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Today in Precalculus

Today in Precalculus. Go over homework Notes: Parabolas Completing the square Writing the equation given the graph Applications Homework. Example 1. Prove that the graph of y 2 + 2y – 8x – 7 = 0 is a parabola y 2 + 2y = 8x + 7 y 2 + 2y + 1 = 8x + 7 + 1 y 2 + 2y + 1 = 8x + 8

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Today in Precalculus

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  1. Today in Precalculus • Go over homework • Notes: Parabolas • Completing the square • Writing the equation given the graph • Applications • Homework

  2. Example 1 Prove that the graph of y2 + 2y – 8x – 7 = 0 is a parabola y2 + 2y = 8x + 7 y2 + 2y + 1 = 8x + 7 + 1 y2 + 2y + 1 = 8x + 8 (y+1)2 = 8 (x + 1) Standard form for a parabola

  3. Example 1 (y+1)2 = 8 (x + 1) Graph: Opens to the right vertex (-1, -1) 4p = 8, p = 2 focal length =2 focal width = 8 focus: (1, -1) directrix: x = -3 axis: y = -1

  4. Example 2 Prove that the graph of x2 – 2x + 3y + 7 = 0 is a parabola x2 – 2x = -3y – 7 x2 – 2x + 1 = -3y – 7 + 1 x2 – 2x + 1 = -3y – 6 (x – 1)2 = -3(y + 2) Standard form for a parabola

  5. Example 2 (x – 1)2 = -3(y + 2) Graph: Opens downward vertex (1, - 2) p = -3/4 focal length = -3/4 focal width = 3 focus: (1, -11/4) directrix: y = -5/4 axis: x = 1

  6. Writing an equation of a parabola Find vertex, substitute into general form vertex: (-2, -2) (y+2)2=4p(x+2) Find another point and substitute for x and y (1,-5) (-5 + 2)2 = 4p(1 + 2) Solve for p 9 = 12p p = ¾ (y+2)2=4(¾)(x+2) (y+2)2=3(x+2)

  7. Writing an equation of a parabola vertex: (-4, -1) (x + 4)2=4p(y+1) point (2,-4) (2 + 4)2 = 4p(-4 + 1) 36 = -12p p = -3 (x + 4)2=4(-3)(y + 1) (x+4)2=-12(y+1)

  8. Writing an equation of a parabola vertex: (2, 1) (y – 1)2=4p(x – 2) point (-1,4) (4 – 1)2 = 4p(-1 – 2) 9 = -12p p = -3/4 (y – 1)2=4(-3/4)(x – 2) (y – 1)2=-3(x – 2)

  9. Application Example 1 Let the focus F have coordinates (0, p) and the vertex be at (0,0) equation: x2 = 4py Because the reflector is 3ft across and 1 ft deep, the points (1.5, 1) and (-1.5, 1) must lie on the parabola. (1.5)2 = 4p(1) 2.25 = 4p p = .5625ft = 6.75in So the microphone should be placed inside the reflector along its axis and 6.75 inches from its vertex.

  10. Application Example 2 Let the roadway be the x-axis. Then the vertex is at (500,25) So the equation is: (x – 500)2 = 4p(y – 25) Points (0, 150) and (1000, 150) also on the parabola.

  11. Application Example 2 So the equation is: (0 – 500)2 = 4p(150 – 25) 250,000 = 500p p = 500 (x – 500)2 = 2000(y – 25) Equation for the shape of the main cable.

  12. Application Example 2 To find the length of the support cables, solve equation for y Support cables are every 100ft, so starting with x = 100, then 200, 300, etc. Cables are 105ft, 70ft, 45ft, 30ft, 25ft, 30ft, 45ft, 70ft, 105ft.

  13. Homework Page 641: 51-56, 59-63 Quiz:

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