The given problem provides the function $f(x) = -x^2 + 6x + 7$, which models the trajectory of a ball launched by Joshua, where $f(x)$ represents the height of the ball (in feet), and $x$ is the time elapsed (in seconds). The task is to determine the range of this quadratic function.

Solution:

1. The function $f(x) = -x^2 + 6x + 7$ is a quadratic equation with a negative leading coefficient ($-x^2$). This indicates that the parabola opens downwards and has a maximum value.

2. The vertex of a quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. Substituting $a = -1$ and $b = 6$: $x = -\frac{6}{2(-1)} = 3$

3. To find the maximum height of the ball, substitute $x = 3$ into $f(x)$: $f(3) = -(3)^2 + 6(3) + 7 = -9 + 18 + 7 = 16$

   Thus, the maximum height is 16 feet.

4. The range of the function corresponds to all the possible heights of the ball. Since the parabola opens downwards and reaches a maximum height of 16, the range is: $0 \leq y \leq 16$ (The ball cannot have a negative height.)

Answer:

The correct range is $0 \leq y \leq 16$.

Would you like further details or have additional questions?

Related Questions:

1. How can you determine the domain of a quadratic function in this context?
2. Why does the parabola open downward in this problem?
3. How do you calculate the vertex of a quadratic function quickly?
4. What would change in the range if the parabola opened upwards?
5. How can you graph the function $f(x) = -x^2 + 6x + 7$?

Tip:

Always analyze the leading coefficient ($a$) of a quadratic function to determine whether it opens upwards ($a > 0$) or downwards ($a < 0$). This affects whether it has a minimum or maximum value!