The problem provided states:

Given that the maximum value of the function $y = -x^2 + 6x + k$ is 12, find the value of $k$.

Solution:

1. The function $y = -x^2 + 6x + k$ is a quadratic function, and since the coefficient of $x^2$ is negative, the parabola opens downward, meaning it has a maximum value.

2. The formula for the x-coordinate of the vertex of a quadratic function $y = ax^2 + bx + c$ is given by: $x = \frac{-b}{2a}$ In this case, $a = -1$ and $b = 6$, so: $x = \frac{-6}{2(-1)} = 3$

3. The y-coordinate of the vertex represents the maximum value of the function. Substituting $x = 3$ into the equation to find the maximum value of $y$: $y = -(3)^2 + 6(3) + k$ Simplifying: $y = -9 + 18 + k = 9 + k$

4. We're told that the maximum value of $y$ is 12, so: $9 + k = 12$ Solving for $k$: $k = 12 - 9 = 3$

Thus, the value of $k$ is 3.

Would you like any additional details or clarifications?

Related Questions:

1. How do you find the vertex of a quadratic function in general?
2. What happens to the graph of the quadratic function when $k$ changes?
3. How can you determine if a quadratic function has a maximum or minimum value?
4. How would the solution change if the coefficient of $x^2$ were positive?
5. How do you find the x-intercepts of the function $y = -x^2 + 6x + k$?

Tip:

For quadratic functions, the maximum or minimum value always occurs at the vertex, which can be easily found using the formula $x = \frac{-b}{2a}$.