The function f(x)equalsnegative 0.25 (x minus 2 )squared plus 8 models the path of a volleyball when the player is 2 feet from the net. The height of the net is 7 ft comma 4 in. Will the ball go over the net? What if the player moves back so she is 4 feet from the net? Explain.
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Answer:
The ball will go over the net when she's standing 2 feet away from the net, but not at 4 ft from the net
Explanation:
Suppose then when the player is 2ft from the net, we can plug in x = 2:
As 8 feet > 7 ft 4 in, the ball will go over the net
If the player moves back so that she's 4 feet from the net, plug in x = 4:
As 7 ft < 7 ft 4 in, this time the ball will NOT go over the net
Final answer:
The function models the path of a volleyball when the player is 2 feet from the net and will the ball go over the net at different distances.
Explanation:
The function f(x) = -0.25(x-2)^2 + 8 models the path of a volleyball when the player is 2 feet from the net. To determine if the ball will go over the net, we need to compare the height of the ball's path with the height of the net.
First, let's convert the height of the net to feet. The height, 7 ft 4 in, is equivalent to 7.33 feet. Now, substitute x = 2 into the function to find the height of the ball when the player is 2 feet from the net. f(2) = -0.25(2-2)^2 + 8 = 8 ft.
The ball will go over the net because the height of the ball, 8 feet, is greater than the height of the net, 7.33 feet.
If the player moves back so she is 4 feet from the net, substitute x = 4 into the function: f(4) = -0.25(4-2)^2 + 8 = 10 ft. The ball would still go over the net, as the height of the ball, 10 feet, is greater than the net's height.
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