The given quadratic equation for the height $h$ of the toy rocket as a function of time $t$ is:

$h(t) = -16t^2 + 24t + 40$

This represents the height of the rocket at any time $t$, where $t$ is in seconds and $h$ is in feet.

Interpretation of the Vertex:

The general form of a quadratic equation is $h(t) = at^2 + bt + c$. The coordinates of the vertex of a parabola in this form can be found using the formula for the time of the vertex (or maximum height in this case):

$t_{\text{vertex}} = \frac{-b}{2a}$

Here, $a = -16$, $b = 24$, and $c = 40$.

Step 1: Calculate the time at the vertex.

$t_{\text{vertex}} = \frac{-24}{2(-16)} = \frac{-24}{-32} = 0.75 \, \text{seconds}$

So, the vertex occurs at $t = 0.75$ seconds after launch.

Step 2: Calculate the height at the vertex.

Substitute $t = 0.75$ into the height equation:

$h(0.75) = -16(0.75)^2 + 24(0.75) + 40$ $h(0.75) = -16(0.5625) + 18 + 40$ $h(0.75) = -9 + 18 + 40 = 49 \, \text{feet}$

Interpretation:

• The vertex $(0.75, 49)$ tells us that the rocket reaches its maximum height of 49 feet at 0.75 seconds after launch.
• This means that the highest point (or apex) of the rocket's trajectory occurs 0.75 seconds after launch, at which time the rocket is 49 feet above the ground.

Would you like further details on this? Here are 5 follow-up questions to explore:

1. How long is the rocket in the air before it returns to the ground?
2. What is the rocket's height at $t = 1$ second?
3. How can you derive the time it takes for the rocket to hit the ground?
4. What is the initial height of the rocket at $t = 0$?
5. Can we find the time intervals when the rocket is higher than 40 feet?

Tip: The vertex of a quadratic function often represents the maximum or minimum value, depending on the orientation of the parabola. In the case of a negative leading coefficient (like here), the vertex is the maximum.