1. Slide 9- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Quadratic Equations 9.1Introduction to Quadratic Equations 9.2Solving Quadratic Equations by Completing the Square 9.3The Quadratic Formula 9.4Formulas 9.5Applications and Problem Solving 9.6Graphs of Quadratic Functions 9.7 Functions 9

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley INTRODUCTION TO QUADRATIC EQUATIONS Write a quadratic equation in standard form ax 2 + bx + c = 0, a > 0, and determine the coefficients a, b, and c. Solve quadratic equations of the type ax 2 + bx = 0, where b  0, by factoring. Solve quadratic equations of the type ax 2 + bx + c = 0, where b  0 and where c  0, by factoring. Solve applied problems involving quadratic equations. 9.1a b c d

4. Slide 9- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Write a quadratic equation in standard form ax 2 + bx + c = 0, a > 0, and determine the coefficients a, b, and c.a

5. Slide 9- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The following are quadratic equations. They contain polynomials of second degree. 4x 2 + 7x – 5 = 0 3y 2 – y = 9 5a 2 = 8a12m 2 = 144 Quadratic Equation A quadratic equation is an equation equivalent to an equation of the type ax 2 + bx + c = 0, a > 0, where a, b, and c are real-number constants. We say that the preceding is the standard form on a quadratic equation.

6. Slide 9- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Write in standard form and determine a, b, and c. a. 5x 2 + 8x – 3 = 0 The equation is in standard form. 5x 2 + 8x – 3 = 0 a = 5; b = 8; c = –3 b.6y 2 = 5y 6y 2 – 5y = 0 a = 6; b = –5; c = 0

7. Slide 9- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations of the type ax 2 + bx = 0, where b  0, by factoring.b

8. Slide 9- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Solve: 8x 2 + 3x = 0. Solution 8x 2 + 3x = 0 x(8x + 3) = 0 Factoring x = 0 or 8x + 3 = 0 Using the principle of zero products x = 0 or 8x = –3 x = 0 or Check: 8x 2 + 3x = 08x 2 + 3x = 0 8(0) 2 + 3(0) = 0 0 = 0 True The solutions both check.

9. Slide 9- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A quadratic equation of the type ax 2 + bx = 0, where c = 0 and b  0, will always have 0 as one solution and a nonzero number as the other solution.

10. Slide 9- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations of the type ax 2 + bx + c = 0, where b  0 and where c  0, by factoring.c

11. Slide 9- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solve: (y – 7)(y – 2) = 4y – 22 Solution Write the equation in standard form and then try factoring. (y – 7)(y – 2) = 4y – 22 y 2 – 9y + 14 = 4y – 22 Multiplying y 2 – 13y + 36 = 0 Standard form (y – 4)(y – 9) = 0 y – 4 = 0 or y – 9 = 0 y = 4 or y = 9 The solutions are 4 and 9.

12. Slide 9- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems involving quadratic equations.d

13. Slide 9- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D The number of diagonals d in a polygon that has n sides is given by the formula If a polygon has 54 diagonals, how many sides does it have? 1. Familarize. A sketch can help us to become familiar with the problem. We draw a hexagon (6 sides) and count the diagonals. As the formula predicts, for n = 6, there are

14. Slide 9- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 2. Translate. Since the number of diagonals is 54, we substitute 54 for d: 3. Solve. We solve the equation for n, first reversing the equation for convenience.

15. Slide 9- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 4. Check. Since the number of sides cannot be negative,  9 cannot be a solution. 5. State. The polygon has 12 sides.

16. Slide 9- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.1 1. Solve 5x 2 = 30. a)  6 b)  5 c) d)

17. Slide 9- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.1 1. Solve 5x 2 = 30. a)  6 b)  5 c) d)

18. Slide 9- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.1 2. Solve x 2 = 2x + 24. a)  3, 8 b)  4, 8 c)  4, 6 d)  6, 4

19. Slide 9- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.1 2. Solve x 2 = 2x + 24. a)  3, 8 b)  4, 8 c)  4, 6 d)  6, 4

20. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE Solve quadratic equations of the type ax 2 = p. Solve quadratic equations of the type (x + c) 2 = d. Solve quadratic equations by completing the square. Solve certain problems involving quadratic equations of the type ax 2 = p. 9.2a b c d

21. Slide 9- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations of the type ax 2 = p.a

22. Slide 9- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Principle of Square Roots The equation x 2 = d has two real solutions when d > 0. The solutions are The equation x 2 = d has no real-number solution when d < 0. The equation x 2 = 0 has 0 as its only solution.

23. Slide 9- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Solve: x 2 = 25 Solution We use the principle of square roots: x 2 = 25 x = 5 or x =  5 We check mentally that 5 2 = 25 and (  5) 2 = 25. The solutions are 5 and  5.

24. Slide 9- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Solve: a) x 2 = 19 b) 9x 2 = 27 c) 5x 2  60 = 0 Solution a) Check: x 2 = 19 x 2 = 19 19 19 19 = 19 19 = 19 The solutions are

25. Slide 9- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solve b) 9x 2 = 27 c) 5x 2  60 = 0 continued Solution b) The check is left to the student. c)

26. Slide 9- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations of the type (x + c) 2 = d.b

27. Slide 9- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solve: a) (x  3) 3 = 16b) (x + 3) 2 = 5 Solution a) x  3 = 4 or x  3 =  4 x = 7 or x =  1 The solutions are 7 and  1. We leave the check to the student.

28. Slide 9- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued b) (x + 3) 2 = 5 The solutions check and can be written as (read as “negative three plus or minus the square root of five”).

29. Slide 9- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Solve: x 2 + 8x + 16 = 17 Solution x 2 + 8x + 16 = 17 (x + 4) 2 = 17 Sometimes we can factor an equation to express it as a square of a binomial.

30. Slide 9- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations by completing the square.c

31. Slide 9- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Completing the Square To complete the square for an expression like x 2 + bx, we take half of the coefficient of x and square it. Then we add that number, which is (b/2) 2.

32. Slide 9- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Complete the square: a) x 2  16x Solution a) The coefficient of x is  16. Half of  16 is  8 and (  8) 2 = 64 Thus, x 2  16 becomes a perfect-square trinomial when 64 is added: x 2  16x + 64 is the square of x  8 The number 64 completes the square. Check: (x  8) 2 = (x  8)(x  8) = x 2  8x  8x + 64 = x 2  16x + 64

33. Slide 9- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Solve by completing the square. x 2  8x =  15 Solution Take half of  8 and square it to get 16. We add 16 to both sides of the equation. x 2  8x + 16 =  15 + 16 (x  4) 2 = 1 x  4 = 1 or x  4 =  1 x = 5 or x = 3

34. Slide 9- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example G Solve by completing the square. x 2  14x  7 = 0 Solution We have x 2  14x  7 = 0 x 2  14x = 7 x 2  14x + 49 = 7 + 49 (x  7) 2 = 56

35. Slide 9- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example H Solve by completing the square. 3x 2 + 7x + 1 = 0 Solution The coefficient of the x 2 term must be 1. When it is not, we must multiply or divide on both sides to find an equivalent equation with an x 2 coefficient of 1.

36. Slide 9- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued

37. Slide 9- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving by Completing the Square To solve a quadratic equation ax 2 + bx + c = 0 by completing the square: 1. If a ≠ 1, multiply by 1/a so that the x 2 –coefficient is 1. 2. If the x 2 –coefficient is 1, add so that the equation is in the form x 2 + bx = –c, or if step (1) has been applied. 3. Take half of the x-coefficient and square it. Add the result on both sides of the equation. 4. Express the side with the variables as the square of a binomial. 5. Use the principle of square roots and complete the solution.

38. Slide 9- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve certain problems involving quadratic equations of the type ax 2 = p.d

39. Slide 9- 39 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example I The Taipei 101 tower in Taiwan is 1670 feet tall. How long would it take an object to fall to the ground from the top? 1. Familiarize. A formula that fits this situation is s = 16t 2, where s is the distance, in feet, traveled by a body falling freely from rest in t seconds. We know that s is 1670 feet. 2. Translate. We know the distance is 1670 feet and that we need to solve for t. 1670 = 16t 2

40. Slide 9- 40 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 3. Solve. 1670 = 16t 2 4. Check. The number  10.2 cannot be a solution because time cannot be negative. s = 16(10.2) 2 = 16(104.04) = 1664.64 This answer is close. 5. State. It takes about 10.2 seconds for an object to fall to the ground from the top of Taipei 101.

41. Slide 9- 41 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.2 1. Solve: (x + 2) 2 = 7 a) b) c) d)

42. Slide 9- 42 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.2 1. Solve: (x + 2) 2 = 7 a) b) c) d)

43. Slide 9- 43 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.2 2. Solve: x 2 – 8x = –16 a) 8 b)  2 c) 4 d)  4

44. Slide 9- 44 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.2 2. Solve: x 2 – 8x = –16 a) 8 b)  2 c) 4 d)  4

45. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley THE QUADRATIC FORMULA Solve quadratic equations using the quadratic formula. Find approximate solutions of quadratic equations using a calculator. 9.3a b

46. Slide 9- 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve quadratic equations using the quadratic formula.a

47. Slide 9- 47 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Quadratic Formula The solutions of ax 2 + bx + c = 0 are given by

48. Slide 9- 48 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Solve using the quadratic formula. 2x 2 + 9x  5 = 0 Solution We identify a, b, and c and substitute into the quadratic formula: 2x 2 + 9x  5 = 0 a = 2, b = 9, c = –5

49. Slide 9- 49 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued a = 2, b = 9, c =  5 Be sure to write the fraction bar all the way across.

50. Slide 9- 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued

51. Slide 9- 51 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Quadratic Equations To solve a quadratic equation: 1. Check to see if it is in the form ax 2 = p or (x + c) 2 = d. If it is, use the principle of square roots as in Section 9.2 2. If it is not in the form of (1), write it in standard form. ax 2 + bx + c = 0 with a and b nonzero. 3. Then try factoring. 4. If it is not possible to factor or if factoring seems difficult, use the quadratic formula. The solutions of a quadratic equation can always be found using the quadratic formula. They cannot always be found by factoring. (When the radicand b 2 – 4ac  0, the equation has real-number solutions. When b 2 – 4ac < 0, the equation has no real-number solutions.

52. Slide 9- 52 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Solve: w 2 =  12w + 4 Solution We write w 2 =  12w + 4 in standard form, identify a, b, and c, and solve using the quadratic formula: 1w 2 + 12w  4 = 0 a = 1, b = 12, c = –4

53. Slide 9- 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solve: 5x 2  x + 3 = 0 Solution a = 5, b =  1, c = 3 Since the radicand,  59 is negative, there are no real-number solutions.

54. Slide 9- 54 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find approximate solutions of quadratic equations using a calculator.b

55. Slide 9- 55 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Use a calculator to approximate to the nearest tenth the solution to the expression Solution Using a calculator we have:

56. Slide 9- 56 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 1. Solve x 2 – 8x + 6 = 0. a) b) c) d)

57. Slide 9- 57 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 1. Solve x 2 – 8x + 6 = 0. a) b) c) d)

58. Slide 9- 58 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 2. Solve 2x 2 – 7x + 2 = 0. a) b) c) d)

59. Slide 9- 59 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 2. Solve 2x 2 – 7x + 2 = 0. a) b) c) d)

60. Slide 9- 60 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 3. Approximate the solutions of x 2 – 9x + 4 = 0. a)  4.24, 9.424 b) 4.969, 13.031 c) 3.795, 5.207 d) 0.469, 8.531

61. Slide 9- 61 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.3 3. Approximate the solutions of x 2 – 9x + 4 = 0. a)  4.24, 9.424 b) 4.969, 13.031 c) 3.795, 5.207 d) 0.469, 8.531

62. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FORMULAS Solve a formula for a given letter. 9.4a

63. Slide 9- 63 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve a formula for a given letter.a

64. Slide 9- 64 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Formulas Formulas can be linear, rational, radical, or quadratic equations, as well as other types of equations. To solve formulas, we use the same steps that we use to solve equations. Probably the greatest difference is that while the solution of an equation is a number, the solution of a formula is generally a variable expression.

65. Slide 9- 65 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Solve for w: y = bw – cw + 3. Solution y = bw – cw + 3 We want this letter alone on one side. y – 3 = bw – cw Subtracting 3 y – 3 = w(b – c) Collecting like terms Dividing by b – c on both sides

66. Slide 9- 66 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Solve: Solution Multiply each side by the LCD Factor out the GCF

67. Slide 9- 67 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Solve: Solution Square both sides Solve for y.

68. Slide 9- 68 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.4 1. Solve for d. a) b) c) d)

69. Slide 9- 69 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.4 1. Solve for d. a) b) c) d)

70. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley APPLICATIONS AND PROBLEM SOLVING Solve applied problems using quadratic equations. 9.5a

71. Slide 9- 71 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Solve applied problems using quadratic equations.a

72. Slide 9- 72 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A The park board wants to increase the size of the children’s play area. The addition will be in the shape of a rectangle and have an area of 270 square feet. The width of the park is to be 3 feet less than the length. Find the dimensions of the addition. Solution 1. Familarize. We first make a drawing and label it. We let l = length of one side and l – 3 = the width. Area is the length times the width 2. Translate. A = lw 270 = l(l – 3) 270 sq. ft l l  3

73. Slide 9- 73 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 3. Solve. 270 = l(l – 3) 270 = l 2 – 3l 0 = l 2 – 3l – 270 a = 1, b =  3, c =  270

74. Slide 9- 74 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 4. Check. Length cannot be negative, so  15 does not check. A = lw = (18)(15) = 270 5. State. The length is 18 feet and the width is 15 feet.

75. Slide 9- 75 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B The hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs. Solution 1. Familarize. We first make a drawing and label it. We let s = length, in yards, of one leg. Then s + 28 = the length, in yards, of the other leg. 2. Translate. We use the Pythagorean theorem: s 2 + (s + 28) 2 = 52 2 s + 28 s 52

76. Slide 9- 76 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 3. Solve. s 2 + (s + 28) 2 = 52 2 s 2 + s 2 + 56s + 784 = 2704 2s 2 + 56s  1920 = 0 s 2 + 28s  960 = 0 a = 1, b = 28, c =  960

77. Slide 9- 77 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued 4. Check. Length cannot be negative, so  48 does not check. 5. State. One leg is 20 yards and the other leg is 48 yards.

78. Slide 9- 78 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.5 1. The hypotenuse of a right triangle is 51 feet long. One leg is 21 feet shorter than the other. Find the length of the shorter leg. a) 8 feet b) 15 feet c) 24 feet d) 45 feet

79. Slide 9- 79 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.5 1. The hypotenuse of a right triangle is 51 feet long. One leg is 21 feet shorter than the other. Find the length of the shorter leg. a) 8 feet b) 15 feet c) 24 feet d) 45 feet

80. Slide 9- 80 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.5 2. The width of a rectangle is 4 m less than the length. The area is 357 m 2. Find the width. a) 7 m b) 17 m c) 21 m d) 87.25 m

81. Slide 9- 81 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.5 2. The width of a rectangle is 4 m less than the length. The area is 357 m 2. Find the width. a) 7 m b) 17 m c) 21 m d) 87.25 m

82. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley GRAPHS OF QUADRATIC EQUATIONS Graph quadratic equations. Find the x-intercepts of a quadratic equation. 9.6a b

83. Slide 9- 83 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Graph quadratic equations.a

84. Slide 9- 84 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations of the form y = ax 2 + bx + c with a ≠ 0, have graphs that are either cupped upward or downward. These graphs are symmetric with respect to an line of symmetry, as shown below. When folded along its axis, the graph has two halves that match exactly. The graphs of quadratic equations are called parabolas. The point at which the equation crosses its axis of symmetry is called the vertex. Graphs of Quadratic Equations Line of symmetry

85. Slide 9- 85 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Graph y = 2x 2 Solution Complete a table of values. After several ordered pairs are found, we plot them and connect them with a smooth curve. (2, 8)82 (1, 2)21 (0, 0)00 (  1, 2) 2 11 (  2, 8) 8 22 (x, y) y = 2x 2 x y The vertex is (0, 0) and the axis of symmetry is the y-axis.

86. Slide 9- 86 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example B Graph Solution We select numbers for x, find the corresponding y values, plot the resulting ordered pairs, and connect them with a smooth curve. (4,  4) 44 4 (2,  1)  1 2 (0, 0)00 (  2,  1) 11 22 (  4,  4) 44 44 (x, y)x y

87. Slide 9- 87 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Finding the Vertex For a parabola given by the quadratic equation y = ax 2 + bx + c 1. The x-coordinate of the vertex is The line of symmetry is x = –b/(2a). 2. The second coordinate of the vertex is found by substituting the x-coordinate into the equation and computing y.

88. Slide 9- 88 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C Graph: x 2  6x + 1 Solution We want to plot the vertex and some other points on either side of the vertex. We substitute 3 for x to find the y-coordinate of the vertex: (3) 2  6(3) + 1 = 9  18 + 1 =  8 The vertex is (3,  8). The axis of symmetry is x = 3.

89. Slide 9- 89 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued x y (x, y) x 2  6x + 1 01(0, 1) 1 44(1,  4) 2 77(2,  7) 3 88(3,  8) 4 77(4,  7) 5 44(5,  4) 61(6, 1)

90. Slide 9- 90 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Tips for Graphing Quadratic Equations 1. Graphs of quadratic equations y = ax 2 + bx + c are all parabolas. They are smooth cup-shaped symmetric curves, with no sharp points or kinks in them. 2. Find the vertex and the line of symmetry. 3. The graph of y = ax 2 + bx + c opens up if a > 0. It opens down if a < 0. 4. Find the y-intercept. It occurs when x = 0, and it is easy to compute.

91. Slide 9- 91 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Graph: 2x 2 + 4x + 1 Solution We want to plot the vertex and some other points on either side of the vertex. We substitute  1 for x to find the y-coordinate of the vertex: 2(  1) 2 + 4(  1) + 1 = 2  4 + 1 =  1 The vertex is (  1,  1). The axis of symmetry is x =  1.

92. Slide 9- 92 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued x y (x, y) 2x 2 + 4x + 1 33 7 (  3, 7) 22 1 (  2, 1) 11 11(  1,  1) 01(0, 1) 17(1, 7) Vertex

93. Slide 9- 93 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Find the x-intercepts of a quadratic equation.b

94. Slide 9- 94 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Find all y- and x-intercepts of the graph y = x 2 + 2x  8 Solution To find the y-intercept, we replace x with 0 and solve for y: y = 0 2 + 2(0)  8 y =  8 The y-intercept is (0,  8). To find the x-intercept(s), we replace y with 0 and solve for x: 0 = (x + 4)(x  2) x + 4 = 0 or x  2 = 0 x =  4 or x = 2

95. Slide 9- 95 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued y-intercept (0,  8) x-intercepts  4 and 2 x-intercept y-intercept x-intercept

96. Slide 9- 96 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.6 1. Graph y = – x 2 – x + 6 a)b) c)d)

97. Slide 9- 97 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.6 1. Graph y = – x 2 – x + 6 a)b) c)d)

98. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTIONS Determine whether a correspondence is a function. Given a function described by an equation, find function values (outputs) for specified values (inputs). Draw a graph of a function. Determine whether a graph is that of a function. Solve applied problems involving functions and their graphs. 9.7a b c d e

99. Slide 9- 99 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether a correspondence is a function.a

100. Slide 9- 100 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Functions appear regularly in magazines and newspapers although they are not usually referred to as such. Ordered pairs form a correspondence from one set of numbers to another. For example: Each student in a college, there corresponds his or her student ID number. To each item in a store, there corresponds its price. The first set is called the domain and the second set is called the range. Given a member of the domain, there is just one member of the range to which it corresponds. This kind of correspondence is called a function.

101. Slide 9- 101 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example A Determine whether or not the correspondence is a function. a) Domain Range x 5 y 7 z 9 Solution This is a function because each member of the domain is matched to just one member of the range.

102. Slide 9- 102 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley continued b) Domain Range 27 4 6969 811 c) Domain Range New York Giants Jets ChicagoBears MiamiDolphins This is a function because each member of the domain is matched to just one member of the range. This is NOT a function because one member of the domain is matched to more than one member of the range.

103. Slide 9- 103 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function, Domain, and Range A function is a correspondence between a first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to exactly one member of the range.

104. Slide 9- 104 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relation A relation is a correspondence between a first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to at least one member of the range.

105. Slide 9- 105 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Given a function described by an equation, find function values (outputs) for specified values (inputs).b

106. Slide 9- 106 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For the function given by f(x) = x + 3, find each of the following. a) f(9)b) f(  4)c) f(0) Solution a) f(9) = 9 + 3, or 12 b) f(  4) =  4 + 3, or  1 c) f(0) = 0 + 3, or 3 Example B

107. Slide 9- 107 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example C For the function f(y) = 3y 2 + y, find each of the following. a) y(2)b) y(  3) Solution a) f(2) = 3  2 2 + 2 = 3  4 + 2 = 14 b) f(  3) = 3(  3) 2 + (  3) = 3  9  3 = 24

108. Slide 9- 108 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Draw a graph of a function.c

109. Slide 9- 109 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Functions To graph a function, we usually calculate ordered pairs of the form (x, y) or (x, f(x)), plot them, and connect the points. The symbols y and f(x) are often used interchangeably when working with functions and their graphs.

110. Slide 9- 110 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example D Graph f(x) = 3  x 2 Solution xf(x)f(x) 22 11 11 2 03 12 2 11

111. Slide 9- 111 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example E Graph f(x) = |x| + 1 Solution xf(x)f(x) 33 4 22 3 11 2 01 12 23 34

112. Slide 9- 112 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Objective Determine whether a graph is that of a function.d

113. Slide 9- 113 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Vertical-Line Test A graph represents a function if it is impossible to draw a vertical line that intersects the graph more than once.

114. Slide 9- 114 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example F Determine whether each of the following is a graph of a function. a) The graph is that of a function because it is not possible to draw a vertical line that crosses the graph more than once. b) The graph is NOT that of a function because it is possible to draw a vertical line that crosses the graph more than once.

115. Slide 9- 115 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley a) The graph is NOT that of a function because it is possible to draw a vertical line that crosses the graph more than once. Example G Determine whether each of the following is a graph of a function. b) The graph is that of a function because it is not possible to draw a vertical line that crosses the graph more than once.

116. Slide 9- 116 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.7 1. Graph h(x) = – |x| a)b) c)d)

117. Slide 9- 117 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.7 1. Graph h(x) = – |x| a)b) c)d)

118. Slide 9- 118 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.7 2. Which of the following is the graph of a function? a)b) c)d)

119. Slide 9- 119 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Section 9.7 2. Which of the following is the graph of a function? a)b) c)d)