The height in meters of a rocket after t seconds can be modeled by
h(t) = -4.9(t - 4)2 + 80
a) You can see this as the equation of a parabola in vertex form: y = a(x - h)2 + k
So the vertex is (4, 80) , and it is a maximum.
The maximum height is 80 meters , and it occurs when t = 4 seconds.
Also you can see that any possible value other than 0 for -4.9(t - 4)2 will take away from the 80 .
b) The y-intercept occurs when t = 0 .
h(0) = -4.9(0 - 4)2 + 80
h(0) = 1.6
So the y-intercept is 1.6 .
It is the height of the rocket at 0 seconds – the height when it was launched.
c) Let's find t such that h(t) = 60 .
-4.9(t - 4)2 + 80 = 60 Subtract 80 from both sides
-4.9(t - 4)2 = -20 Divide both sides by -4.9 .
(t - 4)2 = 200/49 Take the ± square root of both sides.
t - 4 = ±√200 / 7 Add 4 to both sides.
t = ±√200 / 7 + 4
t ≈ 6.02 and t ≈ 1.98
Any value for t that falls between 1.98 and 6.02 will cause h to be greater than 60 .
6.02 - 1.98 = 4.04 . So it will be above 60 meters for about 4.04 seconds.
d) average rate of change = change in height / change in time
avg rate of change = [ h(5.5) - h(2.5) ] / [ 5.5 - 2.5 ]
= [ ( -4.9(5.5 - 4)2 + 80 ) - ( -4.9(2.5 - 4)2 + 80 ) ] / [ 5.5 - 2.5 ]
= [ ( -4.9(1.5)2 + 80 ) - ( -4.9(-1.5)2 + 80 ) ] / [ 3 ]
Notice that the two terms in the numerator will be the same, so their difference is 0 .
= [ 0 ] / [ 3 ]
= 0
e) This one is done the same as the last problem.
avg rate of change = [ h(6) - h(3) ] / [6 - 3]
= [ ( -4.9(6 - 4)2 + 80 ) - ( -4.9(3 - 4)2 + 80 ) ] / [ 6 - 3 ]
Can you finish it from here?
f) The rocket hits the ground when h(t) = 0 . That is..when
-4.9(t - 4)2 + 80 = 0
All you need to do is solve this equation for t . If you get 8 (or something a little bigger than 8) as a solution, then Perry is right...if not, Perry is wrong. 
