Let's break down the problem based on the image provided:

We have a function that models the height of a ball thrown vertically upward:

$h(t) = -5t^2 + 12t + 2.1$ where:

• $h(t)$ is the height in meters,
• $t$ is the time in seconds.

Part a) Finding the Vertex:

The vertex of a quadratic function $h(t) = at^2 + bt + c$ occurs at the time $t = -\frac{b}{2a}$. Here, the coefficients are:

• $a = -5$,
• $b = 12$.

Using the formula for the vertex: $t_{\text{vertex}} = -\frac{12}{2(-5)} = \frac{12}{10} = 1.2 \text{ seconds}.$

Now, plug $t = 1.2$ into the equation for height to find the maximum height: $h(1.2) = -5(1.2)^2 + 12(1.2) + 2.1 = -5(1.44) + 14.4 + 2.1 = -7.2 + 14.4 + 2.1 = 9.3 \text{ meters}.$

Thus, the vertex is at $t = 1.2$ seconds, and the maximum height is 9.3 meters.

Part b) Time in the Air:

The ball hits the ground when $h(t) = 0$. We solve the equation:

$-5t^2 + 12t + 2.1 = 0$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -5$, $b = 12$, and $c = 2.1$. Let's compute the discriminant first:

$\Delta = 12^2 - 4(-5)(2.1) = 144 + 42 = 186$

Now, solve for $t$:

$t = \frac{-12 \pm \sqrt{186}}{-10} = \frac{-12 \pm 13.64}{-10}$

This gives two solutions:

$t_1 = \frac{-12 + 13.64}{-10} = -0.164 \quad (\text{Not feasible, negative time})$ $t_2 = \frac{-12 - 13.64}{-10} = 2.664 \text{ seconds}.$

Thus, the ball is in the air for approximately 2.7 seconds.

Part c) Graphing the Function:

The graph of this function would be a parabola opening downward (since the coefficient of $t^2$ is negative). Important points to plot are:

• The initial height at $t = 0$: $h(0) = 2.1$ meters.
• The vertex at $t = 1.2$ seconds, $h(1.2) = 9.3$ meters.
• The time when the ball hits the ground, at $t = 2.7$ seconds, $h(2.7) = 0$ meters.

Would you like more details on any step or further explanations?

Five follow-up questions:

1. How do we derive the quadratic formula?
2. Can you explain how to interpret the vertex of a parabola in a physical context?
3. Why is the discriminant important in solving quadratic equations?
4. How would the problem change if the ball were thrown with a different initial velocity?
5. What is the significance of the negative solution for time in part b)?

Tip:

Always check your work by substituting key points back into the original equation to verify accuracy, especially with quadratic functions.