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Complex Numbers

2. Complex Numbers. ALGEBRA 2 LESSON 5-6. 1. Simplify – –169. 2. Find |–1 + i | Simplify each expression. (–7 i )(–3 i ) 5. (–3 + 4 i ) – (–3 – 8 i ) 7. Solve x 2 + 2 = 0. –13 i. –21. 12 i. 1 2. ±2 i. Translating Parabolas. ALGEBRA 2 LESSON 5-3.

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Complex Numbers

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  1. 2 Complex Numbers ALGEBRA 2 LESSON 5-6 • 1. Simplify – –169. • 2. Find |–1 + i| • Simplify each expression. • (–7i)(–3i) • 5. (–3 + 4i) – (–3 – 8i) • 7. Solve x2 + 2 = 0. –13i –21 12i 1 2 ±2i

  2. Translating Parabolas ALGEBRA 2 LESSON 5-3 (For help, go to Lesson2-6.) Identify the parent function of each function. Then graph the function by translating the parent function. 1.y = –x + 2 2.y = |3x| + 2 3.y = –|x + 1| – 1 4.y = 2x, 2 units down 5.y = x, 4 units up, 1 unit right Write an equation for each translation. 5-3

  3. Translating Parabolas ALGEBRA 2 LESSON 5-3 Solutions 1.y = –x + 2 parent function: y = –x translate 2 units up 3.y = –|x + 1| – 1 parent function: y = –x translate 1 unit left and one unit down 2.y = |3x| + 2 parent function: y = |3x| translate 2 units up 4.y = 2x, 2 units down: y = 2x – 2 5.y = x, 4 units up and 1 unit right: y = (x – 1) + 4, or y = x + 3 5-3

  4. 25. –5 + 3i 26. –9 – i 27. 3 + 2i 28. 4 – 7i 29. 6 + 3i 30. 1 – 7i 31. 7 + 4i 32. –2 – 3i 33. 10 + 6i 34. –7 – 10i 35. 10 36. 26 – 7i 37. 9 + 58i 38. 9 – 23i 39. –36 40. 65 + 72i 41. ± 5i 42. ± 43. ± 44. ± i 7 45. ±6i 46. ± 47. –i, –1 – i, i 48. –2i, –4 – 2i, 12 + 14i 49. 1 – i, 1 – 3i, –7 – 7i 50. ± i 65 51. ±7i 8i 3 3 i 15 3 i 2 2 Complex Numbers ALGEBRA 2 LESSON 5-6 5-6

  5. Complex Numbers ALGEBRA 2 LESSON 5-6 54. a.A: –5, B: 3 + 2i, C: 2 – i, D: 3i, E: –6 – 4i, F: –1 + 5i b. 5, –3 – 2i, –2 + i, –3i, 6 + 4i, 1 – 5i 55. a. Check students’ work. b. a circle with radius 10 and center at the origin 56. –5, 5 57. 288i 58. –1 + 5i 59. 10 – 4i 60. 8 – 2i 61. 11 – 5i 62. 6 + 10i 63. 7 – i 64. 10 + 11i 65. –27 + 8i 66. –13 + I 5-6

  6. Properties of Parabolas • The graph of y = ax2 + bx + c is a parabola when a ≠ 0. • When a > 0, the parabola opens up. When a < 0, the parabola opens down. • The axis of symmetry is the line x = from completing the square • The x-coordinate of the vertex is The y-coordinate of the vertex is the value of y when x = , or y = f( ) • The y-intercept is (0, c).

  7. b 2a 2 2(– ) x = – = – = 4 Find the x-coordinate of the vertex. 1 4 1 4 y = – (4)2 + 2(4) – 3 = 1 Find the y-coordinate of the vertex. Graph the vertex and the axis of symmetry x = 4. Graph two points on one side of the axis of symmetry, such as (6, 0) and (8, –3). Then graph corresponding points (2, 0) and (0, –3). The maximum value of the function is 1. Properties of Parabolas ALGEBRA 2 LESSON 5-2 1 4 Graph y = – x2 + 2x – 3. What is the maximum value of the function? Since a < 0, the graph of the function opens down, and the vertex represents the maximum value. Find the coordinates of the vertex. 5-2

  8. Relate: revenue equals price times number of get-away packages sold Define: Line R = revenue. Let p = price of a get-away package. Let –0.12p + 60 = number of a get-away packages sold. Write: R = p ( –0.12p + 60 ) = –0.12p2 + 60pWrite in standard form. Properties of Parabolas ALGEBRA 2 LESSON 5-2 The number of weekend get-away packages a hotel can sell is modeled by –0.12p + 60, where p is the price of a get-away package. What price will maximize revenue? What is the maximum revenue? 5-2

  9. b 2a 60 2(–0.12) p = – = – = 250 Find p at the vertex. Properties of Parabolas ALGEBRA 2 LESSON 5-2 (continued) Find the maximum value of the function. Since a < 0, the graph of the function opens down, and the vertex represents a maximum value. R = –0.12(250)2 + 60(250) Evaluate R for p = 250 = 7500 Simplify. A price of $250 will maximize revenue. The maximum revenue is $7500. 5-2

  10. Properties of Parabolas • Vertex form of a quadratic function The vertex form of a quadratic function is y = a(x − h)2 + k. The coordinates of the vertex of the parabola are (h, k).

  11. Properties of Parabolas • Each function in the first column is written in standard form, in the vertex form in the second column. Check that each function is equivalent. • Patterns Copy and complete the table. • Compare the values of and h in each row. Write a formula to show the relationship between and h.

  12. Vertex form • The graph of y = a(x − h)2 + k is the graph of y = ax2 translated h units horizontally and k units vertically. • When h is positive the graph shifts right; when h is negative the graph shifts left. • When k is positive the graph shifts up; when k is negative the graph shifts down. • The vertex is (h, k), and the axis of symmetry is the line x = h.

  13. 2 3 The graph of y = (x + 1)2 – 2 is a translation of the graph of the parent function y = x2. 2 3 Step 1: Graph the vertex (–1, –2). Draw the axis of symmetry x = –1. Step 2: Find another point. When x = 2, y = (2 + 1)2 – 2 = 4. Graph (2, 4). 2 3 Step 3: Graph the point corresponding to (2, 4). It is three units to the left of the axis of symmetry at (–4, 4). Step 4: Sketch the curve. Translating Parabolas ALGEBRA 2 LESSON 5-3 2 3 Graph y = (x + 1)2 – 2. You can graph it by translating the parent function or by finding the vertex and the axis of symmetry. 5-3

  14. b 2a x = – Find the x-coordinate of the vertex. Substitute for a and b. (–70) 2(–7) = – = –5 y = –7 (–5)2 – 70(–5) – 169 Find the y-coordinate of the vertex. = 6 y = a(x – h)2 + k Write in vertex form. = –7(x – (–5))2 + 6 Substitute for a, h and k. = –7(x + 5)2 + 6 Translating Parabolas ALGEBRA 2 LESSON 5-3 Write y = –7x2 – 70x – 169 in vertex form. The vertex is at (–5, 6). The vertex form of the function is y = –7(x + 5)2 + 6. 5-3

  15. Assignment 41 • Page 244 22, 24, 29 • Page 251 2,4, 24, 28, 52

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