If a > 0 then the parabola opens up. - PowerPoint PPT Presentation

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If a > 0 then the parabola opens up.

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  1. maximum minimum 3.3 Factored Form of a Quadratic Relation All quadratic equations can be modeled in the form: y = a(x – s)(x – t) provided a¹ 0. If a < 0 then the parabola opens down. If a > 0 then the parabola opens up.

  2. The axis of symmetry is at s t The vertex is at y = a(x – s)(x – t) The zeros are at x = s and x = t.

  3. a) Determine the zeros of the parabola. The zeros are x = 0 and x = 3. Example 1: Given:y = x(x – 3) Let y = 0 0 = x(x – 3)

  4. c) Determine the vertex of the parabola. Substitute into the b) Determine the equation of the axis of symmetry. original equation. Given:y = x(x – 3) The axis of symmetry is halfway between the zeros. x =

  5. Given:y = x(x – 3)

  6. 20 – w = 0 Example 2: Determine the zeros and the axis of symmetry of the following: a) A = w(20 – w) 20 = w w = 0 The zeros are at 0 and 20 The axis of symmetry is halfway between 0 and 20. w = 10

  7. 4x – 8= 0 Determine the zeros and the axis of symmetry of the following: b) y = 2x(4x – 8) 4x = 8 2x = 0 x = 0 x = 2 The zeros are at 0 and 2 The axis of symmetry is halfway between 0 and 2. x = 1

  8. The x-coordinate of the vertex is at x = (1, – 4) Example 3 Sketch the graph of: y = (x + 1)(x – 3) The zeros are at – 1 and 3. Substitute x = 1 into the original equation to find the vertex. y = (1 + 1)(1 – 3) y = – 4

  9. Example: Sketch the graph of: y = – (4 – x)(8 – x) The zeros are at 4 and 8. The axis of symmetry is at x = 6 Substitute x = 6 in the equation to determine the y value of the vertex. y = – (4 – 6)(8 – 6) y = –(– 2)(2) y = 4 The vertex is at (6, 4).

  10. Homework • Pg 155 #1, 2, 4, 5, 7, 18

  11. Example 4 Determine the equation of the parabola whose x-intercepts are 2 and – 3 and whose y-intercept is 6. sub s = 2 and t = – 3 y = a(x – s)(x – t) To find a, substitute (0,6) into the equation. y = a(x – 2)(x + 3) 6 = a(0 – 2)(0 + 3) 6 = a(– 2)(3) 6 = a(– 6) y = – (x – 2)(x + 3) – 1 = a

  12. 15 m 150 m b) Determine the height of the ball when it was 20 m from the hole. A person made a golf shot that went over a tree and into the cup. The tree was 15 m high and the distance from the green was 150 m. a) Determine an equation that traces the path of the ball.

  13. (75,15) 15 m (0, 0) 150 m (150, 0) y = a(x – s)(x – t) sub s = 0, t = 150 y = a(x – 0)(x – 150) sub point (75, 15) 15 = a(75 – 0)(75 – 150) 15 = a(75)(–75)

  14. (75,15) (0, 0) (150, 0) y = a(x – s)(x – t) s = 0, t = 150 This is the equation of the flight of the golf ball.

  15. (75,15) (130,y) (0, 0) (150, 0) 130 m 20 m Substitute x = 130 into the equation. y = 6.9 b) Determine the height of the golf ball when it was 20 m from the hole. The height was 6.9 m.

  16. Retailing Problem: Maximizing CD Revenue. Max operates a store that sells CDs. All CDs sell for $20 each. In order to increase revenue, he decides to increase his price. He knows that over the last six months he has sold an average of 280 Cd’s per day at $20 each. Market research indicates that for every $0.50 increase in price, daily sales will drop by five units. What unit price will maximize Max’s daily profit? ?

  17. Let x represent the number of price increases. Let 5x represent each loss of sales. Recall: Revenue = Price × number of units sold Price = 20 + 0.50x Number of CDs sold = 280 – 5x

  18. Equation: Revenue = price × number of CDs sold 20 + 0.50x = 0 280 – 5x = 0 The axis of symmetry is at: = 8 = $5760.00 is the maximum revenue Revenue = (20 + 0.50x)(280 – 5x) Determine the zeros: 20 = – 0.50x 280 = 5x – 40 = x 56 = x The zeros are at – 40 and 56 Substitute x = 8 into the equation. Revenue = (20 + 0.50×8)(280 – 5×8) = (24.00)(240)

  19. $160 $140 $120 $100 $80 $60 Increase in profit $40 $20 0 $7.50 $2.50 $5.00 Increase in price His maximum revenue will be $5760.00. The price of each CD will be $24.00 and the number of sales will be 240 per day.

  20. Example: Sketch the graph of: y = – (4 – x)(8 – x) The zeros are at 4 and 8. The axis of symmetry is at x = 6 Substitute x = 6 in the equation to determine the y value of the vertex. y = – (4 – 6)(8 – 6) y = –(– 2)(2) y = 4 The vertex is at (6, 4).

  21. Example The x-intercepts are –1 and – 3 and the y-intercept is –3. Determine the equation of the parabola. y = a(x + 1)(x + 3) sub (0, – 3) into the equation –3 = a(0 + 3)(0 + 1) –3 = 3a –1 = a equationy = – (x + 1)(x + 3)

  22. Bridges The second span of the Bluewater Bridge Sarnia, Ontario is supported by a pair of steel parabolic arches. The arches are set in the concrete foundations that are on opposite sides of the St. Clair River 281 m apart. The top of each arch rises 71 m above the river. Determine and algebraic expression that models the arch.

  23. (0, 71) (–140.5,0) (140.5,0) The distances from the center to the zeros is 140.5 m so the zeros are at (–140.5,0) and (140.5,0). The vertex is at (0, 71).

  24. y = a(x – s)(x – t) s = –140.5 and t = 140.5 y = a( x + 140.5)(x – 140.5) To solve for a, substitute (0, 71) 71 = a(0 + 140.5)(0 – 140.5) 71 = a(–140.5)(140.5) 71 = a(–19740.25) – 0.004 = a Equation:y = – 0.004(x – 140.5)(x + 140.5)

  25. Homework • Pg 156 #3, 6, 8 – 16 • Workbook pg 43 and 44