1 / 24

Vertex Form

Vertex Form. Monday, April 14 th. Vertex form. What is the vertex (max or min point) of the function: y = –5x 2 + 4? Maximum = (0, 4) Minimum = (0, 4) Maximum = (4, 0) Minimum = (4, 0). Vertex form. What is the vertex (max or min point) of the function: y = –5x 2 + 4? Maximum = (0, 4)

ardice
Télécharger la présentation

Vertex Form

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Vertex Form Monday, April 14th

  2. Vertex form What is the vertex (max or min point) of the function: y = –5x2 + 4? • Maximum = (0, 4) • Minimum = (0, 4) • Maximum = (4, 0) • Minimum = (4, 0)

  3. Vertex form What is the vertex (max or min point) of the function: y = –5x2 + 4? • Maximum = (0, 4) • Minimum = (0, 4) • Maximum = (4, 0) • Minimum = (4, 0)

  4. Vertex form What is the vertex (max or min point) of the function: y = 3x2 + 4? • Maximum = (0, 4) • Minimum = (0, 4) • Maximum = (4, 0) • Minimum = (4, 0)

  5. Vertex form What is the vertex (max or min point) of the function: y = 3x2 + 4? • Maximum = (0, 4) • Minimum = (0, 4) • Maximum = (4, 0) • Minimum = (4, 0)

  6. Vertex form y = a(x)2 + k Vertex is (0, k)

  7. Vertex form y = a(x – h)2 + k Vertex is (h, k)

  8. Vertex form y = -0.5(x – 2)2 + 3 Vertex is (2, 3)

  9. Vertex form What is the vertex (max or min point) of the function: y = 3(x – 5)2 + 4? • Maximum = (-5, 4) • Maximum = (3, 4) • Minimum = (-5, 4) • Minimum = (5, 4)

  10. Vertex form What is the vertex (max or min point) of the function: y = 3(x – 5)2 + 4? • Maximum = (-5, 4) • Maximum = (3, 4) • Minimum = (-5, 4) • Minimum = (5, 4)

  11. Vertex form What is the vertex (max or min point) of the function: y = 2(x + 3)2 + 4? • Maximum = (-3, 4) • Maximum = (3, 4) • Minimum = (-3, 4) • Minimum = (3, 4)

  12. Vertex form What is the vertex (max or min point) of the function: y = 2(x + 3)2 + 4? • Maximum = (-3, 4) • Maximum = (3, 4) • Minimum = (-3, 4) • Minimum = (3, 4)

  13. Vertex form What is the vertex (max or min point) of the function: y = –½(x + 7)2 + 4? • Maximum = (-7, 4) • Maximum = (7, 4) • Minimum = (-7, 4) • Minimum = (7, 4)

  14. Vertex form What is the vertex (max or min point) of the function: y = –½(x + 7)2 + 4? • Maximum = (-7, 4) • Maximum = (7, 4) • Minimum = (-7, 4) • Minimum = (7, 4)

  15. Ms. Breimer leaps from her stair to the ground during her epic dash to teach you! MsBreimer reaches a max height of 2.5m after 0.32s. At 0s, she stands on the 2m high stair case. What is her height above the ground as a function of time?

  16. MsBreimer reaches a max height of 2.5m after 0.32s. At 0s, she stands on the 2m high stair case. y (0.32, 2.5) y = a(t – h)2 + k y = a(t – 0.32)2 + 2.5 (0, 2) What is a? Plug in a point! (0, 2) 2 = a(0 – 0.32)2 + 2.5 2 = a(0.1024) + 2.5 t 2 – 2.5 = a(0.1024) a = –4.9

  17. If MsBreimer reaches a max height of 2.5m after 0.32s. At 0s, she stands on the 2m high stair case. y (0.32, 2.5) The equation for Ms. Breimers height above the ground as a function of time is: y = –4.9(t – 0.32)2 + 2.5 t

  18. If MsBreimer reaches a max height of 2.5m after 0.32s. At 0s, she stands on the 2m high stair case. y (0.32, 2.5) y = –4.9(t – 0.32)2 + 2.5 From this equation, we can figure out if Ms. Breimer is on earth or another planet, what her initial jump velocity was, and much more! t

  19. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = Ax2 + bx + c

  20. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = –4.9t2 + 3.1t + 2

  21. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = –4.9t2 + 3.1t + 2 Compare this to the equation for height: y2 = ½gt2 + v1t + y1 y2 = the height above the ground after some time, t g = the acceleration due to gravity (this is –9.8m/s2 on earth and –1.6m/s2 on the moon) v2 = the initial velocity y1 = the initial height off the ground

  22. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = –4.9t2 + 3.1t + 2 Compare this to the equation for height: y2 = ½gt2 + v1t + y1 ½g = -4.9 Ms. Breimer is on a planet with earth’s gravity g = -9.8

  23. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = –4.9t2 + 3.1t + 2 Compare this to the equation for height: y2 = ½gt2 + v1t + y1 v1 = 3.1 Ms. Breimer jumped with an initial velocity of 3.1m/s

  24. Using the quadratic equation y = –4.9(t – 0.32)2 + 2.5 Put this in standard format: y = –4.9t2 + 3.1t + 2 Compare this to the equation for height: y2 = ½gt2 + v1t + y1 y1 = 2 Our staircase is 2m high

More Related