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.