Please help with this question. A rocket leaves a launcher at a height of 6 feet off the ground with an initial velocity of 128 feet per second. The equation describing the rocket's height after t seconds is h(t)=-16t^2+128t+6. Find the maximum height reached by the rocket and how many seconds it takes for the rocket to reach that height

Given:

The equation describing the rocket's height after t seconds is

To find:

The maximum height and time when it is maximum.

Solution:

We have,

Leading coefficient is -16, which is negative. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

If a quadratic function is , then

In the given function, a=-16, b=128 and c=6.

Putting x=4 in the given function, we get

So, vertex of the function is (4,262).

Therefore, maximum reached by the rocket is 262 feet it 4 seconds.

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

step-by-step explanation:

no solution.