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anonymous

  • one year ago

A building has an entry the shape of a parabolic arch 96 ft high and 18 ft wide at the base as shown below. A parabola opening down with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is ninety six feet and its width from left to right is eighteen feet. Find an equation for the parabola if the vertex is put at the origin of the coordinate system.

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  1. mathstudent55
    • one year ago
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    |dw:1436844571604:dw|

  2. anonymous
    • one year ago
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    ok

  3. anonymous
    • one year ago
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    whats do i do now?

  4. mathstudent55
    • one year ago
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    \(y = ax^2 + bx + c\) Use each of the three points in the equation above to come up with three equations. Then solve the equations simultaneously for a, b, and c.

  5. anonymous
    • one year ago
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    how do i use that formula u gave me?

  6. anonymous
    • one year ago
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    because its at the origin, and its pointing down, the y would be the height

  7. mathstudent55
    • one year ago
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    \(0 = a\cdot 0^2 + b \cdot 0 + c\) c = 0

  8. anonymous
    • one year ago
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    so y= -96, a=0?

  9. anonymous
    • one year ago
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    o ok

  10. mathstudent55
    • one year ago
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    That was for point (0, 0). We have c = 0.

  11. anonymous
    • one year ago
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    ok

  12. anonymous
    • one year ago
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    So y = 96, since it's axis of symmetry is the origin, the x's on either side would be -9 and 9 SO we have now two points on the parabola. (9,96) and (-9,96)

  13. anonymous
    • one year ago
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    (96 = a*-9^2 + b*-9 + 0) (96 = a*9^2 + b*9 + 0) these would be the two equations?

  14. mathstudent55
    • one year ago
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    Now let's try the point (9, -96) \(y = ax^2 + bx\) There is no c bec we already know c is zero. \(-96 = a(9^2) + b(9) \) \(81a + 9b = -96\) Here is one equation.

  15. anonymous
    • one year ago
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    2 equations? there will be only one equation

  16. mathstudent55
    • one year ago
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    Now we use the point (-9, -96) to get the other equation. \(-96 = a(-9)^2 + b(-9) \) \(-96 = 81a - 9b\) \(81a - 9b = -96\) Here is the other equation. Since now we only have the two variables a and b, all we need is 2 equations. We can now find a and b. We need to solve the system of equations below: \(81a + 9b = -96\) \(81a - 9b = -96\)

  17. anonymous
    • one year ago
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    The vertex form, as you probably know, is: y = a(x - h)^2 + k where (h, k) is the vertex and that is point (0, 0), so your equation becomes: y = a(x - 0)^2 + 0 or y = ax^2 to find "a", you have to realize that (9, -96) is a point, so: -96 = a(9^2) = 81a a = -96/81 = -32/27 so, y = (-32/27)x^2 vertex form: y = (-32/27)(x - 0)^2 + 0

  18. anonymous
    • one year ago
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    im confused now?

  19. anonymous
    • one year ago
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    from there you can just expand the vertex form to get the standard form

  20. mathstudent55
    • one year ago
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    Add the equations to eliminate b: \(162a = -192\) \(a = -\dfrac{192}{162} = -\dfrac{32}{27} \) Now we need b.

  21. anonymous
    • one year ago
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    I have no idea what math student is doing, we do not need2 equations!!!

  22. anonymous
    • one year ago
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    ok can i get run by this step by step because im so lost

  23. anonymous
    • one year ago
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    using vertex form first then expanding to get standard form is quite simple. Follow my previous long post

  24. anonymous
    • one year ago
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    The vertex form, as you probably know, is: y = a(x - h)^2 + k where (h, k) is the vertex and that is point (0, 0), so your equation becomes: y = a(x - 0)^2 + 0 or y = ax^2 to find "a", you have to realize that (9, -96) is a point, so: -96 = a(9^2) = 81a a = -96/81 = -32/27 so, y = (-32/27)x^2 vertex form: y = (-32/27)(x - 0)^2 + 0 Then multiply it out to get the standard form.

  25. anonymous
    • one year ago
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    OH i now see what math student is doing. Albeit a bit lengthy, but works

  26. mathstudent55
    • one year ago
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    \(81a + 9b = -96\) \(81\left( -\dfrac{32}{27} \right) + 9b = -96\) \(-2592 + 243b = -2592\) \(243b = 0\) \(b = 0\) Now we can write the equation of the parabola as: \(y = ax^2 + bx + c\) \(y = -\dfrac{32}{27}x^2 \)

  27. anonymous
    • one year ago
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    so from y = (-32/27)(x - 0)^2 + 0 the standard form would be y = (-32/27)x^2?

  28. anonymous
    • one year ago
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    Essentially the same thing, but using vertex method then expanding takes fewer time (just a hint for next time)

  29. anonymous
    • one year ago
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    Yes chris

  30. anonymous
    • one year ago
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    so that would be my equation for the parabola if the vertex is put at the origin of the coordinate system?

  31. anonymous
    • one year ago
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    yes, that is what we both got.

  32. anonymous
    • one year ago
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    perfect thank you very much

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