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Holt McDougal Algebra 2 2-6 The Quadratic Formula Warm Up Write each function in standard form. Evaluate b 2 – 4ac for the given values of the valuables. 1. f(x) = (x – 4) 2 + 3 2. g(x) = 2(x + 6) 2 – 11 f(x) = x 2 – 8x + 19 g(x) = 2x 2 + 24x + 61 4. a = 1, b = 3, c = –3 3. a = 2, b = 7, c = 5 9 21
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Solve quadratic equations using the Quadratic Formula. Classify roots using the discriminant. Objectives
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Holt McDougal Algebra 2 2-6 The Quadratic Formula
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Holt McDougal Algebra 2 2-6 The Quadratic Formula The symmetry of a quadratic function is evident in the last step,. These two zeros are the same distance,, away from the axis of symmetry,,with one zero on either side of the vertex.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Find the zeros of f(x)= 2x 2 – 16x + 27 using the Quadratic Formula. Example 1: Quadratic Functions with Real Zeros Set f(x) = 0. Write the Quadratic Formula. Substitute 2 for a, –16 for b, and 27 for c. Simplify. Write in simplest form. 2x 2 – 16x + 27 = 0
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Find the zeros of f(x) = 4x 2 + 3x + 2 using the Quadratic Formula. Example 2: Quadratic Functions with Complex Zeros Set f(x) = 0. Write the Quadratic Formula. Substitute 4 for a, 3 for b, and 2 for c. Simplify. Write in terms of i. f(x)= 4x 2 + 3x + 2
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Holt McDougal Algebra 2 2-6 The Quadratic Formula The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Find the type and number of solutions for the equation. Example 3A: Analyzing Quadratic Equations by Using the Discriminant x 2 + 36 = 12x x 2 – 12x + 36 = 0 b 2 – 4ac (–12) 2 – 4(1)(36) 144 – 144 = 0 b 2 – 4ac = 0 The equation has one distinct real solution.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Find the type and number of solutions for the equation. Example 3B: Analyzing Quadratic Equations by Using the Discriminant x 2 + 40 = 12x x 2 – 12x + 40 = 0 b 2 – 4ac (–12) 2 – 4(1)(40) 144 – 160 = –16 b 2 –4ac < 0 The equation has two distinct nonreal complex solutions.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Find the type and number of solutions for the equation. Example 3C: Analyzing Quadratic Equations by Using the Discriminant x 2 + 30 = 12x x 2 – 12x + 30 = 0 b 2 – 4ac (–12) 2 – 4(1)(30) 144 – 120 = 24 b 2 – 4ac > 0 The equation has two distinct real solutions.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula The graph shows related functions. Notice that the number of real solutions for the equation can be changed by changing the value of the constant c.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula An athlete on a track team throws a shot put. The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t 2 + 24.6t + 6.5. The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t. To the nearest foot, how far does the shot put land from the athlete? Example 4: Sports Application
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Step 1 Use the first equation to determine how long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t. y = –16t 2 + 24.6t + 6.5 Example 4 Continued Set y equal to 0. 0 = –16t 2 + 24.6t + 6.5 Use the Quadratic Formula. Substitute –16 for a, 24.6 for b, and 6.5 for c.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Example 4 Continued Simplify. The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Step 2 Find the horizontal distance that the shot put will have traveled in this time. x = 29.3t Example 4 Continued Substitute 1.77 for t. Simplify. x ≈ 29.3(1.77) x ≈ 51.86 x ≈ 52 The shot put will have traveled a horizontal distance of about 52 feet.
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Check Use substitution to check that the shot put hits the ground after about 1.77 seconds. Example 4 Continued The height is approximately equal to 0 when t = 1.77. y = –16t 2 + 24.6t + 6.5 y ≈ –16(1.77) 2 + 24.6(1.77) + 6.5 y ≈ –50.13 + 43.54 + 6.5 y ≈ –0.09
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Properties of Solving Quadratic Equations
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Properties of Solving Quadratic Equations
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Group Work Group work problems
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Holt McDougal Algebra 2 2-6 The Quadratic Formula Homework Section 2-6 in the workbook Workbook page 81: 1 - 10
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