The problem involves Amaya serving a volleyball, with the ball's height modeled by the quadratic equation:

$y = -0.02(x - 18)^2 + 12$

Where:

• $x$ is the horizontal distance (in feet) from Amaya.
• $y$ is the height (in feet) of the ball.

Let’s analyze the two parts of the question:

Part (a): How far is the ball from Amaya when it is at its maximum height? Explain.

The equation provided is in vertex form: $y = a(x - h)^2 + k$ Where $(h, k)$ is the vertex of the parabola. In this case:

• $h = 18$, which represents the horizontal distance from Amaya where the ball reaches its maximum height.
• $k = 12$, which represents the maximum height of the ball.

So, the ball is 18 feet away from Amaya when it reaches its maximum height.

Part (b): Describe how you would find the ball’s height when it crosses the net at $x = 30$.

To find the height of the ball when it crosses the net, substitute $x = 30$ into the equation for $y$:

$y = -0.02(30 - 18)^2 + 12$

First, calculate $(30 - 18)^2$: $(30 - 18) = 12$ $12^2 = 144$

Now substitute into the equation: $y = -0.02(144) + 12$ $y = -2.88 + 12$ $y = 9.12$

Thus, the ball is approximately 9.12 feet high when it crosses the net at $x = 30$.

Would you like more details on any step? Here are 5 related questions for practice:

1. How do you determine the vertex of a quadratic equation in standard form?
2. What does the value of $a$ in the equation $y = a(x - h)^2 + k$ tell you about the parabola's direction?
3. How do you find the horizontal distance where the ball is at a specific height, such as 10 feet?
4. Can you explain why the maximum height occurs at $x = h$ in vertex form?
5. How would the trajectory change if the value of $a$ were less negative (e.g., $a = -0.01$)?

Tip:

To quickly find the maximum or minimum of a parabola, use the vertex form of the equation—it directly gives the vertex!