part of my regents packet

Question

Let h(t) = 16t ^2 + 64t + 80 represent the height of an object above the ground after t seconds. Determine the number of seconds it takes to achieve its maximum height. Justify your answer. State the time interval, in seconds, during which the height of the object decreases. Explain your reasoning.

Answer 1

The equation should be -16t^2 +64t +80. The maximum height will be at the vertex

The x coordinate of the vertex is at -b/2a which in this case is -64/-32 = 2

When t=2 h(t)= -16(2)^2 +64(2) + 80= -64 + 128 +80 = 144

The height of the object will be decreasing from the time it reaches its maximum height until it hits the ground. At this point h(t) will =0 so solve -16t^2 +64t +80 = 0 Factor out 16 to start

-16( t^2 -4t -5)=0

t^2 -4t -5 = 0

(t-5)(t+1)=0 t=5 or -1 ( can discard -1) so object hits the ground at t=5 which means height is decreasing in the interval between 2 and 5 seconds