Arocket is launched upward so that its distance, in feet, above the ground after t seconds is represented by the function below. what is its maximum height? h(t) = -14t^2 +420t a) 13,020 ft b) 5684 ft c) 3150 ft d) 1740 ft e) none

The maximun height that the rocket will reach is 3150 ft.

Step-by-step explanation:

This problem requieres that we maximize the function. For this we need the function to be of the form:

We already have that covered, as the height of the rocket is represented by:

We start calculating the derivative of h(t), wich is:

We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.

So we find the value in wich the derivative equals zero:

This means that 15 seconds is the time at wich the rocket will reach it's maximum height.

We replace this value in the original equation to solve the problem:

With this we can conclude that the maximum height the rocket will reach is 3150 ft.

60

step-by-step explanation:

the image is given below

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this