A ball is being tossed across a field.The path is a parabola, and the equation of the path is f(x)=-4x^2+8x. what is the highest the ball will be above the flat field and for what value of x?

Expert Answers

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The equation of the path followed by the ball is given by y = -4x^2 + 8x where x represents the horizontal coordinate and y is the corresponding height.

The graph of the path followed by the ball is:

The ball is at its highest point when the curvature of the graph changes. At this point the first derivative of the function f(x) = -4x^2 + 8x is equal to 0

f'(x) = -8x + 8

f'(x) = 0

=> -8x + 8 = 0

=> x = 1

At x = 1, f(x) = -4 + 8 = 4

The ball is at the highest point when x = 1 and its height at this point is equal to 4.

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