For a quadratic $h(t) = at^2 + bt + c$, the maximum (if the parabola opens downward) occurs at the vertex. Recall the formula for the $t$-coordinate of the vertex: $t = -\frac{b}{2a}$.

Once you know the value of $t$ at the vertex, plug that $t$ back into $h(t)$ to get the rocket’s maximum height.

In Desmos, type h(t) = -t^2 + 6t + 7 (you can use x instead of t if you prefer: y = -x^2 + 6x + 7). This will graph the rocket’s height over time.

Click or tap on the top of the parabola, or use Desmos’s built-in feature by typing maximum(h) (or using the menu on the graph). The point that appears is the vertex of the parabola; the y-coordinate of this point is the rocket’s maximum height.

The height is modeled by the quadratic function

Because the coefficient of $t^2$ is negative ($-1$), the parabola opens downward. That means the rocket’s maximum height occurs at the vertex of this parabola.

So the task is to find the vertex’s $y$-value (the maximum value of $h(t)$).

For a quadratic in the form $h(t) = at^2 + bt + c$, the $t$-coordinate of the vertex is given by

Here $a = -1$, $b = 6$, and $c = 7$. Substitute $a$ and $b$ into the formula:

Simplify the fraction to find the time when the rocket reaches its highest point.

First simplify the expression for $t$:

So the rocket reaches its maximum height at $t = 3$ seconds. Now find the height at that time by substituting $t = 3$ into $h(t)$:

Therefore, the maximum height the rocket reaches is 16 meters.