1 / 59

Graphs

This text explores the modeling of a volleyball's path and the height analysis of a soccer ball using equations and graphs. It includes determining if a player can reach the ball, identifying the greatest height of the ball, and analyzing the path and height of a jump. The text also discusses zips on a tent and the length of poles used for support.

neilr
Télécharger la présentation

Graphs

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. Graphs NCEA Excellence

  2. 2007

  3. Question 5c • Jake and Ryan are playing volleyball. • Jake is 4.0 metres from the net on one side and Ryan is 1.5 metres from the net on the other • side, as shown in the diagram. • When the ball is hit from one player to the other, the path of the ball can be modelled by a parabola.

  4. Question 5C • The height of the ball when it leaves Jake is 1 metre above the ground. • The ball reaches its maximum height of 3 metres above the ground when it is directly above the net. • When Ryan jumps, he can reach to a height of 2.2 metres. • Form an equation to model the path of the ball and use it to determine whether or not Ryan could reach it when he jumps.

  5. X = -4, y = 1

  6. X = 1.5, y = ? He will not be able to reach the ball.

  7. Question 5a • John and Richard were playing with a soccer ball. • The graph shows the height of the ball above the ground, during one kick from John, J, towards • Richard, R. • The height of the ball above the ground is y metres. • The horizontal distance of the ball from John is x metres. • The graph has the equation y = 0.1(9 –x)(x + 1).

  8. y = 0.1(9 –x)(x + 1)

  9. y = 0.1(9 –x)(x + 1) • Write down the value of the y-intercept and explain what it means in this situation.

  10. y = 0.1(9 –x)(x + 1) • Write down the value of the y-intercept and explain what it means in this situation. • X = 0, y = 0.9 • This is the initial height of the ball when it is kicked by John.

  11. y = 0.1(9 –x)(x + 1) • What is the greatest height of the ball above the ground?

  12. y = 0.1(9 –x)(x + 1) • What is the greatest height of the ball above the ground? • When x = 4 • Y = 2.5 m 4 9 -1

  13. Draw the graph • y = –x(x – 1) – 2 • = –x2 + x – 2

  14. y = –x(x – 1) – 2= –x2 + x – 2 You can either substitute in values or draw the graph of y = –x(x – 1) moved down 2

  15. Draw the graph • y = 2x2– 8

  16. y = 2x2– 8

  17. Draw the graph • 2y + 3x = 6

  18. 2y + 3x = 6

  19. Write the equations

  20. 2006

  21. Question 5 Part of the dirt cycle track is shown on the graph below. • The first section, AB, is a straight line and the second section, BCD, is part of a parabola.

  22. The equation of the second section, BCD, is • It begins at B, a horizontal distance of 40 metres from the start. • How high is the start of this second section, B, above ground-level?

  23. X = 40, y = 4

  24. The point A is 6 metres above the ground. What is the gradient of the first section AB?

  25. How high above ground-level is the lowest point, C?

  26. In the graph below, the dotted line DEF shows the path of a rider performing a jump. • The rider leaves the track at D, jumps to maximum height at E, and lands on a platform at F. • This jump can also be modelled by a parabola. • E is 8 metres above the ground and a horizontal distance of 60 metres from the start. • B and D are both the same height above the ground. • F is a horizontal distance of 72 metres from the start. • Calculate the height of the platform at F.

  27. X = 50, y = 4

  28. X = 72, y = ?

  29. Draw the graph • y = x2 − 6x

  30. y = x2 − 6x = x(x - 6)

  31. Draw the graph • y = 3 − 2x − x2 • = (1 − x)(3 + x)

  32. y = 3 − 2x − x2= (1 − x)(3 + x)

  33. Draw the graph • 4y − 2x = 8

  34. 4y − 2x = 8

  35. Write the equations

  36. 2005

  37. Question 5 • Mere buys a small tent for her little sister for Christmas. • The top part is modelled by a parabola. • There are three zips: BE, DE, and EF. • The equation of the frame of the top part of the tent is • where y is the height in cm and x is the distance from the centre line in cm.

  38. What is the height of the top of the tent?

  39. What is the height of the top of the tent? • 80 cm

  40. What is the length of the horizontal zip, EF?

  41. What is the length of the horizontal zip, EF? • Y = 0 • X = 40cm

  42. The length of AC is 32 cm. • Find the length of zip BE. • x = 16 • X = 67.2 cm

  43. Two poles, PR and QS help keep the tent upright. • They are 25 cm from the centre-line of the tent. • The bottom part of the tent is also modelled by a parabola. • It cuts the vertical axis at y = –20. • The length FG = 10 cm. • Find the length of the pole PR.

More Related