Ja'Von kicks a soccer ball into the air. The function f(x) = –16(x – 2)2 + 64 represents the height of the ball, in feet, as a function of time, x, in seconds. What is the maximum height the ball reaches?

The maximum height the ball reaches is 64 feet

Step-by-step explanation:

The given function is f(x) = -16·(x - 2)² + 64

From the equation, the path described by the ball is an inverted n-shaped parabola

The maximum height is therefore, the tip of the parabola

At the maximum height, the slope = 0 because the tip is momentarily flat

Since the slope (y₂ - y₁)/(x₂ - x₁)  = The derivative Δy/Δx, we find the derivative of the function and equate it to zero to find the coordinates at the  maximum height

Δy/Δx = dy/dx = d(-16·(x - 2)² + 64)/dx = -32·(x - 2) = -32·x + 64

To check if it is a maximum, we have;

d²y/dx² = d(-32·x + 64)/dx = -32 which is negative, indicating that the slope is reducing and we at the maximum point of the slope

Therefore for the maximum height Δy/Δx = dy/dx = -32·x + 64 = 0

64 = 32·x

x = 64/32 = 2 seconds

We now have the x-value at the slope, the f(x) value, is therefore;

f(2) = -16·(2 - 2)² + 64 = 64 feet

Therefore, the maximum height the ball reaches is 64 feet.

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