12. (1 point) The height y (in feet) of a ball thrown by a child

is
y = −
1
16
x
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) To find the height of the ball when it leaves the child's hand, we need to find the value of y when x is 0, as the ball is thrown horizontally from that point.

Substituting x = 0 into the equation, we have:

y = -1/16(0)^2 + 4(0) + 5
y = -1/16(0) + 0 + 5
y = 0 + 0 + 5
y = 5

Therefore, the ball is 5 feet high when it leaves the child's hand.

(b) To find the maximum height of the ball, we need to find the vertex of the quadratic equation. The vertex form of a quadratic equation is given by y = a(x-h)^2 + k, where (h, k) is the vertex.

The equation y = -1/16x^2 + 4x + 5 can be rewritten in vertex form as:

y = -1/16(x^2 - 64x) + 5

Completing the square, we have:

y = -1/16(x^2 - 64x + 1024) + 5 + 1/16(1024)
y = -1/16(x - 32)^2 + 5 + 64
y = -1/16(x - 32)^2 + 69

The vertex of this parabola is (32, 69), so the maximum height of the ball is 69 feet.

(c) To find the distance from the child where the ball strikes the ground, we need to find the x-coordinate when y is 0.

0 = -1/16x^2 + 4x + 5

Multiplying both sides by 16 to eliminate the fraction, we have:

0 = -x^2 + 64x + 80

Rearranging the equation, we get:

x^2 - 64x - 80 = 0

Using the quadratic formula to solve for x, we have:

x = (-(-64) ± √((-64)^2 - 4(1)(-80))) / (2(1))
x = (64 ± √(4096 + 320)) / 2
x = (64 ± √4416) / 2
x = (64 ± 66.41) / 2

x ≈ -1.205 or x ≈ 65.205

Since the distance from the child cannot be negative, the ball strikes the ground approximately 65.205 feet away from the child.

To answer these questions, we can use the given equation for the height of the ball as a function of the horizontal distance. Let's break down the steps to solve each part:

(a) How high is the ball when it leaves the child’s hand?
To find the height of the ball when it leaves the child's hand, we need to determine the value of y when x is 0 (because x represents the horizontal distance from the point of throw). Let's substitute x = 0 into the equation and solve for y:
y = -(1/16)(0)^2 + 4(0) + 5
y = -0 + 0 + 5
y = 5

Therefore, the ball is 5 feet high when it leaves the child's hand.

(b) What is the maximum height of the ball?
To find the maximum height of the ball, we need to determine the vertex of the parabolic equation. The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula x = -b/(2a).

In our equation, a = -1/16 and b = 4. Let's substitute these values into the formula:
x = -4 / (2*(-1/16))
x = -4 / (-1/8)
x = -4 * (-8/1)
x = 32

Now that we have the x-coordinate of the vertex, we can substitute it back into the equation to find the y-coordinate:
y = -(1/16)(32)^2 + 4(32) + 5
y = -(1/16)(1024) + 128 + 5
y = -64 + 128 + 5
y = 69

Therefore, the maximum height of the ball is 69 feet.

(c) How far from the child does the ball strike the ground?
To find where the ball strikes the ground, we need to determine the value of x when y is 0 (since the height is 0 at ground level). Let's substitute y = 0 into the equation and solve for x:
0 = -(1/16)x^2 + 4x + 5
This is a quadratic equation that we can solve by factoring, completing the square, or using the quadratic formula. Factoring might not be straightforward in this case, so let's use the quadratic formula:

x = [-4 ± sqrt(4^2 - 4(-1/16)(5))] / (2*(-1/16))
x = [-4 ± sqrt(16 + 5/4)] / (-1/8)
x = [-4 ± sqrt(64 + 5) / (-1/8)
x = [-4 ± sqrt(69)] / (-1/8)

Since we're looking for a positive distance, we can disregard the negative solution:
x = [-4 + sqrt(69)] / (-1/8)
x = -8 / (1/8) + sqrt(69) / (1/8)
x = -8 * 8 + 8 * sqrt(69)
x = -64 + 8 * sqrt(69)
x ≈ -0.897 feet

Therefore, the ball strikes the ground approximately 0.897 feet away from the child.

(a) y(0) = ?

(b)
this is just a parabola. You know the vertex is at x = -b/2a

(c) y=0 at x=?

c'mon - don't forget your Algebra I now that you're taking calculus...