Slide 2 / 222. Algebra II. Quadratic Functions

Transcription

1 Slide 1 / 222

2 Slide 2 / 222 Algebra II Quadratic Functions

3 Slide 3 / 222 Table of Contents Key Terms Explain Characteristics of Quadratic Functions Combining Transformations (review) Graph Quadratic Functions Solve Quadratic Equations by Graphing Solve Quadratic Equations by Factoring Application of Zero Product Property Solve Quadratic Equations Using Square Roots Solve Quadratic Equations by Completing the Square Solve Quadratic Equations using the Quadratic Formula The Discriminant Vertex Form More Application Problems using Quadratics click on the topic to go to that section

4 Slide 4 / 222 Key Terms Return to Table of Contents

5 Slide 5 / 222 Key Terms Quadratic Equation: An equation that can be written in the standard form ax2 + bx + c = 0. Where a, b and c are real numbers and a does not equal 0. Quadratic Function: Any function that can be written in the form y = ax2 + bx + c. Where a, b and c are real numbers and a does not equal 0.

6 Slide 6 / 222 Key Terms Parabola: The curve result of graphing a quadratic equation

7 Slide 7 / 222 Key Terms Zero(s) of a Function: An x value that makes the function equal zero. Also called a "root," "solution" or "x-intercept"

8 Slide 8 / 222 Key Terms Vertex: The highest or lowest point on a parabola. 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

9 Slide 9 / 222 Key Terms Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves

10 Slide 10 / 222 Explain Characteristics of Quadratic Functions Return to Table of Contents

11 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

12 Slide 12 / 222 Characteristics of Quadratics 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.

13 Slide 13 / 222 Standard Form Write 2x2 = x + 4 in standard form: Practice writing quadratic equations in standard form: (Simplify if possible.)

14 Slide 14 / 222 Standard Form Write 3x = -x2 + 7 in standard form, if possible:

15 Slide 15 / 222 Standard Form Write 6x2-6x = 12 in standard form and simplify, if possible:

16 Slide 16 / 222 Standard Form Write 3x - 2 = 5x in standard form:

17 Slide 17 / 222 Standard Form 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.

18 Slide 18 / 222 Graph The parabola will open upward if a > 0 or downward if a < 0. Teacher Notes When graphed, a quadratic function will make the shape of a parabola.

19 Slide 19 / 222 Domain The domain of a quadratic function is all real numbers.

20 Slide 20 / 222 Range To determine the range of a quadratic function, ask yourself two questions: Is the vertex a minimum or maximum? What is the y-value of the vertex? If the vertex is a minimum, then the range is all real numbers greater than or equal to the y-value of the vertex. The range of this quadratic is

21 Slide 21 / 222 Range 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

22 Slide 22 / 222 Axis of Symmetry 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

23 Slide 23 / 222 X-Intercepts 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

24 Slide 24 / If a parabola opens downward, the vertex is the highest value on the parabola. False True

25 Slide 25 / 222 A a>0 B a<0 C a=0 2 If a parabola opens upward then...

26 Slide 26 / 222 A discriminant B quadratic equation C axis of symmetry D vertex E maximum 3 The vertical line that divides a parabola into two symmetrical halves is called...

27 Slide 27 / Which of the following shows a quadratic equation correctly written in standard form? A B D E C

28 Slide 28 / What is the equation for the axis of symmetry for the quadratic function A B D E C

29 Slide 29 / What is the domain of the quadratic function below? A B D C

30 Slide 30 / 222 What is the range of the quadratic function below?

31 Teacher Notes Slide 31 / 222 Combining Transformations (REVIEW) Return to Table of Contents

32 Slide 32 / 222 Combining Transformations To combine transformations, follow order of operations: horizontal reflection if b is negative horizontal stretch of 1/b if b < 1 horizontal shrink of 1/b if b > 1 horizontal slide of c vertical reflection if a is negative vertical stretch of a if a > 1 vertical shrink of a if a < 1 vertical slide of d

33 Slide 33 / 222 Combining Transformations Graph y = 2f(.5x+1) - 2 Let the graph of f(x) be

34 Slide 34 / 222 Combining Transformations Graph y =(-1/2 )f(2x + 1) + 2 Let the graph of f(x) be

35 Slide 35 / 222 Combining Transformations Graph y = 3f(-.5x - 2) + 1 Let the graph of f(x) be

36 Slide 36 / 222 Combining Transformations Let the graph of f(x) be Graph y = (-1/2)f(-x + 2) +1

37 Slide 37 / 222 Graph the Transformation Consider the graph y = x2 and the rules for stretches and shrinks, Teacher Notes Graph

38 Slide 38 / 222 A C B D 7 Given the graph of h(x), which of the following graphs is y = 2h(-x+1) - 3?

39 Slide 39 / Given the graph of h(x), which of the following graphs is y = -0.5h(2x - 1) + 2? B A C D

40 Slide 40 / 222 Graph Quadratic Functions Return to Table of Contents

41 Slide 41 / 222 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

42 Slide 42 / 222 Graphing Task: Graph y = 3x2 6x + 1 Step 1: Find the Axis of Symmetry Recall the Formula: a=3 b = -6 x = - (- 6) = 6 = 1 2(3) 6 Therefore, the axis of symmetry is x = 1.

43 Slide 43 / 222 Graphing Task: Graph y = 3x2 6x + 1 Step 2: To find the vertex, substitute equation and find y. y = 3x2-6x + 1 y = 3(1)2 + -6(1) + 1 y=3-6+1 y = -2 Vertex = (1, -2) for x in the

44 Slide 44 / 222 Graphing Task: Graph y = 3x2 6x + 1 Step 3: Find the y-intercept. The y-intercept occurs when x = 0, so substitute zero for x in the equation. y = 3x2-6x + 1 y = 3(0)2 + -6(0) + 1 y=0-0+1 y=1 y intercept = (0, 1)

45 Slide 45 / 222 Graphing Task: Graph y = 3x2 6x + 1 Step 4: Plot an additional point. Choose an x-value to substitute into the function. Using x = -1 y = 3x2-6x + 1 y = 3(-1)2 + -6(-1) + 1 y=3+6+1 y = 10 point = (-1, 10)

46 Slide 46 / 222 Graphing Task: Graph y = 3x2 6x + 1 Step 5: Using the axis of symmetry, reflect the points to get the other half of the parabola. Connect with a smooth curve.

47 Slide 47 / 222 A B C D x=1 x = -1 x=2 x = -3 9 What is the axis of symmetry for y = x 2 + 2x - 3 (Step 1)?

48 Slide 48 / 222 A (-1, -4) B (1, -4) C (-1, -6) D (1, -6) 10 What is the vertex for y = x 2 + 2x - 3 (Step 2)?

49 Slide 49 / 222 A (0, -3) B (0, 3) 11 What is the y-intercept for y = x 2 + 2x - 3 (Step 3)?

50 Slide 50 / 222 Graph Practice: Graph

51 Slide 51 / 222 Graph Practice: Graph

52 Slide 52 / 222 Graph Practice: Graph

53 Slide 53 / 222 Solve Quadratic Equations by Graphing Return to Table of Contents

54 Slide 54 / 222 Solve by Graphing 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.

55 Slide 55 / 222 Solve by Graphing How many zeros do the parabolas have? What are the values of the zeros? No zeroes click 2 zeroes; x = -1 and x=3 click click 1 zero; x=1

56 Slide 56 / 222 Vocabulary 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. When writing a quadratic function as its related quadratic equation, you replace y with 0. So y = 0. y = ax2 + bx + c Quadratic Function 0 = ax2 + bx + c ax2 + bx + c = 0 Quadratic Equation

57 Slide 57 / 222 Solve by Graphing 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.

58 Slide 58 / 222 Solve by Graphing Step 1 - Write the Related Function 2x2-18 = 0 2x2-18 = y y = 2x2 + 0x - 18

59 Slide 59 / 222 Solve by Graphing Step 2 - Graph the Function 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 other 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. Hint y = 2x2 + 0x 18

60 Slide 60 / 222 Solve by Graphing Step 2 - Graph the Function y = 2x2 + 0x 18 x=0 (3,0) (2,-10) (0,-18)

61 Slide 61 / 222 Solve by Graphing Step 3 - Find the zeros y = 2x2 + 0x 18 x=0 The zeros appear to be 3 and -3. (3,0) (2,-10) (0,-18)

62 Slide 62 / 222 Solve by Graphing Step 3 - Find the zeros y = 2x2 + 0x 18 Substitute 3 and -3 for x in the quadratic equation. Check 2x2 18 = 0 2(3)2 18 = 0 2(9) - 18 = = 0 0=0 2(-3)2 18 = 0 2(9) - 18 = = 0 0=0 The zeros are 3 and -3.

63 Slide 63 / Solve the equation by graphing the related function and identifying the zeros. -12x + 18 = -2x2 A y = -2x2-12x + 18 B y = 2x2-12x - 18 C y = -2x2 + 12x - 18 Step 1: Which of these is the related function?

64 Slide 64 / What is the axis of symmetry? y = -2x2 + 12x - 18 x = -3 B x=3 C x=4 D x = -5 Formula: -b 2a A

65 Slide 65 / y = -2x2 + 12x - 18 A (3,0) B (-3,0) C (4,0) D (-5,0) What is the vertex?

66 Slide 66 / y = -2x2 + 12x - 18 A (0,0) B (0, 18) C (0, -18) D (0, 12) What is the y-intercept?

67 Slide 67 / If two other points are (5, -8) and (4, -2), what does the graph of y = -2x2 + 12x - 18 look like? B C D A

68 Slide 68 / 222 y = -2x2 + 12x - 18 Find the zero(s) A -18 B 4 C 3 D -8 17

69 Teacher Notes Slide 69 / 222 Solve Quadratic Equations by Factoring Return to Table of Contents

70 Slide 70 / 222 Solve by Factoring 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

71 Slide 71 / 222 Solve by Factoring 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.

72 Slide 72 / 222 Solve by Factoring 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)

73 Slide 73 / 222 Solve by Factoring Example To factor x2 + 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 Practice: To factor a quadratic trinomial of the form x2 + bx + c, find two factors of c whose sum is b.

74 Slide 74 / 222 Solve by Factoring Practice: Factors of -12 Sum 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.)

75 Slide 75 / 222 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?? x? = 0 Teacher Notes Zero Product Property

76 Slide 76 / 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

77 Slide 77 / 222 Solve by Factoring 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.

78 Slide 78 / 222 Example: Solve x = 12x Remember: The equation has to be written in standard form (ax2 + bx + c). Solve by Factoring

79 Slide 79 / 222 A x = -7 F x=3 B x = -5 G x=5 C x = -3 H x=6 D x = -2 I x=7 E x=2 J x = Solve

80 Slide 80 / 222 A m = -7 F m=3 B m = -5 G m=5 C m = -3 H m=6 D m = -2 I m=7 E m=2 J m = Solve

81 Slide 81 / 222 A h = -12 F h=3 B h = -4 G h=4 C h = -3 H h=6 D h = -2 I h=8 E h=2 J h = Solve

82 Slide 82 / 222 A d = -7 F d=3 B d = -5 G d=5 C d = -3 H d=6 D d = -2 I d=7 E d=0 J d = Solve

83 Slide 83 / 222 Berry Method to Factor Example: Solve When a does not equal 1, check first for a GCF, then use the Berry Method. 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.

84 Slide 84 / 222 Berry Method to Factor Solve Use the Berry Method. a = 8, b = 2, c = -3 Step 1 Step 2-4 and 6 are factors of -24 that add to +2 Step 3 Step 4 Discard common factors

85 Slide 85 / 222 Berry Method to Factor Solve Use the Zero Product Rule to solve.

86 Slide 86 / 222 Berry Method to Factor Solve Use the Berry Method. a = 4, b = -15, c = -25

87 Slide 87 / 222 Berry Method to Factor Solve Use the Zero Product Rule to solve.

88 Slide 88 / 222 Solve Berry Method to Factor

89 Slide 89 / 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

90 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

91 Slide 91 / 222 Hint: Two consecutive integers can be expressed as x and x + 1. Two consecutive even integers can be expressed as x and x The product of two consecutive even integers is 48. Find the smaller of the two integers.

92 Slide 92 / 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?

93 Slide 93 / 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 of 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 keystudents features and graph the function. type their answers here Problem is from: Click link for exact lesson.

94 Slide 94 / 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

95 Slide 95 / 222 Teacher Notes Solve Quadratic Equations Using Square Roots Return to Table of Contents

96 Slide 96 / 222 Solve Using Square Roots Consider the following quadratic: One option is to factor to solve for x. Because there is no "bx" term... another method to solve is using the square roots method. To solve, move the constant, "c" to the other side of the equation and take the square root of each side. IMPORTANT!! When taking the square root, you MUST consider both the positive and negative answer. Both 82 and (-8)2 equal 64.

97 Slide 97 / 222 Solve Using Square Roots You can solve a quadratic equation by the square root method if you can write it in the form: If x and c are algebraic expressions, then: This can also be written as:

98 Slide 98 / 222 Solve Using Square Roots What if x2 has a coefficient other than 1? Example: Solve 4x2 = 20 using the square roots method.

99 Slide 99 / When you take the square root of a real number, True False your answer will always be positive.

100 Slide 100 / 222 A 4 B 2 C -2 D 26 E If x2 = 16, then x =

101 Slide 101 / 222 using the square root method. A 5 E B 20 F -5 C 4 2 G -4 D -2 H Solve

102 Slide 102 / 222 A 4 B 2 C -2 D 26 E If y2 = 4, then y =

103 Slide 103 / 222 A B C D E 30 If 8j2 = 96, then j =

104 Slide 104 / If 4h2-10= 30, then h = B C D E A

105 Slide 105 / If (3g - 9)2 + 7= 43, then g = A C D E B

106 Slide 106 / 222 Solve Using Square Roots Challenge: Solve (2x - 1)² = 20 using the square root method.

107 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. Teacher Notes Slide 107 / 222

108 Slide 108 / A rock is dropped from a 1000 foot tower. The height of 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

109 Slide 109 / 222 Solving Quadratic Equations by Completing the Square Return to Table of Contents

110 Slide 110 / 222 Completing the Square 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.

111 Slide 111 / 222 Completing the Square Consider the following square. If its side length is (x + 3) then the area of the square would be? Notice all 3 expressions are equivalent. In the standard form, b = 6 and c = 9. That is, (b/2)2 = c.

112 Slide 112 / 222 Completing the Square Form a perfect square trinomial with lead coefficient of 1 Find the value that completes the square. x2 + bx +c where c = (b/2)2

113 Slide 113 / Find (b/2)2 if b = 14.

114 Slide 114 / Find (b/2)2 if b = 10.

115 Slide 115 / Find (b/2)2 if b = -12.

116 Slide 116 / Complete the square to form a perfect square trinomial. x2 + 18x +?

117 Slide 117 / Complete the square to form a perfect square trinomial. x2-6x +?

118 Slide 118 / 222 Completing the Square 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.

119 Slide 119 / 222 Completing the Square Let's look at an example to solve: x2 + 14x -15 = 0 Step 1 - Rewrite Equation 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? Step 2 - Find (b/2)2

120 Slide 120 / 222 Completing the Square Let's look at an example to solve: x2-2x - 2 = 0 Step 1 - Rewrite Equation 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? Step 2 - Find (b/2)2

121 Slide 121 / Solve the following by completing the square : A -5 B -2 C -1 D 5 E 2 x2 + 6x = -5

122 Slide 122 / Solve the following by completing the square : A -10 B -2 C -1 D 10 E 2 x2-8x = 20

123 Slide 123 / Solve the following by completing the square : A -6 B C 0 D 6 E Hint: Look for GCF -36x = 3x

124 Slide 124 / 222 Completing the Square Challenge: 3x2-10x = -3 Step 1 - Rewrite Equation *Note: There is no GCF to factor out like the previous example. 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 Step 2 - Find (b/2)2

125 Slide 125 / 222 Completing the Square Challenge: 4x2-17x + 4 = 0 Step 1 - Rewrite Equation *Note: There is no GCF to factor out. 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 Step 2 - Find (b/2)2

126 Slide 126 / 222 Completing the Square Challenge: *Note: There is no GCF to factor out. -6x2-25x - 25 = 0

127 Slide 127 / Solve the following by completing the square : B C D E A

128 Slide 128 / 222 Solve Quadratic Equations by Using the Quadratic Formula Return to Table of Contents

129 Slide 129 / 222 Solving Quadratics At this point you have learned how to solve quadratic equations by: graphing factoring using square roots and completing the square 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. Today we will be given a tool to solve ANY quadratic equation, and it ALWAYS works!

130 Slide 130 / 222 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 Teacher Notes Now try Completing the Square on the standard form of a quadratic equation.

131 Slide 131 / 222 Completing the Square Step 1 - Rewrite Equation and factor out a Steps 2 and 3 - Find (b/2)2, Add the result to both sides, simplify Step 4 - Factor & Simplify Step 5 - Take Square Root of both sides Step 6 - Solve for x

132 Slide 132 / 222 The solutions of ax2 + bx + c = 0, where a 0, are: Teacher Notes The Quadratic Formula

133 Slide 133 / 222 Quadratic Formula 2 Example 1: Solve 2x + 3x - 5 = 0 a = 2 b = 3 c = -5 How can you check your answers? Once you identify the values of a, b, and c, simply substitute into the quadratic formula and simplify as much as possible.

134 Slide 134 / 222 Quadratic Formula Example 2: Solve: 2x = x2-3 Identify a, b, and c, then substitue into the formula and simplify. Don't forget to check your results! To use the Quadratic Formula, the equation must be in standard form (ax2 + bx +c = 0).

135 Slide 135 / Solve the following equation using the quadratic formula: A -5 F 1 B -4 G 2 C -3 H 3 D -2 I 4 E -1 J 5

136 Slide 136 / 222 A -5 F 1 B -4 G 2 C -3 H 3 D -2 I 4 E -1 J 5 45 Solve the following equation using the quadratic formula:

137 Slide 137 / 222 A -5 F 1.5 B -4 G 2 C -3 H 3 D -2 I 4 E -1.5 J 5 46 Solve the following equation using the quadratic formula:

138 Slide 138 / 222 Quadratic Formula Solve using the quadratic formula, and simplify the result. x2-2x - 4 = 0 Example 3:

139 Slide 139 / Find the larger solution to the equation.

140 Slide 140 / Find the smaller solution to the equation.

141 Slide 141 / 222 Which Method Factoring Quadratic Formula Completing the Square Graphing Teacher Notes Work in small groups to solve the quadratic equation using the following different methods.

142 Slide 142 / 222 Which Method Factoring Quadratic Formula Completing the Square Graphing Teacher Notes Work in small groups to solve the quadratic equation using the following different methods.

143 Slide 143 / 222 The Discriminant Return to Table of Contents

144 Slide 144 / 222 Solutions Recall what it means to have 0, 1, or 2 solutions/zeros/roots 2 real solutions no real solutions 1 real solution

145 Slide 145 / 222 The Discriminant 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.

146 Slide 146 / 222 The Discriminant Discriminant: the radicand in the Quadratic Formula (the piece of the equation under the radical sign). Note, it does NOT include the radical sign. Quadratic Formula: Discriminant:

147 Slide 147 / 222 The Discriminant 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.

148 Slide 148 / 222 The Discriminant Teacher Notes Examine the following graphs and determine the relationship between the discriminant of a quadratic and its graph. Discriminant Discriminant Discriminant

149 Slide 149 / 222 The Discriminant CONCLUSION: 2 reveal click If bto - 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

150 Slide 150 / What is the value of the discriminant of 2x2-2x + 3 = 0?

151 Slide 151 / 222 A 0 B 1 C 2 50 Use the discriminant to find the number of solutions for 2x2-2x + 3 = 0

152 Slide 152 / What is the value of the discriminant of x2-8x + 4 = 0?

153 Slide 153 / 222 A 0 B 1 C 2 52 Use the discriminant to find the number of solutions for x2-8x + 4 = 0

154 Slide 154 / 222 Vertex Form Return to Table of Contents

155 Slide 155 / 222 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. A quadratic equation in vertex form:

156 Slide 156 / 222 Vertex Form 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:

157 Slide 157 / 222 Vertex Form Find the vertex, direction of openness and the axis of symmetry for each. B. C. A.

158 Slide 158 / 222 Vertex Form Find the vertex, direction of openness and the axis of symmetry for each. E. D.

159 Slide 159 / Find the vertex for the graph of A B D C

160 Slide 160 / 222 A up B down C left D right 54 Find the direction of opening for the graph of

161 Slide 161 / Find the axis of symmetry for the graph of

162 Slide 162 / Find the vertex: A C D B

163 Slide 163 / 222 A up B down C left D right 57 Give the direction of opening for the graph of

164 Slide 164 / Give the axis of symmetry for the graph of

165 Slide 165 / Identify the vertex of B C D A

166 Slide 166 / 222 A up B down C left D right 60 Give the direction of openness of

167 Slide 167 / The axis of symmetry for the graph of is.

168 Slide 168 / Find the vertex of A C D B

169 Slide 169 / 222 A up B down C left D right 63 Find the direction of opening for

170 Slide 170 / Find the axis of symmetry for the graph of

171 Slide 171 / 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.

172 Slide 172 / 222 Standard Form to Vertex Form Example: Rewrite the following quadratic in vertex form, and name the vertex.

173 Slide 173 / 222 Standard Form to Vertex Form Practice: Rewrite the following quadratic in vertex form, and name the vertex.

174 Slide 174 / What is the vertex form of: A B D C

175 Slide 175 / What is the vertex form of: A C D B

176 Slide 176 / What is the vertex form of: A C D B

177 Slide 177 / 222 Comparing Functions 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.

178 Slide 178 / 222 Two Functions Write two different quadratic equations whose graphs have vertices at (3.5, -7).

179 Slide 179 / 222 Standard Form to Vertex Form 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.

180 Slide 180 / 222 Standard Form to Vertex Form Rewrite the following quadratic in vertex form, and name the vertex.

181 Slide 181 / 222 Standard Form to Vertex Form Rewrite the following quadratic in vertex form, and name the vertex.

182 Slide 182 / What is the vertex form of: A B D C

183 Slide 183 / What is the vertex form of: A C D B

184 Slide 184 / What is the vertex form of: A C D B

185 Slide 185 / 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 of the parabola is the "turning point" and is on the axis of symmetry.

186 Slide 186 / 222 Focus and Directrix of a Parabola Every point on the parabola is the same distance from the directrix and the focus. L1=L2 L1 L2 Focus Directrix Axis of Symmetry

187 Slide 187 / 222 Parabolas The parts are the same for all parabolas,regardless of the direction in which they open. y=ax2+bx+c x=ay2+by+c Axis of Symmetry Focus Focus Vertex Directrix Axis of Symmetry Vertex Directrix

188 Slide 188 / 222 Parabolas Parabolas can open up or down (vertical parabolas), OR left or right (horizontal parabolas). (Horizontal parabolas are not functions.) Parabolas that open left or right have the general equation and the vertex form of where (h,k) is the vertex.

189 Slide 189 / 222 Parabola Summary General Form y= ax2 + bx + c x= ay2 +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 x=h y=k (h, k) (h, k) Axis of Symmetry Vertex Directrix Focus

190 Slide 190 / 222 What is the vertex of A (-3, 2) B (-3, -2) C (2, 3) D (-2, -3)? 71

191 Slide 191 / 222 What is the vertex of A (3, 2) B (-3, -2) C (2, 3) D (-2, -3)? 72

192 Slide 192 / 222 What is the vertex of A (3, 2) B (-3, -2) C (2, 3) D (2, -3)? 73

193 Slide 193 / 222 Converting from General Form to Vertex Form

194 Slide 194 / 222 What value completes the square?? 74

195 Slide 195 / 222 What value should be added to balance the equation? 16? 75

196 Slide 196 / 222 What is the y-coordinate of the vertex? 76

197 Slide 197 / 222 What is the x-coordinate of the vertex? 77

198 Slide 198 / 222 What is the vertex of y= x2-8x +21? A (4, 5) B (-4, 5) C (-5, 4) D (5, 4) 78

199 Slide 199 / 222 What is the equation of the directrix for the following equation? A y=2 B y = -4 C x=3 D x = -5 79

200 Slide 200 / 222 Where is the focus for the following equation? A (-3, 5) B (3, 5) C (5, 3) D (5, -3) 80

201 Slide 201 / 222 What is the equation of the directrix for the following equation? A y = -2.5 B y = -1.5 C x = -3.5 D x =

202 Slide 202 / What is the equation of the parabola with vertex (2,3) and directrix y = 4? A y = 4(x - 2)2 + 3 C x = 4(y - 2)2 + 3 D x = 1/4(y - 2)2 + 3 B y = -1/4(x - 2)2 + 3

203 Slide 203 / 222 More Application Problems Using Quadratics Return to Table of Contents

204 Slide 204 / 222 Quadratic Functions in the Real World Click on the bike to learn more.

205 Slide 205 / 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 Business:Interest Rate Height of a Projectile Distances

206 Slide 206 / 222 Quadratic Equations and Applications 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.

207 Slide 207 / 222 Applications The product of two consecutive negative integers is What are the numbers?

208 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?

209 Slide 209 / 222 Applications The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers.

210 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

211 Slide 211 / The product of a number and a number 3 more than the original is 418. What is the smallest value the original number can be?

212 Slide 212 / The product of two consecutive positive even integers is 168. Find the larger of the numbers.

213 85 Two cars left an intersection at the same time, one 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

214 Slide 214 / 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.

215 Slide 215 / In the accompanying diagram, the width of the inner Application Problemsis represented by x - 3 and its length by rectangle 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 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. Problem is from: Click link for exact lesson. Use the the next page for space to solve.

216 Slide 216 / 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.

217 Slide 217 / 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

218 Slide 218 / 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? x x x x

219 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 of 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 Note: This question has two correct approaches and two different correct solutions. Can you find them both? Hint: Start by factoring. Problem is from: Click link for exact lesson. Slide 219 / 222

220 Slide 220 / Part A Input your answer for h = From PARCC sample test 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.

221 Slide 221 / Part A Input your answer for k = From PARCC sample test 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.

222 Slide 222 / 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