Question

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.

Answer 1

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 $f(x) = -0.25(x-2)^2 + 8$ then when the player is 2ft from the net, we can plug in x = 2:

$f(2) = -0.25(2-2)^2 + 8 = 0 + 8ft$

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:

$f(4) = -0.25(4 - 2)^2 + 8 = -1 + 8 = 7 ft$

As 7 ft < 7 ft 4 in, this time the ball will NOT go over the net