trig

the height of a projectile is modeled by the equation y=-2x^2+38x+10, where x is a time, in seconds, and y is height, in feet, During what interval of time, to the nearest tenth of a second, is the projectile at least 125 feet above ground

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asked by nate
  1. y=-2x^2+38x+10
    we look for y>125
    First we solve for points where y=125:
    125=-2x^2+38x+10
    -2x^2+38x-115=0
    Solving for x, we find
    x=3.78 or x=15.22
    Thus the projectile is over 125' between t=3.78 s and 15.22 s.

    Note: a parabola with the leading coefficient negative (-2x²) is concave downwards.

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  2. graph h vs t.

    I don't think it ever gets that high.

  3. It's a little confusing.

    The original equation for h is
    h(x)=-2x^2+38x+10 (Blue line)

    I subtracted 125 to find the roots when the projectile exceeds h=125 (red line)
    h(x)=-2x^2+38x+10-125, or
    h(x)=-2x^2+38x-115

    See:
    http://img337.imageshack.us/img337/4594/1292008237.png

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