# 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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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.

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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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