algebra
posted by Anonymous .
an arrow is shot into the air with an initial velocity of 96 feet per second. the height in feet of the arrow t seconds after it was shot into the air is given by the function h(x)=16t^2+96t. find the maximum height of the arrow. what is the answer

h(x) = 16t^2 + 96t.
The parabola opens downward. Therefore,
the max. point is the vertex.
t(V) = b/2a = 96 / 32 = 3s = The value of t at the vertex.
h=16*3^2 + 96*3 = 144 + 288 = 144Ft
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