To find the value of $k$ such that the line $y = -x + 3$ is tangent to the graph of the function $f(x) = -x^2 - x + k$, we need to ensure that the line and the curve intersect at exactly one point, and at that point, their slopes are equal.

Step 1: Set up the equations

The line is given by:

$y = -x + 3$

The function is given by:

$f(x) = -x^2 - x + k$

For tangency, these two expressions for $y$ must be equal at some point:

$-x^2 - x + k = -x + 3$

Step 2: Simplify the equation

Move all terms to one side of the equation:

$-x^2 - x + k - (-x + 3) = 0$

Simplify it:

$-x^2 + k - 3 = 0$

Or:

$-x^2 = 3 - k$

Step 3: Find the slope

Let’s differentiate $f(x) = -x^2 - x + k$ to find the slope of the tangent line at the point of tangency:

$f'(x) = -2x - 1$

Since the line is tangent, the slope of the line must equal the derivative of the function at the point of tangency. So:

$-2x - 1 = -1$

Now, let’s solve for $x$:

$-2x - 1 = -1 \implies$