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Let's work through the problem step by step based on the provided equation for the path of the basketball, \( y = -(x - 5)^2 + 25 \), where \( y \) is the height in meters and \( x \) is the horizontal distance in meters.

### (a) Is the turning point a maximum or a minimum?

The given equation is in the form of a **quadratic** equation. Since the coefficient of \( (x - 5)^2 \) is negative, the parabola opens **downwards**, indicating that the turning point is a **maximum**.

### (b) What are the coordinates of the turning point?

The equation \( y = -(x - 5)^2 + 25 \) is written in **vertex form**, \( y = a(x - h)^2 + k \), where the vertex (turning point) is at \( (h, k) \).

From the equation:
- \( h = 5 \) (the value of \( x \) that makes \( (x - 5) = 0 \)),
- \( k = 25 \).

Thus, the turning point is **(5, 25)**.

### (c) What is the y-intercept?

To find the y-intercept, set \( x = 0 \) and solve for \( y \):
\[
y = -(0 - 5)^2 + 25 = -25 + 25 = 0
\]
So, the y-intercept is **(0, 0)**.

### (d) What is the maximum height of the ball?

The maximum height corresponds to the y-value of the turning point. From part (b), the y-coordinate of the turning point is **25 meters**, so the maximum height of the ball is **25 meters**.

### (e) What is the height of the ball at these horizontal distances?

We substitute the given values of \( x \) into the equation \( y = -(x - 5)^2 + 25 \) to find the height \( y \).

- For \( x = 3 \):
  \[
  y = -(3 - 5)^2 + 25 = -(2)^2 + 25 = -4 + 25 = 21
  \]
  The height at \( x = 3 \) is **21 meters**.

- For \( x = 7 \):
  \[
  y = -(7 - 5)^2 + 25 = -(2)^2 + 25 = -4 + 25 = 21
  \]
  The height at \( x = 7 \) is **21 meters**.

- For \( x = 10 \):
  \[
  y = -(10 - 5)^2 + 25 = -(5)^2 + 25 = -25 + 25 = 0
  \]
  The height at \( x = 10 \) is **0 meters**.

---

Would you like more details on any part of the solution? Here are five questions you might find useful for extending this knowledge:

1. How would the equation change if the maximum height occurred at a different horizontal distance?
2. Can you derive the quadratic equation from its factored form, starting from the vertex form?
3. What happens to the shape of the parabola if the coefficient of \( (x - 5)^2 \) is positive instead of negative?
4. How can you find the axis of symmetry for any quadratic equation in vertex form?
5. If the ball reached a height of 15 meters, at what horizontal distance would this occur?

### Tip:
When dealing with quadratic equations, the vertex form \( y = a(x - h)^2 + k \) makes it easy to identify the turning point and whether the parabola opens upwards or downwards!

The path of a basketball is given by y = -(x - 5)^2 + 25, where y meters is the height and x meters is the horizontal distance. The questions are: a) Is the turning point a maximum or a minimum? b) What are the coordinates of the turning point? c) What is the y-intercept? d) What is the maximum height of the ball? e) What is the height of the ball at these horizontal distances: x = 3, x = 7, and x = 10?

This problem explores a quadratic equation describing the path of a basketball. Learn to find the turning point, y-intercept, maximum height, and heights at specific horizontal distances. Suitable for students in Grades 9-10 studying quadratic functions.

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