HELP! Algebra 2 Quadratics!?

Question

A rocket is shot straight into the air at an initial velocity of 96 ft./sec. The height of the rocket after x seconds is represented by the formula h = 96x – 16x2. What is the maximum height the rocket reaches?
6 ft.
144 ft.
180 ft.
240 ft.

Answer 1

equation:
h = 96 x - 16(x^2)
first derivation of length or height is velocity
second derivation of length or height is acceleration
first derivation:
h'<=> 96 - 2.16x = 0
h'<=> 96 - 32x = 0
h'<=> - 32x = -96
so - 32x = -96
- 32x =-96
x = 3 sec
back to first equation and substitute x by 3
h = 96x – 16x2.
<=> 96(3) – 16(3^2).
<=> 144 ft without initial velocity
but u have initial velocity of 96 ft./sec
first derivation:
h'<=> (96 - 2.16x) + 96 = 0
h'<=> 192 - 32x = 0
h'<=> - 32x = -192
so - 32x = -192
x = 6sec
back to first equation and substitute x by 6
h = 96x – 16x2.
<=> 96(6) – 16(6^2).
<=> 0 ft
so may be it mean in 6 second your rocket will fall down.
and a peak only 144 ft....

i'm sorry .......... if this answer make u unsatisfied

Answer 2

Velocity = dh/dx = 96 - 32x
At, Maximum height, velocity becomes 0,
So set the velocity = 0 and find time(x),
0 = 96 - 32x
x = 3 seconds...........At maximum height.

Maximum height = h(3) = 96(3) - 16(9)
maximum height = 144 ft

Answer 3

This amounts to finding the vertex of the parabola.

h = -16x² + 96x
h = -16(x² - 6x)

Complete the square.

h -16*9 = -16(x² - 6x + 9)
h - 144 = -16(x - 3)²

The maximum height is 144 ft.
It occurs at x = 3 seconds.

Answer 4

If you're in algebra 2 you should be able to understand this. I was in Alg. 1 & we had quite a few problems such as this.

Fill in 'h' with the given amount.
Move 16x^2 to the other side, so all amounts are positive.
divide.
You should be able to get it from there.