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1 Algebra I Quadratics 2015-11-04 www.njctl.org 2 Table of - PowerPoint PPT Presentation

1 Algebra I Quadratics 2015-11-04 www.njctl.org 2 Table of Contents Key Terms Click on the topic to go to that section Characteristics of Quadratic Equations Transforming Quadratic Equations Graphing Quadratic Equations


  1. 15 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function. 7 y = – – x 2 5 Answer up, wider A up, narrower B down, wider C down, narrower D 40

  2. What Does "C" Do? What does "c" do in y = ax 2 + bx + c ? y = x 2 + 6 y = x 2 + 3 y = x 2 y = x 2 – 2 y = x 2 – 5 y = x 2 – 9 41

  3. What Does "C" Do? What does "c" do in y = ax 2 + bx + c ? "c" moves the graph up or down the same value as "c." "c" is the y- intercept. 42

  4. 16 Without graphing, what is the y- intercept of the the given parabola? y = x 2 + 17 Answer 43

  5. 17 Without graphing, what is the y- intercept of the the given parabola? y = –x 2 –6 Answer 44

  6. 18 Without graphing, what is the y- intercept of the the given parabola? y = –3x 2 + 13x – 9 Answer 45

  7. 19 Without graphing, what is the y- intercept of the the given parabola? y = 2x 2 + 5x Answer 46

  8. 20 Choose all that apply to the following quadratic: f(x) = –.7x 2 –4 opens up y-intercept of y = –4 A A Answer opens down y-intercept of y = –2 B B wider than parent function y-intercept of y = 0 C C narrower than parent y-intercept of y = 2 D D function y-intercept of y = 4 E y-intercept of y = 6 F 47

  9. 21 Choose all that apply to the following quadratic: 4 f(x) = – – x 2 –6x 3 y-intercept of y = –4 E opens up A Answer opens down y-intercept of y = –2 B F wider than parent function y-intercept of y = 0 C G narrower than parent function D y-intercept of y = 2 H y-intercept of y = 4 I y-intercept of y = 6 J 48

  10. 22 The diagram below shows the graph of y = − x 2 − c. Answer Which diagram shows the graph of y = x 2 − c ? D C A B From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011. 49

  11. Graphing Quadratic Equations Return to Table of Contents 50

  12. Graph by Following Six Steps: Step 1 - Find Axis of Symmetry Step 2 - Find Vertex Step 3 - Find Y intercept Step 4 - Find two more points Step 5 - Partially graph Step 6 - Reflect 51

  13. Axis of Symmetry Step 1 - Find Axis of Symmetry What is the Axis of Symmetry? Teacher Notes Axis of Symmetry 52

  14. Step 1 - Find Axis of Symmetry Graph y = 3x 2 – 6x + 1 –b x = –– Formula: 2a a = 3 b = –6 x = – (–6) = 6 = 1 2(3) 6 The axis of symmetry is x = 1. 53

  15. Step 2 - Find Vertex Step 2 - Find the vertex by substituting the value of x (the axis of symmetry) into the equation to get y. y = 3x 2 – 6x + 1 a = 3, b = –6 and c = 1 y = 3(1) 2 + –6(1) + 1 y = 3 – 6 + 1 y = –2 Vertex = (1,–2) 54

  16. Step 3 - Find y intercept What is the y-intercept? Teacher Notes y- intercept 55

  17. Step 3 - Find y intercept Graph y = 3x 2 – 6x + 1 The y- intercept is always the c value, because x = 0. y = ax 2 + bx + c c = 1 y = 3x 2 – 6x + 1 The y-intercept is 1 and the graph passes through (0,1). 56

  18. Step 4 - Find Two More Points Graph y = 3x 2 – 6x + 1 Find two more points on the parabola. Choose different values of x y = 3x 2 – 6x + 1 and plug in to find points. y = 3(–1) 2 – 6(–1) + 1 Let's pick x = –1 and x = –2 y = 3 + 6 + 1 y = 10 (–1,10) 57

  19. Step 4 - Find Two More Points (continued) Graph y = 3x 2 – 6x + 1 y = 3x 2 – 6x + 1 y = 3(–2) 2 – 6(–2) + 1 y = 3(4) + 12 + 1 y = 25 (–2, 25) 58

  20. Step 5 - Graph the Axis of Symmetry Graph the axis of symmetry, the vertex, the point containing the y-intercept and two other points. 59

  21. Step 6 - Reflect the Points Reflect the points across the axis of symmetry. Connect the points with a smooth curve. (4,25) 60

  22. 23 What is the axis of symmetry for y = x 2 + 2x - 3 (Step 1)? Answer 61

  23. 24 What is the vertex for y = x 2 + 2x - 3 (Step 2)? A (-1, -4) Answer B (1, -4) C (-1, 4) 62

  24. 25 What is the y-intercept for y = x 2 + 2x - 3 (Step 3)? A -3 B 3 Answer 63

  25. Graph y= x 2 + 2x – 3 axis of symmetry = –1 vertex = –1, –4 y intercept = –3 2 other points (step 4) (1,0) (2,5) Partially graph (step 5) Reflect (step 6) 64

  26. Graph y = 2x 2 – 6x + 4 65

  27. Graph y = –x 2 – 4x + 5 66

  28. Graph y = 3x 2 – 7 67

  29. Solve Quadratic Equations by Graphing Return to Table of Contents 68

  30. Find the Zeros One way to solve a quadratic equation in standard form is find the zeros by graphing. A zero is the point at which the parabola intersects the x-axis. A quadratic may have one, two or no zeros. 69

  31. Find the Zeros How many zeros do the parabolas have? What are the values of the zeros? No zeroes click click 2 zeroes; click 1 zero; (doesn't cross x = -1 and x=3 x=1 the "x" axis) 70

  32. Review To solve a quadratic equation by graphing follow the 6 step process we already learned. Step 1 - Find Axis of Symmetry Step 2 - Find Vertex Step 3 - Find Y intercept Step 4 - Find two more points Step 5 - Partially graph Step 6 - Reflect 71

  33. 26 Solve the equation by graphing. –12x + 18 = –2x 2 Which of these is in standard form? Answer A y = –2x 2 – 12x + 18 B y = 2x 2 – 12x + 18 C y = –2x 2 + 12x – 18 72

  34. 27 What is the axis of symmetry? y = –2x 2 + 12x – 18 A –3 Answer B 3 C 4 D –5 73

  35. 28 y = –2x 2 + 12x – 18 What is the vertex? A (3,0) B (–3,0) Answer C (4,0) D (–5,0) 74

  36. 29 y = –2x 2 + 12x – 18 What is the y- intercept? (0, 0) A (0, 18) B Answer (0, –18) C (0, 12) D 75

  37. If two other points are (5, –8) and (4 ,–2),what does 30 the graph of y = –2x 2 + 12x – 18 look like? B A Answer C D 76

  38. 31 y = –2x 2 + 12x – 18 What is(are) the zero(s)? A –18 click for graph of answer B 4 Answer C 3 D –8 77

  39. Solve Quadratic Equations by Factoring Return to Table of Contents 78

  40. Solving Quadratic Equations by Factoring Review of factoring - To factor a quadratic trinomial of the form x 2 + bx + c, find two factors of c whose sum is b. Example - To factor x 2 + 9x + 18, look for factors whose sum is 9. Factors of 18 Sum 1 and 18 19 2 and 9 11 x 2 + 9x + 18 = (x + 3)(x + 6) 3 and 6 9 79

  41. Solving Quadratic Equations by Factoring When c is positive, it's factors have the same sign. The sign of b tells you whether the factors are positive or negative. When b is positive, the factors are positive. When b is negative, the factors are negative. 80

  42. Solving Quadratic Equations by Factoring Remember the FOIL method for multiplying binomials 1. Multiply the First terms (x + 3)(x + 2) x x = x 2 2. Multiply the Outer terms (x + 3)(x + 2) x 2 = 2x 3. Multiply the Inner terms (x + 3)(x + 2) 3 x = 3x 4. Multiply the Last terms (x + 3)(x + 2) 3 2 = 6 (x + 3)(x + 2) = x 2 + 2x + 3x + 6 = x 2 + 5x + 6 I F O L 81

  43. 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. Numbers Algebra 3(0) = 0 If ab = 0, 4(0) = 0 Then a = 0 or b = 0 82

  44. Zero Product Property Example 1: Solve x 2 + 4x – 12 = 0 Use "FUSE" ! (x + 6) (x – 2) = 0 Factor the trinomial using the FOIL method. x + 6 = 0 or x – 2 = 0 Use the Zero property –6 –6 + 2 +2 x = –6 x = 2 –6 2 + 4(–6) – 12 = 0 Substitue found value into original –6 2 + (–24) – 12 = 0 equation 36 – 24 – 12 = 0 0 = 0 or Equal - problem 2 2 + 4(2) – 12 = 0 solved! The 4 + 8 – 12 = 0 solutions are -6 and 2. 0 = 0 83

  45. Zero Product Property The equation has to be Example 2: Solve x 2 + 36 = 12x –12x –12x written in standard form (ax 2 + bx + c). So subtract 12x from both sides. x 2 – 12x + 36 = 0 (x – 6)(x – 6) = 0 Factor the trinomial using the FOIL x – 6 = 0 method. +6 +6 x = 6 Use the Zero property Substitue found 6 2 + 36 = 12(6) value into original equation 36 + 36 = 72 72 = 72 Equal - problem solved! 84

  46. Zero Product Property Example 3: Solve x 2 – 16x + 48= 0 Factor the trinomial using the (x – 4)(x – 12) = 0 FOIL method. Use the Zero x – 4 = 0 x –12 = 0 property +4 +4 +12 +12 x = 4 x = 12 4 2 – 16(4) + 48 = 0 Substitue found 16 – 64 + 48 = 0 value into original –48+48 = 0 equation 0 = 0 12 2 – 16(12) + 48 = 0 144 –192 + 48 = 0 –48 + 48 = 0 0 = 0 –48 Equal - problem solved! 85

  47. 32 Solve x 2 – 5x + 6 = 0 –7 3 A F –5 5 B G Answer –3 6 C H –2 7 D I 2 15 E J 86

  48. 33 Solve m 2 + 10m + 25 = 0 –7 3 A F Answer –5 5 B G 6 –3 H C –2 7 D I 15 J 2 E 87

  49. 34 Solve h 2 – h = 12 3 –12 F A 4 Answer G –4 B –3 6 C H –2 D 8 I 2 12 E J 88

  50. 35 Solve d 2 – 35d = 2d 0 –7 F A 5 –5 G B Answer 6 –3 H C 35 7 D I 37 12 J E 89

  51. 36 Solve 8y 2 + 2y = 3 3 / 4 – 3 / 4 F A 1 / 2 G – 1 / 2 B Answer 4 / 3 H – 4 / 3 C –2 D –3 I 3 J 2 E 90

  52. 37 Which equation has roots of − 3 and 5? A x 2 + 2x − 15 = 0 B x 2 − 2x − 15 = 0 Answer C x 2 + 2x + 15 = 0 D x 2 − 2x + 15 = 0 91

  53. Solve Quadratic Equations Using Square Roots Return to Table of Contents 92

  54. Square Root Method You can solve a quadratic equation by the square root method if you can write it in the form: x ² = c If x and c are algebraic expressions, then: x = c or x = – c √ √ written as: x = ± c √ 93

  55. Square Root Method Solve for z: z ² = 49 z = ± 49 √ z = ±7 The solution set is 7 and –7 94

  56. Square Root Method A quadratic equation of the form x 2 = c can be solved using the Square Root Property. Example: Solve 4x 2 = 20 4x 2 = 20 Divide both sides 4 4 by 4 to isolate x ² x 2 = 5 √ 5 5 The solution set is and – √ 5 x = ± √ 95

  57. Square Root Method Solve 5x ² = 20 using the square root method: 5x 2 = 20 5 5 x 2 = 4 x = or x = – √ 4 √ 4 x = ± 2 96

  58. Square Root Method Solve (2x – 1) ² = 20 using the square root method. click click or 2x – 1 = 20 √ 2x – 1 = – 20 √ 2x – 1 = (4)(5) 2x – 1 = – (4)(5) √ √ 2x – 1 = 2 5 2x – 1 = –2 5 √ √ 2x = 1 + 2 5 √ 2x = 1 – 2 5 √ √ 1 + 2 5 1 – 2 5 √ x = 2 x = 2 solution: x = 1 ± 2 5 √ click 2 97

  59. 38 When you take the square root of a real number, your answer will always be positive. True Answer False 98

  60. 39 If x 2 = 16, then x = A 4 B 2 Answer C –2 D 26 E –4 99

  61. 40 If y 2 = 4, then y = A 4 B 2 Answer C –2 D 26 E –4 100

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