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Review Quadratics

Review Quadratics. Monday , April 28th . Test. We have a request to have the test on Thursday – would that make things difficult for anyone? . Review: Developing Quadratic Equations.

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Review Quadratics

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  1. ReviewQuadratics Monday, April 28th

  2. Test • We have a request to have the test on Thursday – would that make things difficult for anyone?

  3. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. a) What is the equation for Amy’s jump?

  4. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. What is the equation for Amy’s jump? y = A(x – h)2 + k

  5. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. What is the equation for Amy’s jump? y = A(x – h)2 + k y = A(x – 1.5)2 + 2

  6. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. What is the equation for Amy’s jump? y = A(x – h)2 + k y = A(x – 1.5)2 + 2 How could we get A?

  7. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. What is the equation for Amy’s jump? y = A(x – h)2 + k y = A(x – 1.5)2 + 2 How could we get A? Plug in the point from where she jumps, (0, 0)

  8. Review: Developing Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. What is the equation for Amy’s jump? y = A(x – h)2 + k y = A(x – 1.5)2 + 2 0 = A(0 – 1.5)2 + 2 How could we get A? Plug in the point from where she jumps, (0, 0)

  9. Review: Using Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. b) What is Amy’s height off the ground when she is 1m away from the place where she leaped off the ground?

  10. Review: Using Quadratic Equations In a high jump competition, Amy jumps over the bar reaching a maximum height of 2m above the ground when she is 1.5m away from the place where she leaped off the ground. c) What horizontal distance does Amy cover by the time she lands on the mat which is 0.5m off the ground?

  11. In Your Teams James throws a velociraptoregg a horizontal distance of 50m beforeitsplatters on the ground. He notes that the eggreaches a maximum height of 8m in the middle of thistrajectory. Hisonlyhopeatappeasing the angryvelociraptor parents is to show hisskillatfiguring out the equation for thisquadratic. Help James!

  12. FactoredForm Find the x-intercepts of the followingquadratic by factoring: y = x2 + 4x – 32 y = (x + 8)(x – 4) 0 = (x + 8)(x – 4) 0 = x + 8  x = -8, or (-8, 0) 0 = x – 4  x = 4, or (4, 0)

  13. FactoredForm Find the x-intercepts of the followingquadratic by factoring: y = –3x2 – 14x - 8 y = (–3x – 12)(–3x – 2) / –3 0 = (x + 4)(–3x – 2) 0 = x + 4 x = -4, or (-4, 0) 0 = –3x – 2 x = -2/3, or (-2/3, 0)

  14. Vertex Form Put the followingquadraticinto vertex form, and find the vertex: y = x2 – 6x + 5 y = x2 – 6x + 9 – 9 + 5 y = (x – 3)2 – 9 + 5 y = (x – 3)2 – 4 Vertex = (3, -4)

  15. Vertex Form Put the followingquadraticinto vertex form, and find the vertex: y = –2x2 – 16x + 10 y = –2[x2 + 8x – 5] y = –2[x2+ 8x + 16 – 16 – 5] y = –2[(x + 4)2 – 16 – 5] y = –2[(x + 4)2 – 21] y = –2(x + 4)2 + 42 The vertex is (-4, 42)

  16. Using Symmetry Use the shortest strategy to find the vertex of this quadratic: y = (x – 2)(x + 8) The x-intercepts are at 2 and -8 Half-way between these intercepts is the vertex at x = -3. At this point: y = (-3 – 2)(-3 + 8) = (-5)(5) = -25 Therefore, the vertex is (-3, -25)

  17. Jeopardy Review of Quadratics

  18. Homework • Page 255 #15, 16, 17, 19 – 21

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