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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.

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Graphs

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  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.

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