Graphing Quadratic Functions in Standard Form (Vertex Form) — Transcript

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1. 0:00Welcome to a video on graphing quadratic  functions in standard form. In a previous video,
2. 0:06we graphed quadratic functions in general form or  the form of f of x equals a times x squared plus
3. 0:12b times x plus c. So we should know that the graph  of this is a parabola. But if we take general form
4. 0:16and we complete the square on the right side,  we'll have standard form that looks like this,
5. 0:21where f of x is equal to a times the  quantity x minus h squared plus k.
6. 0:26What's nice about standard form is we can tell a  lot about the graph just by looking at the form
7. 0:31of the function. If a is positive or greater than  zero, the parabola opens up, and if it's negative
8. 0:38or less than zero, it opens down. Also, the  vertex of the parabola occurs at the point h, k,
9. 0:45and the vertical line passing through the vertex  is the axis of the parabola or axis of symmetry.
10. 0:52One of the things we have to be very careful about  is identifying the correct sign of the x and y
11. 0:56coordinates of the vertex, which are identified  by h and k. But it has to be in the form of the
12. 1:02quantity x minus h squared plus k. What that means  is if we had x minus h, we'd use a positive value
13. 1:11for h, and if it was x plus h, we'd use a negative  value of h. However, for k, if it's plus k,
14. 1:19we use a positive k, and if it's minus k,  we're going to use a negative value for k.
15. 1:24So for example, if we have y equals the quantity  x minus three squared minus one, the x-coordinate
16. 1:30of the vertex would be positive three, and  the y-coordinate would be negative one. So
17. 1:36you can think that the x-coordinate  is the opposite of what you see,
18. 1:39but the y-coordinate is exactly the same. So  if we have y equals the quantity x plus three
19. 1:46squared plus five, the x-coordinate  would be negative three in this case,
20. 1:50but the y-coordinate would be positive five. The  real reason why this is negative three is because
21. 1:55it actually has to be written in the form of x  minus a negative three squared plus five, but
22. 2:05as you already know, we normally write  minus a negative as plus a positive.
23. 2:10So we should be very careful when we're  identifying the signs of the x and y
24. 2:13coordinates for the vertex. Let's go ahead and  take a look at a graph of a quadratic function
25. 2:18in standard form. What we're going to do here  is leave h and k equal to zero and change the
26. 2:24value of a. Notice when a is equal to one, the  parabola opens up. As we increase the value of
27. 2:30a, it still opens up, but it is stretched  vertically. As we decrease the value of a,
28. 2:37it is stretched horizontally, but as soon as  a is negative, the parabola opens downward.
29. 2:46Let's go ahead and put the value of a back  to one and change the value of h. Remember,
30. 2:53h is the x-coordinate of the  vertex. So as you might expect,
31. 2:57changing the value of h shifts the parabola  left or right. Notice that when h is positive
32. 3:05four point seven, the equation is in the form of  x minus four point seven. And when h is negative,
33. 3:15the equation is in the form x plus the value  because they’re really subtracting a negative.
34. 3:23Let's see what happens when we change k. Remember,
35. 3:25k is the y-coordinate of the  vertex. So if we increase k,
36. 3:29it shifts the parabola up. If we decrease k, or if  k is negative, the parabola is shifted downward.
37. 3:39Let's go ahead and take a look at a couple  examples. Let's say we want to graph this
38. 3:44quadratic function by finding the vertex, axis of  symmetry, and the intercepts. Well, we should be
39. 3:50able to tell that the value of a would be one,  so that tells us the parabola opens up. Next,
40. 3:57h and k would be the coordinates of the vertex.  So the value of h would be positive three, and
41. 4:03the value of k would be negative four, so  we have three negative four for the vertex.
42. 4:10Remember, the axis of symmetry is  a line passing through the vertex,
43. 4:14so the equation of the axis  must be x equals three. Again,
44. 4:19we found all of this useful information  just by looking at the form of the function.
45. 4:23Let's go ahead and review how to  find the intercepts. Remember,
46. 4:26to find the y-intercept, we're supposed to  set x equal to zero and solve for y. So we
47. 4:31would have y equals zero minus three squared  minus four. That would be negative three minus
48. 4:41four, which is negative seven. So the  y-intercept would be zero negative seven.
49. 4:46And then lastly, in order  to find the x-intercepts,
50. 4:50we're supposed to set y equal to  zero and solve for x. So we would
51. 4:53have zero equals the quantity x minus  three squared minus four. So add four to
52. 5:00both sides and then square root both sides of  the equation. Don't forget the plus or minus.
53. 5:06So on the left, we have plus or minus two equals x
54. 5:11minus three. Add three to both sides,  and we’d have three plus or minus two.
55. 5:17So we have two x-intercepts. The  first one would be three plus two,
56. 5:22which would be five, and the second x-intercept  would be three minus two, which would be one.
57. 5:28That should be plenty of information  to make a nice graph. Let's go ahead
58. 5:31and sketch it. We'll plot the vertex  at three negative four. Let's go ahead
59. 5:39and sketch the axis of symmetry in next. The  y-intercept is zero negative seven. Here now,
60. 5:48because we have the axis of symmetry,  we know there must be a mirror image of
61. 5:51this point three units to the right of the  axis of symmetry here. That'll be helpful.
62. 5:57And then lastly, we have our two  x-intercepts, five and one. Again,
63. 6:04that looks good because they're symmetrical  across the axis of symmetry. So now we can
64. 6:08just make a nice parabola passing through these  five points. Let's go ahead and try another one.
65. 6:16Okay, we notice that it's in standard form. a  is equal to negative two, so this parabola opens
66. 6:22downward. The vertex is h, k, so since we have x  plus one, the x-coordinate would be negative one,
67. 6:30and we have plus eight on the end, so the  y-coordinate would be eight, and therefore,
68. 6:35the equation of the axis must  be x equals negative one.
69. 6:41Now, it doesn't ask us to find the x and y  intercepts; it just says to find at least two
70. 6:46additional points. I'm going to go ahead and  at least find the y-intercept. So to find the
71. 6:50y-intercept, we set x equal to zero and solve  for y. We have y equals negative two times the
72. 6:56quantity of zero plus one squared plus  eight. That would be one squared plus eight,
73. 7:04which would be negative two plus eight, that would  be six. So our y-intercept would be zero six.
74. 7:08Let's go ahead and sketch this information  and see what we have. The vertex is
75. 7:12negative one eight, the axis of symmetry is the  vertical line passing through that point, and we
76. 7:20have the y-intercept of zero six here. There must  be a mirror image of this point one unit to the
77. 7:27left of the axis of symmetry there, and we do  have some good information to make a nice graph.
78. 7:32But let's go ahead and find the x-intercepts  again just to review. Remember, to find the
79. 7:37x-intercepts, we set y equal to zero. So we’d have  zero equals negative two times the quantity of x
80. 7:44plus one squared plus eight. So we're  going to subtract eight and then divide
81. 7:49by negative two. That would give us four  equals the quantity x plus one squared.
82. 7:55Now we square root both sides of the equation.  We have our plus or minus there. Looks like we
83. 8:00have plus or minus two equals x plus one, so we  subtract one on both sides, and we have negative
84. 8:07one plus or minus two. So the first x-intercept  would be negative one plus two, which would be
85. 8:15one zero, and the second intercept would be negative  one minus two, which would be negative three zero.
86. 8:21So we were pretty fortunate here; these came  out very nice. Let's go ahead and plot those
87. 8:24and complete our graph. We have one, and notice  they are two units to the right of the axis,
88. 8:31and negative three is two units to the left  of the axis, which is a good indication that
89. 8:36we're doing this correctly. Let's go ahead  and sketch our graph, and there it is.
90. 8:44Okay, I hope you found this video  helpful. Thank you for watching!