if the line y=-x+3 is tangent to the graph of f(x)=-x^2-x+k what is k

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 \( x \):

\[
-2x - 1 = -1 \implies
\]

If the line y = -x + 3 is tangent to the graph of f(x) = -x^2 - x + k, what is k?

Find the value of k such that the line y = -x + 3 is tangent to the quadratic function f(x) = -x^2 - x + k. This problem involves algebraic concepts like solving quadratic equations and applying the condition for tangency. Suitable for students in Grades 9-12.

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