Algebra II Quadratic Functions 2014-10-14 www.njctl.org Slide 3 / - - PDF document

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Algebra II

Quadratic Functions

www.njctl.org 2014-10-14

Slide 2 / 222 Table of Contents

Explain Characteristics of Quadratic Functions Graph Quadratic Functions Solve Quadratic Equations by Graphing Solve Quadratic Equations by Factoring Solve Quadratic Equations Using Square Roots

Key Terms Application of Zero Product Property

Solve Quadratic Equations by Completing the Square Solve Quadratic Equations using the Quadratic Formula The Discriminant Combining Transformations (review) Vertex Form More Application Problems using Quadratics

click on the topic to go to that section

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Key Terms

Return to Table of Contents

Slide 4 / 222 Slide 5 / 222 Slide 6 / 222

Zero(s) of a Function: An x value that makes the function equal zero. Also called a "root," "solution" or "x-intercept"

Key Terms Slide 7 / 222

Minimum Value: The y-value of the vertex if a > 0 and the parabola opens upward Maximum Value: The y-value of the vertex if a < 0 and the parabola opens downward Vertex: The highest or lowest point on a parabola.

Key Terms Slide 8 / 222

Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves

Key Terms Slide 9 / 222

Explain Characteristics

• f Quadratic

Functions

Return to Table of Contents

Slide 10 / 222

Remember: A quadratic equation is any equation that can be written in the form ax2 + bx + c =0 Where a, b, and c are real numbers and a ≠ 0 Question 1: Is a quadratic equation? Question 2: Is a quadratic equation?

Characteristics of Quadratics Slide 11 / 222

Remember: A quadratic equation is any equation that can be written in the form ax2 + bx + c =0 Where a, b, and c are real numbers and a ≠ 0 Question 1: Is a quadratic equation? Question 2: Is a quadratic equation?

Characteristics of Quadratics

Answer Question 1: Yes, can be rewritten as 2x2-x-4=0 Question 2: No, there is no x

2 term.

Slide 11 (Answer) / 222

The form ax2 + bx + c = 0 is called the standard form of a quadratic equation. The standard form is not unique. For example, x2 - x + 1 = 0 can also be written -x2 + x - 1 = 0. Also, 4x2 - 2x + 2 = 0 can be written 2x2 - x + 1 = 0.

Characteristics of Quadratics Slide 12 / 222

Practice writing quadratic equations in standard form: (Simplify if possible.) Write 2x2 = x + 4 in standard form:

Standard Form Slide 13 / 222

Practice writing quadratic equations in standard form: (Simplify if possible.) Write 2x2 = x + 4 in standard form:

Standard Form

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Answer

2x2 - x - 4 = 0

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Write 3x = -x2 + 7 in standard form, if possible:

Standard Form Slide 14 / 222

Write 3x = -x2 + 7 in standard form, if possible:

Standard Form

Answer

x2 + 3x - 7 = 0

Slide 14 (Answer) / 222

Write 6x2 - 6x = 12 in standard form and simplify, if possible:

Standard Form Slide 15 / 222

Write 6x2 - 6x = 12 in standard form and simplify, if possible:

Standard Form

Answer

x2 - x - 2 = 0

Slide 15 (Answer) / 222

Write 3x - 2 = 5x in standard form:

Standard Form Slide 16 / 222

Write 3x - 2 = 5x in standard form:

Standard Form

Answer

Not a quadratic equation

Slide 16 (Answer) / 222

Similar to Quadratic Equations, the standard form of a Quadratic Function is y = ax2 + bx + c, where a ≠ 0. Notice, a can be positive or negative.

Standard Form Slide 17 / 222

When graphed, a quadratic function will make the shape of a parabola. The parabola will open upward if a > 0 or downward if a < 0.

Graph Slide 18 / 222

When graphed, a quadratic function will make the shape of a parabola. The parabola will open upward if a > 0 or downward if a < 0.

Graph

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Teacher Notes Ask students which color parabola has an a>0 (PINK) and which color has a<0 (BLUE)

Slide 18 (Answer) / 222

The domain of a quadratic function is all real numbers.

Domain Slide 19 / 222 Slide 20 / 222

If the vertex is a maximum, then the range is all real numbers less than or equal to the y-value of the vertex. The range of this quadratic is

Range Slide 21 / 222

An axis of symmetry (also known as a line of symmetry) will divide the parabola into mirror images. The line of symmetry is always a vertical line of the form

Axis of Symmetry Slide 22 / 222

The x-intercepts are the points at which a parabola intersects the x-axis. These points are also known as zeros, roots or solutions and solution sets. Each quadratic function will have 0, 1, or 2 or real solutions. 2 real solutions no real solutions 1 real solution

X-Intercepts Slide 23 / 222

1 If a parabola opens downward, the vertex is the highest value on the parabola. True False

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1 If a parabola opens downward, the vertex is the highest value on the parabola. True False

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Answer

True

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2 If a parabola opens upward then...

A

a > 0

B

a < 0

C

a = 0

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2 If a parabola opens upward then...

A

a > 0

B

a < 0

C

a = 0

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Answer

A

Slide 25 (Answer) / 222

3 The vertical line that divides a parabola into two symmetrical halves is called...

A

discriminant

B

quadratic equation

C

axis of symmetry

D

vertex

E

maximum

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3 The vertical line that divides a parabola into two symmetrical halves is called...

A

discriminant

B

quadratic equation

C

axis of symmetry

D

vertex

E

maximum

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Answer

C

Slide 26 (Answer) / 222 Slide 27 / 222

Slide 27 (Answer) / 222 Slide 28 / 222 Slide 28 (Answer) / 222

Slide 29 / 222 Slide 29 (Answer) / 222

What is the range of the quadratic function below?

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Slide 30 (Answer) / 222

Combining Transformations (REVIEW)

Return to Table of Contents

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Combining Transformations (REVIEW)

Return to Table of Contents

Teacher Notes

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This section is intended as review for students to recall transformations and make the connection to vertex form of quadratics.

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Let the graph of f(x) be Graph y = 2f(.5x+1) - 2

Combining Transformations Slide 33 / 222

Let the graph of f(x) be Graph y = 2f(.5x+1) - 2

Combining Transformations

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Answer

horizontal stretch of 2 horizontal shift left 2 vertical stretch of 2 vertical shift down 2

Slide 33 (Answer) / 222

Let the graph of f(x) be Graph y =(-

1/2 )f(2x + 1) + 2

Combining Transformations Slide 34 / 222

Let the graph of f(x) be Graph y =(-

1/2 )f(2x + 1) + 2

Combining Transformations

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Answer

horizontal stretch 1/2 horizontal shift left 1/2 vertical reflection over x-axis vertical stretch of 1/2 vertical shift up 2

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Let the graph of f(x) be Graph y = 3f(-.5x - 2) + 1

Combining Transformations

Answer

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Let the graph of f(x) be Graph y = (-1/2)f(-x + 2) +1

Combining Transformations Slide 36 / 222

Let the graph of f(x) be Graph y = (-1/2)f(-x + 2) +1

Combining Transformations

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Answer

horizontal reflection over y-axis horizontal shift right 2 vertical reflection over x-axis vertical stretch of 1/2 vertical shift up 1

Slide 36 (Answer) / 222

Consider the graph y = x 2 and the rules for stretches and shrinks, Graph

Graph the Transformation Slide 37 / 222

Consider the graph y = x 2 and the rules for stretches and shrinks, Graph

Graph the Transformation

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Teacher Notes When you reach these slides and students graph the transformation, discuss the location of the vertex and how that relates to the equation.

Slide 37 (Answer) / 222

7 Given the graph of h(x), which of the following graphs is y = 2h(-x+1) - 3?

A B C D

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7 Given the graph of h(x), which of the following graphs is y = 2h(-x+1) - 3?

A B C D

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Answer

B

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8 Given the graph of h(x), which of the following graphs is y = -0.5h(2x - 1) + 2?

A B C D

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8 Given the graph of h(x), which of the following graphs is y = -0.5h(2x - 1) + 2?

A B C D

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Answer

D

Slide 39 (Answer) / 222

Graph Quadratic Functions

Return to Table

• f Contents

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Graph by Following Five Steps:

Step 1 - Find Axis of Symmetry Step 2 - Find Vertex Step 3 - Find y-intercept Step 4 - Locate another point Step 5 - Reflect and Connect

Slide 41 / 222 Slide 42 / 222 Slide 43 / 222

Slide 44 / 222 Slide 45 / 222 Slide 46 / 222

9 What is the axis of symmetry for y = x 2 + 2x - 3 (Step 1)?

A

x = 1

B

x = -1

C

x = 2

D

x = -3

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9 What is the axis of symmetry for y = x 2 + 2x - 3 (Step 1)?

A

x = 1

B

x = -1

C

x = 2

D

x = -3

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Answer

B

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10 What is the vertex for y = x 2 + 2x - 3 (Step 2)?

A

(-1, -4)

B

(1, -4)

C

(-1, -6)

D

(1, -6)

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10 What is the vertex for y = x 2 + 2x - 3 (Step 2)?

A

(-1, -4)

B

(1, -4)

C

(-1, -6)

D

(1, -6)

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Answer

A

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11 What is the y-intercept for y = x 2 + 2x - 3 (Step 3)?

A

(0 , -3)

B

(0 , 3)

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11 What is the y-intercept for y = x 2 + 2x - 3 (Step 3)?

A

(0 , -3)

B

(0 , 3)

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Answer

A

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Practice: Graph

Graph Slide 50 / 222

Practice: Graph

Graph

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Answer

Slide 50 (Answer) / 222

Practice: Graph

Graph Slide 51 / 222

Practice: Graph

Graph

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Answer

Slide 51 (Answer) / 222

Practice: Graph

Graph Slide 52 / 222

Practice: Graph

Graph

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Answer

Slide 52 (Answer) / 222

Solve Quadratic Equations by Graphing

Return to Table of Contents

Slide 53 / 222

When asked to solve a quadratic equation, there are several ways to do so. One way to solve a quadratic equation in standard form is to find the zeros of the related function by graphing. A zero is the point at which the parabola intersects the x-axis. A quadratic function may have one, two or no zeros.

Solve by Graphing Slide 54 / 222

How many zeros do the parabolas have? What are the values of the zeros? No zeroes 2 zeroes; x = -1 and x=3 1 zero; x=1

click click click

Solve by Graphing Slide 55 / 222

Every quadratic function has a related quadratic equation. A quadratic equation is used to find the zeroes of a quadratic

• function. When a function intersects the x-axis its y-value is

zero. y = ax2 + bx + c Quadratic Function 0 = ax2 + bx + c ax2 + bx + c = 0 Quadratic Equation When writing a quadratic function as its related quadratic equation, you replace y with 0. So y = 0.

Vocabulary Slide 56 / 222

One way to solve a quadratic equation in standard form is to find the zeros or x-intercepts of the related function. Solve a quadratic equation by graphing: Step 1 - Write the related function. Step 2 - Graph the related function. Step 3 - Find the zeros (or x-intercepts) of the related function.

Solve by Graphing Slide 57 / 222

2x2 - 18 = 0 2x2 - 18 = y y = 2x2 + 0x - 18 Step 1 - Write the Related Function

Solve by Graphing Slide 58 / 222

Use the same five-step process for graphing The axis of symmetry is x = 0. The vertex is (0, -18). The y-intercept is (0, -18). Since the vertex is the y-intercept, locate two

• ther points by substituting values for x. We'll

use (2,-10) and (3,0) Graph these points and use reflection across the axis of symmetry. Connect all points with a smooth curve. Step 2 - Graph the Function y = 2x2 + 0x – 18

Solve by Graphing Slide 59 / 222

Use the same five-step process for graphing The axis of symmetry is x = 0. The vertex is (0, -18). The y-intercept is (0, -18). Since the vertex is the y-intercept, locate two

• ther points by substituting values for x. We'll

use (2,-10) and (3,0) Graph these points and use reflection across the axis of symmetry. Connect all points with a smooth curve. Step 2 - Graph the Function y = 2x2 + 0x – 18

Solve by Graphing

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Hint

Vertex

The vertex is found by substituting the x-coordinate of the Axis of Symmetry into the equation and solving for y.

Y I n t e r c e p t

The point where the line passes through the y-axis. This occurs when the x-value is 0.

Two Points Find two more points on the same side of the Axis of Symmetry as the point containing the y-intercept.

Slide 59 (Answer) / 222

(3,0) # x = 0 (2,-10) # (0,-18) # # # Step 2 - Graph the Function y = 2x2 + 0x – 18

Solve by Graphing Slide 60 / 222

The zeros appear to be 3 and -3.

(3,0)

#

x = 0 (2,-10)

#

(0,-18)

# # #

Step 3 - Find the zeros y = 2x2 + 0x – 18

Solve by Graphing Slide 61 / 222

The zeros are 3 and -3. Substitute 3 and -3 for x in the quadratic equation. Check 2x2 – 18 = 0 2(3)2 – 18 = 0 2(9) - 18 = 0 18 - 18 = 0 0 = 0

#

Step 3 - Find the zeros

y = 2x2 + 0x – 18 2(-3)2 – 18 = 0 2(9) - 18 = 0 18 - 18 = 0 0 = 0

#

Solve by Graphing Slide 62 / 222

12

A B C

y = -2x2 + 12x - 18 Solve the equation by graphing the related function and identifying the zeros.

• 12x + 18 = -2x2

Step 1: Which of these is the related function? y = 2x2 - 12x - 18 y = -2x2 - 12x + 18

Slide 63 / 222

12

A B C

y = -2x2 + 12x - 18 Solve the equation by graphing the related function and identifying the zeros.

• 12x + 18 = -2x2

Step 1: Which of these is the related function? y = 2x2 - 12x - 18 y = -2x2 - 12x + 18

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Answer

C Slide 63 (Answer) / 222

13 What is the axis of symmetry? Formula: -b 2a y = -2x2 + 12x - 18

A

x = -3

B

x = 3

C

x = 4

D

x = -5

Slide 64 / 222

13 What is the axis of symmetry? Formula: -b 2a y = -2x2 + 12x - 18

A

x = -3

B

x = 3

C

x = 4

D

x = -5

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Answer

B Slide 64 (Answer) / 222

14 What is the vertex? y = -2x2 + 12x - 18

A

(3,0)

B

(-3,0)

C

(4,0)

D

(-5,0)

Slide 65 / 222

14 What is the vertex? y = -2x2 + 12x - 18

A

(3,0)

B

(-3,0)

C

(4,0)

D

(-5,0)

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Answer

A Slide 65 (Answer) / 222

15 What is the y-intercept? y = -2x2 + 12x - 18

A

(0,0)

B

(0, 18)

C

(0, -18)

D

(0, 12)

Slide 66 / 222

15 What is the y-intercept? y = -2x2 + 12x - 18

A

(0,0)

B

(0, 18)

C

(0, -18)

D

(0, 12)

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Answer

C Slide 66 (Answer) / 222

16

A B C D

If two other points are (5, -8) and (4, -2), what does the graph of y = -2x2 + 12x - 18 look like?

Slide 67 / 222

16

A B C D

If two other points are (5, -8) and (4, -2), what does the graph of y = -2x2 + 12x - 18 look like?

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Answer

C Slide 67 (Answer) / 222

17 Find the zero(s) y = -2x2 + 12x - 18

A

• 18

B

4

C

3

D

• 8

Slide 68 / 222

17 Find the zero(s) y = -2x2 + 12x - 18

A

• 18

B

4

C

3

D

• 8

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Answer

C Slide 68 (Answer) / 222

Solve Quadratic Equations by Factoring

Return to Table of Contents

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Solve Quadratic Equations by Factoring

Return to Table of Contents

Teacher Notes

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Please Note: This is not meant to be the students first lesson on

• factoring. The emphasis here is

factoring followed by using the Zero Product Property to find the zeros

• f the quadratic. There are various

ways to teach factoring. It is advantageous to show students multiple methods as they can then choose which works best for them.

Slide 69 (Answer) / 222

In addition to graphing, there are additional ways to find the zeros or x-intercepts of a quadratic. This section will explore solving quadratics using the method of factoring. A complete review of factoring can be found in the Fundamental Skills of Algebra (Supplemental Review) Unit.

Fundamental Skills of Algebra (Supplemental Review) Click for Link

Solve by Factoring Slide 70 / 222

Review of factoring - Factoring is simply rewriting an expression in an equivalent form which uses multiplication. To factor a quadratic, ensure that you have the quadratic in standard form: ax2+bx+c=0 Tips for factoring quadratics: · Check for a GCF (Greatest Common Factor). · Check to see if the quadratic is a Difference of Squares or other special binomial product.

Solve by Factoring Slide 71 / 222

Examples: Quadratics with a GCF: 3x2 + 6x in factored form is 3x(x + 2) Quadratics using Difference of Squares: x2 - 64 in factored form is (x + 8)(x - 8) Additional Quadratic Trinomials: x2 - 12x +27 in factored form is (x - 9)(x - 3) 2x2 - x - 6 in factored form is (2x + 3)(x - 2)

Solve by Factoring Slide 72 / 222

Practice: To factor a quadratic trinomial of the form x

2 + bx + c, find two factors

• f c whose sum is b.

Example - To factor x

2 + 9x + 18, look for factors of 18 whose sum is 9.

(In other words, find 2 numbers that multiply to 18 but also add to 9.) Factors of 18 Sum

Solve by Factoring Slide 73 / 222

Practice: To factor a quadratic trinomial of the form x

2 + bx + c, find two factors

• f c whose sum is b.

Example - To factor x

2 + 9x + 18, look for factors of 18 whose sum is 9.

(In other words, find 2 numbers that multiply to 18 but also add to 9.) Factors of 18 Sum

Solve by Factoring

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Answer Factors of 18 Sum 1 and 18 19 2 and 9 11 3 and 6 9 x 2 + 9x + 18 = (x + 3)(x + 6)

Slide 73 (Answer) / 222

Practice: Factor x2 + 4x - 12, look for factors of -12 whose sum is 4. (in other words, find 2 numbers that multiply to -12 but also add to 4.) Factors of -12 Sum

Solve by Factoring Slide 74 / 222

Practice: Factor x2 + 4x - 12, look for factors of -12 whose sum is 4. (in other words, find 2 numbers that multiply to -12 but also add to 4.) Factors of -12 Sum

Solve by Factoring

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Answer Factors of -12 Sum

• 1 and 12 11
• 2 and 6 4
• 3 and 4 1

1 and -12 -11 2 and -6 -4 3 and -4 -1 x2 + 4x - 12 = (x - 2)(x + 6)

Slide 74 (Answer) / 222 Zero Product Property

= 0

x

? ?

Imagine this: If 2 numbers must be placed in the boxes and you know that when you multiply these you get ZERO, what must be true?

Slide 75 / 222

Zero Product Property

= 0

x

? ?

Imagine this: If 2 numbers must be placed in the boxes and you know that when you multiply these you get ZERO, what must be true?

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Teacher Notes

This slide is meant to encourage discussion among students. Hopefully, most will respond and realize that one or both of the numbers in the boxes MUST be

• zero. This will help with their

understanding of the Zero Product Property using binomials.

Slide 75 (Answer) / 222 Zero Product Property

For all real numbers a and b, if the product of two quantities equals zero, at least one of the quantities equals zero.

If a b = 0 then a = 0 or b = 0 Slide 76 / 222

Example: Solve x2 + 4x - 12 = 0

• 1. Factor the trinomial.
• 2. Using the Zero Product Property, set each factor equal to zero.
• 3. Solve each simple equation.

Now... combining the 2 ideas of factoring with the Zero Product Property, we are able to solve for the x-intercepts (zeros) of the quadratic.

Solve by Factoring Slide 77 / 222

Example: Solve x2 + 4x - 12 = 0

• 1. Factor the trinomial.
• 2. Using the Zero Product Property, set each factor equal to zero.
• 3. Solve each simple equation.

Now... combining the 2 ideas of factoring with the Zero Product Property, we are able to solve for the x-intercepts (zeros) of the quadratic.

Solve by Factoring

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Answer

(x + 6) (x - 2) = 0 x + 6 = 0 or x - 2 = 0 x = -6 x = 2 x2 + 4x - 12 = 0

Slide 77 (Answer) / 222

Example: Solve x2 + 36 = 12x Remember: The equation has to be written in standard form (ax2 + bx + c).

Solve by Factoring Slide 78 / 222

Example: Solve x2 + 36 = 12x Remember: The equation has to be written in standard form (ax2 + bx + c).

Solve by Factoring

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Answer

x2 + 36 = 12x

• 12 x -12x

x2 - 12x + 36 = 0 (x - 6)(x - 6) = 0 x - 6 = 0 +6 +6 x = 6

Slide 78 (Answer) / 222

18 Solve

A

x = -7

B

x = -5

C

x = -3

D

x = -2

E

x = 2

F

x = 3

G

x = 5

H

x = 6

I

x = 7

J

x = 15

Slide 79 / 222

18 Solve

A

x = -7

B

x = -5

C

x = -3

D

x = -2

E

x = 2

F

x = 3

G

x = 5

H

x = 6

I

x = 7

J

x = 15

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Answer

E, F Slide 79 (Answer) / 222

19 Solve

A

m = -7

B

m = -5

C

m = -3

D

m = -2

E

m = 2

F

m = 3

G

m = 5

H

m = 6

I

m = 7

J

m = 15

Slide 80 / 222

19 Solve

A

m = -7

B

m = -5

C

m = -3

D

m = -2

E

m = 2

F

m = 3

G

m = 5

H

m = 6

I

m = 7

J

m = 15

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Answer

B Slide 80 (Answer) / 222

20 Solve

A

h = -12

B

h = -4

C

h = -3

D

h = -2

E

h = 2

F

h = 3

G

h = 4

H

h = 6

I

h = 8

J

h = 12

Slide 81 / 222

20 Solve

A

h = -12

B

h = -4

C

h = -3

D

h = -2

E

h = 2

F

h = 3

G

h = 4

H

h = 6

I

h = 8

J

h = 12

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Answer

C, G Slide 81 (Answer) / 222

21 Solve

A

d = -7

B

d = -5

C

d = -3

D

d = -2

E

d = 0

F

d = 3

G

d = 5

H

d = 6

I

d = 7

J

d = 37

Slide 82 / 222

21 Solve

A

d = -7

B

d = -5

C

d = -3

D

d = -2

E

d = 0

F

d = 3

G

d = 5

H

d = 6

I

d = 7

J

d = 37

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Answer

E, J Slide 82 (Answer) / 222

Berry Method to factor Step 1: Calculate ac. Step 2: Find a pair of numbers m and n, whose product is ac, and whose sum is b. Step 3: Create the product . Step 4: From each binomial in step 3, factor out and discard any common factor. The result is the factored form. Example: Solve When a does not equal 1, check first for a GCF, then use the Berry Method.

Berry Method to Factor Slide 83 / 222

Use the Berry Method. a = 8, b = 2, c = -3

• 4 and 6 are factors of -24 that add to +2

Solve Step 1 Step 2 Step 3 Step 4 Discard common factors

Berry Method to Factor Slide 84 / 222

Use the Zero Product Rule to solve. Solve

Berry Method to Factor Slide 85 / 222

Solve Use the Berry Method. a = 4, b = -15, c = -25

Berry Method to Factor Slide 86 / 222

Use the Zero Product Rule to solve. Solve

Berry Method to Factor Slide 87 / 222

Solve

Berry Method to Factor Slide 88 / 222

Solve

Berry Method to Factor

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Answer

• r

Slide 88 (Answer) / 222

Application of the Zero Product Property

In addition to finding the x-intercepts of quadratic equations, the Zero Product Property can also be used to solve real world application problems. Return to Table of Contents

Slide 89 / 222

Example: A garden has a length of (x+7) feet and a width of (x +3) feet. The total area of the garden is 396 sq. ft. Find the width of the garden.

Application Slide 90 / 222

Example: A garden has a length of (x+7) feet and a width of (x +3) feet. The total area of the garden is 396 sq. ft. Find the width of the garden.

Application

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Answer

(x+7)(x+3) = 396 x2 + 10x + 21 = 396 x2 + 10x - 375 = 0 (x-15)(x+25)=0 x=15 x=-25 x=15 feet width = 15+3 = 18 feet

Slide 90 (Answer) / 222

22 The product of two consecutive even integers is

• 48. Find the smaller of the two integers.

Hint: Two consecutive integers can be expressed as x and x + 1. Two consecutive even integers can be expressed as x and x + 2.

Slide 91 / 222

22 The product of two consecutive even integers is

• 48. Find the smaller of the two integers.

Hint: Two consecutive integers can be expressed as x and x + 1. Two consecutive even integers can be expressed as x and x + 2.

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Answer x(x+2)=48 x2 + 2x = 48 x2 +2x - 48 = 0 (x - 6)(x + 8) = 0 x = 6 x=-8 x = 6, 1st integer x + 2 = 8, 2nd integer OR x = -8, 1st integer x + 2 = -6, 2nd integer

Slide 91 (Answer) / 222

23 The width of a rectangular swimming pool is 10 feet less than its length. The surface area of the pool is 600 square feet. What is the pool's width?

Slide 92 / 222

23 The width of a rectangular swimming pool is 10 feet less than its length. The surface area of the pool is 600 square feet. What is the pool's width?

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Answer L(L - 10)=600 L2 - 10L = 600 L2 - 10L - 600 = 0 (L - 30)(L + 20) = 0 L= 30 L= -20 Length = 30 feet Width = L- 10 = 20 feet

Slide 92 (Answer) / 222

24 A science class designed a ball launcher and tested it by shooting a tennis ball straight up from the top of a 15-story building. They determined that the motion

• f the ball could be described by the

function: , where t represents the time the ball is in the air in seconds and h(t) represents the height, in feet, of the ball above the ground at time t. What is the maximum height of the ball? At what time will the ball hit the ground? Find all key features and graph the function.

Students type their answers here

Problem is from: Click link for exact lesson.

Slide 93 / 222

24 A science class designed a ball launcher and tested it by shooting a tennis ball straight up from the top of a 15-story building. They determined that the motion

• f the ball could be described by the

function: , where t represents the time the ball is in the air in seconds and h(t) represents the height, in feet, of the ball above the ground at time t. What is the maximum height of the ball? At what time will the ball hit the ground? Find all key features and graph the function.

Students type their answers here

Problem is from: Click link for exact lesson.

[This object is a pull tab]

Answer

Slide 93 (Answer) / 222

25 A ball is thrown upward from the surface of Mars with an initial velocity of 60 ft/sec. What is the ball's maximum height above the surface before it starts falling back to the surface? Graph the function. The equation for "projectile motion" on Mars is:

Students type their answers here

Slide 94 / 222

25 A ball is thrown upward from the surface of Mars with an initial velocity of 60 ft/sec. What is the ball's maximum height above the surface before it starts falling back to the surface? Graph the function. The equation for "projectile motion" on Mars is:

Students type their answers here

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Answer Maximum Height = 138.5 ft

Slide 94 (Answer) / 222

Solve Quadratic Equations Using Square Roots

Return to Table of Contents

Slide 95 / 222

Solve Quadratic Equations Using Square Roots

Return to Table of Contents

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Teacher Notes Note: In this section it is important to keep reminding students that when they take the square root of both sides of an equation, they must consider both the positive and negative answers. Many mistakes when solving these questions result in students only using the positive square root.

Slide 95 (Answer) / 222 Slide 96 / 222 Slide 97 / 222

What if x2 has a coefficient other than 1? Example: Solve 4x2 = 20 using the square roots method.

Solve Using Square Roots Slide 98 / 222

What if x2 has a coefficient other than 1? Example: Solve 4x2 = 20 using the square roots method.

Solve Using Square Roots

Answer

x = ±√5 4x 2 = 20 4 4 x 2 = 5

Slide 98 (Answer) / 222

26 When you take the square root of a real number, your answer will always be positive. True False

Slide 99 / 222

26 When you take the square root of a real number, your answer will always be positive. True False

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Answer

False

Slide 99 (Answer) / 222

27 If x2 = 16, then x =

A

4

B

2

C

• 2

D

26

E

• 4

Slide 100 / 222

27 If x2 = 16, then x =

A

4

B

2

C

• 2

D

26

E

• 4

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Answer

A E

Slide 100 (Answer) / 222

28 Solve using the square root method.

A 5 B

20

C

4

D

• 2

E

• 5

F

2

G

• 4

H

• 20

Slide 101 / 222

28 Solve using the square root method.

A 5 B

20

C

4

D

• 2

E

• 5

F

2

G

• 4

H

• 20

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Answer

D F

Slide 101 (Answer) / 222

29 If y2 = 4, then y =

A

4

B

2

C

• 2

D

26

E

• 4

Slide 102 / 222

29 If y2 = 4, then y =

A

4

B

2

C

• 2

D

26

E

• 4

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Answer

B C

Slide 102 (Answer) / 222 Slide 103 / 222 Slide 103 (Answer) / 222

Slide 104 / 222 Slide 104 (Answer) / 222

32 If (3g - 9)2 + 7= 43, then g =

A B C D E

Slide 105 / 222

32 If (3g - 9)2 + 7= 43, then g =

A B C D E

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Answer

A D

Slide 105 (Answer) / 222

Challenge: Solve (2x - 1)² = 20 using the square root method.

Solve Using Square Roots Slide 106 / 222

Challenge: Solve (2x - 1)² = 20 using the square root method.

Solve Using Square Roots

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Answer

Slide 106 (Answer) / 222

33 A physics teacher put a ball at the top of a ramp and let it roll toward the floor. The class determined that the height of the ball could be represented by the equation, ,where the height, h, is measured in feet from the ground and time, t, in

• seconds. Determine the time it takes the ball to

reach the floor.

Students type their answers here

Problem is from: Click link for exact lesson.

Slide 107 / 222

33 A physics teacher put a ball at the top of a ramp and let it roll toward the floor. The class determined that the height of the ball could be represented by the equation, ,where the height, h, is measured in feet from the ground and time, t, in

• seconds. Determine the time it takes the ball to

reach the floor.

Students type their answers here

Problem is from: Click link for exact lesson.

[This object is a pull tab]

Teacher Notes Use the lesson on Engage NY to show two separate methods for discussion with the students.

Answer: 1/2 second

Slide 107 (Answer) / 222

34 A rock is dropped from a 1000 foot tower. The height

• f the rock as a function of time can be modeled by

the equation: . How long does it take for the rock to reach the ground?

Students type their answers here Answer

Slide 108 / 222

Solving Quadratic Equations by Completing the Square

Return to Table of Contents

Slide 109 / 222

In algebra, "Completing the Square" is a technique for changing a quadratic expression from standard form: ax2 + bx + c to the vertex/graphing form: a(x + h)2 + k. It can also be used as a method for solving quadratic equations.

Completing the Square Slide 110 / 222 Slide 111 / 222

Slide 112 / 222 Slide 112 (Answer) / 222

35 Find (b/2)2 if b = 14.

Slide 113 / 222

35 Find (b/2)2 if b = 14.

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Answer

49 Slide 113 (Answer) / 222

36 Find (b/2)2 if b = 10.

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36 Find (b/2)2 if b = 10.

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Answer

25 Slide 114 (Answer) / 222

37 Find (b/2)2 if b = -12.

Slide 115 / 222

37 Find (b/2)2 if b = -12.

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Answer

36 Slide 115 (Answer) / 222

38 Complete the square to form a perfect square trinomial. x2 + 18x + ?

Slide 116 / 222

38 Complete the square to form a perfect square trinomial. x2 + 18x + ?

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Answer

81 Slide 116 (Answer) / 222

39 Complete the square to form a perfect square trinomial. x2 - 6x + ?

Slide 117 / 222

39 Complete the square to form a perfect square trinomial. x2 - 6x + ?

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Answer

9 Slide 117 (Answer) / 222

Step 1 - Write the equation in the form x2 + bx = c. Step 2 - Find (b ÷ 2)2. Step 3 - Complete the square by adding (b ÷ 2)2 to both sides of the equation. Step 4 - Factor the perfect square trinomial. Step 5 - Take the square root of both sides. Step 6 - Write two equations, using both the positive and negative square root and solve each equation.

Completing the Square Slide 118 / 222

Let's look at an example to solve: x2 + 14x -15 = 0 Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve How can you check your solutions?

Completing the Square Slide 119 / 222

Let's look at an example to solve: x2 + 14x -15 = 0 Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve How can you check your solutions?

Completing the Square

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Answer

STEP 1: x2 + 14x = 15 STEP 2: (14 ÷ 2)2 = 49 STEP 3: x2 + 14x + 49 = 15 + 49 STEP 4: (x + 7)2 = 64 STEP 5: x + 7 = ±8 STEP 6: x + 7 = 8 or x + 7 = -8 SOLUTION: x = 1 or x = -15

Slide 119 (Answer) / 222

Let's look at an example to solve: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve How can you check your solutions?

Completing the Square

x2 - 2x - 2 = 0

Slide 120 / 222

Let's look at an example to solve: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve How can you check your solutions?

Completing the Square

x2 - 2x - 2 = 0

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Answer

STEP 1: x2 - 2x = 2 STEP 2: (-2 ÷ 2)2 = (-1)2 = 1 STEP 3: x2 - 2x + 1 = 2 + 1 STEP 4: (x - 1)2 = 3 STEP 5: x - 1 = ± √3 STEP 6: x - 1 = √3 or x - 1 = -√3 SOLUTION: x = 1 + √3 or x = 1 - √3

Slide 120 (Answer) / 222

40 Solve the following by completing the square : x2 + 6x = -5

A

• 5

B

• 2

C

• 1

D

5

E

2

Slide 121 / 222

40 Solve the following by completing the square : x2 + 6x = -5

A

• 5

B

• 2

C

• 1

D

5

E

2

Answer

A C

Slide 121 (Answer) / 222

41 Solve the following by completing the square : x2 - 8x = 20

A

• 10

B

• 2

C

• 1

D

10

E

2

Slide 122 / 222

41 Solve the following by completing the square : x2 - 8x = 20

A

• 10

B

• 2

C

• 1

D

10

E

2

Answer

B D

Slide 122 (Answer) / 222

Slide 123 / 222 Slide 123 (Answer) / 222

Challenge: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

Completing the Square

3x2 - 10x = -3

*Note: There is no GCF to factor out like the previous example.

Slide 124 / 222

Challenge: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

Completing the Square

3x2 - 10x = -3

*Note: There is no GCF to factor out like the previous example.

Slide 124 (Answer) / 222

Challenge: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

Completing the Square

4x2 - 17x + 4 = 0

*Note: There is no GCF to factor out.

Slide 125 / 222

Challenge: Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

Completing the Square

4x2 - 17x + 4 = 0

*Note: There is no GCF to factor out.

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Answer

Slide 125 (Answer) / 222

Challenge:

Completing the Square

• 6x2 - 25x - 25 = 0

*Note: There is no GCF to factor out.

Slide 126 / 222 Slide 126 (Answer) / 222

43 Solve the following by completing the square :

A B C D E

Slide 127 / 222

43 Solve the following by completing the square :

A B C D E

Answer

A B

Slide 127 (Answer) / 222

Return to Table of Contents

Solve Quadratic Equations by Using the Quadratic Formula

Slide 128 / 222

At this point you have learned how to solve quadratic equations by: Today we will be given a tool to solve ANY quadratic equation, and it ALWAYS works! Many quadratic equations may be solved using these methods. Though completing the square works for any quadratic equation, it can be cumbersome to repeatedly use the algorithm. · graphing · factoring · using square roots and · completing the square

Solving Quadratics Slide 129 / 222

Now try Completing the Square on the standard form of a quadratic equation.

Completing the Square

Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

Slide 130 / 222

Now try Completing the Square on the standard form of a quadratic equation.

Completing the Square

Step 1 - Rewrite Equation Step 2 - Find (b/2)2 Step 3 - Add the result to both sides Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Write 2 Equations & Solve

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Teacher Notes

see next slide for solution

Slide 130 (Answer) / 222

Steps 2 and 3 - Find (b/2)2 , Add the result to both sides, simplify Step 1 - Rewrite Equation and factor out a Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Solve for x

Completing the Square Slide 131 / 222

Slide 132 / 222 Slide 132 (Answer) / 222

Example 1: Solve 2x2 + 3x - 5 = 0 a = 2 b = 3 c = -5 Once you identify the values of a, b, and c, simply substitute into the quadratic formula and simplify as much as possible. How can you check your answers?

Quadratic Formula Slide 133 / 222

Slide 133 (Answer) / 222

Example 2: Solve: 2x = x2 - 3

To use the Quadratic Formula, the equation must be in standard form (ax2 + bx +c = 0). Identify a, b, and c, then substitue into the formula and simplify. Don't forget to check your results!

Quadratic Formula Slide 134 / 222

Example 2: Solve: 2x = x2 - 3

To use the Quadratic Formula, the equation must be in standard form (ax2 + bx +c = 0). Identify a, b, and c, then substitue into the formula and simplify. Don't forget to check your results!

Quadratic Formula

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Answer

Slide 134 (Answer) / 222

44 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1

F

1

G

2

H

3

I

4

J

5

Slide 135 / 222

44 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1

F

1

G

2

H

3

I

4

J

5

Answer

F I

Slide 135 (Answer) / 222

45 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1

F

1

G

2

H

3

I

4

J

5

Slide 136 / 222

45 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1

F

1

G

2

H

3

I

4

J

5

Answer

B J

Slide 136 (Answer) / 222

46 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1.5

F

1.5

G

2

H

3

I

4

J

5

Slide 137 / 222

46 Solve the following equation using the quadratic formula:

A

• 5

B

• 4

C

• 3

D

• 2

E

• 1.5

F

1.5

G

2

H

3

I

4

J

5

Answer

F I

Slide 137 (Answer) / 222

Example 3:

Solve using the quadratic formula, and simplify the result. x

2 - 2x - 4 = 0

Quadratic Formula Slide 138 / 222

Example 3:

Solve using the quadratic formula, and simplify the result. x

2 - 2x - 4 = 0

Quadratic Formula

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Answer

Slide 138 (Answer) / 222

47 Find the larger solution to the equation.

Slide 139 / 222

47 Find the larger solution to the equation.

Answer

Slide 139 (Answer) / 222

48 Find the smaller solution to the equation.

Slide 140 / 222 Slide 140 (Answer) / 222

Work in small groups to solve the quadratic equation using the following different methods. Factoring Quadratic Formula Completing the Square Graphing

Which Method Slide 141 / 222

Work in small groups to solve the quadratic equation using the following different methods. Factoring Quadratic Formula Completing the Square Graphing

Which Method

Teacher Notes

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Have students compare and contrast which method is better to use for this problem. *Answer: x = -3

• r

x = .5

Slide 141 (Answer) / 222

Factoring Quadratic Formula Completing the Square Graphing Work in small groups to solve the quadratic equation using the following different methods.

Which Method Slide 142 / 222

Factoring Quadratic Formula Completing the Square Graphing Work in small groups to solve the quadratic equation using the following different methods.

Which Method

Teacher Notes

[This object is a teacher notes pull tab]

Have students compare and contrast which method is better to use for this problem. *Answer: x = -3

• r

x = .5

Slide 142 (Answer) / 222

The Discriminant

Return to Table of Contents

Slide 143 / 222

2 real solutions no real solutions 1 real solution Recall what it means to have 0, 1, or 2 solutions/zeros/roots

Solutions Slide 144 / 222

At times, it is not necessary to solve for the zeros or roots of a quadratic function, but simply to know how many roots exist (zero, one, or two). The quickest way to determine how many solutions a quadratic has, algebraically, is to calculate what's called the discriminant. It may look familiar, as the discriminant is a part of the quadratic formula.

The Discriminant Slide 145 / 222 Slide 146 / 222

Other important tips before practice: · The square root of a positive number has two solutions. · The square root of zero is 0. · The square root of a negative number has no real solution.

The Discriminant Slide 147 / 222

Slide 148 / 222 Slide 148 (Answer) / 222

If b2 - 4ac > 0 (POSITIVE) the quadratic has two real solutions If b2 - 4ac = 0 (ZERO) the quadratic has one real solution If b2 - 4ac < 0 (NEGATIVE) the quadratic has no real solutions CONCLUSION:

The Discriminant

click to reveal

Slide 149 / 222

49 What is the value of the discriminant of 2x2 - 2x + 3 = 0 ?

Slide 150 / 222

49 What is the value of the discriminant of 2x2 - 2x + 3 = 0 ?

Answer

• 20

Slide 150 (Answer) / 222

50 Use the discriminant to find the number of solutions for 2x2 - 2x + 3 = 0 A B 1 C 2

Slide 151 / 222

50 Use the discriminant to find the number of solutions for 2x2 - 2x + 3 = 0 A B 1 C 2

Answer

A

Slide 151 (Answer) / 222

51 What is the value of the discriminant of x2 - 8x + 4 = 0 ?

Slide 152 / 222

51 What is the value of the discriminant of x2 - 8x + 4 = 0 ?

Answer

48

Slide 152 (Answer) / 222

52 Use the discriminant to find the number

• f solutions for x2 - 8x + 4 = 0

A B 1 C 2

Slide 153 / 222

52 Use the discriminant to find the number

• f solutions for x2 - 8x + 4 = 0

A B 1 C 2

Answer

C

Slide 153 (Answer) / 222

Vertex Form

Return to Table of Contents

Slide 154 / 222

Vertex Form

A quadratic equation in vertex form: So far, we have been using quadratics in standard form. However, sometimes when graphing, it is more useful to write them in Vertex Form.

Slide 155 / 222

Vertex Form shows the location of the vertex (h , k). The a still tells the direction of opening. And the axis of symmetry is x = h. Example: Find the vertex, direction of opening and the axis of symmetry for the graph of: A quadratic function written in vertex form:

Vertex Form Slide 156 / 222

Vertex Form shows the location of the vertex (h , k). The a still tells the direction of opening. And the axis of symmetry is x = h. Example: Find the vertex, direction of opening and the axis of symmetry for the graph of: A quadratic function written in vertex form:

Vertex Form

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Answer

Vertex: (6 , 4) Direction of openness is down Axis of Symmetry: x = 6

Slide 156 (Answer) / 222

Find the vertex, direction of openness and the axis of symmetry for each. A. B. C.

Vertex Form Slide 157 / 222

Find the vertex, direction of openness and the axis of symmetry for each. A. B. C.

Vertex Form

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A. B. C.

Slide 157 (Answer) / 222

D. E.

Vertex Form

Find the vertex, direction of openness and the axis of symmetry for each.

Slide 158 / 222

D. E.

Vertex Form

Find the vertex, direction of openness and the axis of symmetry for each.

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D. E.

Slide 158 (Answer) / 222

53 Find the vertex for the graph of

A B C D

Slide 159 / 222

53 Find the vertex for the graph of

A B C D

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Answer

D

Slide 159 (Answer) / 222

54 Find the direction of opening for the graph of

A

up

B

down

C

left

D

right

Slide 160 / 222

54 Find the direction of opening for the graph of

A

up

B

down

C

left

D

right

[This object is a pull tab]

Answer

B

Slide 160 (Answer) / 222 Slide 161 / 222

Slide 161 (Answer) / 222 Slide 162 / 222 Slide 162 (Answer) / 222

57 Give the direction of opening for the graph of

A

up

B

down

C

left

D

right

Slide 163 / 222

57 Give the direction of opening for the graph of

A

up

B

down

C

left

D

right

[This object is a pull tab]

Answer

A

Slide 163 (Answer) / 222

58 Give the axis of symmetry for the graph of

Slide 164 / 222

58 Give the axis of symmetry for the graph of

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Answer

x = -9

Slide 164 (Answer) / 222 Slide 165 / 222 Slide 165 (Answer) / 222

60 Give the direction of openness of

A

up

B

down

C

left

D

right

Slide 166 / 222

60 Give the direction of openness of

A

up

B

down

C

left

D

right

[This object is a pull tab]

Answer

A

Slide 166 (Answer) / 222

61 The axis of symmetry for the graph of is ______ .

Slide 167 / 222

61 The axis of symmetry for the graph of is ______ .

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Answer

x = 0

Slide 167 (Answer) / 222 Slide 168 / 222 Slide 168 (Answer) / 222

Slide 169 / 222 Slide 169 (Answer) / 222 Slide 170 / 222

Slide 170 (Answer) / 222 Converting from Standard Form to Vertex Form

To convert from standard form to vertex form, we need to recall the method for completing the square. Step 1 - Write the equation in the form y = x2 + bx + ___ + c - ___ Step 2 - Find (b ÷ 2)2 Step 3 - Write the result from Step 2 in the first blank and in the second blank. Step 4 - Rewrite the first three terms as a perfect square.

Slide 171 / 222 Slide 172 / 222

Slide 172 (Answer) / 222 Slide 173 / 222 Slide 173 (Answer) / 222

65 What is the vertex form of:

A B C D

Slide 174 / 222

65 What is the vertex form of:

A B C D

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Answer

A

Slide 174 (Answer) / 222 Slide 175 / 222

Slide 175 (Answer) / 222 Slide 176 / 222 Slide 176 (Answer) / 222

Compare the following functions based on information from the

• equations. What do the graphs have in common? How are

they different? Sketch both graphs to confirm your conclusions.

Comparing Functions Slide 177 / 222

Compare the following functions based on information from the

• equations. What do the graphs have in common? How are

they different? Sketch both graphs to confirm your conclusions.

Comparing Functions

[This object is a pull tab]

Answer

Both quadratic equation have their vertices at (3, 4). However the graph of f(x) has less vertical stretch than the graph of g(x), and the graph of f(x) opens down, whereas the graph of g(x) opens up. f(x) g(x)

Slide 177 (Answer) / 222

Write two different quadratic equations whose graphs have vertices at (3.5, -7).

Two Functions Slide 178 / 222

What if "a" does not equal 1? Step 1 - Write the equation in the form y = ax2 + bx +__+ c - __ Step 2 - Factor: y = a(x2 + (b/a)x +__)+ c - __ Step 3 - Find (b/a ÷ 2)2 Step 4 - Put your answer from Step 3 in the first blank and multiply Step 3 by a to fill in the second blank. Step 5 - Write trinomial as perfect square.

Standard Form to Vertex Form Slide 179 / 222 Slide 180 / 222 Slide 180 (Answer) / 222

Slide 181 / 222 Slide 181 (Answer) / 222 Slide 182 / 222

Slide 182 (Answer) / 222 Slide 183 / 222 Slide 183 (Answer) / 222

Slide 184 / 222 Slide 184 (Answer) / 222 Geometric Definition of a Parabola

A parabola is a locus* of points equidistant from a fixed point, the focus, and a fixed line, the directrix.

*locus is just a fancy word for set.

Every parabola is symmetric with respect to a line through the focus and perpendicular to the

• directrix. The vertex
• f the parabola is the

"turning point" and is

• n the axis of

symmetry.

Slide 185 / 222

Axis of Symmetry Directrix Focus L1 L2 L1=L2

Focus and Directrix of a Parabola

Every point on the parabola is the same distance from the directrix and the focus.

Slide 186 / 222 Parabolas

The parts are the same for all parabolas,regardless of the direction in which they open. Directrix Axis of Symmetry Vertex Focus y=ax2+bx+c Vertex Focus Directrix Axis of Symmetry x=ay2+by+c

Slide 187 / 222 Slide 188 / 222

General Form y= ax

2 + bx + c

x= ay

2 +by + c

Vertex Form y= a(x - h)

2 +k

x= a(y - k)

2 + h

Opens a>0 opens up a<0 opens down a>0 opens to the right a<0 opens to the left Axis of Symmetry x = h y = k Vertex (h , k) (h , k) Directrix Focus

Parabola Summary Slide 189 / 222

71 What is the vertex of ?

A

(-3, 2)

B

(-3, -2)

C

(2, 3)

D

(-2, -3)

Slide 190 / 222

71 What is the vertex of ?

A

(-3, 2)

B

(-3, -2)

C

(2, 3)

D

(-2, -3)

Answer

A

Slide 190 (Answer) / 222

72 What is the vertex of ?

A

(3, 2)

B

(-3, -2)

C

(2, 3)

D

(-2, -3)

Slide 191 / 222

72 What is the vertex of ?

A

(3, 2)

B

(-3, -2)

C

(2, 3)

D

(-2, -3)

Answer

B

Slide 191 (Answer) / 222 Slide 192 / 222

Slide 192 (Answer) / 222 Slide 193 / 222 Slide 193 (Answer) / 222

Slide 194 / 222 Slide 194 (Answer) / 222 Slide 195 / 222

Slide 195 (Answer) / 222 Slide 196 / 222 Slide 196 (Answer) / 222

Slide 197 / 222 Slide 197 (Answer) / 222

78 What is the vertex of y= x

2 - 8x +21?

A

(4, 5)

B

(-4, 5)

C

(-5, 4)

D

(5, 4)

Slide 198 / 222

78 What is the vertex of y= x

2 - 8x +21?

A

(4, 5)

B

(-4, 5)

C

(-5, 4)

D

(5, 4)

[This object is a pull tab]

Answer

A

Slide 198 (Answer) / 222

79 What is the equation of the directrix for the following equation?

A

y = 2

B

y = -4

C

x = 3

D

x = -5

Slide 199 / 222

79 What is the equation of the directrix for the following equation?

A

y = 2

B

y = -4

C

x = 3

D

x = -5

Answer

C Slide 199 (Answer) / 222

80 Where is the focus for the following equation?

A

(-3 , 5)

B

(3 , 5)

C

(5 , 3)

D

(5 , -3)

Slide 200 / 222

80 Where is the focus for the following equation?

A

(-3 , 5)

B

(3 , 5)

C

(5 , 3)

D

(5 , -3)

Answer

C Slide 200 (Answer) / 222 Slide 201 / 222

Slide 201 (Answer) / 222

82 What is the equation of the parabola with vertex (2,3) and directrix y = 4?

A y = 4(x - 2)2 + 3

y = -1/4(x - 2)2 + 3 x = 4(y - 2)2 + 3 x = 1/4(y - 2)2 + 3

B C D

Slide 202 / 222

82 What is the equation of the parabola with vertex (2,3) and directrix y = 4?

A y = 4(x - 2)2 + 3

y = -1/4(x - 2)2 + 3 x = 4(y - 2)2 + 3 x = 1/4(y - 2)2 + 3

B C D

[This object is a pull tab]

Answer

B Slide 202 (Answer) / 222

More Application Problems Using Quadratics

Return to Table of Contents

Slide 203 / 222 Quadratic Functions in the Real World

Click on the bike to learn more.

Slide 204 / 222 Quadratic Equations and Applications

A sampling of applied problems that lend themselves to being solved by quadratic equations: Number Reasoning Free Falling Objects Geometry: Dimensions Distances Business:Interest Rate Height of a Projectile

Slide 205 / 222

PLEASE KEEP THIS IN MIND. When solving applied problems that lead to quadratic equations, you might get a solution that does not satisfy the physical constraints of the problem. For example, if x represents a width of a garden and the two solutions of the quadratic equations are -9 and 1, the value -9 is rejected since a width must be a positive number. We call this an extraneous solution.

Quadratic Equations and Applications Slide 206 / 222

The product of two consecutive negative integers is 1122. What are the numbers?

Applications Slide 207 / 222

The product of two consecutive negative integers is 1122. What are the numbers?

Applications

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Answer

Leading Questions... · What does it mean for numbers to be consecutive? · What can you use to represent an unknown number? *Let them use variable of their choice. Let n = first number Therefore n + 1 = second number n(n + 1) = 1122 n2 + n = 1122 n2 + n - 1122 = 0 (n + 34)(n - 33) = 0 n = -34 and n = 33. The solution is either -34 and -33 or 33 and 34, since the direction ask for negative integers -34 and -33 are the correct pair.

Slide 207 (Answer) / 222

Applications

Two cars left an intersection at the same time, one heading north and one heading west. Some time later, they were exactly 100 miles apart. The head headed north had gone 20 miles farther than the car headed west. How far had each car traveled?

Slide 208 / 222 Applications

Two cars left an intersection at the same time, one heading north and one heading west. Some time later, they were exactly 100 miles apart. The head headed north had gone 20 miles farther than the car headed west. How far had each car traveled?

[This object is a pull tab]

Answer

x+20 100 x

Use a2 + b2 = c2 to solve x2 + (x+20) 2 = 1002 x2 + x2 + 40x + 400 = 10,000 2x2 + 40x - 9600 = 0 2(x2 +20x - 4800) = 0 x2 + 20x - 4800 = 0 x = 60 or x = -80 Since the distance cannot be negative, discard the negative solution. The distances are 60 miles (Westbound Car) and 60 + 20 = 80 miles (Northbound Car). Let x = the distance traveled by westbound car Therfore (x + 20) = the distance traveled by northbound car

*Encourage students to draw a picture! Prompt by asking what relationship they know about sides of a right triangle.

Slide 208 (Answer) / 222

The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers.

Applications Slide 209 / 222

The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers.

Applications

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Answer

Let n = first number Therefore n + 2 = second number n(n + 2) = 4[n + (n + 2)] - 1 n2 + 2n = 4[2n + 2] - 1 n2 + 2n = 8n + 8 - 1 n2 + 2n = 8n + 7 n2 - 6n - 7 = 0 (n - 7)(n + 1) = 0 n = 7 and n = -1 Test 7 and 9 Test -1 and 1

7 x 9 = 4[7 + (7 + 2)] - 1 63 = 4(16) - 1 63 = 64 -1 63 = 63 (-1) x 1 = 4[-1 + (-1 + 2)] - 1

• 1 = 4[-1 + 1] - 1
• 1 = 4(0) - 1
• 1 = -1

We get two sets of answers.

*This example will challenge students to determine which solution they use. Encourage them to test both and see if both are acceptable. Often, students are inclined to simply exclude the negative answer, but for this example, both solutions work!

Slide 209 (Answer) / 222

The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

Applications Slide 210 / 222

The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.

Applications

[This object is a pull tab]

Answer

x x + 6

Let x = width Therefore x + 6 = length x( x + 6) = 91 x2 + 6x = 91 x2 + 6x - 91 = 0 (x - 7)(x + 13) = 0 x = 7 or x = -13 Since the length cannot be negative... The width is 7 and the length is 13. *Encourage students to draw a picture! Ask students what relationship they know about length and width of a rectangle and area of a rectangle.

Slide 210 (Answer) / 222

83 The product of a number and a number 3 more than the original is 418. What is the smallest value the

• riginal number can be?

Slide 211 / 222

83 The product of a number and a number 3 more than the original is 418. What is the smallest value the

• riginal number can be?

[This object is a pull tab]

Answer

Slide 211 (Answer) / 222

84 The product of two consecutive positive even integers is 168. Find the larger of the numbers.

Slide 212 / 222

84 The product of two consecutive positive even integers is 168. Find the larger of the numbers.

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Answer

Larger Number = 14

Slide 212 (Answer) / 222

85 Two cars left an intersection at the same time,

• ne heading north and the other heading east.

Some time later they were 200 miles apart. If the car heading east traveled 40 miles farther, how far did the northbound car go?

Slide 213 / 222 Slide 213 (Answer) / 222

86 A square's length is increased by 6 units and its width is increased by 4 units. The result of this transformation is a rectangle with an area of 195 square units. Find the area of the original square.

Slide 214 / 222 Slide 214 (Answer) / 222

87 In the accompanying diagram, the width of the inner rectangle is represented by x - 3 and its length by x + 3. The width of the outer rectangle is represented by 3x + 4 and the length by 3x - 4. Express the area of the pink shaded region as a polynomial in terms of x.

Students type their answers here

Problem is from: Click link for exact lesson.

Step 1: Write an expression to represent the area of the larger rectangle. Step 2: Write an expression to represent the area of the smaller rectangle. Step 3: Subtract the polynomial to get your final answer.

Use the the next page for space to solve.

Application Problems Application Problems

Slide 215 / 222

Step 1: Write an expression to represent the area of the larger rectangle. Step 2: Write an expression to represent the area of the smaller rectangle. Step 3: Subtract the polynomial to get your final answer.

Answer

Slide 216 / 222

88 A large painting in the style of Rubens is 3 ft. longer than it is wide. If the wooden frame is 12 in. wide, the area of the picture and frame is 208 ft2 , find the dimensions of the painting. (Draw a diagram.)

Students type their answers here Answer

Slide 217 / 222

89 The rectangular picture frame below is the same width all the way around. The photo it surrounds measures 17" by 11". The area of the frame and photo combined is 315 sq. in. What is the length of the outer frame?

17

x x

11

x x

Answer

Slide 218 / 222

90 Two mathematicians are neighbors. Each owns a separate rectangular plot of land that shares a boundary and have the same dimensions. They agree that each has an area of square

• units. One mathematician sells his plot to the other.

The other wants to put a fence around the perimeter

• f his new combined plot of land. How many linear

units of fencing will he need? Write your answer as an expression of x.

Students type their answers here

Problem is from: Click link for exact lesson.

Note: This question has two correct approaches and two different correct solutions. Can you find them both? Hint: Start by factoring.

Slide 219 / 222

90 Two mathematicians are neighbors. Each owns a separate rectangular plot of land that shares a boundary and have the same dimensions. They agree that each has an area of square

• units. One mathematician sells his plot to the other.

The other wants to put a fence around the perimeter

• f his new combined plot of land. How many linear

units of fencing will he need? Write your answer as an expression of x.

Students type their answers here

Problem is from: Click link for exact lesson.

Note: This question has two correct approaches and two different correct solutions. Can you find them both? Hint: Start by factoring.

[This object is a pull tab]

Answer *Selecting which boundary is common will affect the solution because the length of the common side is not included when finding the perimeter of the combined plot. 8x + 6

• r

10x + 6

Slide 219 (Answer) / 222

91 Part A An expression is given. x2 - 8x + 21 Determine the values of h and k that make the expression (x - h)2 + k equivalent to the given expression. Input your answer for h =

From PARCC sample test

Slide 220 / 222

91 Part A An expression is given. x2 - 8x + 21 Determine the values of h and k that make the expression (x - h)2 + k equivalent to the given expression. Input your answer for h =

From PARCC sample test

[This object is a pull tab]

Answer h = 4

Slide 220 (Answer) / 222

92 Part A An expression is given. x2 - 8x + 21 Determine the values of h and k that make the expression (x - h)2 + k equivalent to the given expression. Input your answer for k =

From PARCC sample test

Slide 221 / 222

92 Part A An expression is given. x2 - 8x + 21 Determine the values of h and k that make the expression (x - h)2 + k equivalent to the given expression. Input your answer for k =

From PARCC sample test

[This object is a pull tab]

Answer k = 5

Slide 221 (Answer) / 222

93 Part B An equation is given. x2 - 8x + 21 = (x - 4)2 + 3x - 16 Find the one value of x that is a solution to the given equation.

From PARCC sample test

Slide 222 / 222

93 Part B An equation is given. x2 - 8x + 21 = (x - 4)2 + 3x - 16 Find the one value of x that is a solution to the given equation.

From PARCC sample test

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Answer 7

Slide 222 (Answer) / 222