3. Lesson 1 Contents Example 1Graph a Quadratic Function Example 2Axis of Symmetry, y-Intercept, and Vertex Example 3Maximum or Minimum Value Example 4Find a Maximum Value

4. Quadratic Functions Quadratic Functions A quadratic function has the form f (x) = ax 2 + bx + c where a, b and c are real numbers and a is not equal to 0. The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downward and vary in "width" or "steepness", but they all have the same basic "U" shape.

5. Axis of Symmetry All parabolas are symmetric with respect to a line called the axis of symmetry. A parabola intersects its axis of symmetry at a point called the vertex of the parabola (think of this point as the point where the curve begins to turn around.) The equation of the axis of symmetry is The x-coordinate of the vertex is.

6. More on Quadratic Functions The domain of this function is the set of all real numbers. The y intercept of the graph of f is given by f(0) = c. The x intercepts are found by solving the equation ax 2 + bx + c = 0

7. Example 1-1a First, choose integer values for x. Then, evaluate the function for each x value. Graph the resulting coordinate pairs and connect the points with a smooth curve. Graph by making a table of values. Answer: (1, 3) 31 (0, –1) –10 (–1, –3) –3–1 (–2, –3) –3–2 (–3, –1) –1–3 (x, y) f (x)x

8. Example 1-1a

9. Example 1-1b 167214 f(x)f(x) 210–1–2 x Answer: Graph by making a table of values.

10. Try These Graph the following function by making a table of values. 1. f(x) = 2x 2 2. f(x) = 2x 2 – 4

11. Example 1-2a Consider the quadratic function Find the y -intercept, the equation of the axis of symmetry, and the x -coordinate of the vertex. Begin by rearranging the terms of the function so that the quadratic term is first, the linear term is second and the constant term is last. Then identify a, b, and c. So,and The y -intercept is 2.

12. Example 1-2a You can find the equation of the axis of symmetry using a and b. Answer:The y -intercept is 2. The equation of the axis of symmetry is x = 2.Therefore, the x -coordinate of the vertex is 2. Equation of the axis of symmetry Simplify.

13. Example 1-2a Make a table of values that includes the vertex. Choose some values for x that are less than 2 and some that are greater than 2. This ensures that points on either side of the axis of symmetry are graphed. Vertex Answer: (4, 2) 24 (3, –1) –13 (2, –2) –22 (1, –1 ) –11 (0, 2) 20 (x, f(x)) f(x)f(x)x

14. Answer: Then graph the points from your table connecting them with a smooth curve. Graph the vertex and the y -intercept. Example 1-2a Use this information to graph the function. As a check, draw the axis of symmetry,, as a dashed line. (0, 2) The graph of the function should be symmetric about this line. (2, –2)

15. Consider the quadratic function a. Find the y -intercept, the equation of the axis of symmetry, and the x - coordinate of the vertex. b. Make a table of values that includes the vertex. Example 1-2b Answer: y -intercept: 3 ; axis of symmetry: x -coordinate: 3 –2–5–6–5–23f(x)f(x) 543210x Answer:

16. Example 1-2b c. Use this information to graph the function. Answer:

17. Maximum and Minimum Values When you have a quadratic function in the form, f(x) = ax 2 + bx + c if a > 0, then the parabola opens up and therefore has a minimum but no maximum if a < 0, then the parabola opens down and therefore has a maximum but no minimum.

18. Example 1-3a Consider the function Determine whether the function has a maximum or a minimum value. Answer:Since the graph opens down and the function has a maximum value. For this function,

19. Example 1-3a State the maximum or minimum value of the function. The maximum value of this function is the y -coordinate of the vertex. Answer:The maximum value of the function is 4. The x -coordinate of the vertex is Find the y -coordinate of the vertex by evaluating the function for Original function

20. Consider the function a. Determine whether the function has a maximum or a minimum value. b. State the maximum or minimum value of the function. Example 1-3b Answer: minimum Answer: –5

21. Try These Determine whether the function has a maximum or a minimum value. State the maximum or minimum value. 1. f(x) = x 2 – 8x + 2 2. f(x) = 3 – x 2 – 6x

22. End of Lesson 1

23. Assignment P 291 #20, 24, 36, 38

24. Lesson 2 Contents Example 1Two Real Solutions Example 2One Real Solution Example 3No Real Solution Example 4Estimate Roots Example 5Write and Solve an Equation

25. To Solve Quadratic Equations by Graphing: The basic idea behind solving by graphing is that you can graph the equation and “SEE” the solutions. The solutions are the points where the graph crosses the x-axis, also known as the x-intercepts. So you can look at the x - intercepts of the graph to find the solutions to the equation.

26. Here is an example: Solve x 2 – 8x + 15 = 0 by using the following graph. The graph crosses the x-axis at 3 and 5. The solution is x = 3, 5

27. The number of real solutions is at most two. Quadratic Solutions No solutionsOne solutionTwo solutions x = 3 x = -2, x = 2

28. Example 2-1a Solveby graphing. Graph the related quadratic function The equation of the axis of symmetry is Make a table using x values around Then graph each point. x–101234 f (x)f (x)0–4–6 –40

29. Example 2-1a From the table and the graph, we can see that the zeroes of the function are –1 and 4. Answer: The solutions of the equation are –1 and 4.

30. Example 2-1a CheckCheck the solutions by substituting each solution into the original equation to see if it is satisfied.

31. Example 2-1b Answer: –3 and 1 Solveby graphing.

32. Example 2-2a Solveby graphing. Write the equation in form. Add 4 to each side. Graph the related quadratic function x01234 f(x)f(x)41014

33. Example 2-2a Notice that the graph has only one x - intercept, 2. Answer: The equation’s only solution is 2.

34. Example 2-2b Solveby graphing. Answer: 3

35. Example 2-4a Solveby graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. The equation of the axis of symmetry of the related function is x0123456 f(x)3–2–5–6–5–23

36. Example 2-4a The x -intercepts of the graph are between 0 and 1 and between 5 and 6. Answer:One solution is between 0 and 1 and the other is between 5 and 6.

37. Example 2-4b Solveby graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. Answer:between 0 and 1 and between 3 and 4

38. Video of Example http://www.teachertube.com/view_video.php ?viewkey=65e4db9defb2cfc74f65 http://www.teachertube.com/view_video.php ?viewkey=65e4db9defb2cfc74f65

39. Example f(x) = x 2 - 4 Ex 1: Choose the correct solution. A. -2 and 2. l -4 l 2 l -2 l No solution Answer: A

40. Example: f(x) = 2x - x 2 A. x = 2 B. x = 0 C. x = 1 D. x = 0 and 2 E. No solution Ex 2: Choose the correct solution. Answer: D

41. Try These P 297-298 #14, 15, 22, 24

42. End of Lesson 2

43. Assignment P 297-298 #18, 19, 26, 34

44. Lesson 3 Contents Example 1Two Roots Example 2Double Root Example 3Greatest Common Factor Example 4Write an Equation Given Roots

45. Video on Solving Equations by Factoring http://www.phschool.com/atschool/ academy123/html/bbapplet_wl- problem-431091.html

46. Solving Quadratic Equations by Factoring 1.Be sure your polynomial is in standard form and equals 0. 2.Factor the polynomial. 3.Set each factor equal to 0. 4.Solve the equations from step 3. Example: Solve x 2 + 9x + 20 = 0 by factoring. 1.x 2 + 9x + 20 = 0 is already in standard form and equals 0. 2.(x + 5)(x + 4) = 0 Factor the polynomial. 3.x + 5 = 0 and x + 4 = 0 Set each factor equal to 0. 4.x = -5 x = -4 Solve the equations from step 3.

47. Example 3-1a Solveby factoring. Answer: The solution set is {0, –4}. Original equation Add 4x to each side. Factor the binomial. Solve the second equation. Zero Product Property or

48. Example 3-1a Check Substitute 0 and –4 in for x in the original equation.

49. Example 3-1a Solveby factoring. Original equation Subtract 5x and 2 from each side. Factor the trinomial. Zero Product Property or Solve each equation. Answer: The solution set is Check each solution.

50. Solve each equation by factoring. a. b. Example 3-1b Answer: {0, 3} Answer:

51. Example 3-2a Answer: The solution set is {3}. Solveby factoring. Original equation Add 9 to each side. Factor. Zero Product Property or Solve each equation.

52. Example 3-2a Check The graph of the related function, intersects the x -axis only once. Since the zero of the function is 3, the solution of the related equation is 3.

53. Example 3-2b Answer: {–5} Solveby factoring.

54. Example 3-3a Multiple-Choice Test Item What is the positive solution of the equation ? A –3 B 5 C 6 D 7 Read the Test Item You are asked to find the positive solution of the given quadratic equation. This implies that the equation also has a solution that is not positive. Since a quadratic equation can either have one, two, or no solutions, we should expect this equation to have two solutions.

55. Example 3-3a Solve the Test Item Answer: D Both solutions, –3 and 7, are listed among the answer choices. However, the question asks for the positive solution, 7. Original equation Factor. Divide each side by 2. Factor. or Zero Product Property Solve each equation.

56. Example 3-3b Multiple-Choice Test Item What is the positive solution of the equation ? A 5 B – 5 C 2 D 6 Answer: C

57. End of Lesson 3

58. Assignment P 304 #14, 16, 18, 22

59. Lesson 4 Contents Example 1Equation with Rational Roots Example 2Equation with Irrational Roots Example 3Complete the Square Example 4Solve an Equation by Completing the Square Example 5Equation with a  1 Example 6Equation with Complex Solutions

60. Perfect Square Trinomial How to recognize a Perfect Square Trinomial: First be sure that the trinomial is in descending order such as 4x 2 + 12x + 9. 1.It has three terms. 2.The first and last terms are perfect squares and are positive. 3.The middle term is twice the product of the first and last term.

61. Example of perfect square trinomial Example: 9x 2 + 30x + 25 Since 9x 2 and 25 are perfect squares and are positive, lets check the middle term. The square root of 9x 2 and 25 are 3x and 5 respectively. Now, find the product and double it. (3x)(5) = 15x so double that and get 30x, your middle term. Therefore, 9x 2 + 30x + 25 is a perfect square trinomial.

62. Completing the Square Video http://www.phschool.com/atschool/academy123/htm l/bbapplet_wl-problem-431076.html http://revver.com/video/440223/completing-the- square/

63. Example 4-1a Solve by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Subtract 7 from each side. Solve each equation. Answer: The solution set is {–15, 1}. Write as two equations. or

64. Example 4-1b Solve by using the Square Root Property. Answer: {3, 13}

65. Example 4-2a Solve by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Add 5 to each side. Use a calculator. Write as two equations. or

66. Example 4-2a Answer: The exact solutions of this equation are and The approximate solutions are 1.5 and 8.5. Check these results by finding and graphing the related quadratic function. Original equation Subtract 12 from each side. Related quadratic function

67. Example 4-2b Solve by using the Square Root Property. Answer:

68. Example 4-3a Find the value of c that makesa perfect square. Then write the trinomial as a perfect square. Step 1Find one half of 16. Step 2Square the result of Step 1. Step 3Add the result of Step 2 to Answer:The trinomial can be written as

69. Example 4-3b Answer: 9 ; (x + 3) 2 Find the value of c that makes a perfect square. Then write the trinomial as a perfect square.

70. Example 4-4a Solve by completing the square. Write the left side as a perfect square by factoring. Since add 4 to each side. Rewrite so the left side is of the form Notice that is not a perfect square.

71. Example 4-4a Answer: The solution set is {–6, 2}. Square Root Property Subtract 2 from each side. Solve each equation. Write as two equations. or

72. Example 4-4b Answer: {–6, 1} Solve by completing the square.

73. Example 4-5a Solve by completing the square. Divide by the coefficient of the quadratic term, 3. Notice that is not a perfect square. Add to each side. Since add to each side.

74. Example 4-5a Write the left side as a perfect square by factoring. Simplify the right side. Square Root Property Add to each side.

75. Example 4-5a Solve each equation. Answer: The solution set is Write as two equations. or

76. Example 4-5b Solve by completing the square. Answer:

77. Example 4-6a Solve by completing the square. Notice that is not a perfect square. Rewrite so the left side is of the form Since add 1 to each side. Write the left side as a perfect square by factoring.

78. Example 4-6a Square Root Property Subtract 1 from each side. Answer: The solution set is Notice that these are imaginary solutions.

79. Example 4-6a Check A graph of the related function shows that the equation has no real solutions since the graph has no x -intercepts. Imaginary solutions must be checked algebraically by substituting them in the original equation.

80. Example 4-6b Solve by completing the square. Answer:

81. End of Lesson 4