2. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.
3. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex.
4. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex.
5. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,
6. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,
7. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,then the reflection of the point across the center line is alsoon the parabola.
8. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,then the reflection of the point across the center line is alsoon the parabola.
9. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,then the reflection of the point across the center line is alsoon the parabola. There is exactly one parabola that goesthrough these three points.
10. Graphs of Quadratic EquationsThe graphs of the equations of the formy = ax2 + bx + c and x = ay2 + bx + care parabolas.Each parabola has a vertex and the center line that containsthe vertex. Suppose we know another point on the parabola,then the reflection of the point across the center line is alsoon the parabola. There is exactly one parabola that goesthrough these three points.
11. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.
12. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.
13. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.The x-intercept is obtained by setting y = 0 and solve for x.
14. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.The x-intercept is obtained by setting y = 0 and solve for x.The graphs of y = ax2 + bx = c are up-down parabolas.
15. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.The x-intercept is obtained by setting y = 0 and solve for x.The graphs of y = ax2 + bx = c are up-down parabolas.If a > 0, the parabola opens up. a>0
16. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.The x-intercept is obtained by setting y = 0 and solve for x.The graphs of y = ax2 + bx = c are up-down parabolas.If a > 0, the parabola opens up.If a < 0, the parabola opens down. a>0 a<0 y = x2 y = –x2
17. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptsand the y-intercept.The y-intercept is obtained by setting x = 0 and solve for y.The x-intercept is obtained by setting y = 0 and solve for x.The graphs of y = ax2 + bx = c are up-down parabolas.If a > 0, the parabola opens up.If a < 0, the parabola opens down. a>0 a<0Vertex Formula (up-down parabolas) The x-coordinate ofthe vertex of the parabola y = ax2 + bx + c is at x = -b . 2a
18. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.
19. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a
20. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.
21. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.
22. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.
23. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15
24. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 –(2)The vertex is at x =2(–1) = 1
25. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 –(2)The vertex is at x =2(–1) = 1y = 16.
26. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16.
27. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) =1y = 16.The y-intercept is at (0, 15)
28. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) =1y = 16. (0, 15)The y-intercept is at (0, 15)
29. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15).
30. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15).
31. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw,
32. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:
33. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0
34. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0
35. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5
36. Graphs of Quadratic EquationsFollowing are the steps to graph the parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.Example A. Graph y = –x2 + 2x + 15 (1, 16) –(2)The vertex is at x =2(–1) = 1y = 16. (0, 15) (2, 15)The y-intercept is at (0, 15)Plot its reflection (2, 15)Draw, set y = 0 to get x-int: (-3, 0) (5, 0)–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5
37. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.
38. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opensto the right. x = y2
39. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opens If a<0, the parabola opento the right. to the left. x = y2 x = –y2
40. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opens If a<0, the parabola opento the right. to the left. x = y2 x = –y2Each sideway parabola is symmetric to a horizontal centerline.
41. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opens If a<0, the parabola opento the right. to the left. x = y2 x = –y2Each sideway parabola is symmetric to a horizontal centerline. The vertex of the parabola is on this line.
42. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opens If a<0, the parabola opento the right. to the left. x = y2 x = –y2Each sideway parabola is symmetric to a horizontal centerline. The vertex of the parabola is on this line. If we know thelocation of the vertex and another point on the parabola, theparabola is completely determined.
43. Graphs of Quadratic EquationsThe graphs of the equationsx = ay2 + by + care parabolas that open sideway.If a>0, the parabola opens If a<0, the parabola opento the right. to the left. x = y2 x = –y2Each sideway parabola is symmetric to a horizontal centerline. The vertex of the parabola is on this line. If we know thelocation of the vertex and another point on the parabola, theparabola is completely determined. The vertex formula is thesame as before except its for the y coordinate.
44. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2a
45. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2aFollowing are the steps to graph the parabola x = ay2 + by + c.
46. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2aFollowing are the steps to graph the parabola x = ay2 + by + c. –b 2a• Set y = in the equation to find the x coordinate of the vertex.
47. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2aFollowing are the steps to graph the parabola x = ay2 + by + c. –b 2a• Set y = in the equation to find the x coordinate of the vertex.2. Find another point; use the x intercept (c, 0) if its not the vertex.
48. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2aFollowing are the steps to graph the parabola x = ay2 + by + c. –b 2a• Set y = in the equation to find the x coordinate of the vertex.2. Find another point; use the x intercept (c, 0) if its not the vertex.3. Locate the reflection of the point across the horizontal center line. These three points form the tip of the parabola. Trace the parabola.
49. Graphs of Quadratic EquationsVertex Formula (sideway parabolas)The y coordinate of the vertex of the parabolax = ay2 + by + cis at y = –b . 2aFollowing are the steps to graph the parabola x = ay2 + by + c. –b 2a• Set y = in the equation to find the x coordinate of the vertex.2. Find another point; use the x intercept (c, 0) if its not the vertex.3. Locate the reflection of the point across the horizontal center line. These three points form the tip of the parabola. Trace the parabola.4. Set x = 0 to find the y intercept.
50. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15
51. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1
52. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16
53. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).
54. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15or (15, 0).
55. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15or (15, 0). (16, 1)
56. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15or (15, 0). (16, 1) (15, 0)
57. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15or (15, 0). (16, 1)Plot its reflection. (15, 0)
58. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2). (15, 0)
59. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2). (15, 0)Draw.
60. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2) (15, 0)Draw.
61. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2) (15, 0)Draw. Get y-int:–y2 + 2y + 15 = 0
62. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2) (15, 0)Draw. Get y-int:–y2 + 2y + 15 = 0y2 – 2y – 15 = 0(y – 5) (y + 3) = 0y = 5, -3
63. Graphs of Quadratic EquationsExample B. Graph x = –y2 + 2y + 15 –(2)Vertex: set y = 2(–1) =1 then x = –(1)2 + 2(1) + 15 = 16so v = (16, 1).Another point:Set y = 0 then x = 15 (15, 2)or (15, 0). (16, 1)Plot its reflection.Its (15, 2) (15, 0)Draw. Get y-int: (0, -3)–y2 + 2y + 15 = 0y2 – 2y – 15 = 0(y – 5) (y + 3) = 0y = 5, -3
64. Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-interceptand the y-intercept.The y-intercept is (o, c) obtained by setting x = 0.The x-intercept is obtained by setting y = 0 and solve theequation 0 = ax2 + bx + c which may or may not have realnumber solutions. Hence there might not be any x-intercept.Following are the steps to graph a parabola y = ax2 + bx + c.3. Set x = -b in the equation to find the vertex. 2a2. Find another point, use the y-intercept (0, c) if possible.3. Locate its reflection across the center line. These three points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.The graph of y = ax2 + bx = c are up-down parabolas.If a > 0, the parabola opens up.If a < 0, the parabola opens down.
65. Graphs of Quadratic Equations Exercise A. Graph the following parabolas by finding and plotting the vertex point and intercepts. 1. y = x2 – 2x – 3 and x = y2 – 2y – 3 2. y = –x2 + 2x + 3 and x = –y2 + 2y + 3 3. y = x2 + 2x – 3 and x = y2 + 2y – 3 4. y = –x2 – 2x + 3 and y = –x2 – 2x + 3 5. x = y2 – 2y – 8 6. x = –y2 + 2y – 5 7. x = y2 + 4y + 3 8. x = –y2 – 3y – 4 9. x = y + 2y – 8 2 10. x = –y2 – 2y + 811. x = – y2 – 6y + 27 12. x = y2 – 6y + 60
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