https://file.aiquickdraw.com/mathbot/file/1726483365155edk4a9ne.jpg

Let's start by analyzing the problem and graph you provided in the image. 

### Part D: General Form of the Equation of the Parabola

The graph displays a parabola that opens to the right, indicating it is in the form:

\[
(y - k)^2 = 4p(x - h)
\]

where \((h, k)\) is the vertex of the parabola and \(p\) is the distance from the vertex to the focus (or directrix).

From the graph:
- The vertex is at \((0, 0)\), since the parabola is centered at the origin.
- The focus seems to be located at \((2, 0)\), which means \(p = 2\).

Therefore, the equation of the parabola in standard form is:

\[
y^2 = 8x
\]

Now, to convert this to the general form, we expand and rearrange:

\[
y^2 = 8x \quad \Rightarrow \quad y^2 - 8x = 0
\]

This is the **general form** of the equation of the parabola.

### Part E.6: Ball Trajectory Problem

Given:
- The maximum height the ball reaches is 5 meters.
- The total horizontal distance is 10 meters (since the ball lands back at 10).

The equation of the ball’s height as a function of its horizontal distance \(x\) will take the form of a downward-opening parabola:

\[
y = ax^2 + bx + c
\]

To find the equation, use the following points:
- The ball starts and lands at \((0, 0)\) and \((10, 0)\), meaning when \(x = 0\) or \(x = 10\), \(y = 0\).
- The maximum height occurs at \(x = 5\), so \(y = 5\) when \(x = 5\).

Let me calculate the exact equation for this.The equation that gives the ball's height \( y \) as a function of its horizontal distance \( x \) is:

\[
y = -\frac{1}{5}x^2 + 2x
\]

### Part E.6.b: Height at \( x = 4 \)

Next, let's calculate the height of the ball when \( x = 4 \).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually compute the height at \(x = 4\) by plugging the value into the equation:

\[
y = -\frac{1}{5}x^2 + 2x
\]

Substitute \(x = 4\):

\[
y = -\frac{1}{5}(4)^2 + 2(4) = -\frac{16}{5} + 8 = -3.2 + 8 = 4.8
\]

So, the height at \(x = 4\) is **4.8 meters**.

Would you like more details or have any questions about this?

### Here are some related questions to expand on this:
1. How can we derive the equation for a different maximum height or distance?
2. What is the vertex form of a parabola and how is it useful?
3. How would the equation change if the ball had a different starting position?
4. How do we calculate the focus and directrix for a given parabola?
5. What are the applications of parabolas in physics, specifically projectile motion?

**Tip**: The vertex of a parabola always lies halfway between the roots (x-intercepts) if the parabola is symmetric.

What is the general form of the equation of the parabola and solve projectile motion problems involving a ball's trajectory?

Learn how to find the general form of a parabola and solve projectile motion problems. This example explores the equation of a parabola, calculating the height of a ball thrown in a parabolic path, and determining key points such as the vertex. Suitable for Grades 9-12 students studying algebra and quadratic functions.

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