The height of an object off the ground, h (in feet) t seconds after it is launched into the air is given by

h (t) = - 16t2 + 128t, 0 ≤ t ≤ 8.

Find the average rate of change of h over the interval

[4, 8].

Answer: - 64

Step-by-step explanation:

h (t) = - 16t² + 128t

dh/dt = - 32t + 128

At t = 4,

dh/dt = - 16 (4) ² + 128 = 0

At t = 4, dh/dt = 0

At t = 8,

dh/dt = - 16 (8) ² + 128 = - 128

At t = 8, dh/dt = - 128

The rate of change of h is constant, the average rate change is

(0 - 128) / 2 = - 64