The line x + y = k, where k is a constant, is a tangent line
to the graph of the function f(x) = 2x - 5x + 2. What is the
value of K
(AI -1
(B) 2 (C) -2 (d)-4

We are given the function \( f(x) = 2x^2 - 5x + 2 \) and the line \( x + y = k \), which is tangent to the graph of the function. To find the value of \( k \), we'll follow these steps:

### Step 1: Express the line equation in slope-intercept form
The given line is \( x + y = k \), which can be rewritten as:
\[
y = -x + k
\]
So the slope of the line is \( -1 \).

### Step 2: Find the derivative of \( f(x) \)
The slope of the tangent line to the function at any point is given by the derivative of \( f(x) \). First, let's find the derivative of \( f(x) = 2x^2 - 5x + 2 \):
\[
f'(x) = \frac{d}{dx}(2x^2 - 5x + 2) = 4x - 5
\]
So the slope of the tangent line at any point \( x \) is \( 4x - 5 \).

### Step 3: Set the slopes equal
Since the line is tangent to the graph of the function, the slope of the line must equal the slope of the tangent to the curve at the point of tangency. Therefore, we set the slope of the line equal to the derivative of \( f(x) \):
\[
4x - 5 = -1
\]
Solving for \( x \):
\[
4x = 4 \implies x = 1
\]

### Step 4: Find the y-coordinate of the point of tangency
Now that we have \( x = 1 \), substitute it into the original function \( f(x) \) to find the corresponding \( y \)-coordinate:
\[
f(1) = 2(1)^2 - 5(1) + 2 = 2 - 5 + 2 = -1
\]
So the point of tangency is \( (1, -1) \).

### Step 5: Find the value of \( k \)
At the point \( (1, -1) \), the point lies on the line \( y = -x + k \). Substituting \( x = 1 \) and \( y = -1 \) into this equation:
\[
-1 = -(1) + k
\]
\[
-1 = -1 + k
\]
\[
k = 0
\]

Thus, the value of \( k \) is \( \boxed{0} \), but this option isn't in the provided list. There might be a mistake in either the problem or the options provided.

Would you like further clarification on this, or would you like to explore a related concept?

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Here are 5 questions that expand on this concept:
1. How do we find the derivative of any quadratic function?
2. What does the slope of a tangent line represent in calculus?
3. How do we solve a system of equations involving a function and a line?
4. Why is it important that the line is tangent to the graph, and not just intersecting it?
5. How can we solve similar problems for functions other than quadratics?

**Tip**: The slope of a tangent line can always be found by differentiating the function!

The line x + y = k, where k is a constant, is a tangent line to the graph of the function f(x) = 2x^2 - 5x + 2. What is the value of k? (A) -1 (B) 2 (C) -2 (D) -4

Learn how to find the value of k for the tangent line x + y = k to the quadratic function f(x) = 2x^2 - 5x + 2. This problem involves algebraic concepts, differentiation, and solving for tangent lines. Suitable for students in Grades 10-12.

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