n
n + __21
or
–n – __21
n2
Alternatively, by cutting a –n-frame and 2 black tile in halves and
adding 1 ⁄ 4 of a black tile to a collection for n 2 – 5n + 6, the
resulting collection can be formed into a square with edge n – 2 1 ⁄2.
If the original collection has value 0, the square array has value 1 ⁄ 4
and its edge has value 1 ⁄ 2 or – 1 ⁄ 2 . If n – 2 1 ⁄ 2 = 1⁄ 2 then n = 3 and if
n – 2 1 ⁄ 2 = – 1 ⁄ 2 , then n = 2.
f) Beginning with a collection for n 2 + n, if one cuts the n–frame
in halves and adds 1 ⁄ 4 of a black tile, a square array with edge
n + 1 ⁄ 2 can be formed. This square array has value 6 1 ⁄ 4 and its edge
has value 21⁄ 2 or –2 1⁄ 2. If n + 1⁄2 = 2 1⁄2 , then n = 2; if n + 1⁄ 2 = – 21 ⁄ 2,
then n = –3.
n
__
2
n 2 + n + 1⁄ 4 = ( n + 1⁄ 2 ) 2 or ( –n – 1 ⁄ 2 ) 2
g) A collection for n 2 + 3n – 10 contains a n 2 -mat, 3
n-frames and 10 red tile. If this collection is arranged as
shown on the left and 2 n-frames and 2 –n-frames are
added, the value of the collection is unchanged and the
resulting collection can be formed into an array with
edges n – 2 and n + 5. This array has value 0 if one of the
edges has value 0, that is if n = 2 or n = –5.
n+5
Here are other equations you might have students solve:
(n – 4)(n + 2) = 0; n 2 + 4n – 5 = 0; n 2 + 6n = –8;
n 2 = 7n – 6; 2n 2 – 2n = 112. If you create others, be sure
they have integer solutions (in later lessons students
explore sequences with nonintegral arrangement numbers).
n2 + 3n – 10
n–2
ALGEBRA THROUGH VISUAL PATTERNS | 201
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
8 Place a transparency of Focus
8
Master 10.2 on the overhead, revealing the top half only. Tell the students that the arrangement shown
is the nth arrangement of an extended sequence of counting piece
arrangements. Ask them to form the
–3rd through 3rd arrangements
(with edges) of this sequence. Then
distribute a copy of Focus Master
10.3 to each student and have the
students write a formula for v(n),
construct its graph (see completed
graph below), and record their observations about the graph.
LINEAR & QUADRATIC EQUATIONS
Arrangements numbered –3 through 3 are shown on the
bottom half of Focus Master 10.2.
Here are three possibilities for v(n): v(n) = n 2 – 2n – 3;
v(n) = (n + 1)(n – 3); and v(n) = (–n – 1)(–n + 3).
Edge pieces help to illustrate the latter two formulas:
oooo oooo oooo
oooo oooo oooo
LESSON 10
FOCUS BLACKLINE MASTER 10.2
oooo oooo oooo
o
oooo oooo oooo
o
o
o
The coordinate points associated with these arrangements are
shown on the completed graph on the left.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
Here are some observations made about the graph:
FOCUS BLACKLINE MASTER 10.3
v (n )
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.
32
30
28
The graph is symmetric about the vertical line that passes through
n = 1. That is, if the graph is folded along the vertical line that goes
through n = 1, the points to the right of the fold coincide with those to
the left of the fold.
26
24
22
20
18
16
14
12
The point (1,–4) is a turning point where the graph stops falling and
starts to rise (looking from left to right).
10
8
6
4
2
n
–8 –7 –6 –5 –4 –3 –2 –1
–2
1
2
3
4
5
6
7
The smallest value for v(n) is –4. It occurs when n = 1.
8
–4
( n + 1)(n – 3)
v(n) = ______________________
202 | ALGEBRA THROUGH VISUAL PATTERNS
When n is greater than 1, as n increases so does v(n). When n is less
than 1, as n decreases v(n) increases.
The domain of v(n) = n 2 – 2n – 3 is the integers and its range is the
set of integers greater than or equal to –4.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
9 Place a copy of Focus Master 10.4
on the overhead, and tell the students that Arrangements I and II are
the nth arrangements of two different extended sequences. Ask them
to write formulas for v 1 (n), the
value of the nth arrangement of the
first sequence and v 2 (n), the value of
the nth arrangement of the second
sequence.
LINEAR & QUADRATIC EQUATIONS
9
Here are possible formulas for the given nth arrangements:
v 1 (n) = 6n – 2
v 2 (n) = n 2 + 7n – 8
Other formulas are possible. For example, v 2 (n) = (n + 8)(n – 1);
edge pieces may help the students see this formula:
LESSON 10
FOCUS BLACKLINE MASTER 10.4
n+8
n–1
I
II
10 Give each student a copy of
Focus Master 10.5 and ask them to
do the following:
a) record their formulas for v 1 (n)
and v 2 (n);
b) graph v 1(n) and v 2(n) on the coordinate grid, indicating the points on
the graph of v1(n) with an x and those
on the graph of v 2 (n) with an o (see
completed graph on the next page);
10 The completed graphs are shown on a copy of Focus Master
10.5 on the next page.
On the next page are some observations about the graphs. If
students don’t bring these up, you might prompt discussion by
posing questions such as: What points, if any, do the 2 graphs have
in common? What do the common points on the graphs tell about
the 2 sequences of counting piece arrangements? When is v 1 (n)
greater than v 2 (n)? How do the shapes of the graphs compare?
How could you change the equation of v 1 so that it doesn’t
intersect v 2 ? Are v 1 and v 2 functions? How do the domains and
ranges of v 1 and v 2 compare?, etc. If you wish, you may use the
discussion as an opportunity to introduce inequality notation.
c) record observations about the
graphs and the relationships between them.
Discuss their observations.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 203
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.5
v (n )
36
x
32
24
Note: you might have the students form the 2nd arrangement of
each sequence and verify that they have the same value. Likewise
for the –3rd arrangements.
x
20
x
16
12
8
4
x
x
n
1
2
3
4
5
6
–4
x –8
x
continued
Two points, (2,10) and (–3,–20), are on both graphs. This tells us that
the 2nd arrangements of the 2 sequences have the same value and
the –3rd arrangements also have the same value. It also tells us the
equation 6n – 2 = n 2 + 7n – 8 has 2 solutions, n = 2 and n = –3.
x
28
–11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1
10
–12
–16
When n is between –3 and 2, the values of the arrangements in
Sequence I are greater than the values of the arrangements in
Sequence II. Using inequality notation: if –3 < n < 2, then v1(n) > v2(n).
–20
x
–24
–28
6n – 2
x: v 1(n) = ______________________
n 2 + 7n – 8
o: v 2(n) = ______________________
Observations:
Observations:
When n is less than –3 and when n is greater than 2, the value of
Sequence II is greater than the value of Sequence I. Using inequality
notation, if n < –3 or n > 2, then v 2(n) > v 1 (n).
The graph of v 1 (n) follows the path of a straight line, and the graph of
v 2 (n) follows the path of a U-shaped curve.
The graph of Sequence I always rises as values of n increase from left
to right. Looking at the graph of Sequence II from left to right, the
graph 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 midway
between 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 (–3 1 ⁄ 2 ,–20 1 ⁄ 4 ). Since the curve is symmetric , we think the turning point is half way between n = – 3 and n = – 4,
or at n = – 3 1 ⁄ 2 . We found the y-coordinate of the turning point by
finding (3 1 ⁄ 2 ) 2 + 7(–3 1 ⁄ 2 ) – 8 = – 20 1 ⁄ 4 .
A function such as v 1 (n) whose graph lies on a straight line is
called a linear function; and a function such as v 2 (n) whose graph
lies on a parabola (i.e., a U-shaped curve) is called a quadratic
function.
204 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
11 Point out to the students that
in Action 10 they used information
from their graphs to find the solutions of the equation 6n – 2 = n 2 +
7n – 8. Then ask the students to
solve this equation by a method
that doesn’t involve graphing. Discuss the methods the students use.
11 You might ask the students for their views on the advantages
and disadvantages of graphical versus non-graphical methods. (In
Lesson 13 students are introduced to graphing calculators, at
which time their views may change.)
Students will use a variety of ways to solve this equation. Some of
these ways are shown below.
6n – 2
added
a) One can view n 2 + 7n – 8 as an n + 8 by n – 1
array whose value equals 6n – 2. When 6n – 2 is
removed from this array, the remaining portion must
have value 0. If this portion is rearranged and an nstrip and a –n-strip is added to it, its value is unchanged and the result is an n + 3 by n – 2 array
whose value is 0. Hence, one of the dimensions of
the array must have value 0. Thus,
n = –3 or n = 2.
0
(n + 3)(n – 2) = 0
b) Alternatively, removing an n-strip and a –n-strip from the
arrangement of value 0, and then adding 6 black tile to it, results
in an n + 1 by n array whose value is 6, that is, an array whose
dimensions 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.
0
n2 + n – 6 = 0
(n + 1)n = 6
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 205
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
11 continued
c) One can also find a solution by completing the square. By
taking the above n + 1 by n array whose value is 6 and cutting the
n-strip in half and adding 1 ⁄ 4 of a black tile, one can form a square
of dimension n + 1 ⁄ 2 whose value is 6 1 ⁄ 4 or 25 ⁄ 4 . Hence, n + 1 ⁄ 2 = 5 ⁄ 2
or n + 1 ⁄ 2 = – 5 ⁄ 2 and, as before, one has n = 2 or n = –3.
(n
1 2
+ __
2 )
=
1
6 __
4
One might record the above actions using algebraic symbols as
follows:
n2
n 2 + 7n – 8
+ 7n – 8 – (6n – 2)
n2 + n – 6
n2 + n
2
n + n + 1⁄ 4
(n + 1 ⁄ 2 ) 2
n + 1⁄ 2
n
n
=
=
=
=
=
=
=
=
=
6n – 2
0
0
6
6 + 1 ⁄ 4 = 25 ⁄
25 ⁄ 4
± 5⁄ 2
– 1⁄ 2 ± 5⁄2
2 or n = –3
4
d) In the above completing the square procedures, one can
avoid cutting pieces by introducing an appropriate multiplication in the procedure. Once one has determined that a
collection of 1 n 2 -mat and 1 n-strip has value 6, then a
collection of 4 n 2 -mats and 4 n-strips has value 24, so that a
collection of 4 n 2 -mats and 4 n-strips and 1 black tile has
value 25. This collection can be formed into a square of
side 2n + 1. Hence, (2n + 1) 2 = 25, whence
2n + 1 = 5 or 2n + 1 = –5 and, as before, n = 2 or n = –3.
n2 + n = 6
4n2 + 4n = 24
(2n + 1)2 = 25
206 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
12 Give each student coordinate
grid paper (see Appendix) and a copy
of Focus Master 10.6. Have them
carry out the instructions. When
the students are finished, invite
volunteers to share their questions,
observations, and conjectures.
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.6
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 v 1(n) and v 2 (n) for n from –3 to 3. Then graph v 1(n) and v 2(n) on
the same coordinate axes.
12 The activities on Focus Master 10.6 could be completed as
homework and then discussed in class.
This action is intended to acquaint students with the graphs of
constant, linear, and quadratic graphs. These are considered in
greater detail in subsequent lessons.
A constant function is a function whose range consists of a single
quantity such as v 1 (n) in pair 2 where the value for every n is –7.
Both v 1 (n) and v 2 (n) in pair 1 are examples of linear functions. In
general, v(n) is a linear function if v(n) is of the form an + b where
a is not zero. Both v 1 (n) and v 2 (n) in pair 3 and v 2 (n) in pair 2 are
examples of quadratic functions. In general, v(n) is a quadratic
function if v(n) is of the form an 2 + bn + c where a is not zero.
b) Determine when v 1(n) = v 2(n).
Pair 1
v 1(n) = –3n + 2
v 2(n) = 4n – 12
A table of values for v 1 (n) and v 2 (n) is useful for generating and
organizing ordered pairs to plot. Following is a table for Pair 1.
Pair 2
v 2(n) = –n 2 – 2n + 8
v 1(n) = –7
n
–3
–2
–1
0
1
2
3
Pair 3
v 1(n) =
n2
+2
v 2(n) =
–n 2
+4
2 Record your general observations and conjectures about graphing constant, linear, and
quadratic functions.
v 1 (n)
11
8
5
2
–1
–4
–7
v 2 (n)
–24
–20
–16
–12
–8
–4
–0
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 207
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS
COMMENTS
ACTIONS
12 continued
In each of the graphs shown here, points on the graph of v 1 (n) are
indicated by an o and those on the graph of v 2 (n) by an x.
Pair 1
v (n )
12
11
x
One can determine when v 1 (n) = v 2 (n) for Pair 1 and Pair 3 from
the graphs. However, not all solutions of v 1 (n) = v 2 (n) in Pair 2 are
apparent from the portion of the graph shown for that pair. The
students may deduce by symmetry that, in addition to being equal
when n = 3, v 1 (n) and v 2 (n)are also equal when n = –5. The students can also find this solution by extending their graph to the
left or through the use of Algebra Pieces.
10
9
x
8
7
6
x5
4
3
x
2
1
–3 –2 –1
–1
Pair 2
1 2
x
o3
n
12
–2
x
o
10
–6
–7
–9
–10
–11
o
–12
–13
–14
–15
x
x
5
4
3
10
9
8
o
x
1 2 3
7
6
x
n
x3 o
o
x
2o
–2
–3
–18
–4
–19
–5
–3 –2 –1
–1
–6
–2
1
x
x
o oo–7o o o o
–22
–8
–23
–9
–24
–10
o
5
4
–17
–21
o
11
x
2
1
–3 –2 –1
–1
The graph of a constant function lies
along a horizontal line, that of a
linear function along a non-horizontal line, and that of a quadratic
function along a U-shaped curve.
12
o
7
6
o–16
o –20
o
o
v (n )
x9
x 8x
–5
–8
Pair 3
11
–3
–4
Following are some conjectures
the students might make about
the graphs:
v (n )
x
1 2 3
–3
–4
x
–5
–6
x
n
A graph that is linear rises from left
to right if the coefficient of n is
positive and falls if the coefficient is
negative.
For an equation whose graph follows
a linear path, the constant moves the
graph up or down from the horizontal axis. The coefficient of the nterm determines the “steepness” of
the graph.
If the coefficient of the n 2 -term is negative the U opens down (the U
would spill water). If the coefficient of the n 2 -term is positive the U
opens up (the U would hold water).
A line and a U-shaped curve can only intersect in 0, 1, or 2 points; 2
different lines can intersect in 0 or 1 point; 2 different U-shaped
curves can intersect in 0, 1, or 2 points.
208 | ALGEBRA THROUGH VISUAL PATTERNS
LINEAR & QUADRATIC EQUATIONS
LESSON 10
…
–3
–2
–1
0
1
2
3
…
FOCUS BLACKLINE MASTER 10.1
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 209
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.2
…
…
–3
–2
210 | ALGEBRA THROUGH VISUAL PATTERNS
–1
0
1
2
3
© THE MATH LEARNING CENTER
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.3
v (n )
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
n
–8 –7 –6 –5 –4 –3 –2 –1
–2
1
2
3
4
5
6
7
8
–4
v(n) = ______________________
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 211
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.4
I
212 | ALGEBRA THROUGH VISUAL PATTERNS
II
© THE MATH LEARNING CENTER
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.5
v (n )
36
32
28
24
20
16
12
8
4
n
–11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1
1
2
3
4
5
6
–4
–8
–12
–16
–20
–24
–28
x: v 1(n) = ______________________
o: v 2(n) = ______________________
Observations:
Observations:
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 213
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOCUS BLACKLINE MASTER 10.6
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 v 1 (n) and v 2(n) for n from –3 to 3. Then graph v 1(n) and v 2(n) on
the same coordinate axes.
b) Determine when v 1(n) = v 2 (n).
Pair 1
v 1(n) = –3n + 2
v 2(n) = 4n – 12
Pair 2
v 1(n) = –7
v 2(n) = –n 2 – 2n + 8
Pair 3
v 1(n) = n 2 + 2
v 2(n) = –n 2 + 4
2 Record your general observations and conjectures about graphing constant, linear, and
quadratic functions.
214 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
LINEAR & QUADRATIC EQUATIONS
LESSON 10
FOLLOW-UP BLACKLINE MASTER 10
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) n 2 + 2n = 35
b) (n – 5)(n – 1) = 21
c) 2n 2 – 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 horizontal 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.
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 215
LINEAR & QUADRATIC EQUATIONS
LESSON 10
ANSWERS TO FOLLOW-UP 10
1
a)
n2
36
n2 + 2n + 1
2
a) One possibility:
v(n) = –n + 2
7
o
o
o
o
oo
o
ooo oooo
ooo oooo
ooo
5
4
3
–4 –3 –2 –1
–1
–2
n
1 2 3 4
2
1
–3
(n – 3) 2 = 25
n – 3 = 5 or –5
n = 8 or –2
o oo
o
7
6
1
(n – 5)(n – 1) = 21
n 2 – 6n + 5 = 21
n 2 – 6n + 9 = 25
Rearrange and
add 4 black:
oo
10
9
8
3
2
(n – 1)(n – 5)
oo
v (n)
11
6
5
4
21
oooo oooo
b) One possibility:
v(n) = 3n + 2
v (n )
o
b)
n 2 + 2n = 35
+ 2n + 1 = 36
(n + 1) 2 = 36
n + 1 = 6 or –6
n = 5 or –7
–3 –2 –1
–1
–2
n
–3
–4
–5
oo
–6
–7
oooo
oo
25
oo
c) One possibility:
v(n) =n 2 – 3
oooo
oo
(n – 3)2 = 25
d) One possibility:
v(n) = –2n 2
v (n )
v (n )
oo
oooo oooo
oo
oooo
c)
oo
6
oooo oooo
oo
oooo
oo
2n2 – 6n
oo
oo
oo
o
Double, rearrange,
and add 9 black:
oo
oo
2n 2
81
4
–3 –2 –1
–1
–2
3
–3
2
–4
–5
5
36
oo
1 2 3
– 6n = 36
4n 2 – 12n = 72
2
4n – 12n + 9 = 81
(2n – 3) 2 = 81
2n – 3 = 9 or –9
2n = 12 or –6
n = 6 or –3
1
–3 –2 –1
–1
–2
–3
3
n
–6
–7
–8
–9
n
1 2 3
1 2 3
One possibility: v 1 (n) = n + 2
v 2 (n) = 2n – 5
v (n)
ooo
x
ooo
ooo
13
(2n – 3)2
12
x o
o
11
10
9
o
x
o
o x
8
7
6
o
o
5
4
3
2
1
x
o
o
x
x
–1
–2
1 2 3
–3
x
x
4 5 6
7 8 9
n
o: v1(n) = n + 2
x: v2(n) = 2n – 5
216 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
COMPLETE SEQUENCES
LESSON 11
THE BIG IDEA
Complete sequences of arrangements, which have an
arrangement corresponding to every point of a number line, are introduced. The values of these arrangements serve as a vehicle for studying linear and quadratic graphs and equations.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
An extended sequence of
arrangements is augmented to
provide an arrangement
corresponding to every point
on the number line.
Graphs of complete sequences
of arrangements are constructed, leading to a discussion
of lines and parabolas. The
coordinates of various points
of graphs are determined by
observation or by using Algebra Pieces or sketches to solve
the appropriate equations.
Students graph straight lines
and write their equations. They
find the intercepts, turning
points, and points of intersections of intersecting parabolas,
and solve linear and quadratic
equations.
Materials
Start-Up Masters 11.1-11.2;
1 copy of each per student
and 1 transparency of each.
Materials
Follow-Up 11, 1 copy per
student.
Coordinate grid paper (see
Appendix).
Materials
Algebra Pieces for the
overhead.
Algebra pieces for each
student.
Focus Masters 11.1-11.2,
1 copy of each per student
and 1 transparency of each.
Focus Masters 11.3-11.4,
1 transparency of each.
Coordinate grid paper (see
Appendix), 10 or 11 sheets
per student and 1 transparency.
Algebra pieces for the
overhead.
ALGEBRA THROUGH VISUAL PATTERNS | 217
TEACHER NOTES
218 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLETE SEQUENCES
LESSON 11
START-UP
Overview
An extended sequence of arrangements is augmented to provide an
arrangement corresponding to
every point on the number line.
ACTIONS
1 Distribute a copy of Start-Up
Master 11.1 to each student. Show
them the Algebra Piece Arrangement of 2 n-frames and 1 red tile
shown below. Tell them it is the nth
arrangement of an extended sequence of counting piece arrangements. Ask them to sketch, in Section A of Start-Up Master 11.1, the
–3rd through 3rd arrangements of
the sequence, representing counting
pieces by grid squares.
2 Distribute a copy of Start-Up
Master 11.2 to each student. For
the sequence introduced in Action 1,
ask the students to record a formula for v(n) in the space provided
and then construct a graph of v(n).
Materials
Start-Up Masters 11.111.2; 1 copy of each per
student and 1 transparency of each.
Algebra Pieces for the
overhead.
COMMENTS
1
The students can use red and black pens or markers, if available, to fill in squares to represent red and black tile. In the
sketches shown below, the hatched squares represent red tile.
–3
–2
arrangement number
–1
0
1
2
3
2
The completed graph is shown below. Some students may have
other expressions for v(n) that are equivalent to the one shown.
COMPLETE SEQUENCES
LESSON 11
START-UP BLACKLINE MASTER 11.2
v (n )
11
10
9
8
7
6
5
4
3
2
1
n
–8 –7 –6 –5 –4 –3 –2 –1
1
1
2
3
4
5
6
7
8
–2
3
–4
5
–6
–7
8
–9
10
–11
v(n) =
2n – 1
ALGEBRA THROUGH VISUAL PATTERNS | 219
COMPLETE SEQUENCES
LESSON 11
START-UP
ACTIONS
3 Mention to the students that
there is no point on the graph for
n = 1 1 ⁄ 2 since there is no 1 1 ⁄ 2 th
arrangement. Ask the students to
imagine that the sequence has been
augmented to contain such an
arrangement. Ask them to sketch, in
Section B of Start-Up Master 11.2,
what they think the 1 1 ⁄ 2th arrangement looks like. Have them compute
the value of that arrangement and
add the corresponding point to their
graph. Discuss their ideas and reasoning. Repeat for n = 3 1 ⁄ 2 and
n = –2 3⁄ 4 in Sections C and D of
Start-Up Master 11.2.
COMMENTS
3
Below on the left is a sketch of a 1 1 ⁄ 2 th arrangement, based on
the pattern of the arrangements in the original sequence. Its net
value is 2. Thus, (1 1 ⁄ 2 ,2) is the point on the graph corresponding
to this arrangement. Similarly, (3 1 ⁄ 2 ,6) is the point associated with
the 3 1 ⁄ 2 th arrangement.
B
C
v (1 1 ⁄ 2 ) = 2(1 1 ⁄ 2 ) – 1 = 2
The net value of the –2 3 ⁄ 4 th arrangement is –6 1 ⁄ 2 , as illustrated
below. Its corresponding point on the graph is (–2 3 ⁄ 4 ,–6 1 ⁄ 2 ).
D
220 | ALGEBRA THROUGH VISUAL PATTERNS
v (3 1 ⁄ 2 ) = 2(3 1 ⁄ 2 ) – 1 = 6
COMPLETE SEQUENCES
LESSON 11
START-UP
ACTIONS
4 Have the students do the following on Start-Up Master 11.1 and
11.2:
a) On Start-Up Master 11.2 choose
some point between two integers
on the positive part of the n-axis
and suppose it represents the positive number P. Then choose some
point between two integers on the
negative part of the n-axis and
suppose it represents the negative
number Q.
b) In Sections E and F of Start-Up
Master 11.1, sketch, respectively, the
Pth and Qth arrangements and
determine the values of these
arrangements.
c) Add the points associated with
the Pth and Qth arrangements to
the graph in Start-Up Master 11.2.
Discuss the methods the students
use to carry out these procedures.
COMMENTS
4
The students choice of location for P and Q will vary.
Shown below are sketches of the Pth and Qth arrangements for the
choices of P and Q shown on the graph at the left below. The corresponding points (P,2P – 1) and (Q,2Q – 1) are shown on the graph.
v (P) = 2P – 1
v (Q ) = 2 Q – 1
The location of the points on the graph can be determined by
measuring. For example, one can mark off on the edge of a piece
of 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 these
two lengths will be the distance of the point (P,2P – 1) above the
n-axis. This is illustrated below. Some of the students may locate
the 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.
COMPLETE SEQUENCES
LESSON 11
START-UP BLACKLINE MASTER 11.2
v (n )
11
10
9
8
7
6
5
4
3
2
1
Q
P
n
–8 –7 –6 –5 –4 –3 –2 –1
1
1
2
3
4
5
6
7
8
–2
–3
–4
5
–6
–7
–8
9
10
–11
v(n) =
2n – 1
ALGEBRA THROUGH VISUAL PATTERNS | 221
COMPLETE SEQUENCES
LESSON 11
START-UP
ACTIONS
5 Ask the students to imagine that
the sequence of arrangements
discussed in Actions 1-4 has been
augmented so there is an arrangement corresponding to every point
on the n-axis. Ask them how they
could show this on their graphs.
Discuss.
Introduce the terms complete sequence of arrangements and real
number.
COMMENTS
5
The resulting graph is a straight line, only a portion of which
shows in the graph the students have constructed. The actual
graph extends indefinitely in both directions, as indicated by the
arrowheads.
A sequence of arrangements which has an arrangement corresponding to every point of a coordinate axis will be referred to
as a complete sequence of arrangements. The set of numbers
corresponding to the points of a coordinate axis is called the set
of real numbers. (Every real number may be represented by a
decimal, which may or may not terminate.) Thus a complete
sequence of arrangements is a sequence which has an arrangement for every real number.
COMPLETE SEQUENCES
LESSON 11
START-UP BLACKLINE MASTER 11.2
v (n )
11
10
9
8
7
6
5
4
3
2
1
Q
P
n
–8 –7 –6 –5 –4 –3 –2 –1
1
2
–3
–4
–5
6
–7
–8
–9
10
–11
v(n) =
222 | ALGEBRA THROUGH VISUAL PATTERNS
2n – 1
1
2
3
4
5
6
7
8
COMPLETE SEQUENCES
LESSON 11
START-UP
6
The students may suggest a variety of ways to represent the
xth arrangement. It might be represented by a sketch. For example, the 1st sketch below consists of 2 unshaded strips, each
labeled x, and a single red counting piece. It is understood that
each 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 positive, the strip is filled with black; if x is negative it is filled with
red; and if x is 0, it is empty. The 2nd sketch uses edge pieces to
represent 2x as a rectangle with edge values 2 and x.
x
x
2
2x – 1
x
2x – 1
Alternatively, Algebra Pieces might be used to form a representation. The frames are to be thought of as x-frames rather than
n-frames; that is, each frame represents a strip of counting pieces
or partial pieces whose value is x.
2x – 1
Henceforth, the letter x will be used when referring to a generic
arrangement in a complete sequence of arrangements. (With this
convention, the horizontal axis in the graph shown in Action 5
would be labeled x, the vertical axis labeled v(x) and the formula
written as v(x) = 2x – 1.) The letter n will be used when referring
to a generic arrangement in a sequence of arrangements corresponding to integers, e.g., an extended sequence of arrangements.
When arrangement
numbers are real
numbers:
n - frame
–n - frame
oooo
When arrangement
numbers are integers:
oooo
generic point x on the horizontal
axis in Action 5 is selected and to
create a representation of the xth
arrangement. Discuss their representations.
oooo
6 Ask the students to suppose a
COMMENTS
oooo
ACTIONS
x - frame
–x - frame
The above usage follows the customary, but not universal, practice
of using letters like x, y, and z to represent quantities that can
take on any value, integral or not, (i.e., continuous variables) and
using letters like k, m, and n to represent quantities that have
integral values (i.e., discrete variables). The choice of a letter to
represent a generic arrangement is arbitrary. For example, one
might refer to the zth arrangement and write v(z) = 2z – 1. In
this case, if frames were used to represent the zth arrangement,
they would be referred to as z-frames or –z-frames and have
values z or –z, respectively.
ALGEBRA THROUGH VISUAL PATTERNS | 223
TEACHER NOTES
224 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLETE SEQUENCES
LESSON 11
START-UP BLACKLINE MASTER 11.1
A
B
C
D
E
F
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 225
COMPLETE SEQUENCES
LESSON 11
START-UP BLACKLINE MASTER 11.2
v (n )
11
10
9
8
7
6
5
4
3
2
1
n
–8 –7 –6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
7
8
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
v(n) =
226 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
COMPLETE SEQUENCES
LESSON 11
FOCUS
Overview
Graphs of complete sequences of
arrangements are constructed,
leading to a discussion of lines and
parabolas. The coordinates of various points of graphs are determined
by observation or by using Algebra
Pieces or sketches to solve the
appropriate equations.
Materials
Algebra pieces for each
student.
Coordinate grid paper
(see Appendix), 10 or 11
sheets per student and
1 transparency.
Algebra pieces for the
overhead.
Focus Masters 11.3-11.4,
1 transparency of each.
COMMENTS
ACTIONS
1 Show the students the following
oooo oooo oooo
xth arrangement from a complete
sequence of arrangements. Ask them
to write a formula for v(x). Then
give a sheet of coordinate grid
paper to each student and have
them construct a graph consisting
of all points (x,y) such that y = v(x).
oooo oooo oooo
Focus Masters 11.1-11.2,
1 copy of each per student and 1 transparency
of each.
1
You may want to clarify that the partial counting piece is 1 ⁄ 2 of
a black counting piece, so y = v(x) = 7 ⁄ 2 – 3x. Note that, in the
graph shown below, the horizontal axis is labeled x. The vertical
axis is labeled y, it could also be labeled v(x).
Some students may write y = 7 ⁄ 2 + 3(–x). You may want to discuss
why 7 ⁄ 2 – 3x and 7 ⁄ 2 + 3(–x) are equivalent expressions. One way
to see this is to note that a collection of 3 –x-frames is the
opposite of a collection of 3 x-frames, that is 3(–x) = –(3x), and
adding the opposite of a value is equivalent to subtracting that
value. Thus, 7 ⁄ 2 + 3(–x) = 7 ⁄ 2 + (–(3x)) = 7 ⁄ 2 – 3x.
y
11
10
9
8
7
6
5
4
3
2
1
–8 –7 –6 –5 –4 –3 –2 –1
1
1
2 3 4 5 6 7 8
x
2
–3
–4
–5
–6
7
8
9
–10
–11
y = 7⁄2 – 3x
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 227
COMPLETE SEQUENCES
LESSON 11
FOCUS
ACTIONS
COMMENTS
1
continued
Notice that for this graph, the values for x included in the graph
are all of the real numbers and the values for y are all of the real
numbers. Hence, the domain and range of y = 7 ⁄ 2 – 3x are the real
numbers.
(x, y) is on the graph in Action 1 and
that y = 8. Have them determine
what x equals. Ask for volunteers to
describe the methods they use.
Discuss.
The students might see from their graph in Action 1, if drawn
accurately, that if the point (x, y) is on their graph and y = 8, then
x = –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 been
obtained by graphing the function y = 7 ⁄ 2 – 3x and determining
the point on this graph for which y = 8.
Some fractions can be avoided by doubling the collection of
pieces representing v(x) as shown in the figure below. In this
figure, we have 2v(x) = 16, hence the 6 –x-frames have a total
value 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 .
oooo oooo oooo oooo oooo oooo
Repeat if y is: a) –9, b) –14.2, c) 100.
2
oooo oooo oooo oooo oooo oooo
2 Tell the students that the point
2 v ( x ) = 7 + 6(– x ) = 16
Symbolically, the steps in this method of determining x could be
recorded as follows:
7 ⁄ 2 + 3(–x) = 8
7 + 6(–x) = 16
6(–x) = 9
2(–x) = 3
2x = –3
x = – 3⁄ 2
a) When y = –9, it is difficult to determine the exact value of x
from 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 a
total value of 25, that is 6x = 25. Hence, x = 25 ⁄ 6 = 4 1 ⁄ 6 .
b) Proceeding as in b), 6x = 35.4 and x = 5.9.
c) Again proceeding as in b), 6x = –193 and x = –32 1 ⁄ 6 .
228 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
3 Place a transparency of Focus
3
Master 11.1 on the overhead and
give a copy to each group. Tell the
students this is a graph of y = v(x)
for a certain complete sequence of
arrangements. Ask them to write a
formula for v(x) in the space provided on the bottom of Focus Master 11.1 and to sketch or construct
an Algebra Piece representation of
the xth arrangement. Discuss.
COMPLETE SEQUENCES
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 expression 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 sketch
of the xth arrangement is shown on the right.
x
x
x
x
LESSON 11
FOCUS BLACKLINE MASTER 11.1
y
v(x ) = 4x + 2
v(x ) = 4x + 2
22
20
18
16
14
12
10
8
6
4
2
1
–8 –7 –6 –5 –4 –3 –2 –1
2
3
4
5
6
7
8
x
–2
–4
–6
–8
–10
–12
–14
–16
–18
–20
–22
v(n) = __________________________
4 Discuss the students’ ideas about
ways the numbers, 4 and 2, in the
formula for v(x) relate to its graph.
Introduce the terms coefficient,
slope, intercept, and slope-intercept
form of the equation of a line.
4
Focus Master 11.2, shown on the next page, may be useful for
discussion. The number 4 in the product 4x is called the coefficient
of 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, an
increase of 4). This rate of change, that is, the change in y for each
unit increase in x, is called the slope of the line. (If y had decreased as x increased, the change, and hence the slope, would
have been negative.)
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 229
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.2
10
4
continued
The constant 2 is the value of y when x is 0 or, to put it another
way, the y-coordinate of the point where the line crosses the yaxis. This value is called the y-intercept of the line.
increase of 4
8
6
1
increase of 4
4
A line may also cross the x-axis. If it does, the value of y is zero at
the point of intersection with the x–axis and the value of x at that
point is called the x-intercept. For y = 4x + 2, the x-intercept is – 1⁄ 2 .
y - intercept
2
1
–2
1
–1
2
slope
–2
x - intercept
The equation y = 4x + 2 is said to be in slope-intercept form where
the 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 line
whose slope is a and y-intercept is b.
y = 4x + 2
5 Give each student a sheet of
coordinate grid paper. Tell them that
the graph of y = v(x) for a certain
complete sequence of counting piece
arrangements is a straight line which
passes through the points (–2,10)
and (4,–8). Ask them to graph and
find a formula for v(x). Then have them
construct an Algebra Piece representation or draw a sketch of the
xth arrangement. Discuss.
5
Students will need to locate axes and then scale them. In the
graph shown below, the x-axis is scaled so that each subdivision
represents 1 unit and the y-axis is scaled so that each subdivision
represents 2 units. The graph will appear differently for other
scales, but will still be a straight line.
v (x )
18
16
14
12
10
8
6
4
2
–8
–7
–6
–5
–4
–3
–2
1
–1
2
–4
6
8
–10
12
–14
–16
18
230 | ALGEBRA THROUGH VISUAL PATTERNS
x
2
3
4
5
6
7
8
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
v (x )
If the points (–2,10) and (4,–8) are located on a graph and
a straightedge is used to draw a line through them, one
sees that the y-intercept of the line is 4.
18
16
14
The slope of the line is –3 since y values decrease by 3 as
x values increase by 1. One way to determine this is to
note the y values decrease 18 units (from 10 to –8) as the
x values increase 6 units (from –2 to 4), which is equivalent to a y-decrease of 3 units for every 1 unit x-increase.
(–2, 10)
10
8
6
4
decrease
of 18
2
–2
1
–1
x
2
3
4
5
6
7
8
–2
–4
(4, –8)
–6
Since the line has slope –3 and y-intercept 4, y = v(x) =
–3x + 4. The students may use other methods to find a
formula for v(x).
8
–14
–16
–18
increase of 6
slope =
–18 ⁄ 6
= –3
–x
–x
–x
3
oooo oooo oooo
–12
oooo oooo oooo
Shown below are various sketches and Algebra Piece
representations of y = –3x + 4, some of which include
edge pieces. In the sketches, a numeral alongside the edge
of a rectangle denotes the value of the edge. Note that
the value of an edge may differ from its length—the length
of an edge is always positive or zero, while the value of an
edge can be positive, negative, or zero. If the value of an
edge is positive, the value of the edge and its length are
the same. If the value of an edge is negative, the value of
the edge and its length are opposites. For example, if the
value of an edge is –3, its length is 3.
–10
oooo oooo oooo
–3
o
–4
oooo oooo oooo
–5
o
–6
oooo oooo oooo
–7
oooo oooo oooo
–8
–3x
–x
4
–3
–3x
4
x
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 231
LESSON 11
COMPLETE SEQUENCES
FOCUS
ACTIONS
6 Repeat Action 5 for the points
(–2,–8) and (4,7).
COMMENTS
6
In the graph shown below, the x-axis is scaled so that each
subdivision represents 1 ⁄ 2 unit and the y-axis is scaled so that
each subdivision represents 1 unit. The graph’s appearance will
differ for other scales, but will still be a straight line.
v (x )
8
7
6
5
4
3
increase
of 15
2
1
–3
–2
–1
1
2
3
4
x
–1
–2
–3
–4
–5
–6
–7
–8
increase
of 6
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 15
as x values increase by 6. This is equivalent to a y-increase of 2 1 ⁄ 2
for 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 showing
that it provides the correct values for v(–2) and v(4), namely –8
and 7.
On the left below is an Algebra Piece representation in which an
x-frame has been cut in half. On the right is a sketch in which the
values of edges and regions are shown.
5
__
2
1
v (x ) = 2 __
x–3
2
232 | ALGEBRA THROUGH VISUAL PATTERNS
5
__
x
2
x
5
v (x) = __
x–3
2
–3
COMPLETE SEQUENCES
LESSON 11
FOCUS
of coordinate grid paper. Ask them
to construct or sketch an Algebra
Piece representation of the xth
arrangement of a complete sequence
of arrangements for which the graph
of v(x) is a straight line that satisfies
the conditions in a) below. Invite
several volunteers to sketch their
xth arrangements at the overhead
and then have the students graph
v(x) and verify that the conditions in
a) are met. Discuss relationships
among the Algebra Piece representations and the equations and graphs
that represent them. Repeat for b)-h).
a) y-intercept is 4
7
The collection of all equations that satisfy a) form a family of
lines whose y-intercept is 4. Similarly, the collections of equations
satisfying b) and e) form families of lines. This idea is investigated
further in Lesson 12. The conditions given in c), d), f), and g) each
produce a unique line.
a) If the graph of v(x) has y-intercept 4, then v(0) = 4. This will be
the case if the “constant” part of the arrangements has value 4.
Shown below are 2 possibilities.
v(x) = x + 4
oooo oooo
7 Give each student 2 or 3 sheets
COMMENTS
oooo oooo
ACTIONS
v ( x ) = –2 x + 4
The graphs of the above equations have y-intercept 4 and slopes
1 and –2, respectively.
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
b) If the graph of v(x) is a straight line whose slope is 3, then the
values 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 obtained
by adding a constant to this expression, e.g., v(x) = 3x + 2. The
difference in the graphs of these 2 expressions is that the y-intercept
of the 1st is 0 and that of the 2nd is 2.
f) slope is – 4⁄ 6 and y-intercept is –3
v (x ) = 3x
v(x) = 3x + 2
c) If the graph of v(x) is a straight line and both the slope and
y-intercept are given, there is only one possibility. In this case,
v(x) = –2x + 3, or an equivalent expression.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 233
COMPLETE SEQUENCES
LESSON 11
FOCUS
ACTIONS
COMMENTS
7 continued
d) 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.
y
(x1,b)
(x2,b)
(0,0)
y=b
x
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 3
units for every run of 4 units. Hence, any line whose equation of
the 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 may
point 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 be
rewritten as 4x + 6y = –18. The latter equation is commonly
referred to as the standard form of the equation. In general, an
equation ax + by = c, where a, b, and c are integers, is in standard
form. Notice the variables are on one side of the equation and the
constant is on the other. It is not as easy to “see” the slope and
intercept of the graph of a linear equation in standard form as
compared to an equation in slope-intercept form.
8 Give each student a sheet of
8
a) Shown below is an Algebra Piece representation, with edge pieces.
coordinate grid paper.
a) Ask the students to build or
sketch an Algebra Piece representation of the xth arrangement of a
complete sequence of arrangements
for which v 1 (x) = (2 – x)(3 + x).
b) Ask the students to predict what
the graph of v 1 (x) looks like. Discuss
their predictions and then ask them
to draw the graph. Ask the students
for their observations.
234 | ALGEBRA THROUGH VISUAL PATTERNS
o
oooo oooo oooo
o
oooo oooo oooo
v 1( x ) = (2 – x )(3 + x )
COMPLETE SEQUENCES
LESSON 11
FOCUS
ACTIONS
COMMENTS
b) Since v 1 (0) = 6, the y-intercept is 6. The students may observe
that the x-intercepts are 2 and –3 since v 1 (x) = 0 for those values
of x. If x is in the interval between the x-intercepts, both factors
of v 1 (x) are positive and v 1 (x) is positive (i.e., for –3 < x < 2, v 1 (x)
> 0). Outside this interval, one factor of v 1 (x) is positive and the
other is negative, so v 1 (x) is negative (i.e., for x < –3 and x > 2,
v 1 (x) < 0).
An alternate form for v 1 (x) is –x 2 – x + 6, as can be seen from the
Algebra Piece representation shown above.
In the graph of y = v 1 (x) shown below, every subdivision of the
x-axis represents 1 ⁄ 2 unit and every subdivision of the y-axis
represents 1 unit. The shape of the graph will vary for other
scalings of the axes; however, regardless of the scaling, the graph
is a parabola that is symmetric about the vertical line x = – 1 ⁄ 2 and
opens downward.
y
7
6
5
4
3
2
1
–4
3
–2
–1
1
2
3
4
x
–1
–2
–3
–4
–5
–6
–7
–8
y = v 1 ( x ) = (2 – x )(3 + x )
Some students may find a few points on the graph and connect
these points with straight line segments. If so, you might have
them find more points on the graph to help show that it is
rounded rather than angular.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 235
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
9 Ask the students to graph v2(x) = x
9
If y = x and y = (2 – x)(3 + x) are graphed on the same coordinate system, it appears that the two graphs intersect when x is
about 1.7 and –3.7.
on the same coordinate axes as
their sketch of v 1 (x) = (2 – x)(3 + x).
Have them find where the graphs
intersect. Discuss.
y
7
6
y = (2 – x)(3 + x)
5
4
3
2
y=x
1
–4
3
–2
–1
1
2
3
4
x
–1
–2
–3
–4
–5
–6
–7
–8
oooo oooo
oooo
oooo oooo
oooo
The exact values of x where the 2 graphs intersect can be found
by determining when v 1 (x) = v 2 (x), i.e., when (2 – x)(3 + x) = x.
This is the case if the value of the circled portion of the Algebra
Piece representation for v(x), shown at the left, is 0. If the value
of 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 x 2 -mat and
the 2 x-strips has value 6. If one black counting piece is added to
this collection, it has value 7 and can be arranged into a square
with edge (x + 1). Thus, x + 1 equals 7 or – 7 , so x = –1 + 7
or x = –1 – 7 . Since 7 ≈ 2.65, x ≈ 1.65 or x ≈ –3.65.
x+1
x+1
236 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
10
10 Give each student a sheet of
coordinate grid paper. Tell them that
the value of the xth arrangement of
a certain complete sequence of
arrangements is v(x) where v(x) =
1 ⁄ 2 x 2 – x – 8. Ask the students to
construct a graph of v(x) and to find
the graph’s x- and y-intercepts and
the coordinates of its “turning”
point. Discuss.
The students may find it helpful to construct a table of values.
The graph of v(x) is shown below.
y
7
6
5
4
3
2
1
–7
–6
–5
–4
–3
–2
–1
1
2
3
4
5
6
7
8
x
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
y = v ( x ) = 1 ⁄ 2x 2 – x – 8
oooo
oooo
Since v(0) = –8, the y-intercept is –8.
1
oooo
16
oooo
x 2 – 2 x + 1 = ( x – 1) 2 = 17
The x-intercepts occur when v(x) = 0, that is when 1 ⁄ 2 x 2 – x – 8 =
0. This will be the case if 1 ⁄ 2 x 2 – x = 8, or, doubling, if x 2 – 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, which
is x = 1. Since v(1) = – 17 ⁄ 2 , the coordinates of the turning point
are (1,– 17 ⁄ 2 ).
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 237
COMPLETE SEQUENCES
LESSON 11
FOCUS
ACTIONS
11 Give each student a sheet of
coordinate grid paper. Show the
students the following pair of expressions:
v 1(x) = –4x – 3
v 2(x) = 1⁄ 2x 2 + 3x – 27 ⁄ 2
COMMENTS
11
a) A transparency of the completed graph can be made from
Focus Master 11.3.
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.3
y
24
Tell the students these are the
values of the xth arrangements of
two complete sequences of arrangements. Ask the students to do the
following:
22
v 2( x ) = –4x – 3 20
18
16
14
12
v 1( x ) = 1⁄ 2x 2 + 3x – 27⁄ 2
10
8
6
a) graph v 1(x) and v 2 (x) on the same
coordinate axes,
4
2
13 12
11
10
9
8
7
6
5
4
3
2
1
1
2
3
4
5
6
x
2
b) determine the x- and y-intercepts
of the graphs,
4
6
8
10
12
c) determine the values of x for
which v 1 (x) = v 2 (x).
14
16
18
20
b) For v 1 (x): x-intercept is – 3 ⁄ 4 ; y-intercept is –3. For v 2 (x):
x-intercepts are –9 and 3; y-intercept is – 27 ⁄ 2 .
c) If –4x – 3 = 1 ⁄ 2 x 2 + 3x – 27⁄ 2, then, doubling both arrangements,
–8x – 6 = x 2 + 6x – 27 and, adding 8x + 27 to both arrangements,
21 = x 2 + 14x. Hence, the square shown below has value 70. Thus
x + 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 the
graphs of v 1 (x) and v 2 (x) which occurs at the latter value of x is
beyond the scope of the graph shown above.
x
x
x
x
49
x
x
x
x x x x x x x
x2
(x +
238 | ALGEBRA THROUGH VISUAL PATTERNS
7)2
=
x2
x+7
+ 14x + 49 = 21 + 49 = 70
COMPLETE SEQUENCES
LESSON 11
FOCUS
v 1(x) = (x – 6)(x + 2)
v 2(x) = –x 2 + 10x – 16
12
a) A transparency of the completed graph can be made from
Focus Master 11.4.
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.4
y
18
16
14
v 1( x ) = ( x – 6)( x + 2)
12
10
8
6
4
2
5
4
3
2
1
1
2
3
4
5
6
7
9
10
x
2
4
6
8
10
12
14
16
v 2( x ) = – x 2 + 10x – 16
18
b) Since a product is 0 if, and only if, one of its factors is 0, v1(x) = 0
when x – 6 = 0 or x + 2 = 0, that is, if x = 6 or x = –2. Thus the
x-intercepts of the graph of v 1 (x) are 6 and –2. Its y-intercept is
v 1 (0), which is –12.
The x-intercepts of v 2 (x) occur when v 2 (x) = 0, that is when an
arrangement for –x 2 + 10x – 16 has value 0. Thus –x 2 + 10x = 16
or, taking opposites, x 2 – 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 v 2 (0), which is –16.
oooo oooo oooo oooo oooo
12 Repeat Action 11 for the pair:
COMMENTS
oooo oooo oooo oooo oooo
ACTIONS
oooo oooo oooo oooo oooo
–16
oooo oooo oooo oooo oooo
x–5
( x – 5) 2 = x 2 – 10 x + 25 = –16 + 25 = 9
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 239
LESSON 11
COMPLETE SEQUENCES
FOCUS
COMMENTS
ACTIONS
12 continued
c) The values, v 1 (x) and v 2 (x), are equal if the two arrangements
shown below have the same value.
2
x
x
x
x2
x– x– x– x– x– x–
x
–6
–12
–x 2
–16
x x x x x x x x x x
v2(x) = –x2 + 10x – 16
v1(x) = (x – 6)(x + 2)
= x2 – 4x – 12
Adding x 2 – 10x + 12 to both of these arrangements, one has 2x 2
– 14x = –4. or, halving, x 2 – 7x = –2. Completing the square, as
shown below, gives (x – 7 ⁄ 2 ) 2 = –2 + 49 ⁄ 4 = 41 ⁄ 4 .
x
–1
2
x–x–x–
x2
49
4
x
x
x–
x–
x–
–7
2
x
–1
2
–7
2
(x – 72 )2 = (x2 – 7x) + 494
= –2
+ 494 =
41
4
Thus, x – 7 ⁄ 2 = 41⁄ 2 or – 41 ⁄ 2 . So x = (7 + 41) ⁄ 2 or x = (7 –
the help of a calculator, one has x ≈ 6.70 or x ≈ 0.30.
240 | ALGEBRA THROUGH VISUAL PATTERNS
41) ⁄ 2 . With
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
–7x
–7x
49
x
x2
x2
–7x
x
x2
x2
–7x
x
x
–7
(2x – 7)2 = (4x2 – 28x) + 49
=
–8
+ 49 = 41
13 (Optional) a) Have the students
find x if 2x 2 + 5x –10 = 0. Ask for
volunteers to show their solutions.
Discuss the methods the students
use. Repeat for 3x 2 + 5x – 10 = 0.
b) (Challenge) Ask the students to
find all values of x for which ax 2 +
bx + c = 0 where a, b, and c are
arbitrary integers.
You might ask the students to record the steps in their solution
using algebraic notation. For example, the steps in the last solution might be recorded as follows:
(x – 6)(x + 2) = –x 2 + 10x – 16
x 2 – 4x – 12 = –x 2 + 10x – 16
2
2
x – 4x – 12 + (x – 10x + 12) = –x2 + 10x – 16 + (x2 – 10x + 12)
2x 2 – 14x = –4
4x 2 – 28x = –8
2
4x – 28x + 49 = –8 + 49
(2x – 7) 2 = 41
2x – 7 = ± 41
2x = 7 ± 41
x = (7 ± 41)⁄ 2
13 a) You can ask the volunteers to use Algebra Pieces or
sketches to illustrate their thinking.
If 2x 2 + 5x – 10 = 0, then 2x 2 + 5x = 10 and x 2 + 5 ⁄ 2 x = 5. Completing 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 .
5
4
x
5
4
25
16
5
4
x2
x
x
–7
Alternatively, instead of halving 2x 2 – 14x, one could double it, to
obtain 4x 2 – 28x = –8. Then, completing the square, as shown on
the left, one has (2x – 7) 2 = 41. Whence 2x – 7 = 41 or – 41
and, as before, x = (7 + 41) ⁄ 2 or x = (7 – 41) ⁄ 2 .
x
ACTIONS
5
4
(x + 54 )2 = (x2 + 52 x) + 25
16
=
5
+
25
16
=
105
16
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 241
COMPLETE SEQUENCES
LESSON 11
FOCUS
COMMENTS
ACTIONS
13
20x
5
4x
25
16x2
20x
4x
5
continued
The introduction of fractions can be delayed by multiplying the
original equation by an amount so that the coefficient of x 2 is a
perfect square and the number of x-strips can be divided evenly
along the sides of the resulting square. This means that the
number of x-strips must be divisible by twice the length of the
resulting square. This can be accomplished by multiplying the
original equation by 4 times the original coefficient of x 2 , in this
case 4 x 2 or 8. Increasing amounts by a factor of 8, the equation
2x 2 + 5x = 10 becomes 16x 2 + 40x = 80. Then completing the
square 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 .
2
(4x + 5) = (16x2 + 40x) + 25
=
80
+ 25 = 105
If 3x 2 + 5x – 10 = 0, then 3x 2 + 5x = 10. Multiplying by 4 times 3,
or 12, this becomes 36x 2 + 60x = 120. Completing the square, as
shown, one has (6x + 5) 2 = 120 + 25 = 145. Hence, 6x + 5 = 145
or – 145 and x = (5 + 145 ) ⁄ 6 or (5 – 145 ) ⁄ 6 .
5
30x
25
6x
36x2
30x
6x
5
2
b
2abx
(6x + 5) =
=
b2
(36x2
+ 60x) + 25
120
+ 25 = 145
b) If ax 2 + bx + c = 0 , then ax 2 + bx = –c. Multiply by 4a: 4a 2 x 2 +
4abx = –4ac. Complete the square: (2ax + b) 2 = 4a 2 x 2 + 4abx + b 2
= –4ac + b 2 .
Thus: 2ax + b = b2 − 4ac or – b2 − 4ac
2ax
4a 2x 2
2abx
2ax = – b +
x=
(–b +
2
b2 − 4ac or –b –
b − 4ac ) ⁄ 2a
or
(–b –
b2 − 4ac
2
b − 4ac ) ⁄ 2a .
This result is known as the quadratic formula. You may want to ask
the students to use the formula to solve the equations of part a).
b
2ax
2
2
2
(2ax + b) = (4a x + 4abx) + b
=
–4ac
+ b2
2
= b – 4ac
2
242 | ALGEBRA THROUGH VISUAL PATTERNS
If b 2 – 4ac < 0, the solutions are complex numbers. These are
discussed in Lesson 14.
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.1
y
22
20
18
16
14
12
10
8
6
4
2
1
–8 –7 –6 –5 –4 –3 –2 –1
2
3
4
5
6
7
8
x
–2
–4
–6
–8
–10
–12
–14
–16
–18
–20
–22
v(n) = __________________________
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 243
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.2
10
increase of 4
8
6
1
increase of 4
4
y - intercept
2
1
–2
1
–1
2
slope
–2
x - intercept
y = 4x + 2
244 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.3
y
24
22
v 2( x ) = –4x – 3 20
18
16
14
v 1( x ) =
1⁄ 2x 2
+ 3x –
12
27⁄ 2
10
8
6
4
2
–13 –12 –11
–10 –9
–8
–7
–6
–5
–4
–3
–2
–1
1
2
3
4
5
6
x
–2
–4
–6
–8
– 10
– 12
– 14
– 16
–18
–20
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 245
COMPLETE SEQUENCES
LESSON 11
FOCUS BLACKLINE MASTER 11.4
y
18
16
14
v 1( x ) = ( x – 6)( x + 2)
12
10
8
6
4
2
–5
–4
–3
–2
–1
1
2
3
4
5
6
7
8
9
10
x
–2
–4
–6
–8
– 10
– 12
– 14
– 16
–18
246 | ALGEBRA THROUGH VISUAL PATTERNS
v 2( x ) = – x 2 + 10x – 16
© THE MATH LEARNING CENTER
COMPLETE SEQUENCES
LESSON 11
FOLLOW-UP BLACKLINE MASTER 11
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)
x
x
x
x
___
31⁄2
___
31
v(x)
17
___
200
90 1 ⁄ 2
b)
–x
–x
x
v(x)
55
___
___ –230
61 1 ⁄ 4 –121
___ 178
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 complete sequence of arrangements. Graph a) and b) on the same coordinate system. Find all xintercepts, y-intercepts, the coordinates of the turning points, and the values of x for which
v 1(x) = v 2 (x). Explain how you arrived at your answers.
a) v 1(x) = (x + 4)(x – 3)
b) v 2(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) x 2 – x – 12 = –x 2 + x
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 247
COMPLETE SEQUENCES
LESSON 11
ANSWERS TO FOLLOW-UP 11
1
a) v(x) = 3x – 2 1/ 2, so:
if v(x) = 17 then 3x = 17 + 2 1 / 2 = 18 + 1 1 / 2 , and
x = 6 1/ 2;
if x = –31/2, then v(x) = 3x – 21/2 = –101/2 – 221/2 = –13;
if v(x) = 200, then 3x = 200 + 2 1 / 2 = 201 + 1 1 / 2 and
x = 67 1 / 2 .
3
a) x-intercepts: 3 and –4; y-intercept: –12; turning
point: (– 1/ 2, –12 1/ 4)
b) x-intercepts: 2 and –5; y-intercept: –10; turning
point: (– 3/ 2, 12 1/ 4)
y
b) v(x) = –2x + 1 1/ 2 , so:
if x = 55, then v(x) = –110 + 1 1 / 2 = –108 1 / 2 ;
if v(x) = –230, then –2x = –230 –1 1 / 2 , so x = 115 +
3 / 4 = 115 3 / 4 ;
if v(x) = 178, then –2x = 178 – 1 1 / 2, so x = –89 + 3 / 4
= –88 1/ 4.
22
20
18
v 2( x ) = (2 – x )(5 + x )
16
14
12
10
2
y
8
6
11
4
10
2
x
9
–8 –7 –6 –5 –4 –3 –2 –1
8
2
3
4
5
6
7
8
–6
a) y = 3x – 2
6
5
–8
4
–10
v 1( x ) = ( x + 4)(x – 3)
3
–12
–14
2
–16
1
x
–1
1
–4
7
–8 –7 –6 –5 –4 –3 –2 –1
–2
1
2
3
4
5
6
7
8
–18
–20
–22
–2
–3
d) y = – 3
–4
–5
c) y = –x + 3
–6
–7
–8
b) y = 4x – 1
–9
–10
–11
248 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
COMPLETE SEQUENCES
LESSON 11
ANSWERS TO FOLLOW-UP 11 (CONT.)
3
continued
If v 1 (x) = v 2 (x), adding x 2 + 3x + 12 to both arrangements, one has 2x 2 + 4x = 22, or, halving, x 2 + 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 ≈ 2.46
or –4.46.
–12
x2
x– x– x–
x
–3
x
x
2
–x
10
–x
–x ⁄ 2
x
x
x
8
2
–x 2
x
Doubling both arrangements:
v 1 (x) = (x + 4)(x – 3)
= x 2 + x – 12
–x
7x
6
Adding x 2 + 3x + 12 to both arrangements:
x
x
x
x
10
x
x
x
x
x
x
x
6
v 2 (x) = (2 – x)(5 + x)
= –x 2 – 3x + 10
x
x
x
x
x
x
6x + 10
Adding x – 10 to both:
5
5
3x + 5
+8
–x + 16
x– x– x– x– x–
x
a)
16
x
x
x
x
4
4
∴ x = 6⁄ 7
In symbols:
x2
x2
22
–x ⁄ 2 + 8
–x + 16
–x + 16 + (x – 10)
6
x
=
=
=
=
=
3x + 5
6x + 10
6x + 10 + (x – 10)
7x
6⁄ 7
22
2x 2 + 4x
Halving each arrangement:
x
x
x2
11
x2
+ 2x
11
Completing the square:
1
x
x
1
x2
x
12
x
1
(x + 1) 2 = x 2 + 2x + 1 = 11 + 1 = 12
continued
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 249
COMPLETE SEQUENCES
LESSON 11
ANSWERS TO FOLLOW-UP 11 (CONT.)
4
x
x
x
x
b)
4
4
–12
c)
x2
x
–x 2
–x
–12
x2
x– x– x–
x
–x
x
x 2 – x – 12
3
–x 2 + x
Add x 2 – x + 12 to each arrangement:
–3
(x + 4)(x – 3) = x 2 + x – 12
–x + 3
x2
x2
–x –x
12
Add x + 12 to both arrangements:
x
x
Double both arrangements:
x2
15
x2
x2
–x –x
x2
x2
–x –x
x2 + 2x
Complete the square:
1
x
1
x
x2
x
16
x
Complete the square:
–1
1
24
–x
–x
(x + 1) 2 = x 2 + 2x + 1 = 15 + 1 = 16
x + 1 = 4 or –4
x = 3 or –5
1
x2
x2
–x
x2
x2
–x
2x
In symbols:
(x + 4)(x – 3)
x 2 + x – 12
2
x + x – 12 + (x + 12)
x 2 + 2x
2
x + 2x + 1
(x + 1) 2
x+1
x
x
=
=
=
=
=
=
=
=
=
–x + 3
–x + 3
–x + 3 + (x + 12)
15
15 + 1
16
±4
–1 ± 4
3, – 5
2x
–1
(2x – 1) 2 = 4x 2 – 4x + 1 = 24 + 1 = 25
2x – 1 = 5 or –5
x = 3 or –2
In symbols:
x2
250 | ALGEBRA THROUGH VISUAL PATTERNS
25
x 2 – x – 12
– x – 12 + (x 2 – x + 12)
2x 2 – 2x
4x 2 – 4x
2
4x – 4x + 1
(2x – 1) 2
2x – 1
2x
x
=
=
=
=
=
=
=
=
=
–x 2 + x
–x 2 + x + (x 2 – x + 12)
12
24
25
25
±5
6, – 4
3, – 2
© THE MATH LEARNING CENTER
SKETCHING SOLUTIONS
LESSON 12
THE BIG IDEA
Sketches that show the essential features of a mathematical situation are a valuable aid in problem solving.
Many word problems found in beginning algebra
courses are readily solved with the help of an appropriate sketch or diagram.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
Students draw sketches that
illustrate geometric and
numerical situations. They use
their sketches to help them
answer questions about these
situations.
Students use sketches and
diagrams to model mathematical situations. They reason
from their sketches to solve
problems and find solutions of
equations.
Students use sketches to solve
problems. They record their
thought processes using algebraic symbols and equations.
Materials
Materials
Focus Master 12.1, 1 transparency.
Focus Master 12.2, 1 copy
per student.
Focus Master 12.3 (2 pages),
1 copy per student.
Materials
None, other than paper and
pencil.
Follow-Up 12, 1 copy per
student.
ALGEBRA THROUGH VISUAL PATTERNS | 251
TEACHER NOTES
252 | ALGEBRA THROUGH VISUAL PATTERNS
SKETCHING SOLUTIONS
LESSON 12
START-UP
Overview
Students draw sketches that illustrate geometric and numerical
situations. They use their sketches
to help them answer questions
about these situations.
ACTIONS
1 Ask the students to draw a sketch
of a rectangle.
Materials
None, other than paper
and pencil.
COMMENTS
1
Students often have difficulty in drawing sketches that disclose
the essential features of a situation. As a first step in developing
their sketching skills, the students are asked to sketch a familiar
figure. If your students have had experience using sketches and
diagrams to solve problems visually, you may want to skip the
Start-Up and proceed to the Focus.
Most students will draw a rectangle that is wider than it is tall. It
is likely their sketches will contain no words or symbols. Generally, it is unnecessary to label a sketch of a rectangle for the
students to recognize what has been drawn.
2 Ask the students to draw and
label a sketch of a rectangle whose
length is 6 inches longer than its
width. Ask for volunteers to show
their sketches. Discuss with the
students whether or not the sketches
adequately convey the information
given, and whether the words and
symbols used are absolutely essential.
2
3 Tell the students the perimeter
of the rectangle in Action 2 is 56
inches. Ask them to determine the
dimensions of the rectangle, Discuss
the methods the students use.
3 The students will arrive at the dimensions in various ways.
Some may observe that if the length is reduced by 6 inches, the
rectangle becomes a square whose perimeter is 44 inches. Hence,
the length of its side is 11 inches. Thus the dimensions of the
original rectangle are 11 and 11 + 6, or 17, inches.
Below are some possible sketches.
6
d
6
d
6
d
d
6
Others may note that the sum of the width and length is half of
56, or 28, and their difference is 6. Hence they may search for
two numbers that differ by 6 and add to 28.
Still others may see that the perimeter consists of 2 segments of
length 6 and 4 segments of unknown but equal length. The sum of
the lengths of these latter segments is 56 – 12 or 44. Hence each
segment is 11, and the dimensions of the rectangle are 11 and
11 + 6, or 17.
ALGEBRA THROUGH VISUAL PATTERNS | 253
SKETCHING SOLUTIONS
LESSON 12
START-UP
COMMENTS
ACTIONS
4 Discuss with the students how
algebraic symbols and equations can
be used to record their thinking in
finding the dimensions of the rectangle in Action 3.
4 The last solution in Comment 3 might be recorded as follows.
Note, that the algebraic equations become a symbolic way of
recording one’s thinking. They are the record of a chain of thought
rather than a series of manipulations carried out according to
prescribed rules.
Person’s Thinking
Algebraic Symbols
The perimeter of 56 consists of 2
segments of length 6 and 4 other
segments of the same length that
we’ll call d.
56 = 2(6) + 4d = 12 + 4d
So, the 4 segments of length d have
a total length of 56 – 12 or 44.
4d = 56 – 12 = 44
Thus, the length of each segment is
44 ÷ 4, or 11.
d = 44 ÷ 4 = 11
Hence, the dimensions of the
rectangle are 11 and 11 + 6.
5 Ask the students to draw a sketch,
using as few words and symbols as
possible, that portrays a rectangle of
unknown dimensions whose length is
4 units longer than 3 times its width.
Ask for volunteers to replicate their
drawings on the overhead. Discuss
whether the drawings adequately
convey the information given about
the rectangle and whether the words
and symbols used are essential.
5 Having the students draw sketches of a situation before a
problem is posed focuses their attention on creating a sketch that
portrays the essential features of the situation.
Below are some possible sketches. Notice that, in the last sketch
shown, the essential information is carried in the symbols and not
the sketch—that is, if the symbolic phrase “3w + 4” is erased, the
distinguishing feature of the rectangle is lost.
x
4
x
6 Tell the students that the perimeter of the rectangle they drew in
Action 2 is 48 inches. Then ask them
to determine the dimensions of the
rectangle. Ask for volunteers to
describe their thinking.
254 | ALGEBRA THROUGH VISUAL PATTERNS
width = d = 11
length = d + 6 = 17
x
x
3w + 4
4
w
6 The students will use various methods to arrive at the dimensions. One way is to note that the perimeter of 48 inches consists
of 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.
SKETCHING SOLUTIONS
LESSON 12
START-UP
ACTIONS
7 Repeat Action 5 for a rectangle
whose length is 5 inches less than
twice its width. Then ask the students to determine the dimensions
of the rectangle if its perimeter is
32 inches. Have several students
describe their thinking in determining the dimensions of the rectangle.
COMMENTS
7
To emphasize the mathematical relationships in the rectangle,
it is helpful to discuss the students’ sketches before telling them
the perimeter of the rectangle. Here is one sketch:
w
w
w
5
The extended rectangle shown above has a perimeter 10 inches
longer than the original rectangle, so its perimeter is 42 inches.
These 42 inches are composed of 6 equal lengths. So each of
these lengths is 7 inches. The width of the original rectangle is
one of these lengths, or 7 inches; the length of the rectangle is 5
inches less than 2 of these lengths, or 9 inches.
8 Ask the students to sketch a
square. Then have them sketch an
equilateral triangle whose sides are
2 units longer than the sides of the
square. Ask the students to reason
from their diagrams to determine
the length of the side of the square
if the square and the triangle have
equal perimeters. Ask for volunteers
to show their sketches and describe
their thinking.
9 Ask the students to draw diagrams
or sketches which represent a
number and that number increased
by 6. Show the various ways in which
students have done this. Then ask
the students to use one of the
sketches to determine what the
numbers are if their sum is 40.
8
The perimeter of the square, in the following drawing, contains
4 segments of length s; that of the equilateral triangle contains 3
segments of length s and 3 of length 2. Thus, the 3 segments of
length 2 must sum to s. So s is 6.
s
2
s
s
s
s
2
s
s
2
9
Since numbers have no particular shape, the students must
invent a way of portraying number. They might do this in a variety
of ways, e.g., as a length or as an area or as a “blob.”
6
1 1
1 1
1 1
Looking at the sketch on the left above, the smaller number is
represented by a line segment and the larger number by two line
segments. The sum of the lengths of these segments is 40. The
small segment has length 6. Hence, the sum of lengths of the
other 2 segments is 34. Since these 2 segments are congruent, the
length of each is 34 ÷ 2, or 17. Hence the 2 numbers are 17 and
17 + 6, or 23.
ALGEBRA THROUGH VISUAL PATTERNS | 255
SKETCHING SOLUTIONS
LESSON 12
START-UP
ACTIONS
10 Ask the students to draw
sketches that represent 2 numbers
such that 4 times the smaller number is 1 less than the larger. Then
ask them to reason from their
sketches to determine the numbers
if their sum is 36. Ask for volunteers
to show their sketches and explain
their reasoning.
256 | ALGEBRA THROUGH VISUAL PATTERNS
COMMENTS
10
If the sum of the 2 numbers is 36, in the sketch shown
below, the sum of the lengths of the 5 congruent segments (1
segment 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 numbers
are 7 and 4(7) + 1, or 29.
smaller number
larger number
1
SKETCHING SOLUTIONS
LESSON 12
FOCUS
Overview
Students use sketches and diagrams
to model mathematical situations.
They reason from their sketches to
solve problems and find solutions of
equations.
Materials
Focus Master 12.1, 1
transparency.
Focus Master 12.3, 1
copy per student.
Focus Master 12.2, 1
copy per student.
COMMENTS
ACTIONS
1 Place a transparency of Focus
1
Master 12.1 on the overhead, revealing only Situation a). Have the
students draw a sketch or diagram
that represents the given information. Ask for volunteers to show
their sketches.
SKETCHING SOLUTIONS
Shown below are 2 possibilities.
2nd
group
5
1st group
43
LESSON 12
FOCUS BLACKLINE MASTER 12.1
5
1st
43
Situations
2nd
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.
2 Ask the students to reason from
their drawing to determine how
many people are in each group. Ask
for several volunteers to describe
their thinking.
2
Looking at the first drawing above, the number of people in
the extended rectangle is 48. This extended rectangle consists of
4 equal regions. Hence, each region contains 12 people. One of
these regions represents the second group; hence, the second
group contains 12 people. The first group is 5 less than the
number in 3 regions, or 31.
You may want to discuss with the students how their thinking
might be represented in a sequence of algebraic statements. For
example, in the above argument, if one lets x represent the
number of people in a region, the above solution might be recorded 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.
ALGEBRA THROUGH VISUAL PATTERNS | 257
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
3 Reveal Situation b) on Focus
3
Master 12.1. Ask the students to
draw a sketch or diagram that
represents the given situation and
then use their drawing to answer an
appropriate question about the
situation. Ask for volunteers to
share their sketches and solutions.
To facilitate sharing, you might have overhead pens and halfsheets of blank transparencies available so students can prepare
their drawings prior to presenting them to the class.
Below is one possibility, in which a sketch is used to answer the
question, “What are the three numbers?”
1st number
2nd number
SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.1
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.
3rd number
Each box represents 112 ÷ 7. or 16. Therefore, the numbers are
32, 16, and 64.
There are 43 people at the meeting.
b) There are 3 numbers.
The 1st number is twice the 2nd number.
Following is an algebraic representation of the visual reasoning
used above:
The 3rd is twice the 1st.
The sum of the 3 numbers is 112.
c) The sum of 2 numbers is 40.
If
Their difference is 14.
d) The sides of square A are 2 inches longer than the sides of square B.
x = 2nd number,
2x = 1st number,
2(2x) = 4x = 3rd number
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.
So,
She has 14 coins altogether.
f) Three particular integers are consecutive.
2x + x + 4x = 112
7x = 112
x = 112 ⁄ 7 = 16.
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.
4 Repeat Action 3 for the remaining situations on Focus Master 12.1.
Therefore, the 3 numbers are 16, 32, and 64.
4
From time to time as students present their solutions, you
may wish to discuss how a student’s thinking might be represented in a series of algebraic statements. However, the emphasis
of this Action is on sketching situations and reasoning from them.
Following are sample questions and diagrams from which answers
to these questions can be deduced.
258 | ALGEBRA THROUGH VISUAL PATTERNS
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
c) What are the numbers?
Solution 2
Solution 1
larger
40
40 = sum
large
small
14
small
26
__
2
difference
40 – 14 = 26
= 13
14
smaller
small
The area, 40, of the shaded region is the sum of
the 2 numbers; the area, 14, of the unshaded region is the difference. The combined area, 54, of
the shaded and unshaded regions is twice the
larger number. Hence, the larger number is 54 ÷ 2,
or 27. The smaller number is 27 – 14, or 13.
13
The smaller number is 26 ÷ 2, or 13.
The larger number is 40 – 13, or 27.
d) What is the length of the small square?
In each of the following, s is the side of the smaller square, B.
Solution 1
s
1
2
4
2
s
Solution 2
1
s
1
s
s+1
1
B
s+1
s+1
1
1
1
The area of the unshaded
border is 48. Hence, the area
of each of the two 2 x s rectangles is (48 – 4) ÷ 2 or 22.
Thus, s is 11.
1
1
1
B
B
1
Solution 3
1
The area of each of the
four 1 x s rectangles is
(48 – 4) ÷ 4, or 11. Thus,
s is 11.
s+1
The area of the unshaded
border is 48. Hence, the area
of each of the four 1 x (s + 1)
rectangles is 48 ÷ 4, or 12,
and s is 11.
e) How many dimes and how many quarters does Melody
have?
$2.75
5¢
5¢
5¢
5¢
5¢
$1.40
no. of quarters no. of dimes
14 coins
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 quarters
and 5 dimes.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 259
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.1
4
continued
f) What are the 3 integers?
Situations
1
1
1st integer
1 1
1 1
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.
2nd integer
3rd
integer
1st
integer
b) There are 3 numbers.
The 1st number is twice the 2nd number.
The 3rd is twice the 1st.
3rd integer
The sum of the 3 numbers is 112.
2nd integer
The area of the shaded rectangle is the product of
the first 2 integers. The area of the unshaded region
is the difference between that product and the square
of the 3rd integer which is given to be 40. So, each
of the 3 unshaded rectangles has area (40 – 4) ÷ 3,
or 12. Thus, the 3 numbers are 12, 13, and 14.
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.
g) How old is Lucille?
ages
now
ages in
6 years
Karen
Lucille
Karen
Lucille
6
6
Comparison of Karen’s age in 6 years with 3
times Lucille’s age in 6 years:
Karen
6
Lucille (3 times)
6 6 6
These have the same value if each box represents two
6’s, or 12. So, Karen is now 48 and Lucille is 12.
5 Write the following on the overhead:
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?
Ask the students to create sketches
from which they can deduce the
answer to the question. After the
students have worked for 10 to 15
260 | ALGEBRA THROUGH VISUAL PATTERNS
5
The intent here is to engage the students in thinking about the
problem and then reflecting on other students’ work. Some
students may not reach a solution before you distribute the Focus
Master; you might give them the option not to examine it until
they are ready.
In Solution 1, a line segment representing the tank is divided into
12 sections, which is a multiple of both 4 and 6. Pump A will fill 2
of 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 2 2 / 5
hour to fill all 12 sections.
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
minutes, distribute a copy of Focus
Master 12.2 to each student, pointing out that these are sketches
various students have made in
arriving at an answer to the question. Discuss with the students their
ideas and questions about the thinking behind the solutions, and how
their sketches compare with those
on Focus Master 12.2.
SKETCHING SOLUTIONS
In Solution 2, it is determined that pumps A and B together can
fill 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 1
hour varies from one part of the sketch to the next, while in
Solution 2, the length of an interval representing 1 hour remains
the same.
The thinking in Solution 3 is similar to that in Solution 1 except
that the tank is represented by a rectangle rather than a line
segment.
LESSON 12
FOCUS BLACKLINE MASTER 12.2
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?
Solution 1
Pump A
1 hour
Pump B
1 hour
Together
1 hour
2
5
1 hour
of 1 hr.
Solution 2
Time to fill 1 tank: Pump A
Pump B
Tanks filled in
12 hours:
Pump A
Pump B
6 hours
4 hours
6 hours
4 hours
6 hours
4 hours
4 hours
Pumps A and B fill 5 tanks in 12 hours.
Solution 3
Pump A fills
in 1 hour.
1
6
tank
Pump B fills
in 1 hour.
1
4
tank
Pump A fills 4 subdivisions in 1 hour.
Pump B fills 6 subdivisions in 1 hour.
Together, they fill 10 subdivisions in
1 hour:
1 hour
1 hour
4
10
hour
6 Give each student a copy of
Focus Master 12.3 (see following
page). Select, or have the students
select, situations from the Focus
Master. Ask the students to pose
mathematical questions about the
selected situations and then draw
diagrams or sketches from which
they can determine the answers to
the questions. Discuss.
6
You might ask for volunteers to state a question and show,
without comment, the diagrams or sketches, appropriately labeled, that they used to arrive at an answer. Then ask the other
students to suggest how the answer was deduced from the
sketches.
Following is a sample question and solution for each of the
situations on the Focus master.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 261
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.3
6
continued
a) How long does it take to empty the tank using only the
2nd drain?
More Situations
Both drains empty
1
__
tank in 1 hour.
3
a A tank has 2 drains of different sizes.
If both drains are used, it takes 3 hours to empty the tank.
1st drain empties
1
__
tank in 1 hour.
7
If only the first drain is used, it takes 7 hours to empty the tank.
Working together, both
drains empty 7 subdivisions in 1 hour. The 1st
drain empties 3, so the
2nd drain empties 4 subdivisions in 1 hour.
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.
2nd drain
Next, Moe walked 1⁄ 2 the remaining distance and stopped to rest at the park bench.
1 hr.
1 hr.
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.
1 hr.
It takes the 2nd
drain 5 1⁄ 4 hours to
empty the tank.
e) A gallon of paint contains 20% red paint and 80% blue paint.
1 hr.
1 hr.
Red paint is added until the mixture contains 50% red paint.
f) Standard quality coffee sells for $18.00 per kg.
1
__
4
Prime quality coffee sells for $24.00 per kg.
hr.
Every Saturday morning Moonman’s Coffee Shop grinds a 40kg batch of a standard/prime
blend to sell for $22.50/kg.
b) How many books does each girl have now?
continued on back
20
SKETCHING SOLUTIONS
LESSON 12
Lisa
FOCUS BLACKLINE MASTER 12.3 (CONT.)
Lisa
Maria
3
30
33
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
The difference in length of the top and bottom
arrows is the number of books Lisa has. Hence,
she has 33 – 20, or 13, books. So, Maria has
20 – 13, or 7, books.
rest is water).
She added enough of a 40% sugar solution to create a 60% solution.
c) How many people are in the room?
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.
Each of the boxes below
contains the same number
of people; 3 of the boxes
contain women and 2 contain men:
Doubling the men gives 4
boxes of men. Adding 6 to the
women (each X is a woman),
gives 6 more than 3 boxes of
women:
The train averaged 75 mph and the bus averaged 60 mph.
W
W
W
M
M
M
M
M
M
W
W
W
XXXXXX
If the number of men and women are equal, the
last box of men must contain 6 men. Thus, all
boxes contain 6 people and, to begin with there
are 18 women and 12 men in the room.
262 | ALGEBRA THROUGH VISUAL PATTERNS
SKETCHING SOLUTIONS
LESSON 12
FOCUS
ACTIONS
COMMENTS
d) How far is it from school to Moe’s home?
Distance from school to home:
school
park
bench
water
1
home
1
__
3
1
total distance
A
B
Segments A and B are equal. Thus, replacing A with B.
1
1
__
3
1
1
__
3
1
total distance
total distance
B
B
The 3 segments of length 1 comprise the other
third of the distance. Hence, the distance from
school to home is 9 miles.
e) How much pure red paint does Jill add?
Solution 1
original gallon
added part
20% =
1
__
5
gal
1
__
5
1
__
5
gal
gal
1
__
5
gal
1
__
5
gal
1
__
5
1
__
5
gal
red
1
__
5
gal
gal
red
The added part is 3⁄ 5 gallon.
The areas of the rectangles below in Sketch I represent the
amount of red paint in 1 gallon of the mixture and x gallons of
added red paint. If the resulting mixture is to be 50% red paint,
the 2 rectangles should be “leveled off” at 50. This will be the
case if, in Sketch II, area A = area B. Since area A is 30 and
area 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.
Solution 2
Sketch I:
Sketch II:
100%
100%
100%
100%
blue
blue
paint
red
added
red
paint
A
red
20%
red paint
in mixture
1
x
gallons
red
red
0%
0%
50%
50%
30%
20%
50%
B
0%
0%
1
x
gallons
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 263
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
6
continued
f) How much prime coffee and how much standard coffee
are needed to produce 40 kg of blend?
Sketch I:
Sketch II:
24
B
22.5
4.5
price 18
per kg
24
22.5 1.5
A
18
value of
premium
value of
standard
0
0
40 kg
10
30
40
The areas of the rectangles in Sketch I represent the values of
the coffees in the blend. If the blend is to sell for $22.50, the 2
rectangles should “level off” at 22.5. This will be the case if, in
Sketch II, area A = area B. Since the height of B is 1⁄ 3 the height
of A, for the areas to be equal, the base of B must be 3 times
the 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 should
be 10 kg of standard coffee and 30 kg of premium coffee.
g) How many of each coin are there?
numbers
of coins
3
75¢
3 30¢
75¢
?
10¢ 5¢
25¢
40¢
values of coins
The heights of the rectangles represent the number
of coins and their bases the values, so the sum of
the areas of the rectangles is the total value of the
collection. 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 is
40¢, its height is 2.40 ÷ .40 = 6. Thus, there are 6
nickels, 9 dimes, and 12 quarters.
264 | ALGEBRA THROUGH VISUAL PATTERNS
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
h) What is the cost of student and adult tickets?
I. Kyle’s sales:
In the sketches A is the cost of an adult ticket
and S is the cost of a student ticket. Vertical dimensions represent the number of tickets sold.
II. Matt’s sales:
number of
tickets sold
15
48
6
A
8
S
A
III. Kyle’s sales increased
by a factor of 1⁄ 3:
7
38
S
IV. Removing II from III:
26 13
20
64
8
A
S
A
S
Increasing Kyle’s sales by a factor of 1⁄ 3 —so Kyle
and Matt have the same number of adult sales—
and removing Matt’s sales from the result, as
shown in sketch IV, shows that 13 student tickets cost $26, so each cost $2. Thus, in sketch I,
the 15 student tickets cost $30, so the 6 adult
tickets cost $18, and each ticket costs $3.
Alternatively, one could quadruple Kyle’s sales
and triple Matt’s sales. Then Kyle would have 24
A sales and 60 S sales for a total of $192, while
Matt would have 24 A sales and 21 S sales for a
total of $114. So the $78 difference in sales is the
result of 39 S tickets. Hence, each S ticket is $2.
i) How much of the 40% solution did the nurse add?
Sketch I:
The areas of the rectangles in Sketch I at the left
represent the amount of sugar in the solutions. If
the resulting solution is to be 60% sugar, the 2
rectangles should “level off” at 60. This will be the
case if, in Sketch II, area A = area B. Since the area
of B is 1200 x 25, or 30,000, and the height of A is
20, for the areas to be equal, the width of A must
be 30,000 ÷ 20, or 1500. Hence, 1500 ml of the 40%
solution should be added.
85%
40%
0%
sugar in
85% solution
sugar in
40% solution
?
milliliters
1200
Sketch II:
85%
B
20%
60%
40%
25%
60%
A
?
milliliters
1200
0%
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 265
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
6
continued
j) How far is it from Gillette to Spearfish?
1
1 __
2
1
1 __
3
Return Trip:
Trip:
(x + 8) mph
distance from
x mph Gillette to Spearfish
1
1 __
hours
2
distance from
Spearfish to
Gillette
8
A
1
1
– 1__
=
1 __
2
3
32⁄3
1
__
6
B
x
1
1 __
hours
3
x⁄6
x
Area B = Area A
x
__
6
32
= __
,
3
x = 64 mph
The areas of the above rectangles represent distances traveled.
Since the distances are the same, the areas are equal. Thus, if
one rectangle is superimposed on the other as shown above,
the areas of rectangles A and B are equal. So, the distance between Gillette and Spearfish is 64 x (1 1⁄ 2) = 96 miles.
k) What were the student’s 3 test scores?
Average score is 78:
Moving 1 point from last
score to each of 1st 2 scores,
so average of 1st 2 scores is 3
greater than 3rd score:
Moving 7 points from 2nd score
to 1st score, makes 1st score 86:
86
79 79 76
78 78 78
72 76
The 2nd and 3rd scores are 72
and 76, respectively.
l) How far did the train travel?
15
speed
75
mph
60
1200 miles
time
(train)
60
time
(bus)
17 hours
266 | ALGEBRA THROUGH VISUAL PATTERNS
time
(train)
1200 – 1020 =
180 miles
1020 miles
17
The distance traveled is represented by the area of the region
in the 1st sketch to the left. This
region can be divided into the 2
rectangles shown in the 2nd
sketch. The area of the lower
rectangle is 1020 miles. Hence,
that of the upper is 180 miles, so
its length is 180 ÷ 15, or 12
hours. Thus, 12 hours of the trip
were by train, and the distance
traveled by train was 75 x 12, or
900, miles.
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
7 Ask the students to sketch a
7
Here is one sketch of the rectangle:
rectangle whose length is 8 units
greater than its width. Then tell
them the area of the rectangle is
1428 and ask them to find its dimensions. Discuss the equations
that have been solved.
x
x
8
One way to determine the dimensions of the rectangle is to find
2 numbers which differ by 8 and whose product is 1428. If the
rectangle were a square, its dimension would be 1428 which is
about 38. Since its not a square, one dimension should be somewhat larger than this and one dimension somewhat smaller. If one
guesses the dimensions are 34 and 42, a check will verify that this
is correct. Making an educated guess and then checking to see if
it is correct would be more difficult if the dimensions were not
integers.
Another way to proceed is by "completing the square," as shown
in the sketches on the left. If the strip of width 8 in the above
sketch 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 corner
missing. 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 original
rectangle are 34 and 34 + 8, or 42.
16
4
x
x+4
1428
x
4
Figure 1
1444
x+4
Figure 2
The equation x(x + 8) = 1428 has been solved.
8 Ask the students to draw sketches
to solve the following equations:
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 segment
of value x augmented by a segment of value –2.
a) x 2 – 4x + 6 = 5
b) x 2 + 9x = 400
a) x 2 – 4x + 6 = 5
c) x(3x – 4) = 4
One can complete the square as shown in the following sequence
of sketches. Notice that, since x 2 – 4x + 6 = 5, then x 2 – 4x = –1
and x 2 – 4x + 4 = 3.
d) 2x(3 – x) = 3
–1
x2
–4x
x
–2x
x2
x
4
–2x
–2
–2
x 2 – 4x + 4 x – 2
x–2
3
x–2
x–2
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 267
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
8 a) continued
If 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) x 2 + 9x = 400
Completing the square gives the sequence of sketches shown below.
9
__
2
400
x2
x
9x
9
__
x
2
81
__
4
x 2 + 9x +
x2
9
__
2 x
= 400 +
x
9
__
2
81
__
4
420.25
81
__
4
9
x + __
2
With the help of a calculator, one finds
x + 4.5
= ± 420.25
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 the
sketches below.
Since
9
1681 = 41, 2x + 9 = ±41 and the result follows.
18x
81
4x 2 + 36x + 81
= 4(x 2 + 9x ) + 81
4x 2
2x
2x
18x
1681
= 4(400) + 81
2x + 9
9
2x + 9
c) x(3x – 4) = 4
In the following sequence, the second rectangular region is
obtained from the first by increasing its height, and therefore its
area, by a factor of 3.
–2
3x
x
3x
12
4
16
12
4
3x
–4
268 | ALGEBRA THROUGH VISUAL PATTERNS
3x
–4
3x
–2
3x – 2
3x – 2 = ±4
3x = 6 or 3x = –2
x = 2 or x = – 2 ⁄ 3
SKETCHING SOLUTIONS
LESSON 12
FOCUS
COMMENTS
ACTIONS
d) 2x(3 – x) = 3
In the following sequence, the second rectangular region is
obtained from the first by changing the value of its base, and
therefore its area, by a factor of –2.
–3
9
2x – 3 = ± 3
2x
2x
3
–x
3
2x
–6
2x
–6
3
–6
2x
2x = 3 ± 3
x = 1⁄ 2(3 ± 3 )
–3
2x – 3
Many sequences of sketches shown in the solutions above, and
elsewhere in this lesson, contain more figures than may be in the
sketches the students draw. In a number of instances, several
figures shown in a sequence of sketches could be combined into a
single figure, especially if an oral presentation is being made
concurrently, or if solutions are being developed for private use
and not for the benefit of a reader.
ALGEBRA THROUGH VISUAL PATTERNS | 269
TEACHER NOTES
270 | ALGEBRA THROUGH VISUAL PATTERNS
SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.1
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.
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.
© THE MATH LEARNING CENTER
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SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.2
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?
Solution 1
Pump A
1 hour
Pump B
1 hour
Together
1 hour
2
__
5
1 hour
of 1 hr.
Solution 2
Time to fill 1 tank: Pump A
Pump B
Tanks filled in
12 hours:
Pump A
Pump B
6 hours
4 hours
6 hours
4 hours
6 hours
4 hours
4 hours
Pumps A and B fill 5 tanks in 12 hours.
Solution 3
Pump A fills
in 1 hour.
1
__
6
tank
Pump B fills
in 1 hour.
1
__
4
tank
Pump A fills 4 subdivisions in 1 hour.
Pump B fills 6 subdivisions in 1 hour.
Together, they fill 10 subdivisions in
1 hour:
1 hour
1 hour
4
__
10
272 | ALGEBRA THROUGH VISUAL PATTERNS
hour
© THE MATH LEARNING CENTER
SKETCHING SOLUTIONS
LESSON 12
FOCUS BLACKLINE MASTER 12.3
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
© THE MATH LEARNING CENTER
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SKETCHING SOLUTIONS
LESSON 12
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.
274 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
SKETCHING SOLUTIONS
LESSON 12
FOLLOW-UP BLACKLINE MASTER 12
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 students 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?
© THE MATH LEARNING CENTER
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SKETCHING SOLUTIONS
LESSON 12
ANSWERS TO FOLLOW-UP 12
1
a) Let the two numbers be x and x + 6.
Sketches:
Algebraic Statements:
The value of the following region is 1476:
x 2 + (x + 6) 2 = 1476
6x
6
36
2x 2 + 6x + 6x + 36 = 1476
x
x2
x2
x
x
6x
6
The value of half of the region is 738:
18
x
x2
x 2 + 6x + 18 = 738
6x
6
x
Rearrange:
9
x+3
9
x2
(x + 3) 2 + 9 = 738
(x + 3) 2 = 729
x + 3 = 27 or –27
x = 24 or x = –30
If x = 24, x + 6 = 30.
If x = –30, x + 6 = –24.
x+3
The two numbers are 24 and 30 or –24 and –30.
continued
276 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
SKETCHING SOLUTIONS
LESSON 12
ANSWERS TO FOLLOW-UP 12 (CONT.)
b) Let the smaller number be x and the larger number be x + 2d.
Sketches:
Algebraic Statements:
The value of the following region is 520:
2d
x
2dx
4d 2
x2
2dx
x2
x
x 2 + (x + 2d) 2 = 520,
x 2 + x 2 + 4dx + 4d 2 = 520 and
2x + 2d = 32,
2d
x
32
The value of half the region is 260:
d2 d2
(x + d) 2 + d 2 = 260,
16 2 + d 2 = 260,
256 + d 2 = 260,
d 2 = 4.
Since d is positive, d = 2.
Thus x = 16 – d = 14 and
x + 2d = 18.
x2
x+d
x 2 + 2dx + 2d 2 = 260 and
x + d = 16,
x+d
16
Since 16 2 = 256, d 2 = 4 and, since d is positive d
= 2, so x = 16 – 2, or 14 and x + 2d = 18. The
numbers are 14 and 18.
c) Let x be the width of the rectangle and x + 2d its
length. Then x + (x + 2d) is half the perimeter, or 46.
Rearranged it forms a 23 x 23 square with a d x d
corner missing;
The following region has value 493:
23 x + d
x
493
d 2 = (23)2 – 493
= 529 – 493
= 36
xd
x2
xd
46
x + 2d
x+d
23
So d 2 = 36. Hence d = 6, so 2x = 46 – 12 = 34. Thus
x = 17 and the dimensions of the rectangle are 17
and 29.
continued
© THE MATH LEARNING CENTER
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SKETCHING SOLUTIONS
LESSON 12
ANSWERS TO FOLLOW-UP 12 (CONT.)
d) In the following sketch, the base of a rectangle
represents time, the height represents rate of travel, so
the area represents distance traveled.
f) In the following, the base of a rectangle represents
the amount of a solution and the height the percent
of acid in the solution. Thus, the area represents the
amount of acid in the solution.
In the following, r is the rate of the slower car and r + 20
is the rate of the faster car. The total area of the region
is 400.
80
20
100%
Slow
Car r
320
r
Fast
Car
5000
50%
40%
3500
1600
40
4
r = 320 ÷ 8 = 40.
Freshmen
g) x is the amount of pure acid to be added.
Seniors
2000
Juniors
1
50
If the heights of the above rectangles are “leveled
off,” 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.
The rates are 40 mph and 60 mph.
e)
70
160
4
3
20%
2000
40%
3500
40%
100%
Sophomores
The above rectangle is divided into 20 equal parts. The
28 sophomores constitute 4 more than 4 of these
parts, so each part represents (28 – 4) ÷ 4, or 6, students.
Hence, there are 6 x 20, or 120, students. (42 are freshman, 28 are sophomores, 27 are juniors, and 23 are
seniors.)
100
x
If the two rectangles are “leveled off ” to a height of
60%, the areas of the two shaded rectangles are
equal. The area of the rectangle on the left is 20(100),
or 2000. Hence the base x of the rectangle on the
right is 2000 ÷ 40, or 50. Thus 50 cc of pure acid
must be added.
continued
278 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
SKETCHING SOLUTIONS
LESSON 12
ANSWERS TO FOLLOW-UP 12 (CONT.)
h)
A 40 x 60 rectangular
garden with an 864 square
foot border of uniform
width, w:
Rearranging the
border:
a
e
a
f
864
g
b
50
864
40
c
b
60
60
Moving 10 feet from the end of rectangles a
and b to the end of rectangles c and d, and then
completing the square:
2500
3364
d
hw
w
w
e
f
c
w
g
w
h
w
d
2w
40
864
2w
2w + 50
50
2w + 50 =
3364 = 58,
so, w = 4 (i.e., the width
of the border is 4 feet).
© THE MATH LEARNING CENTER
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TEACHER NOTES
280 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
THE BIG IDEA
Analyzing the graphs and Algebra Piece representations
of families of linear and quadratic equations prompts
intuitions and conjectures about the general characteristics of linear and quadratic functions. Exploring a
variety 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.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
Students explore the graphing
calculator and discuss their
successes and challenges with
the graphing calculator.
Students use graphing calculators to graph, solve, and evaluate linear and quadratic equations and inequalities. Special
functions of graphing calculators provide information about
graphs of everyday situations.
Students compare the advantages and disadvantages of the
graphing calculator as a tool
for solving and graphing
equations to by-hand graphing,
Algebra Piece, and symbolic
methods, and mental strategies.
Students use graphs to represent and solve equations and
systems of equations. They
write math expressions that
represent graphs of equations
and inequalities. They use
graphs to solve problems
regarding ice cream sales.
Materials
Start-Up Master 13.1, 1 copy
per student.
Graphing calculators, 1 per
student.
Graphing calculator for the
overhead (recommended).
Materials
Focus Master 13.1, 1 transparency.
Focus Masters 13.2-13.5,
1 copy of each per student
and 1 transparency of each.
Algebra Pieces for each
student.
Algebra Pieces for the
overhead.
Graphing calculators,
1 per student.
Graphing calculator for the
overhead (recommended).
Materials
Follow-Up 13, 1 copy per
student.
Coordinate grid paper (see
Appendix), 6 sheets per
student.
ALGEBRA THROUGH VISUAL PATTERNS | 281
TEACHER NOTES
282 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
START-UP
Overview
Students explore the graphing
calculator and discuss their successes and challenges with the
graphing calculator.
ACTIONS
1 Distribute graphing calculators
and manuals. Hand out copies of
Start-Up Master 12.1 (see following
pages) and suggest that they might
use this sheet for ideas on this
Start-Up activity. Then ask students
to privately explore with their
calculators and find at least two
things that they can do with a
graphing calculator that they didn’t
know about previously. Allow 5-10
minutes for this exploration.
Materials
Start-Up Master 13.1,
1 copy per student.
Graphing calculator for
the overhead (recommended).
Graphing calculators, 1
per student.
COMMENTS
1 With their graphing calculators students should be encouraged
to find out how to perform specific functions that interest them.
Providing manuals for reference is recommended.
Students should be encouraged to demonstrate their findings on
the overhead graphing calculator.
Ask students to share with another
student one thing they discovered.
Ask for volunteers to present their
dicoveries at the overhead.
ALGEBRA THROUGH VISUAL PATTERNS | 283
TEACHER NOTES
284 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
START-UP BLACKLINE MASTER 13.1
Individually use your graphing calculator to determine which graphing calculator functions
are clear to you. Use the calculator manual as needed. Some suggestions for investigation
follow.
1 This list below contains functions that you will need to know. Investigate those that
interest you.
ON/OFF
CLEAR the screen
show blank coordinate axes in the calculator viewing screen
move the cursor around a blank coordinate axes
change the viewing WINDOW size
FORMAT the axes
determine the “standard” WINDOW size on my calculator (on many it is –10 ≤ x ≤ 10
and –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 graph
ZOOM in again—and again
ZOOM out on a graph
ZOOM back to the standard window
TRACE the graph of a function to determine the approximate value of the function
at x = 0, x = 19.75, and x = –37.5
TRACE the graph of a function to determine the value of x when y = 75, when y = –75
GRAPH 2 equations on the same coordinate axes.
TRACE to approximate the intersection of 2 graphs
ZOOM and TRACE to improve your approximation
DRAW a horizontal line on coordinate axes and slide the line up and down
DRAW a vertical line on coordinate axes and slide the line left and right
___ view a table of coordinates of 2 equations listed simultaneously
use a table to find when 0 = 5x + 1
clear MEMory
reset defaults in MEMory
continued on back
© THE MATH LEARNING CENTER
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ANALYZING GRAPHS
LESSON 13
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 + 1
clear MEMory
reset defaults in MEMory
solve equations using the “solver” function
use the “maximum” and minimum” functions to find the turning point of a parabola
use the “intersect” function to find the intersection of 2 graphs
use the “zero” function to find the x-intercepts of a graph
use the “value” function to find v(x) for specific values of x
set 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 that
interest you. Share how these functions work with a classmate.
3 List some functions that are not clearly understood.
286 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
ANALYZING GRAPHS
LESSON 13
FOCUS
2 Ask the students to leave the
arrangements formed in Action 1 on
their desks/tables and to imagine
the graph of y = –3x + 5 in enough
detail that they can “see” important
features of the graph. Ask for volunteers to sketch and explain their
ideas at the overhead. Use this as a
context for recalling the terms slope,
x-intercept, and y-intercept, and
how to determine the value of each.
If it hasn’t come up previously, point
out to students that an x-intercept
of a graph is also referred to as a
zero of the equation, since it is a
point where the value of y is zero.
Focus Masters 13.2-13.5,
1 copy of each per
student and 1 transparency of each.
Algebra Pieces for each
student.
Algebra Pieces for the
overhead.
Graphing calculators,
1 per student.
Graphing calculator for
the overhead (recommended).
COMMENTS
1 An Algebra Piece arrangement for the given sequence is shown
below.
–2nd
–1st
0
1st
2nd
…
oooo oooo
…
ooo
oooo
1 Distribute Algebra Pieces to each
student. Write the formula v(x) =
–3x + 5 on the overhead and tell
the students that this formula represents the xth arrangement of a
complete sequence of counting piece
arrangements. Ask the students to
form the –3rd, –2nd, –1st, 0th, 1st,
2nd, 3rd, and xth arrangements of
this sequence. Discuss.
ooo
ACTIONS
Materials
Focus Master 13.1, 1
transparency.
ooo
Overview
Students use graphing calculators to
graph, solve, and evaluate linear and
quadratic equations and inequalities.
Special functions of graphing calculators provide information about
graphs of everyday situations. Students compare the advantages and
disadvantages of the graphing calculator as a tool for solving and graphing equations to by-hand graphing,
Algebra Piece, and symbolic methods,
and mental strategies.
2 In the above sequence, each time the arrangement number
increases by 1, the value of the arrangement decreases by 3.
Hence, the graph of y = –3x + 5 is a line that falls from left to
right 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-axis
between x = 1 and x = 2, and closer to 2.” Others may mentally
solve the equation –3x + 5 = 0 to determine the x-intercept is
x = 5 ⁄ 3 . And, some may use Algebra Piece representations of
y = –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 3
units vertically for every 1 unit of horizontal change (to the
right), or down 1 unit for every 1 ⁄ 3 unit to the right. Hence, since
the y–intercept is at (0,5), one can drop down 3 units and move
to the right 1 unit to locate the point (1,2) and from there move
down 2 units and over 2 ⁄ 3 unit to locate the x-intercept.
ALGEBRA THROUGH VISUAL PATTERNS | 287
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3 If it didn’t come up in Actions 1
or 2, ask the students to determine
how the slope, y-intercept, and
x-intercept of the graph relate to
the arrangements formed in Action 1.
Discuss.
4 Distribute graphing calculators (if
students don’t have them). Ask the
students to graph v(x) = –3x + 5 on
their calculators, and to determine
various methods of using a graphing
calculator to determine a) below.
Discuss their ideas regarding the
advantages and disadvantages of the
graphing calculator methods when
compared to: hand graphing, mental
strategies, and algebraic procedures
(either with Algebra Pieces or with
symbols representing the pieces).
Repeat for b) and c).
a) the x-intercept
3
The slope of a line is the ratio of the difference in the values
of 2 arrangements to the difference in the corresponding arrangement numbers. The value of the y-intercept is the value of the 0th
arrangement. The x-intercept is the number of the arrangement
whose value is 0.
4
If the calculators were used by other classes, students may need
to clear or turn off graphs that were stored in the calculator.
Throughout this lesson students have opportunities to use the
calculator functions that were explored on Start-Up 12.1, and
they are introduced to other functions as needed or appropriate
for use in the activity. The names of functions and menus that are
referenced in this lesson may vary among calculator brands, and
some brands may not have some of the functions. Hence, you may
need to adapt some actions according to the calculators used by
your students.
a) The students should notice that the TRACE function can give a
very close “approximation” for the x-intercept, but not necessarily an exact value. A series of traces and zooms for y = –3x + 5 is
shown below at the left. Each ZOOM obtains a closer approximation of the x-intercept.
b) the y-intercept
Many students may suggest that mentally calculating the x-intercept
(by mentally determining when 0 = –3x + 5) is simple and therefore using the calculator is not needed to compute the x-intercept. And others may note that symbolic procedures
to locate the x -intercept:
are quick and exact for this equation. Two important
purposes of this lesson are for students to: 1) develop
Trace after 1st zoom:
a “tool kit” of options for graphing and solving
equations and 2) develop a sense for the appropriate uses of the available options.
c) the point where x = 49
Using
ZOOM
and
TRACE
1st trace:
X = 1.91
Y = –.74
Trace after 2nd zoom:
X = 1.67
Y = –.02
X = 1.65
Y = –.05
Trace after 3rd zoom:
X = 1.666
288 | ALGEBRA THROUGH VISUAL PATTERNS
Y = –.0033
On many calculators the “zero,” “intersect,” and
“solver” functions are all appropriate for locating
the x-intercept. Students may also use the table
function to locate the x-intercept.
b) Students may feel that mentally evaluating the
equation at x = 0, by substituting 0 for x to get
y = (–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 the
y-intercept.
ANALYZING GRAPHS
LESSON 13
FOCUS
COMMENTS
ACTIONS
c) Students may graph y = –3x + 5 in a standard window and
attempt tracing the graph to determine the y value for x = 49.
However, this is not possible if x = 49 is outside the viewing
window. (Note: in graphing calculators there is a standard default
viewing 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 is
included and so the corresponding y-value also appears (see
example at the left). This requires making a mental estimate of
the y-value when x = 49. For example, one could note that
y = –3(49) + 5 should be a little more than –3(50), and therefore,
set the minimum y at –150.
Use WINDOW to set minimum and maximums:
x min 0
x max 50
y min –150
y max 20
y = –3x + 5
X= 49
Y= –142
The y -value at x = 49.
5 Ask the students to determine
then
49
–49
–49
–49
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If
oooo oooo oooo
Some students may “see” v(x) using Algebra Pieces:
3(–49) + 5
5
As students discuss the advantages and disadvantages of
various techniques, you might encourage them to make connections among the representations they use for these techniques.
For example, a particular counting piece arrangement corresponds to a point on the graph; a point on the graph can be
described by a pair of values, x and y; and the relationship between the values x and y can be described by a general formula
(in this case, y = –3x + 5). Understanding connections among
these mathematical representations empowers students as
algebraic thinkers.
various methods of identifying the
points on y = –3x + 5 where the
following are true. Discuss.
a) y = –75
b) y = 10 2⁄ 3
c) x = –28.75
d) x = 10
The intent here is to have students continue exploring various
techniques for solving and evaluating equations while developing
comfort with the techniques and a sense about their appropriate
uses. Following are some methods that students may suggest. If
computers are available, you might have the students explore the
use of one or more computer graphing utilities.
e) x = 257
a) One way to solve –3x + 5 = –75 is to use Algebra Pieces, as
shown below:
–75
so,
oooo oooo oooo
=
oooo oooo oooo
–3x
+5
oooo oooo oooo
oooo oooo oooo
–3x
=
–80
and thus,
=
80 / 3
= 26 2 / 3
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 289
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5
continued
Instead of Algebra Pieces, one can use symbols representing
Algebra Piece actions, as shown here:
–3x + 5
–3x + 5 – 5
–3x
–3x ⁄ 3
–x
x
y = –3x + 5 1
y = –75
X= 26.38
Y= –75.16
y = –3x + 5
y = –75
X= 26.66
Y= –74.99
= –75
= –75 – 5
= –80
= –80 ⁄ 3
= –80 ⁄ 3
= 80 ⁄ 3
Another possibility is to adjust the window, then graph both
y = –3x + 5 and y = –75 simultaneously on a graphing calculator,
and finally zoom and trace to approximate the x-value where
these graphs intersect. The line y = –75 can be graphed by using
the Y = function and the GRAPH function on the calculator, or by
using the “horizontal” function from the DRAW menu to form a
horizontal line and then slide the horizontal line vertically until it
intersects y = –3x + 5 at y = –75. In the graphs shown at the left
the horizontal axis was set for –10 ≤ x ≤ 50, and the vertical axis
was 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 on
the calculator, and finally select the function “intersect” to
determine 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 graphing calculator to view a table of values for y = –3x + 5. Scroll to
the entry closest to y = –75. Increments in x may need adjustment in order to locate an x-value that produces a y-value closer
to y = –75.
b) x = –1 8 ⁄ 9
c) If
= –28.75,
then –3x + 5 = –3(–28.75) + 5
= 86.25 + 5
= 91.25
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One could also set the window of the graphing calculator to
include x = –28.75 and y = v(–28.75), and then trace and zoom to
approximate v(–28.75) or use a “value” function. To determine a
window that will include v(–28.75), one could set the upper
bound for y at v(–30), which is easy to compute mentally. Another
approach is to locate the y-value that corresponds to x = –28.75
in a table for y = –3x + 5.
d)-e) Methods similar to those used for c) are appropriate for d)
and e).
6 Place a transparency of Focus
Master 13.1 on the overhead, revealing Part a) only. Discuss the students’ ideas. Then reveal and discuss
Parts b)-d).
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.1
I
y = –3x + 5
II
y = –x + 5
6 a) You may need to remind the students to imagine and predict
the graphs at this point, not to draw or use their calculators.
Students may mention differences in the slope, both its steepness
and rise/fall. They may predict that I and IV are mirror images of
each other across the y-axis, as are lines II and III; and many may
point out they all have y-intercept 5.
b) Here are graphs of the 4 equations. You might suggest that
students be sure to determine the equation associated with each
graph on their calculators.
y
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?
5
(0, 5)
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?
x
–5
d) What are similarities and differences among Algebra Piece representations of the xth
5
arrangements of the sequences represented by equations I-IV?
y=x+5
y = 3x + 5
–5
y = –x + 5
y = –3x + 5
c) Some students may describe these as a “family of lines with a
common y-intercept.” Hence, other family members could be
equations of any lines whose y-intercept is 5. Other students may
suggest that the lines must have slopes of +1, –1, +3, or –3. Still
others may say that, for every line in the family, if the line has
slope m, then another family member must have slope –m, and
others may suggest any lines whose slopes are integers and
whose y-intercepts are 5 belong in the family.
continued next page
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6
continued
d) For example, the xth arrangements of the sequences represented by these equations each contain only x-frames and 5 units
or –x-frames and 5 units.
7 Give each student a copy of
7
Focus Master 13.2 and repeat
Action 6 for several of the equation
families listed. Encourage conjectures and generalizations about
relationships between the graph of a
line or parabola and the constant,
coefficients, and variables in the
equation for the line or parabola.
Encourage discussion about the
information revealed by different
forms of an equation (e.g., factored
or expanded forms of a quadratic).
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.2
Students could explore these in groups or individually as
homework. And you might create other families for students to
examine, based on mathematical ideas or relationships you feel
students need to discuss further or based on prior conjectures
that students have made.
The intent here is for students to continue their search for
insights about relationships among equations, their graphs, and
their Algebra Piece representations.
Conjectures that surface may shift discussion in a number of
directions; 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.
y
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,
5
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
5 I
II
y = –3x – 5
II
III y = –3x + 2
IV y = –3x – 2
y = (x – 3)(x – 4)
y = x2 – 6
6 I
II
y = x2 + 6
II
y = (x – 2)(x – 5)
y = 2(x – 2)(x – 5)
III y = –x 2 – 6
III y = –(x – 2)(x – 5)
IV y = –(x 2 – 6)
IV y = –2(x 2 – 7x + 10)
3 I
II
y=
y = ( 1⁄ 4)x 2
III y = –3x 2 – 6
IV y = ( 3⁄ 4)x 2
4 I
II
y = x(x – 3)
y = x 2 – 2x
–5
5
IV y = (x + 1)(x – 2)
2 I
4x 2
x
y = x 2 – 7x + 12
III y = (x + 2)(x + 3)
7 I
II
28x + 8y = 0
y = –3x + 5
y = –3x + 2
y = –3x – 2
y = –3x – 5
–5
7x + 2y = 6
III 14x + 4y = 4
IV 21x + 6y = –12
8 I
II
y = –5x + 2⁄ 3
3y = –15x + 2
III y = x 2 + 2x
III 0 = –5x – y + 2⁄ 3
IV y = x(x + 3)
IV –2 = –15x – 3y
2) Notice that graphs of equations I and II are identical to the
parabola y = x 2 after translating it 6 units down (I) or up (II) the
y-axis. Similarly, graphs of III and IV are translations of the parabola,
y = –x 2 , 6 units down (III) or up (IV) the y-axis. Graphs of II and III
are reflections of each other across the x-axis, as are graphs of I
and IV. Graphs of I and III are reflections of each other across the
line y = –6, while graphs of II and IV are reflections of each other
across the line y = 6.
3) Some may refer to this as a family of quadratic functions
whose graphs are parabolas with vertices are on the y-axis. This is
also true for the quadratics in 2) above.
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4) These are all quadratic functions and, when written in standard
form, the coefficients of the x 2 and x terms are integers and the
constant is 0. When expressed in factored form, all have x as one
factor.
y = x2
y = x 2 + 2x
y = –x 2
y = x 2 + – 3x
Students may be interested in pursuing the effects of adding
various x-terms to the equation y = x 2. One interesting
conjecture that may come up is that the vertices of all the
parabolas of the form y = x 2 + bx, where b is a real number,
lie on the parabola y = –x 2 (see left); and vertices of all
parabolas of the form y = –x 2 + bx lie on the parabola
y = x 2.
5) Some students may call this a family of parabolas that can be
written in the form y = x 2 + bx + c, where b and c are real numbers not equal to zero. Some may notice that the x-intercepts of
a parabola are easy to identify when the equation is in factored
form (assuming it factors). For example, if the equation is of the
form y = (x – r)(x – s), where r and s are real numbers, the
x-intercepts are x = r and x = s.
Notice that when the two factors in equation I are multiplied, the
product is equation II. Hence, equations I and II have identical
graphs, with x-intercepts 3 and 4.
6) Students may describe this as a family of parabolas whose
vertices 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. For
each family member, its reflection across the x-axis is also in the
family. Students may make conjectures about equations for lines
whose 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 is
a reflection across the y-axis? across the line y = x?” Note:
replacing x with –x in an equation creates a reflection across the
y-axis, and exchanging x and y in an equation creates a reflection
across the line y = x; however, students may not reach these
conclusions.
7) These equations are all written in standard linear form (ax + by
= c). Rewriting each equation in slope-intercept form shows that
all 4 lines have the same slope but different y-intercepts. Hence,
this is a family of parallel lines with slope –7 ⁄ 2 . Notice, the standard form of a linear equation does not give away as many
explicit “clues” about its graph as does the slope-intercept form.
8) The graphs of these equations are all identical; hence the
equations are all equivalent. Students may add other equations
equivalent to these, or they may add a set of equivalent equations
that represent another line.
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8 Write the equations y 1 = 2x + 1
8
a) the TRACE and ZOOM functions on
the graphing calculator
a) The diagrams below show a trace and zoom to locate x = 2.97
and x = 3.03 as approximations for one solution. Additional
zooms improve the approximation, suggesting the graphs intersect at x = 3.
and y 2 = x 2 – 2 on the overhead or
board. Point out that these are
referred to as a system of 2 equations in 2 variables. Ask the groups
to solve this system of equations,
using each of the methods listed
below. Discuss.
b) Algebra Pieces (or sketches of
the pieces)
Solving this system of equations means to find the value of x for
which 2x + 1 = x 2 – 2, i.e., to solve the equations simultaneously. In
terms of the graph, solving the system means finding the points of
intersection of the two graphs. In terms of sequences of counting
piece arrangements, solving the system is equivalent to finding the
values of x for which the 2 different xth arrangements of the
sequences represented by the equations have the same value.
Trace:
c) algebraic symbols
y2 = x 2 – 2
d) the calculator “solver” function
y1 = 2x + 1
e) the calculator “intersect” function
f) the calculator
TABLE
function
X= 2.97
Y= 6.87
Zoom and then trace again:
y2 = x 2 – 2
y1 = 2x + 1
X= 3.03
Y= 7.06
A series of zooms and traces of the other intersection point
suggests x = –1 as a solution. One can verify that x = 3 and –1 are
solutions of the system by testing those points in equations for y 1
and y 2. Since v 1 (3) = 2(3) + 1 = 7 = 3 2 – 2 = v 2 (3), and v 1 (–1) =
2(–1) + 1 = –1 = (–1) 2 – 2 = v 2 (–1), the points (3,7) and (–1,–1)
are intersection points of these graphs.
294 | ALGEBRA THROUGH VISUAL PATTERNS
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b) Illustrated below is an Algebra Piece solution.
=
x2 – 2
2x + 1
?
ooo
ooo
=
Adding –2 x + 2 to both
collections produces these
2 collections.
ooo
=
ooo
=
3
ooo
ooo
x2 – 2x
ooo
=
ooo
x 2 – 2 x + 1 or ( x – 1) 2
=
4
“Completing the squares” by adding
1 unit to the upper right hand corner
of each collection produces this diagram. Since the squares are equal in
value, their edges must be equal.
Hence, x – 1 = 2 or x – 1 = –2, so x = 3
or x = –1. Therefore, x = 3 and x = –1
are the x -coordinates of the intersection points of the 2 graphs.
c) One can also imagine the Algebra Piece actions and record
symbolic procedures that represent those actions. For example:
(x 2
x2 – 2
– 2) + (–2 x + 2)
x2 – 2x
2
x – 2x + 1
( x – 1) 2
x–1
so, x
=
=
=
=
=
=
=
2x + 1
(2 x + 1) + (–2 x + 2)
3
3+1
2 2 or (–2) 2
2 or –2
3 or –1
(form the 2 equal collections)
(add –2 x + 2 to both collections)
(simplify)
(add 1 to both collections)
(form squares of each collection)
(take the square root of the value of
each square)
d) One can use the “solver” function to determine when the
difference (2x + 1) – (x 2 – 2) = 0.
e) If students’ calculators do not have this function, the students
may 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-values
from the 2 tables are equal.
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9 Write the following system of
equations on the overhead and ask
the students to determine the
solution(s), if any, to each system,
using the approach of their choice.
Ask students to verify each solution
using a second method, and so that
one of their methods utilizes the
graphing calculator and one does
not. Discuss as needed.
y = 4 + 2x;
y=x+3
9
Allow plenty of time for students to explore and discuss their
approaches and results with their classmates before discussing as
a class. Students may use a hand graph, a calculator graph, Algebra
Piece procedures, symbolic procedures, mental strategies, the
solver or intersect functions on the calculator, tables generated
by hand or by a calculator, or combinations of these; or, they may
invent 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.
Repeat Action 9 for other systems
of equations selected from the
following list.
a) y = 7 – x 2;
y = –7 + x 2
b) y = 3x – 2;
y = 3x + 1
c) y = 2x + 7;
y = 4x 2 – 3x + 2
d) 2x + 3y = 4;
x–y=7
e) y – 2x = 5;
x + 3y = 6
f) y = x 2 + 7;
y=0
g) y = –2x 2 – 1;
y=0
y = 4 + 2x
y=x+3
X= -1.01
Y= 1.97
An algebraic solution based on what students imagine as Algebra
Piece actions might look like the following:
4 + 2x
4+x
1+x
x
=
=
=
=
x+3
3
0
–1
(form the 2 equal collections)
(remove x from both collections)
(remove 3 units from both collections)
(add –1 to both collections)
a) Here is a solution using algebra symbols to represent Algebra
Pieces:
7 – x 2 = –7 + x 2
14 = 2x 2
7 = x2
7 = x or – 7 = x
Since 7 – (± 7 ) 2 = 0 , and –7 + (± 7 ) 2 = 0, the graphs intersect
at the points (– 7 ,0) and ( 7 ,0). Note that these are exact
points of intersection; calculator methods give decimal approximations 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 a
number and produce the same result as subtracting 2 from the
number. Or, students may reason that since these 2 graphs are
different straight lines with the same slope, they are parallel and
hence, never intersect. Zooming out on the calculator graph (see
diagram on following page) can help verify this; however, it is
important to note that, when 2 lines are not parallel, it is possible
to miss an intersection point by not zooming out far enough.
296 | ALGEBRA THROUGH VISUAL PATTERNS
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COMMENTS
y = 3x – 2
y = 3x + 1
Here is a representation using sketches
of Algebra Pieces:
=
=
3x – 2
3x + 1
=
Adding –3 x to each collection above leaves:
–2= 1
But it is not possible that –2 = 1, so there are no solutions to
the system.
c) As shown in the calculator display below, one approximation of
an intersection point is (–.638,5.72). The other intersection point
is outside the window. Changing the window ranges for x and y
enables one to approximate the other intersection point.
y = 4x 2 – 3x + 2
y = 2x + 7
X= –.638
Y= 5.72
Using Algebra Pieces (see following page) to complete the square
for this quadratic equation illustrates that these graphs do not
intersect at a point whose coordinates are whole or rational
numbers. Rather, the coordinates of the points of intersection are
irrational numbers.
continued next page
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9
continued
ooo
ooo
ooo
ooo
ooo
ooo
2x + 7
=
ooo
ooo
ooo
Adding –2 x to both collections in the
above diagram produces the following
collections:
ooo
4x 2 – 3x + 2
ooo
ooo
ooo
ooo
=
ooo
ooo
4 x 2 – 5x + 2
=
7
ooo o
ooo
ooo o
ooo
x
ooo
ooo
–
1
—
4
ooo
Cutting and rearranging the
pieces in the above collection
produce the following (note
that the pieces on the upper
and right edges of the square
are not edge pieces; rather,
they are quartered – x -frames):
1
––
16
9
––
16
o
7
––
16
ooo
ooo
o
of a
unit
left
over
=
ooo
2x
–5
—x
4
(2x – 54– )2 +
7
––
16
9
–– .
So, (2x – 54– )2 = 6 16
298 | ALGEBRA THROUGH VISUAL PATTERNS
=
7
ANALYZING GRAPHS
LESSON 13
FOCUS
ACTIONS
COMMENTS
Subtracting 7 ⁄ 16 from both collections above produces
(2x – 5 ⁄ 4 ) 2 = 6 9 ⁄ 16 = 105 ⁄ 16 , and so 2x – 5 ⁄ 4 = ± 10516 . Thus,
x = 5 4 ± 10516 and the graphs cross at approximately x ≈ 1.906
2
and x ≈ –.656. When x = –.656, 2x + 7 = 4x2 – 3x + 2 ≈ 5.69. When x
≈ 1.906, 2x + 7 = 4x 2 – 3x + 2 ≈ 10.8. Thus, the 2 points of
intersection for these graphs occur at approximately (–.656,5.69)
and (1.906,10.8). These are rational approximations to irrational
coordinates.
This example illustrates the convenience of the graphing calculator
for quickly finding approximate solutions to equations. Recall that
the TRACE function obtained the approximation x = –.638, which is
close to the algebraic approximation of x = –.656. Repeated
zooms and traces would improve the calculator approximation.
d) In order to graph these equations on the calculator, one must
rewrite them in slope intercept form as y = ( –2 ⁄ 3 )x + 4 ⁄ 3 and
y = x – 7. Then one can graph and trace to find the approximate
intersections, 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 produces
the new equation 2x + 3(x – 7) = 4. Hence, 2x + 3x – 21 = 4, so
5x = 25, and thus, x = 5. This is an example of the method called
solving by substitution.
e) Students will need to rewrite these equations in slope intercept form before graphing them on the calculator, using the
“solver” function on the calculator, or solving them using Algebra
Pieces 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 either
of these systems since neither the parabola y = x 2 + 7 nor the
parabola y = –2x 2 – 1 intersects the x-axis (i.e., the line y = 0).
Note that if x 2 + 7 = 0, then x 2 = –7. But there is no real number
whose square is negative. Solutions for equations like these are
discussed in the next lesson, Complex Numbers.
ALGEBRA THROUGH VISUAL PATTERNS | 299
ANALYZING GRAPHS
LESSON 13
FOCUS
COMMENTS
ACTIONS
10 Give each student a copy of
Focus Master 13.3 and ask the them
to carry out the instructions.
Discuss their results. As needed,
clarify graphing calculator procedures and graphing conventions
such as the use of dotted lines and
open/closed circles.
ANALYZING GRAPHS
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 the
domain would be x > 2 and the point (2,2) would not be considered on the graph. Students may not be familiar with this notation, but will probably speculate what it means and you can clarify
as needed.
1 Write a mathematical statement to describe each graph.
2 Recreate each graph on a graphing calculator.
y
y
b)
c)
1
1
1
1
y
e)
x
x
1
y
f)
5
x
15
y
h)
y
i)
x
y
5
1
1
5
x
{
c) The domain and range of this function, y = x, are the real
numbers.
20
x
15
x + 1, x ≥ 0
b) An equation for this function is y = –x + 1, x < 0 .
The domain is all real numbers and the range is the reals ≥ 1.
y
10
5
g)
y
x
1
d)
a) One statement describing this graph is y = 2 for x ≥ 2. This is a
function whose domain is the real numbers ≥ 2, and 2 is the only
number in the range. Note: Students may also point out that this
graph is a ray whose end point is (2,2).
LESSON 13
FOCUS BLACKLINE MASTER 13.3
a)
10 It may be helpful to have the students complete a) and then
discuss before continuing with the others. Encourage students to
write mathematical statements that use as few words as possible,
but so that someone reading their statements could recreate the
graph exactly.
x
d) You may need to tell students that all points in the shaded
region are part of the graph to be described. This graph includes
the set of all ordered pairs, (x,y) for which x and y are real numbers
and y ≥ –10. The fact that the line y = –10 is a solid line implies
the points on the line y = –10 are included in the graph. Refer to
your calculator's manual to determine how to shade a region.
1
2
x
e) The dotted line implies the points on the line are not included
in 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 x
and y are real numbers and y > 15.
f) This is a graph of the function y = –20 for x < 40. The domain is
all 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 = –x 2 + 4.
It can also be described as a graph of the inequality y ≤ –x 2 + 4.
The y-values are all real numbers less than or equal to 4.
h) This is a graph of the function y = 2x 2 for x ≥ 0. The domain
and range are the real numbers ≥ 0.
i) This is a graph of the inequality y > x for all real numbers x.
300 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
FOCUS
COMMENTS
ACTIONS
11 Give each student a copy of
Focus Master 13.4 and have them
carry out the instructions.
ANALYZING GRAPHS
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 up
with the child? What length race would be most fair? etc. What
are equations and graphs that could represent this situation?
LESSON 13
FOCUS BLACKLINE MASTER 13.4
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.
12 If students have difficulty deciding what to explore, you
might suggest something, basing your choice on its mathematical
potential. 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 investigate the above question, such as to make a chart or table of
values, and/or to graph the times and distances traveled by both
runners. A graph is illustrated on the following page. A table can
be produced by hand or by using the table function on the
graphing calculator. Notice that since Marcus runs 20 meters in 5
seconds, then he runs 4 meters in 1 second. Similarly, since
Franko runs 20 meters in 3 seconds, he runs 20⁄3 meters in 1 second.
74
78
82
86
90
94
98
102
…
11
12
13
14
15
16
17
18
Franko
time in distance
seconds in meters
0
0
1
6 2⁄ 3
2
13 1 ⁄ 3
3
20
…
Marcus
time in
distance
seconds in meters
0
30
1
34
2
38
3
42
…
dents’ questions to explore. Discuss
their results.
…
12 Pick one or more of the stu-
11
12
13
14
15
73 1 ⁄ 3
80
86 2 ⁄ 3
93 1 ⁄ 3
100
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 301
ANALYZING GRAPHS
LESSON 13
FOCUS
ACTIONS
COMMENTS
12
continued
160
Distance in Meters
140
120
100
80
60
40
20
–Franko
–Marcus
5
10
Time in Seconds
15
The table and graph show that Franko catches up with Marcus
between 11 and 12 seconds. Also, at 12 seconds, Franko is at the
80 meter mark, and Marcus is at the 78 meter mark, so Franko is
favored to win a 100 meter race.
302 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
FOCUS
ACTIONS
13 If it hasn’t already been suggested, ask the students to write an
equation for y 1, the distance of
Marcus from the starting line x
seconds after the start of the race,
and an equation for y 2 the distance
of Franko from the starting line x
seconds after the start of the race.
Ask for volunteers to share their
equations and have the class determine whether the equations accurately represent the distances. Then
ask the students to use calculator
graphs of the equations to determine
a fair length for the race (or to
verify their results from Action 13).
COMMENTS
13 Students may verify their equations by testing various
numbers of seconds to see if the meters traveled match those in
their tables from Action 13. For example, their equation for y 1
should 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 y 1 = 4x + 30 and y 2 = ( 20 ⁄ 3 )x.
In order to find the race length that makes the 30 meter head
start 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 y 1 = 4x + 30 and y 2 = ( 20 ⁄ 3 )x and use
trace to determine the point of intersection (see diagram below).
1st trace:
x min 0 x max 20
y min 0 y max 150
y1 = 4x + 30
__ )x
y2 = ( 20
3
X= 11.7
Y= 76.9
Trace after several zooms:
X= 11.25
Y= 75
Many students will probably suggest a “fair” race is about 75 meters.
An extension question could be, “How much of a head start does
Marcus need for 100 meters to be the length of a fair race?”
ALGEBRA THROUGH VISUAL PATTERNS | 303
ANALYZING GRAPHS
LESSON 13
FOCUS
COMMENTS
ACTIONS
14 Give each student a copy of
14
Focus Master 13.5 and ask them to
complete Situation 1. Discuss their
results. Then repeat for Situations 2
and 3.
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.5
For each of the 3 Situations shown below, please do the following:
These situations could be explored in class, and/or as a
homework activity. Either way, it is helpful to encourage students
to 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 of
miles for which the cost will be the same for both companies? As
an example, the following discussion addresses the question:
Which company has the better deal?
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
Students could make a table, write and solve a system of equations describing each company’s cost, or graph the equations by
hand or on a graphing calculator and look for the points where
the graphs intersect.
question, and verification that your answer works.
If y 1 is the cost at We Hardly Try, y 2 is the cost at Rent-A-Wreck,
and x is the number of miles driven, equations for the total rental
cost from each company could be represented as follows:
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
y 1 = $10 + .10x
y 2 = .15x
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 – 16t 2.
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 – 16t 2.
A symbolic solution to this system could look like the following:
10 + .10x = .15x
10 + .10x – .10x = .15x – .10x
10 = .05x
10 ( 1 ⁄ .05 ) = .05x ( 1 ⁄ .05)
200 = x
When x = 200, y 1 = 30 = y 2 . Hence, at 200 miles the cost of
driving a car from either company is the same. For less than 200
miles, Rent-A-Wreck is a better deal, while We Hardly Try is a
better deal for more than 200 miles. Traces on a graphing calculator illustrate this:
1st trace:
x min 0 x max 300 (miles)
y min 0 y max 50 (cost)
WHT
WHT
RAW
RAW
X= 201
304 | ALGEBRA THROUGH VISUAL PATTERNS
Trace after several zooms:
Y= 30.1
X= 200
Y= 30
ANALYZING GRAPHS
LESSON 13
FOCUS
COMMENTS
ACTIONS
x min 0 x max 50 (pizzas)
y min 0 y max 200 (dollars)
profit: y = 4.5x
X= 22.3
Y= 100.53
x min 0 x max 50 (pizzas)
y min 0 y max 200 (dollars)
income: y2 = 7x
cost: y1 = 100 + 2.5x
X= 22.8
Y= 157
Situation 2. Examples of questions that may come up include: If
Saucey’s sells 100 pizzas, how much money will they make or lose
that day? How many pizzas do they need to sell in a day to make
a profit? Should they change their pricing? Following is one way of
answering the question: How many pizzas do they need to sell in
a 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’s
sells 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 total
income greater than the daily overhead cost of $100 can be found
by tracing a graph of y = 4.5x to determine the value of x when y
exceeds $100, as shown at the left. Since y exceeds 100 between
x = 22 and x = 23, Saucey must sell at least 23 pizzas per day or
lose money.
Another approach to answering this question is to simultaneously
graph the equation representing the daily cost, y 1 = 100 + 2.50x
(where x is the number of pizzas sold), and the equation for daily
income, y 2 = 7x. The intersection of the graphs is the point at
which Saucey’s breaks even (see graph at the left). This also
indicates Saucey must sell 23 pizzas in order for daily income to
exceed daily cost.
Situation 3. Here are some questions students may pose: At what
time does the golf ball reach its highest point? How long does it
take before the golf ball hits the ground? How high is the elevated
tee? Where did those formulas come from?
seconds
x
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
lower level tee
height in feet
y1 = 80x – 16x 2
0
36
64
84
96
100
96
84
64
36
0
–44
elevated tee
height in feet
y2 = 80x – 16x 2 + 20
20
56
84
104
116
120
116
104
84
56
20
–24
An example of a table generated by the calculator is shown
at the left, where y 1 describes the height of the ball when hit
from the lower tee, and y 2 describes the height of the ball
from the elevated tee. The table suggests the ball reaches a
high point after 2.5 seconds, and then starts back down again,
hitting the ground after 5 seconds (a table with smaller
increments can be used to see if the high point is slightly
more or less than 2.5). Notice the table lists negative heights,
but since the ball stops descending at ground level, those do
not 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 when
the ball hits the ground (see diagram on the next page).
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 305
ANALYZING GRAPHS
LESSON 13
FOCUS
ACTIONS
COMMENTS
14 continued
Notice, in the diagram below, the 2 graphs are the same shape,
the graph of y 2 is a vertical translation 20 units above the graph
of y 1 , and it takes longer for the ball hit from the elevated tee to
hit the ground.
The coefficients of the variables have physical significance. For
example, the “80” means the golf ball rises at a rate of 80 feet/sec
when it is first hit. On the other hand, the “–16” is the effect of
the pull of the earth’s gravity on the golf ball (the force of gravity
is a downward force of 16 feet per second squared (i.e., per x 2 ,
where x is the number of seconds). Gravity slows the height gain
of the ball over time and eventually the ball starts to descend.
The 20 in the elevated tee equation is the height of the ball at
time 0, so the elevated tee is 20 feet above the ground level.
y2
y1
x = 2.55
y1 = 89.96
y2 = 119.96
x=5
y1 = 0
x = 5.24
y2 = –.12
y1 = 80x – 16x 2
y2 = 80x – 16x 2 + 20
A ball hit from the elevated tee stays 20
feet higher, and hits the ground about a
quarter second later. Note: the graphs of
the equations are not the paths of the
balls, but rather give the heights of the
ball for various times.
306 | ALGEBRA THROUGH VISUAL PATTERNS
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.1
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?
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 307
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.2
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
5 I
II
y = –3x – 5
II
y = (x – 3)(x – 4)
y = x 2 – 7x + 12
III y = –3x + 2
III y = (x + 2)(x + 3)
IV y = –3x – 2
IV y = (x + 1)(x – 2)
2 I
y = x2 – 6
6 I
II
y = x2 + 6
II
y = (x – 2)(x – 5)
y = 2(x – 2)(x – 5)
III y = –x 2 – 6
III y = –(x – 2)(x – 5)
IV y = –(x 2 – 6)
IV y = –2(x 2 – 7x + 10)
3 I
II
y = 4x 2
y = ( 1⁄ 4)x 2
7 I
II
28x + 8y = 0
7x + 2y = 6
III y = –3x 2 – 6
III 14x + 4y = 4
IV y = ( –3⁄ 4)x 2
IV 21x + 6y = –12
4 I
II
y = x(x – 3)
y = x 2 – 2x
8 I
II
y = –5x + 2⁄ 3
3y = –15x + 2
III y = x 2 + 2x
III 0 = –5x – y + 2⁄ 3
IV y = x(x + 3)
IV –2 = –15x – 3y
308 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.3
1 Write a mathematical statement to describe each graph.
2 Recreate each graph on a graphing calculator.
a)
y
y
b)
c)
1
1
x
1
1
1
y
e)
y
f)
y
10
5
5
x
15
y
h)
20
x
15
g)
x
x
1
d)
y
y
i)
x
y
5
1
1
5
x
x
1
2
© THE MATH LEARNING CENTER
x
ALGEBRA THROUGH VISUAL PATTERNS | 309
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.4
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.
310 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
ANALYZING GRAPHS
LESSON 13
FOCUS BLACKLINE MASTER 13.5
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 – 16t 2.
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 – 16t 2.
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 311
ANALYZING GRAPHS
LESSON 13
FOLLOW-UP BLACKLINE MASTER 13
1 For each of the following families of 3 equations, graph the equations on 1 coordinate 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.
c) y = x 2 + 3
a) y = 4x – 1
b) y = 3x – 2
y = 4x + 2
y = x⁄ 5 – 2
y = –2x 2 + 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
2y – 3x = 12
d) y = x 2⁄ 2 – x ⁄ 2 + 3
y = –x 2 – 3x + 5
b) y = x 2 – 8x + 18
c) y = 2⁄ x + 3
y = x2 + 4
2x + y = 7
e) y = x 2 – 3x – 2
f) y = –x 2 – x – 2
x 2 + 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.
312 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
ANALYZING GRAPHS
LESSON 13
ANSWERS TO FOLLOW-UP 13
1
For each of the following families of 3 equations,
graph the equations on 1 coordinate axis, and list the
characteristics that make the equations a family.
d) y = 1 ⁄ x + 5
y = 1⁄ x – 4
y = 1⁄ x + 3
a) y = 4x – 1
y = 4x + 2
y = 4x – 5
These equations produce graphs of inverse variations,
each with a vertical asymptote of the y–axis. Horizontal
asymptotes are y = 5, y = –4, and y = 3, respectively.
These equations produce lines that are parallel with
slopes of 4 and cross the x–axis at 1, –2, and 5 respectively.
e) y = 4(x – 2)(x + 5)
y = –3(x – 2)(x + 5)
y = (x – 2)(x + 5)
b) y = 3x – 2
y = x⁄ 5 – 2
y = –6x – 2
These equations produce three intersecting lines that
intersect at (0,–2) with slopes of 3, 1 ⁄ 5 , and –6 respectively.
c) y = x 2 + 3
y = –2x 2 + 3
y = x 2⁄ 4 + 3
These equations produce three parabolas, all with
vertices at (0,3) and a vertical axis of x = 0. The
equation y = –2x 2 + 3 produces a parabola that opens
downward with a maximum vertex at (0,3). The other
two equations produce parabolas that open upward
each with a minimum vertex at (0,3).
These equations produce parabolas that each intersect
the x–axis at (–5,0) and (2,0). The equation
y = –3(x – 2)(x + 5) produces a parabola that opens
downward. The screen shots above illustrate the
minimum and maximum vertices for each equation.
continued
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 313
ANALYZING GRAPHS
ANSWERS TO FOLLOW-UP 13 (CONT.)
LESSON 13
2
For each of the following systems of equations use
your calculator to find the solutions:
d) y = x ⁄ 2 – x⁄ 2 +3
y = –x 2 – 3x + 5
a) 6x + 3y = 5
2y – 3x = 12
There are two points of intersection, or two solutions,
as indicated by the screen shots below:
2
Graph the equations and choose INTERSECT from the
(or appropriate) menu. By repeatedly pressing
ENTER (or its equivalent) you will see the following
screens, with the last being the intersection of the
two equations. Only the INTERSECTION screen is shown
for the remaining problems.
CALC
e) y = x 2 – 3x – 2
x 2 + 3y –18
There are two points of intersection, or two solutions,
as indicated by the screen shots below:
f) y = –x 2 –x –2
y = (x + 1)(x – 2)
b) y = x 2 – 8x + 18
2x + y = 7
There is no intersection
of these two equations,
or no solution.
c) y = 2 ⁄ x + 3
y = x2 + 4
The graphs of these two
equations touch at the
indicated point.
3 Verify your solutions to Problems 2a) and 2b) by
solving each using another
method. Show your
thinking and reasoning.
Answers will vary, but
students might solve
symbolically or they might
use the table function in
their calculator, as shown
for problem 2a:
4 Discuss the advantages and disadvantages of using
the graphing calculator to solve equations. Give
examples to illustrate your ideas.
Answers will vary.
314 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
COMPLEX NUMBERS
LESSON 14
THE BIG IDEA
Complex numbers are introduced to provide square
roots for negative real numbers. Green and yellow
complex number pieces, along with red and black
counting pieces, are used to carry out computations
involving complex numbers.
START-UP
FOLLOW-UP
FOCUS
Overview
Overview
Overview
Green and yellow complex
number pieces are introduced
to provide square roots for
negative numbers.
Students use tile pieces to
perform computations with
complex numbers. They solve
equations involving complex
numbers.
Students perform computations
and solve equations involving
complex numbers. They associate complex numbers with
points on a coordinate grid.
Materials
Materials
Black and red counting pieces,
20 per student.
Follow-Up 14, 1 copy per
student.
Black and red edge pieces,
12 per student.
Counting and complex
number pieces for use at
home.
Green and yellow complex
number pieces, 20 per
student.
Grid paper (see Appendix),
1 sheet.
Materials
Black and red counting pieces,
20 per student.
Black and red edge pieces,
12 per student.
Green and yellow complex
number pieces, 20 per
student.
Green and yellow edge
pieces, 12 per student.
Black and red counting and
edge pieces for the overhead.
Green and yellow complex
number and edge pieces for
the overhead.
Green and yellow edge pieces,
12 per student.
Black and red counting and
edge pieces for the overhead.
Green and yellow complex
number and edge pieces for
the overhead.
ALGEBRA THROUGH VISUAL PATTERNS | 315
TEACHER NOTES
316 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
START-UP
Overview
Green and yellow complex number
pieces are introduced to provide
square roots for negative numbers.
ACTIONS
1 Distribute black and red counting and edge pieces to the students.
Ask them to construct a square
array, with edge pieces, whose value
is 9. Discuss.
Materials
Black and red counting
pieces, 20 per student.
Black and red edge
pieces, 12 per student.
Green and yellow complex number pieces, 20
per student.
Green and yellow edge
pieces, 12 per student.
Black and red counting
and edge pieces for the
overhead.
Green and yellow complex number and edge
pieces for the overhead.
COMMENTS
1
There are two possibilities; an array with black edge pieces or
an array with red edge pieces:
If a square array has two identical, adjacent edges, the value of
that edge is called a square root of the value of the array. Thus 9
has two square roots: 3 and –3. If x is positive, the symbol x is
used to designate the positive square root of x (cf. Start-Up 9).
2 Ask the students to construct a
2
A red array has one black edge and one red edge:
square array, with edge pieces,
whose value is –9. Discuss.
Since a square red array does not have two identical edges, there
is no positive or negative number that is a square root of –9. In
general if x is a negative number, there is no positive or negative
number that is a square root of x.
ALGEBRA THROUGH VISUAL PATTERNS | 317
COMPLEX NUMBERS
LESSON 14
START-UP
ACTIONS
3 Distribute green and yellow
complex counting and edge pieces
to the students. Tell the students
the purpose of these pieces is to
provide square roots for negative
numbers; in particular, if both edges
of an array are green, the array will
be red. Illustrate.
COMMENTS
3
In these materials, colored tile and edge pieces are represented as follows:
black
red
green
yellow
In an array, if a row and column both have green edges, the piece
which lies at their intersection will be red.
4 Discuss the net value of collections of green/yellow and black/red
pieces. Introduce notation and
terminology for these collections.
4
Green and yellow are opposites. Thus, the net value of the
following 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 system
and net values in the green/yellow system, the letter i will be
used to indicate net values in the green/yellow system. Thus, a
collection of 4 black tile has net value 4 while a collection of 4
green 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.
318 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
START-UP
ACTIONS
COMMENTS
Shown below are two other collections and their net values.
3 – 3i
–4
Numbers of the form a + bi are called complex numbers. If a and b
are integers, a + bi is called a complex or Gaussian integer. An
imaginary number is a complex number for which b ≠ 0. A real
number is a complex number for which b = 0, a pure imaginary
number is an imaginary number for which a = 0.
ALGEBRA THROUGH VISUAL PATTERNS | 319
TEACHER NOTES
320 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
FOCUS
Overview
Students use tile pieces to perform
computations with complex numbers. They solve equations involving
complex numbers.
ACTIONS
1 Ask students to use counting
pieces to find the sum and difference of 3 + 2i and 4 – 5i. Discuss.
Materials
Black and red counting
pieces, 20 per student.
Black and red edge
pieces, 12 per student.
Green and yellow complex number pieces, 20
per student.
Green and yellow edge
pieces, 12 per student.
Black and red counting
and edge pieces for the
overhead.
Green and yellow complex number and edge
pieces for the overhead.
COMMENTS
1
The sum (3 + 2i) + (4 – 5i) can be found by combining a
collection whose value is 3 + 2i with a collection whose value is
4 – 5i and then finding the value of the combined collection.
3 + 2i
4 – 5i
7 – 3i
The difference (3 + 2i) – (4 – 5i) can be found by combining a
collection whose value is 3 + 2i with a collection whose value is
the opposite of 4 – 5i and then finding the value of the combined
collection.
Alternatively, the difference can be found by forming a collection
with net value 3 + 2i from which a collection with value 4 – 5i
can be removed:
3 + 2i
(3 + 2 i ) – (4 – 5 i ) = –1 + 7 i
ALGEBRA THROUGH VISUAL PATTERNS | 321
COMPLEX NUMBERS
LESSON 14
FOCUS
ACTIONS
2 Ask the students to form a
counting piece array for which one
edge has value 1 + 2i and the other
edge has value 2 + 3i. Then ask the
students to form an array to find
the product (1 + 2i)(2 + 3i). Discuss.
COMMENTS
2
You may want to remind the students that a piece in the array
will be red if both of its edges are green. Also, that in the complex numbers, 1 maintains its role as a multiplicative identity. This
means that a piece which has a black edge, i. e., an edge whose
value is 1, will have the color of its other edge; in particular a
piece with black and green edges will be green.
An array with the given edges has four sections as shown.
2i
IV
III
1
I
II
2
3i
A tile in section I will be black. A tile in section III will be red
since both of its edges are green. A tile in either section II or IV
will be green since one edge is black and the other green.
The completed array appears below. It’s value is –4 + 7i.
(1 + 2 i )(2 + 3 i ) = –4 + 7 i
322 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
FOCUS
ACTIONS
3 Ask the students to form an array
for the product (1 + 2i)(2 – 3i).
Discuss.
COMMENTS
3
Since tile in sections I, II, and IV of the array shown below
have 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 those
in IV are green.
2i
IV
III
1
I
II
2
–3i
The tile in Section III have one green and one yellow edge. The
students may offer various reasons why these tile are black. One
argument might be that a tile with two green edges is red, and
hence changing one of these edges to its opposite will change the
value of the tile to its opposite.
Another way to see this is to form an array that has one edge
consisting of a single green edge piece and the other edge consisting of 1 green and 1 yellow edge piece. Since this edge has
value 0, the value of the array is 0. Thus, since the tile with both
edges green is red, the other tile, which has 1 green and 1 yellow
edge, must be black.
G
R
B
G
Y
The completed array is shown below.
(1 + 2 i )(2 – 3 i ) = 8 + i
ALGEBRA THROUGH VISUAL PATTERNS | 323
COMPLEX NUMBERS
LESSON 14
FOCUS
COMMENTS
ACTIONS
4 Ask the students to complete the
following table showing the color of
a tile with the given edges.
edge
edge 1
2
B
R
G
Y
B
4
Since a tile 1 with 1 black edge has the color of the other
edge, 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 1
yellow edge is black. Thus the following is known:
edge
edge 1
2
B
R
G
Y
G
Y
B
R
B
B
R
G
R
R
B
Y
G
G
R
Y
Y
B
The remaining entries can be determined by methods similar to
those described in Comment 3. For example, the color of a tile
with 1 red and 1 green edge can be determined
by forming an array in which one edge is green G
G
Y
and the other edge is black and red. The completed table is shown below.
B
R
edge
edge 1
2
B
R
G
Y
B
B
R
G
Y
R
R
B
Y
G
G
G
Y
R
B
Y
Y
G
B
R
Replacing colors by values in the above table produces the
multiplication table shown to the left.
324 | ALGEBRA THROUGH VISUAL PATTERNS
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
COMPLEX NUMBERS
LESSON 14
FOCUS
ACTIONS
5 Ask the students to use number
COMMENTS
5
a) 7 + i
pieces to find the following:
b) 10 + 2i
a) (3 – 2i) + (4 + 3i)
c) –2 – 4i
b) (4 + 6i) – (–6 + 4i)
d)
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)
(2 – 3 i )(4 – 2 i ) = 2 – 16 i
e)
(3 + i )(3 – 2 i ) = 11 – 3 i
f)
(–1 + 3 i )(1+ 3 i ) = –10
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 325
COMPLEX NUMBERS
LESSON 14
FOCUS
COMMENTS
ACTIONS
5
continued
g) If an array whose value is 5i is constructed so one edge has
value 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 number
of 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.
5i
1 + 2i
5 i ÷ (1 + 2 i ) = 2 + i
h) The students will devise various strategies for constructing an
array whose value is 16 + 2i and has an edge whose value is 2 – 3i.
2 – 3i
One way to proceed is to lay out an edge of 2 black and 3 yellow
and then consider which of black or red and which of green or
yellow must be in the other edge. Since the resulting array must
contain at least 16 black tile, colors should be chosen that produce
both black and green tile, with more of the former. As shown on
the left, a selection of black leads to a column of 2 black and
3 yellow and a selection of green leads to a column of 2 green
and 3 black. Two of the former and 4 of the latter will produce an
array whose net value is 16 + 2i, as shown below.
16 + 2i
2 – 3i
(16 + 2 i ) ÷ (2 – 3 i ) = 2 + 4 i
326 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
FOCUS
ACTIONS
6 Ask the students to find all
solutions to the following equations:
a)
x2
= –16
b) x 2 + 6x + 34 = 0
COMMENTS
6
a) A 4 x 4 red square will either have two green edges or two
yellow edges. Hence, x = 4i or x = –4i.
b) Since x 2 + 6x + 9 = (x + 3) 2 (see the figure), x 2 + 6x + 34 =
(x + 3) 2 + 25. Hence, if x 2 + 6x + 34 = 0, then (x + 3) 2 = –25.
Therefore x + 3 is the edge of a 5 x 5 red square. Thus x + 3 = 5i
or x + 3 = –5i. So x = –3 + 5i or x = –3 – 5i.
c) x 2 – 4x + 20 = 10
3
3x
9
x
x2
3x
x
3
c) If x 2 – 4x + 20 = 10 then (x – 2) 2 = x 2 – 4x + 4 = –6. Thus x + 2
is 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 (Optional.) Ask the students to
use counting pieces to find a square
root of 2i. Then ask them how they
might use counting pieces, along
with a scissors, to find a square
root of i. Discuss.
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 above
square 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 ⁄ 2 i is a square root of i. Since
each edge could consist of a red and a yellow edge piece, rather
than a black and a green, – 2 ⁄ 2 – 2 ⁄ 2 i is also a square root of i.
continued next page
ALGEBRA THROUGH VISUAL PATTERNS | 327
COMPLEX NUMBERS
LESSON 14
FOCUS
ACTIONS
COMMENTS
7
continued
Note that no new colors—only scissors—are necessary to obtain
square roots for i, that is, no new colors are needed to solve the
equation x 2 = i. This is an illustration of the Fundamental Theorem
of Algebra: Every polynomial equation with complex numbers as
coefficients has solutions which are complex numbers.
2
1
2
The square shown here has area 2.
Its edges are diagonals of unit squares.
328 | ALGEBRA THROUGH VISUAL PATTERNS
COMPLEX NUMBERS
LESSON 14
FOLLOW-UP BLACKLINE MASTER 14
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) x 2 + 10x + 100 = 0
c) x 2 – 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 correspond, 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 corresponding to points P and Q?
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 329
COMPLEX NUMBERS
LESSON 14
ANSWERS TO FOLLOW-UP 14
1
a) A collection for 3 – 5i consists of 3 black and 5 yellow tile; the opposite of a collection for 4 – 2i consists
of 4 red and 2 green tile. Combining these collections and eliminating pairs of opposite tile yields a collection
of 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
b)
c)
a) x = –1 ± 6i
5
5x
25
x
x2
5x
x
5
Completing the square:
x 2 + 10x + 100 = 0
x 2 + 10x + 25 = –75
(x + 5) 2 = –75 = 25(–3)
x + 5 = ±5 3 i
x = –5 ± 5 3 i
–3
–3x
–3x
9
x
x2
x2
–3x
x
x2
x2
–3x
x
x
–3
330 | ALGEBRA THROUGH VISUAL PATTERNS
Multiplying by 4 and then completing the square:
x 2 – 3x + 9 = 0
2
4x – 12x + 36 = 0
(2x – 3) 2 = –45 = 9(–5)
2x – 3 = ±3 5 i
2x = 3 ± 5 i
x = (3 ± 5 i)⁄ 2
© THE MATH LEARNING CENTER
COMPLEX NUMBERS
LESSON 14
ANSWERS TO FOLLOW-UP 14 (CONT.)
3
Coordinates of S are (11, 1) which is associated with the complex number 11 + i. This is the sum of the
numbers corresponding to P and Q.
5
Q (6,4)
4
3
2
S (11,1)
1
0
–1
1
2
3
4
5
6
7
8
9
10
11
12
–2
–3
P (5,–3)
–4
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 331
TEACHER NOTES
332 | ALGEBRA THROUGH VISUAL PATTERNS
APPENDIX
1 ⁄ 4 -Inch
Grid
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 333
APPENDIX 1-Centimeter Grid
334 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
APPENDIX Coordinate Grids
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 335
APPENDIX Red and Black Counting Pieces (print back-to-back with page 337)
336 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
APPENDIX Back for Red and Black Counting Pieces
© THE MATH LEARNING CENTER
ALGEBRA THROUGH VISUAL PATTERNS | 337
APPENDIX Blank Counting Pieces
338 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER
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
342 | ALGEBRA THROUGH VISUAL PATTERNS
© THE MATH LEARNING CENTER