The given problem involves a quadratic function representing the height $h$ of a toy rocket launched from a platform as a function of time $t$:

$h(t) = -16t^2 + 136t + 72$

This is a quadratic equation of the form $h(t) = at^2 + bt + c$, where:

• $a = -16$
• $b = 136$
• $c = 72$

To find how many seconds have gone by when the rocket is at its highest point, we need to determine the time at which the vertex of this parabola occurs. For a quadratic function, the time $t$ at the vertex can be found using the formula:

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

Substituting the values of $a$ and $b$:

$t = -\frac{136}{2(-16)} = \frac{136}{32} = 4.25 \text{ seconds}$

So, the rocket reaches its highest point 4.25 seconds after launch.

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

Here are 5 related questions:

1. What is the maximum height the rocket reaches?
2. How do we find when the rocket hits the ground after launch?
3. Can you explain how the value of $a$ affects the direction of the parabola?
4. How does the value of $b$ influence the position of the vertex?
5. What is the significance of the constant term $c$ in the quadratic equation?

Tip: In quadratic motion problems, the time to reach the maximum height can always be found using the vertex formula $t = -\frac{b}{2a}$, as this gives the peak of a parabolic path.