The problem involves modeling the motion of a ball thrown upward and outward, with its height described by the quadratic function:

$f(x) = -0.2x^2 + 2.1x + 8$

where $x$ is the ball's horizontal distance from the point where it was thrown. Let's analyze each part of the problem:

Part a: Maximum Height and Horizontal Distance

The maximum height of the ball occurs at the vertex of the parabola. The formula to find the vertex for a quadratic function $f(x) = ax^2 + bx + c$ is given by:

$x = -\frac{b}{2a}$

Here, $a = -0.2$ and $b = 2.1$. Substituting these values:

$x = -\frac{2.1}{2(-0.2)} = \frac{2.1}{0.4} = 5.25 \, \text{feet}$

Now, to find the maximum height, substitute $x = 5.25$ back into the function:

$f(5.25) = -0.2(5.25)^2 + 2.1(5.25) + 8$

Calculating this:

$f(5.25) = -0.2(27.5625) + 11.025 + 8$ $f(5.25) = -5.5125 + 11.025 + 8 = 13.5125 \, \text{feet}$

So, the maximum height is approximately 13.5 feet and it occurs 5.25 feet from the point where the ball was thrown.

Part b: Horizontal Distance When the Ball Hits the Ground

To find the horizontal distance when the ball hits the ground, we solve for $x$ when $f(x) = 0$ (i.e., when the height is zero). We solve the quadratic equation:

$-0.2x^2 + 2.1x + 8 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = -0.2$, $b = 2.1$, and $c = 8$. Substituting these values:

$x = \frac{-2.1 \pm \sqrt{(2.1)^2 - 4(-0.2)(8)}}{2(-0.2)}$ $x = \frac{-2.1 \pm \sqrt{4.41 + 6.4}}{-0.4}$ $x = \frac{-2.1 \pm \sqrt{10.81}}{-0.4}$ $x = \frac{-2.1 \pm 3.29}{-0.4}$

Thus, the two solutions are:

$x_1 = \frac{-2.1 + 3.29}{-0.4} = \frac{1.19}{-0.4} = -2.975 \, \text{(discarded as negative)}$ $x_2 = \frac{-2.1 - 3.29}{-0.4} = \frac{-5.39}{-0.4} = 13.475 \, \text{feet}$

Thus, the ball travels approximately 13.5 feet horizontally before hitting the ground.

Part c: Correct Graph of the Parabolic Path

The graph that correctly models the ball’s motion should have:

• A maximum height around 13.5 feet.
• The ball hitting the ground at approximately 13.5 feet horizontally.

Among the options presented, Graph B best fits this behavior, with the peak at around 13.5 feet and hitting the ground at a similar horizontal distance.

Would you like further details or have any questions about this solution?

Here are 5 related questions to consider:

1. How does changing the coefficient $a$ in a quadratic function affect the parabola's shape?
2. Why does the ball's path follow a parabolic trajectory?
3. What would happen to the ball's maximum height if it were thrown with a greater initial velocity?
4. How can we model the ball's velocity using the derivative of the height function?
5. What physical factors, like air resistance, are neglected in this model?

Tip: For any quadratic equation, the vertex always gives either the maximum or minimum point, depending on the parabola's orientation.