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Math Help - Parabolic Arch

  1. #1
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    Parabolic Arch

    Suppose you wanted to design a parabolic arch (a parabola opening downward) as a trellis in your garden. This would mean that the first term of the quadratic equation, the term with the variable raised to the second power, would have to be negative. An engineering friend suggests that the following quadratic function might be aesthetically pleasing.
    h(x) = -5x2/64 + 4
    This function gives the height of the arch (in feet) at any horizontal distance x (in feet) from the centerline of the arch. The trellis (parabola) could be supported by vertical posts at intervals across the inside of the arch. These posts could be placed at 1, 2, 3, 4, 5, 6, and 7 feet on both sides of the centerline (the axis of the parabola). Therefore, for any distance (x-value) the height of a post supporting the arch at that point could be determined from the above function.

    1) When drawing the graph of this parabola, where is its axis located in the coordinate plane?

    2) What is the maximum height of the arch?

    If someone could help with this, that would be great. I almost have all my questions answered to help prepare me so thanks tons for everyone who has helped.

    Kasey
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  2. #2
    MHF Contributor
    Joined
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    The axis (of symmetry) of a parabola is the line passing through the vertex. In "your" parabola, it is the vertical line passing through the highest point of the arch.

    The maximum height of your parabola
    h(x) = -(5/64)x^2 +4
    is, by using x = -b /2a,
    x = -0 / 2(-5/64) = 0 ----------this means the axis of symmetry is at x=0, which is the y-axis in your graph.
    So,
    h(0) = 0 +4 = 4

    That only means that the maximum height of your parabola is 4 ft only.
    In your graph, the vertex, or maximum height is at point (0,4).
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