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the height of a ball thrown straight up with a velocity of 96 ft/s is given by the quadratic function h(t)= -16t^2 +96t. what is the maximum height the ball reaches?
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to find this take the derivative and set it equal to 0. -32t+96=0 solve t=3 plug back in to the original. = 144
a simple way is to find the line of symmetry and sub it into the original equation t = -96/-32 t =3 max height is when t = 3 sub into the original equation
the line of symmetry comes from the general quadratic formula its just x = -b/2a in this question b = 96 and a = -16 so its -96/(2x-16)
thanks
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