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The equation provided in the question is:

\[
y = -\frac{1}{16}x^2 + 6x + 3
\]

Where \(y\) is the height of the ball and \(x\) is the horizontal distance from the point of throw.

Let's break down each part of the question:

### (a) How high is the ball when it leaves the child's hand?
To find the height when the ball leaves the child's hand, we need to substitute \(x = 0\) into the equation because this is the point where the ball is thrown (initial horizontal distance).

\[
y = -\frac{1}{16}(0)^2 + 6(0) + 3 = 3
\]

So, the ball is **3 feet** high when it leaves the child's hand.

### (b) What is the maximum height of the ball?
The equation is a quadratic equation, and the maximum height will occur at the vertex. The formula to find the \(x\)-coordinate of the vertex for a quadratic equation \(y = ax^2 + bx + c\) is:

\[
x = -\frac{b}{2a}
\]

In this case, \(a = -\frac{1}{16}\) and \(b = 6\), so:

\[
x = -\frac{6}{2 \times -\frac{1}{16}} = 48
\]

Now, substitute \(x = 48\) into the equation to find the maximum height:

\[
y = -\frac{1}{16}(48)^2 + 6(48) + 3
\]
\[
y = -\frac{1}{16}(2304) + 288 + 3 = -144 + 288 + 3 = 147
\]

So, the maximum height of the ball is **147 feet**.

### (c) How far from the child does the ball strike the ground?
The ball strikes the ground when \(y = 0\). We need to solve the equation:

\[
0 = -\frac{1}{16}x^2 + 6x + 3
\]

Multiply the entire equation by 16 to eliminate the fraction:

\[
0 = -x^2 + 96x + 48
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = -1\), \(b = 96\), and \(c = 48\).

\[
x = \frac{-96 \pm \sqrt{96^2 - 4(-1)(48)}}{2(-1)}
\]
\[
x = \frac{-96 \pm \sqrt{9216 + 192}}{-2}
\]
\[
x = \frac{-96 \pm \sqrt{9408}}{-2}
\]
\[
x = \frac{-96 \pm 97.98}{-2}
\]

This gives two solutions for \(x\):

\[
x_1 = \frac{-96 + 97.98}{-2} \approx \frac{1.98}{-2} \approx -0.99 \quad (\text{not valid, since distance cannot be negative})
\]
\[
x_2 = \frac{-96 - 97.98}{-2} \approx \frac{-193.98}{-2} \approx 96.99
\]

So, the ball strikes the ground **about 97 feet** from the child.

---

Do you have any questions or want more details?

Here are 5 related questions:
1. How do you determine the vertex of a quadratic function?
2. Why is there a negative sign in front of the \(x^2\) term?
3. How can you solve a quadratic equation without using the quadratic formula?
4. What is the significance of the discriminant in the quadratic formula?
5. How does the height equation change if the child throws the ball with more force?

**Tip:** Always check the sign of the \(x^2\) term in a quadratic equation to understand whether the parabola opens upwards or downwards (which helps determine if it's a maximum or minimum).

The height y (in feet) of a ball thrown by a child is y = -1/16x^2 + 6x + 3, 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?

Explore how to solve a quadratic function problem involving the height of a ball thrown by a child. Learn how to find the initial height, maximum height, and the distance at which the ball hits the ground using key concepts like the vertex formula and quadratic formula. Suitable for high school students in Grades 9-12.

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