Answer:
(a) 5 ft
(b) 21 ft
(c) 17.165 ft
Step-by-step explanation:
(a) When y = 0, y = ...
y = -1/4·0² +4·0 +5
y = 5
The ball is 5 ft high when it leaves the child's hand.
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(b) For quadratic ax²+bx+c, the x-coordinate of the vertex is ...
x = -b/(2a)
For this height function, we have a=-1/4 and b = 4, so the x-coordinate of the vertex is ...
x = (-4)/(2(-1/4)) = 8
The height when x = 8 is ...
y = (-1/4·8 +4)·8 +5 = 2·8 +5 = 21
The maximum height of the ball is 21 feet.
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(c) The ball will hit the ground where y=0, so we want find x at that point.
0 = -1/4x² +4x +5
0 = -1/4(x² -16x) +5 = -1/4(x² -16x +64) +21 . . . . complete the square
0 = -1/4(x -8)² +21 . . . . . . . write as a square
84 = (x -8)² . . . . . . . . . . . . multiply by -4, add 84
2√21 = x -8 . . . . . . . . . . . take the positive square root
x = 8 + 2√21 ≈ 17.165 . . . . feet
The ball strikes the ground about 17.165 feet from the child.
