To interpret the coordinates of the vertex of the quadratic function $h(t) = -16t^2 + 80t + 96$, we can start by finding the vertex itself.

Step 1: Determine the Vertex

The general form of a quadratic function is $h(t) = at^2 + bt + c$, where the $t$-coordinate (or $x$-coordinate in generic terms) of the vertex can be calculated using: $t = -\frac{b}{2a}$

In this function:

• $a = -16$
• $b = 80$

So, $t = -\frac{80}{2 \cdot (-16)} = \frac{80}{32} = 2.5$

Therefore, the $t$-coordinate of the vertex is 2.5 seconds.

Step 2: Find the $h$-coordinate of the Vertex

To find the height (or $h$-coordinate) at $t = 2.5$ seconds, we substitute $t = 2.5$ back into the function: $h(2.5) = -16(2.5)^2 + 80(2.5) + 96$

Calculating each term:

1. $-16 \cdot (2.5)^2 = -16 \cdot 6.25 = -100$
2. $80 \cdot 2.5 = 200$
3. $96$ (constant term)

Adding them up: $h(2.5) = -100 + 200 + 96 = 196$

Therefore, the height at the vertex is 196 feet.

Interpretation of the Vertex Coordinates

1. The $t$-coordinate (or $x$-coordinate):

   • The $t$-coordinate of the vertex is 2.5 seconds. This represents the time after launch when the rocket reaches its maximum height.

2. The $h$-coordinate (or $y$-coordinate):

   • The $h$-coordinate of the vertex is 196 feet. This represents the maximum height of the rocket after launch.

In summary:

• The rocket reaches its maximum height of 196 feet at 2.5 seconds after being launched.

Would you like further explanation on any part of this solution?

---

Here are some related questions to explore:

1. What does the initial height (96 feet) represent in this context?
2. How would the maximum height change if the initial speed were different?
3. Can we determine the total time the rocket is in the air before it lands?
4. What is the significance of the coefficient of $t^2$ being negative?
5. How would a change in the launch angle affect the height function?

Tip: In projectile motion problems, the vertex often represents the maximum height if the path is a parabolic arc.