The given function is $y = -x^2 - 6x + k$. We need to find the value of $k$ such that the maximum value of this function is 2.

The function $y = -x^2 - 6x + k$ is a downward-opening parabola because the coefficient of $x^2$ is negative. The maximum value of this parabola occurs at its vertex.

The vertex of a parabola given by $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$.

For the function $y = -x^2 - 6x + k$:

• $a = -1$
• $b = -6$

The x-coordinate of the vertex is: $x = -\frac{-6}{2(-1)} = \frac{6}{-2} = -3$

Now, we substitute $x = -3$ back into the function to find the y-coordinate (the maximum value): $y = -(-3)^2 - 6(-3) + k = -9 + 18 + k = 9 + k$

We are given that the maximum value is 2: $9 + k = 2$

Solving for $k$: $k = 2 - 9 = -7$

Therefore, the value of $k$ is $\boxed{-7}$.

Would you like any further details or have any questions? Here are some other questions you might find interesting:

1. How do you find the vertex of a quadratic function?
2. Can you explain the process of completing the square?
3. What are the properties of parabolas in the context of quadratic functions?
4. How does changing the coefficients in a quadratic function affect its graph?
5. What is the importance of the discriminant in quadratic equations?

Tip: When dealing with quadratic functions, always remember the vertex formula for finding maximum or minimum values efficiently.