David kicks a ball off the ground. After t seconds, it’s height, h (in feet), is given by the formula h=-16t^2+80t. How many seconds does the ball reach its maximum height? What is the maximum height reached by the ball?

The ball reaches its maximum height at 2.5 seconds

The maximum height is 100 feet

Step-by-step explanation:

Maximum Value of Functions

We use the derivative of a function to find its maximum or minimum value over a given interval.

Given a function y=f(x), the first derivative criterion establishes if x=a is such that:

f'(a)=0, and f''(a) < 0 then x=a is a maximum of f.

The height h in feet of a ball after t seconds is:

Find the first derivative:

Equate to 0:

Subtract 80:

Divide by -32:

t=2.5 seconds

Find the second derivative:

h''(t)=-32

Since h''(t) is always negative, then

The ball reaches its maximum height at 2.5 seconds

To find the value of the maximum height, substitute t=2.5 into the function:

The maximum height is 100 feet

idk but i found this "the graph of y = f(x-2) is the graph of y = f(x) translated two units rightward.  

so, since the point (9,2) lies on the graph of y = f(x), the point (11,2) lies on the graph of y = f(x-2)."

step-by-step explanation:

link is

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step-by-step explanation: