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    The vertex of a parabola

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Since our projectile travels at a constant rate of 340 ft / sec in the horizontal direction, we have, using the formula D = RT, that our horizontal distance x = 340 t, giving
    t = x / 340
    Our above equation h(t) can now be converted into h(x), as follows

    h(x)=-16*(x/340)^2+340*x/340

    giving

    h(x)=-x^2/7225+x

    What is the ordered pair which constitutes the vertex of our parabolic equation h(x)?
    At what distance from our firing point does our projectile reach maximum height?
    What are the ordered pairs corresponding to the zeros of the above equation h(x)
    (solutions to the equation -x^2/7225+x=0)?
    How far does our projectile travel?

    © BrainMass Inc. brainmass.com December 24, 2021, 9:28 pm ad1c9bdddf
    https://brainmass.com/math/algebra/vertex-parabola-382521

    SOLUTION This solution is FREE courtesy of BrainMass!

    The vertex of a parabola y=ax^2+bx+c is at the point x=-b/(2a). In our case we have a=-1/7225, and b=1, so the vertex is at
    x=-1/(-2/7225) = 7225/2 = 3612.5ft. This is the distance from the firing point, where the projectile reaches its maximal height.

    To find the zeros of the function h(x), solve the equation
    -x^2/7225 + x = 0
    factor out x:
    x(-x/7225 + 1) = 0
    so x=0 or -x/7225 + 1=0, x/7225 = 1, x = 7225ft
    The second number, x=7225ft is the maximal distance our projectile travels.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:28 pm ad1c9bdddf>
    https://brainmass.com/math/algebra/vertex-parabola-382521

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