The height y (in feet) of a ball thrown by a child is
y=−1/14x^2+4x+5
where x is the horizontal distance in feet from the point at which the ball is thrown.

(a) how high is the ball when it leaves the child's hand?


(b) what is the maximum height of the ball?


(c) how far from the child does the ball strike the ground?

  (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.

dont think so

step-by-step explanation:

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.