Answer:
2 seconds, assuming the "?" in the equation is supposed to be ^2 (the exponent of 2): f(h) = -16x^2 +64x + 80
Step-by-step explanation:
You can either graph the functuion or take the first derivative to find the maximum height. The first derivative will give us the slope for a values of x. The slope at the top height of the rocket is zero, since it has stopped and is starting it's way back down.
First Derivative:
f(x) = -16x^2 +64x + 80
f'(x) = -32x + 64
Set this equal to zero and solve for x
0 = -32x + 64
x = 2 seconds.
Use 2 seconds in the original equation to find the height at 2 seconds.
f(2) = -16x^2 +64x + 80
f(2) = -16(2)^2 +64*(2) + 80
f(2) = 144 feet
Graph:
A graph is attached. You can locate the maximum at (2,144)
