Sally launches a rocket off of a platform. The height of the rocket, in meters, can be modeled by the function, h(x) = -5(x - 3)2 + 56, where x is in seconds. o What is the height of the platform? o What is the maximum height the rocket reaches? o How long does it take to reach this height?

Time = 3secs

Max height = 56m

Step-by-step explanation:

Given the height reached by the rocket modeled by the equation;

h(x) = -5(x - 3)² + 56 where;

x is in seconds

The rocket velocity at its maximum height is zero.

Hence dh/dx = 0

dh/dx = 2(-5)(x-3)

dh/dx = -10(x-3)

Since dh/dx = 0

0 = -10(x-3)

0 = -10x + 30

10x = 30

x = 3secs

Hence it takes 3 secs to reach the maximum height.

Get the maximum height reached by the rocket. Substitute x = 3 into the equation given;

Recall that:

h(x) = -5(x - 3)² + 56

If x  = 3

h(3) = -5(3 - 3)² + 56

h(3) = 0 + 56

h(3) = 56

Hence the maximum height that the rocket reaches is 56m

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)

where are the graphs?

step-by-step explanation: