A rocket is launched upward with a velocity of 96 feet per second from the top of a 40 foot stage. What is the maximum height attained by the rocket?

2344 feet is the max height

Step-by-step explanation:

1) we find the function.

the function will be a parabola with a y intercept of 40.

the max height will be at the vertex.

f(x) = ax^2 + bx + 40

2) we find the derivative

f'(x) = 2ax + b

3) since the velocity is 96 at x = 0, we get the function:

f'(x) = 2ax + 96

4) find the x intercept of the velocity function

0 = 2ax + 96

-96 = 2ax

-48/a = x

This is the time in seconds when the max height is reached.

5) we finish the equation f(x)

since the antiderivative of 96 is 96x, we can write the equation:

f(x) = -ax^2 + 96x + 40

a will be negative since the parabola is opening downwards.

6) Since -48/a and -b/2a are both equal, a = 1

f(x) = -x^2 + 96x + 40

7) substitute 48 into x

-48/a = 48

f(48) = -(48^2) + 96(48) + 40

f(48) = -2304  + 4608 + 40

f(48) = 2344

2344 feet is the max height

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

what is

step-by-step explanation: