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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.

A ball is thrown vertically upward with an initial velocity of 9 m/s. Its height, in meters, can be modeled by the function h(t) = -5t^2 + 12t + 2.1, where t is the time in seconds after the ball was thrown. The question asks for graphical determination of the vertex, the time the ball is in the air, and a graph of the function.

Learn how to find the vertex and the time in the air of a ball thrown vertically upward using the quadratic function h(t) = -5t^2 + 12t + 2.1. This problem covers key algebraic concepts like finding the vertex of a parabola and solving using the quadratic formula, suitable for high school students.

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