SOLUTION: An athlete whose event is the shot put releases a shot whose path is shown by the graph to the right is released at an angle of 45°, its height, f(x), in feet, can be modeled by f
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-> SOLUTION: An athlete whose event is the shot put releases a shot whose path is shown by the graph to the right is released at an angle of 45°, its height, f(x), in feet, can be modeled by f
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Question 1167170: An athlete whose event is the shot put releases a shot whose path is shown by the graph to the right is released at an angle of 45°, its height, f(x), in feet, can be modeled by f(x)= -0.02x^2 + 1.0x + 5.6, where x is the shot's horizontal distance, in feet, from its point of release
What is the shot's maximum horizontal distance, to the nearest tenth of a foot, or distance of the throw?
(I've tried using the quadratic formula to solve it but I cannot seem to get the decimal it wants.) Answer by ikleyn(52790) (Show Source):
The quadratic formula has nothing to do with this problem.
It is about finding the maximum of a quadratic function.
Solution
The maximum is where the VERTEX is.
The x-coordinate of the vertex is at x = , where "a" is the coefficient at x^2 and "b" is the coefficient at x.
In your case, a = -0.02, b = 1.0, so the x-coordinate of the vertex is
x = = = 25.
Thus we just found the horizontal coordinate: the maximum will happen at x = 25 feet.
Now, to find the maximum height (y-coordinate), substitute x= 25 into the given formula for the height.
