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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.

Layla launches a toy rocket from a platform. The height of the rocket in feet is given by h = -16t^2 + 136t + 72 where t represents the time in seconds after launch. How many seconds have gone by when the rocket is at its highest point?

Learn how to calculate when a toy rocket reaches its highest point using the quadratic equation h(t) = -16t^2 + 136t + 72. This problem demonstrates key concepts in algebra and projectile motion, suitable for students in Grades 9-11.

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