Download 301a MBF3C Pearson Math 11 Lesson Plans

Lesson Plans - Unit 3 – Quadratic Relations
MBF 3C1
3.1 Modelling Quadratic Relations
BIG Idea: Different representations
Learning Goal: Given a constraint on the perimeter of a swimming pool, create a table
of values and graph and determine the maximum area.
AGENDA
10 mins
30 mins
5 mins
Warm-up, Attendance
p. 99 – Investigate
Wrap-up
Do now
Create a table of values for x = -2, -1, 0, 1, 2, then graph each relation (on the same
graph is ok).
a) y = 2x2
2) y = (x – 1)2
Complete Investigation p. 99
Length of side perpendicular
to shore (m)
0
2
4
6
8
10
12
14
16
Length of
other side (m)
8
7
6
5
4
3
2
1
0
Swimming
area (m2)
0
14
24
30
32
30
24
14
0
Maximizing A Swimming Area
Swimming area
35
30
25
20
15
10
5
0
-5 0
2
4
6
8
10
12
14
16
18
Length of side perp to shore
A length of 10 for the side perpendicular to the shore results in the greatest swimming
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
area. The area is 10m × 3m = 30m2.
Reflect
Yes, you should join the points on the graph to show the trend. A solid line should be
used since any area is possible (decimals included).
The point (3, 30) represents the greatest swimming area. The 3 represents the length in
m of the side perpendicular to the shore and the 30 represents the Swimming area in
m2.
The table shows that it is the greatest area because it has the highest value. The graph
shows that this is the highest area because it is the highest point.
Inquire
Find an equation for the area, A, in terms of the length of the perpendicular side, p.
A = (16 – p)(0.5p) = 8p – 0.5p2
MSIP / Homework: pp. 102 #1, 2, 6
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.2 Graphing Quadratic Relations
Learning goal: Use a graphing calculator to graph a quadratic relation and answer
questions about the graph in context.
AGENDA
5 mins Warm-up + Attendance
35 mins
Inquiry – Graphing Calculator
15 mins
Practice pp. 109-110 #1-2
5 mins
Wrap-up
Do now
p. 105 Inquire – HP-39G
pp. 109-110 #1-2
MSIP / Homework: pp. 109-110 #1-2 (use emulator)
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.3 The Role of h and k in the function y = (x – h)2 + k
Learning goal: Describe and sketch graphs of quadratics in the form y = (x – h)2 + k
AGENDA
10 mins
30 mins
15 mins
5 mins
Warm-up + Attendance
p. 111 Investigate (graphing calculator)
Practice pp. 109-110
Wrap-up
Do now
*Complete p. 111 Inquire w/HP-39G
The Role of k in the Graph of y = x2 + k
The graph of y = x2 + k is a vertical translation of the graph y = x2
When k is positive, the graph is translated up.
When k is negative, the graph is translated down.
e.g.
To graph y = x2 + 4, translate the graph of y = x2 up 4 units.
To graph y = x2 – 2, translate the graph of y = x2 down 2 units.
The Role of h in the Graph of y = (x – h)2
The graph of y = (x – h)2 is a horizontal translation of the graph y = x2
When h is positive, the graph is translated left.
When h is negative, the graph is translated right.
e.g.
To graph y = (x – 3)2, translate the graph of y = x2 right 3 units.
To graph y = (x + 1)2, translate the graph of y = x2 left 1 unit.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Example 1
Describe how the graphs of the following can be obtained from the graph of y = x 2.
Sketch both graphs on the same set of axes.
a) y = x2 – 6
Translate the graph of y = x2 down 6 units
b) y = (x + 3)2
Translate the graph of y = x2 left 3 units
2
c) y = (x – 2.5)
Translate the graph of y = x2 right 2.5 units
d) y = (x + 2)2 – 4
Translate the graph of y = x2 left 2 units and down 4 units
e) y = (x – 1)2 + 2
Translate the graph of y = x2 right 1 unit and up 2 units
MSIP / Homework pp. 114-115 #1, 2, 4, 6, 9
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.4 The role of a in y = ax2
Learning goal: Describe and sketch graphs of quadratics in the form y = ax2
AGENDA
15 mins
30 mins
15 mins
Warm-up + Attendance + Homework check pp. 114-115 #1, 2, 4, 6, 9
The Role of a in the Graph of y = ax2
Wrap-up
Do now
Describe the transformations that must be done to the graph of y = x2 to obtain the
graph of y = (x – 3)2 – 1. Sketch both graphs on the same set of axes.
Translate the graph of y = x2 right 2 units and down 1 unit.
The Role of a in the Graph of y = ax2
The graph of y = ax2 is a vertical dilatation (stretch / compression) of the graph of y = x2
When a is greater than 1 (a > 1), the graph is stretched vertically by a factor of a (gets
thinner).
When a is between 0 and 1 (0 < a < 1), the graph is compressed vertically by a factor of a
(gets wider).
When a is negative, the graph is also reflected across the x-axis (opens down).
Step Pattern of a parabola
The step pattern for the base quadratic equation y = x2 is [1, 3,
5].
To graph the parabola, begin at the vertex and each time you
move over 1 unit (left or right), move up 1, 3 and 5 units.
To find the step pattern for any quadratic, multiply the step
pattern [1, 3, 5] by a.
x
y = x2
-3
9
Step
-5
-2
4
-3
-1
1
-1
0
0
1
1
1
3
2
4
5
3
e.g.
for y = 2x2 the step pattern is 2, 6, 10
9
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
for y = -3x2 the step pattern is -3, -9, -15
Example 2)
Describe the translations that must be done to the graph of y = x2 to obtain the graph of
the following. Use the step pattern to graph each equation on the same graph.
a) y = ½x2
Vertical compression of y = x2 by a factor of ½
Step pattern ½, 1.5, 2.5
b) y = 3x2
Vertical stretch of y = x2 by a factor of 3
Step pattern 3, 9, 15
c) y = -2x2
Vertical stretch of y = x2 by a factor of 2.5, reflection across x-axis
Step pattern -2, -6, -9
MSIP / Homework:
Extra practice: Complete p. 117 Inquire w/ HP-39G
Read Example on p. 119 (Step pattern)
pp. 119-120 #1-4, 7
Lesson Plans - Unit 3 – Quadratic Relations
MBF 3C1
3.5 The Vertex Form of a Quadratic Relation y = a(x – h)2 + k
(Student handout next page)
Learning goal: Describe and sketch graphs of quadratics in the form y = ax2
Do now
Describe the translations that must be done to the graph of y = x2 to obtain the graph of
the following. Use the step pattern to graph each equation on the same graph.
a) y = 2(x – 4)2
b) y = -(x + 3)2 – 5
Translate right 4 units, vertical stretch by a factor of 2.
Translate left 3 units, reflect across x-axis, translate down 5.
Vertex Form of a quadratic equation is y = a(x – h)2 + k. The vertex is (h, k). The value of
a indicates whether the parabola opens up (positive) or down (negative), and whether it
is a vertical stretch (a > 1) or compression (0 < a < 1).
A parabola that opens up has a minimum point, and a parabola that opens down has a
maximum point.
Example 1) Complete the following table for Vertex Form.
Equation
Vertex
Step Pattern
Direction
of Opening
Axis of
Symmetry
Optimal
Value
y = (x – 1)2 – 3
y = -(x + 7)2 + 8
y = -2(x – 3)2 + 4
y = 3(x + 7)2 – 5
Example 2) Sketch the graph of y = ½ (x – 6)2 – 4.
The vertex is (6, -4). a = ½ so the step pattern is [0.5, 1.5, 2.5]. Use these to plot the
graph.
MSIP / Homework:
Extra practice: Complete p. 117 Inquire w/ HP-39G
Read Example on p. 119 (Step pattern)
pp. 119-120 #1-4, 7
pp. 123-4 #1ace, 2bd, 3, 4, 5
Max /
Min?
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.5 The Vertex Form of a Quadratic Relation y = a(x – h)2 + k ( Student Handout)
Learning goal: Describe and sketch graphs of quadratics in the form y = ax2
Do now
Describe the translations that must be done to the graph of y = x2 to obtain the graph of
the following. Use the step pattern to graph each equation on the same graph.
a) y = 2(x – 4)2
Description:
b) y = -(x + 3)2 – 5
Description:
Graph of (a)
Graph of (b)
Vertex Form of a quadratic equation is y = a(x – h)2 + k. The vertex is (h, k). The value of
a indicates whether the parabola opens up (a is positive) or down (a is negative), and
whether it is a vertical stretch (a > 1) or compression (0 < a < 1).
A parabola that opens up has a minimum, and a parabola that opens down has a
maximum. In both cases it is the y-value (second coordinate) of the vertex.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Example 1) Complete the following table for Vertex Form.
Equation
Vertex
Step Pattern
Direction
of Opening
y = (x – 1)2 – 3
y = -(x + 7)2 + 8
y = -2(x – 3)2 + 4
y = 3(x + 7)2 – 5
Example 2) Sketch the graph of y = ½ (x – 6)2 – 4.
MSIP / Homework: pp. 123-4 #1ace, 2bd, 3, 4, 5
Axis of
Symmetry
Optimal
Value
Max or
Min?
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Marshmallow Gun Activity
Learning goal: Write equations in Vertex Form that could model a projectile
marshmallow.
Marshmallow gun demo
Do now
In a small group, write an equation in the form y = a(x – h)2 + k that could model the
path of a marshmallow shot out of a marshmallow gun. Explain your choice of values for
a, h and k. Draw a picture of the marshmallow being shot.
Task 1)
For each of the following equations, explain why they can or cannot be a model for a
marshmallow fired out of a gun.
a)
b)
c)
d)
y = -0.1(x – 10)2 + 5
y = -0.1(x + 5)2 + 10
y = 5(x – 1)2 + 8
y = -3(x – 100)2 + 800
Task 2)
Write another equation that models a marshmallow and another one that does not.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Review
1.1 Given a scenario, complete a table of values and graph quadratic data.
(1.2 Graph a quadratic equation using a graphing calculator and determine key features
of the graph.)
1.3 – 1.4 Given a quadratic function in the one of the following forms:
y = x2 + k
y = (x – h)2
y = ax2
y = a(x – h)2 + k
describe the transformation(s) that must be done to the graph of y = x2 and sketch both
graphs on the same set of axes.
1.5 Given a quadratic function or graph, determine key properties including: direction of
opening, step pattern, vertex, axis of symmetry, max/min?, optimum value
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.6a Multiplying Polynomials Using FOIL
Learning goal: Multiply two binomials using FOIL
Do now
Explain the error in the following:
2(x + 1)(x – 3)
= (2x + 2)(2x – 6)
= 4x2 – 12x + 4x – 12
= 4x2 – 8x – 12
Error:
Polynomials
Recall: A polynomial is a sum or difference of terms e.g. 3, 5x + 3, 2x2 – 5x – 3
3 has one term so it is a monomial
5x + 3 has two terms so it is a binomial
2x2 – 5x – 3 has three terms so it is a trinomial
Multiplying Binomials
To multiply a binomial by a monomial we use the Distributive Property a(b + c)=ab+ac
e.g. 2(x – 3)
= 2(x) + 2(-3)
= 2x + (-6) or 2x – 6
How do we expand (x + 2)(x + 3), i.e. multiply two binomials?
The process for multiplying two binomials is called FOIL, which stands for First Outside Inside
Last.
Steps in FOIL
1.
2.
3.
4.
5.
F = First; multiply the first terms in each bracket
O = Outside; multiply the outside terms
I = Inside; multiply the inside terms
L = Last; multiply the last terms in each bracket
Add these 4 products and collect like terms
FOIL can be represented as: (a + b)(c + d) = ac + ad + bc + bd
Example 1) Expand:
a) (x + 2)(x + 3)
= (x)(x) + (x)(3) + (2)(x) + (2)(3)
= x2 + 3x + 2x + 6
b) 5(2x – 3)(x + 1)
= 5 [ (2x)(x) + (2x)(1) + (-3)(x) + (-3)(1) ]
= 5 [ 2x2 + 2x + -3x + -3 ]
= 5 [ 2x2 – x – 3 ]
= 10x2 – 5x – 15
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Standard Form of a Parabola
Standard Form of a Parabola is y = ax2 + bx + c
To convert from Vertex Form to Standard Form we expand and simplify.
Example 2) Convert y = 3(x – 4)2 + 8 to Standard Form
y = 3(x – 4)2 + 8
= 3(x – 4)(x – 4) + 8
= 3[(x)(x) + (x)(-4) + (-4)(x) + (-4)(-4)] + 8
= 3[x2 + -4x + -4x + 16] + 8
= 3[x2 -8x + 16] + 8
= 3x2 – 24x + 48 + 8
= 3x2 – 24x + 56
MSIP / Homework: pp. 124 – 125 #(1-2)ace, 4-6, 8
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.7 Factoring Trinomials and Technology (OMIT)
3.8 Factoring Polynomials
Learning goal: Use the values of b and c to write a quadratic expression in the form ax 2 +
bx + c in Factored Form (x – s)(x – t)
Do now
Write a quadratic relation in Vertex Form y = a(x – h)2 + k. Give it to a classmate to
convert to Standard Form by expanding and simplifying. Compare the original relation
and the final relation. How are they similar?
Yesterday we revisited FOIL: Expand (x + 6)(x – 2) = x2 + 4x – 12
Consider the numbers in the brackets, 6 and -2.
6 x -2 = -12
6 + -2 = 4
Today we will factor polynomials and write them as the product of two binomials.
Factoring is the opposite of FOIL.
Part A: Factoring trinomials of the form ax2 + bx + c where a = 1
If a trinomial in the form x2 + bx + c can be written in the form (x + s)(x + t)
Then s × t = c and s + t = b i.e. we need 2 numbers that multiply to c and sum to b.
Example 1) Factor the following:
Standard Form
x2 + 2x + 1
x2 + 3x + 2
x2 + 4x + 3
x2 + 5x + 4
x2 – 13x + 36
x2 – 4x – 32
Factored Form
(x + 1)(x + 1)
(x + 2)(x + 1)
(x + 3)(x + 1)
(x + 4)(x + 1)
(x – 9)(x – 4)
(x – 8)(x + 4)
Multiplication is commutative, meaning a×b = b×a, so it doesn’t matter what order you
write the numbers in.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Part B: Factoring trinomials of the form ax2 + bx + c where a is a common factor
Recall: A common factor is a factor of every term in a polynomial. E.g. 3x2 + 12x + 9 has a
common factor 3 since every term is divisible by 3.
If a is a common factor for b and c:
i. factor out a
ii. factor as in Part A.
Example 2) Factor the following:
a) 3x2 + 12x + 9 = 3(x2 + 4x + 3) = 3(x + 3)(x + 1)
b) 5x2 – 5x – 30 = 5(x2 – x – 6) = 5(x + 2)(x – 3)
Extra:
Associativity
(a×b)×c = a×(b×c)
State another operation that is: a) Associative. b) Not associative.
a) Addition. b) Subtraction, Division.
Commutativity
a×b = b×a
State another operation that is: a) Commutative. b) Not commutative.
a) Addition. b) Subtraction, Division
Other patterns
Difference of Squares
x2 – 25 = (x + 5)(x – 5)
16 – 49x2 = (4 + 7x)(4 – 7x)
MSIP / Homework: pp. 137-138 #2-6, 9
Binomial Square
x2 + 6x + 9 = (x + 3)(x + 3)
x2 – 18x + 81 = (x – 9)(x – 9)
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
3.9 The Factored Form of a Quadratic Relation
Do now
Fully Factor the following:
x2 – 13x + 40
-2x2 – 16x – 32
36 – x2
The Factored Form of a Quadratic Relation is y = a(x – r)(x – s). It has roots / x-intercepts
r and s. NOTE: the sign is opposite of what is in the equation!
Why?
The x-intercepts occur when y = 0. If y = 0, i.e. 0 = (x – r)(x – s), then one of (x – r) = 0 or
(x – s) = 0.
x–r=0
x–s=0
x=r
x=s
so the solutions are x = r and x = s.
Example 1) Find the x-intercepts of the equation y = 2(x + 4)(x – 2). What do they tell
you about the graph?
The equation is in the form a(x – s)(x – t) so we can read the roots directly from the
equation. The roots are -4 and 2. This means the graph crosses the x-axis at -4 and 2.
Example 2) What are the roots of y = -(15 – x)(10 + x)? Sketch the graph.
This equation is not in the form y = a(x – s)(x – t), so set each factor equal to 0 and
solve.
15 – x = 0
10 + x = 0
15 – 15 – x = 0 – 15 10 – 10 + x = 0 - 10
-x = - 15
x = -10
x = 15
The roots are -10 and 15.
Example 3) What are the roots of y = 2x(x + 6)?
This relation can be written as y = 2(x – 0)(x + 6). The roots are 0 and -6.
Therefore a factor ‘x’ represents an x-intercept of 0.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Example 4) Consider the following graph.
Write its equation in Vertex Form and
Factored Form. Then, convert one to Standard
Form. How does each form connect to the
graph?
Vertex Form: y = -5(x – 2)2 + 20
Factored Form y = -5(x – 0)(x – 4)
y = -5(x)(x – 4)
y = -5x(x – 4)
MSIP / Homework: pp. 141-142 #1, 2, 3ab, 7
Complete p. 139 Investigate and Reflect
Equation
y = (x + 4)(x – 2)
y = (x + 1)(x – 3)
y = 3(x – 3)(x – 5)
y = -0.5(x – 4)(x + 4)
y = 2x(x + 6)
y = (15 – x)(10 + x)
x-intercepts
-4
2
-1
3
3
5
4
-4
0
-6
15
-10
Clarification of Bullet 4
e.g., for y = (x + 4)(x – 2), x-intercepts -4, 2
Partner 1 Chooses x = -4
Evaluates y = (x + 4)(x – 2) when x = -4
y = (-4 + 4)(-4 – 2)
= (0)(-6)
=0
Partner 2 Chooses 2
Evaluates y = (x + 4)(x – 2) when x = 2
y = (2 + 4)(2 – 2)
= (6)(0)
=0
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
Example
PGA golfer Lyon Forrest drives a golf ball down the fairway with his 3-wood. The ball
follows a path given by the following equation y = -0.02x2 + 2x where y is the height in
m and x is the vertical distance in m.
a) Write the equation in Factored Form.
y = -0.02x2 + 2x
= -0.02 (x2 – 100x)
= -0.02 (x)(x – 100)
b) What are the x-intercepts?
The x-intercepts are 0 and 100.
c) How far did the ball travel?
The ball travelled 100m.
d) Sketch the graph.
Test Review: Read p. 143
Complete pp. 144 – 145 #4 – 8, 11 – 16
Note: congruent = same a value
Lesson Plans - Unit 3 – Quadratic Relations
MBF 3C1
Quadratics Review
Do now
Explain how to factor 3x2 – 3x – 18. Show each step.
1. Factor out the common factor 3 to get 3(x2 – x – 6).
2. Consider the factors of -6: -1 and 6, 1 and -6, 2 and -3, -2 and 3. Find the pair that
sums to -1: 2 and -3.
3. Write x2 – x – 6 as the product of two binomials using the numbers found in #2:
3(x – 3)(x + 2).
Summary of Quadratic Forms
Expand & Simplify
Vertex Form
y = a(x – h)2 + k


Standard Form
y = ax2 + bx + c
(Complete The Square)
Vertex is (h, k)
y-intercept is c
Describing transformations
Sketching the graph
Factor


Factored Form
y = a(x – r)(x – s)
FOIL
Roots / x-intercepts are r, s
NOTE: Completing the Square is not covered in MBF3C
Test emphasis on Expanding and Simplifying, Factored Form
Vertex Form and transformations are a minor focus.
You do not need to memorize the Quadratic forms. You do need to know what each
one is useful for / when it is required.
MBF 3C1
Lesson Plans - Unit 3 – Quadratic Relations
MSIP / Homework BLM 3.4.3
1. Complete the following table
Equation
Vertex
y = (x – 2)2 + 1
y = -(x + 4)2 + 6
y = 4(x – 4)2 – 1
y = 3(x + 7)2 – 4
y = -2(x – 10)2 + 100
y = (x – 4)2 + 15
y = -2(x + 2)2 + 64
y = 5(x – 10)2 – 11
y = 2(x + 3)2 – 3
y = -(x – 20)2 + 10
Equation
Vertex
Direction of
Opening
Step Pattern
Max or Min?
Optimal Value
Axis of Symmetry
(2, 1)
(-4, 6)
(4, -1)
(-7, -4)
(10, 100)
(4, 15)
(-2, 64)
(10, -11)
(-3, -3)
(20, 10)
Step Pattern From
Vertex
1, 3, 5
-1, -3, -5
4, 12, 20
3, 9, 15
-2, -6, -10
1, 3, 5
-2, -6, -10
5, 15, 25
2, 6, 10
-1, -3, -5
Direction of
Opening
Up
Down
Up
Up
Down
Up
Down
Up
Up
Down
y = 3(x – 4)2 - 8
(4, -8)
y = -2(x + 1)2
(-1, 0)
y = -(x + 2)2 + 10
(-2, 10)
Up
Down
Down
3, 9, 15
Min
-8
y=4
-2, -6, -10
Max
0
y = -1
-1, -3, -5
Max
10
y = -2