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The problem presents the quadratic function:

\[
f(x) = -2x^2 + ax + b
\]

You are given the following conditions:
1. The **x-coordinate of the vertex** is 3.
2. The graph of the function is **tangential to the line** \( y = 10 \).
3. You are asked to find the value of \( f(2) \).

### Step 1: Use the Vertex Formula
The formula for the x-coordinate of the vertex for a quadratic equation \( f(x) = Ax^2 + Bx + C \) is:

\[
x = -\frac{B}{2A}
\]

For the given function \( f(x) = -2x^2 + ax + b \), we know:
- \( A = -2 \)
- \( B = a \)

Thus, the x-coordinate of the vertex is:

\[
x = -\frac{a}{2(-2)} = \frac{a}{4}
\]

Since the x-coordinate of the vertex is given as 3, we set:

\[
\frac{a}{4} = 3 \implies a = 12
\]

### Step 2: Use the Tangency Condition
The function is tangential to the line \( y = 10 \). This means that at some point, the value of \( f(x) \) is 10, and the derivative (slope) of the function at that point is 0 because tangency implies that the curve just touches the line without crossing it.

Let's find where the derivative of \( f(x) \) equals 0:

\[
f'(x) = \frac{d}{dx} (-2x^2 + 12x + b) = -4x + 12
\]

Set this equal to 0 to find the x-coordinate where tangency occurs:

\[
-4x + 12 = 0 \implies x = 3
\]

Thus, the function is tangent to the line \( y = 10 \) at \( x = 3 \). We also know that \( f(3) = 10 \), so substitute \( x = 3 \) into the original function:

\[
f(3) = -2(3)^2 + 12(3) + b = 10
\]

Simplifying:

\[
f(3) = -18 + 36 + b = 10 \implies 18 + b = 10 \implies b = -8
\]

### Step 3: Find \( f(2) \)
Now that we know \( a = 12 \) and \( b = -8 \), substitute these values into the original function:

\[
f(x) = -2x^2 + 12x - 8
\]

Find \( f(2) \):

\[
f(2) = -2(2)^2 + 12(2) - 8 = -2(4) + 24 - 8 = -8 + 24 - 8 = 8
\]

### Final Answer:
\[
f(2) = 8
\]

Let me know if you need further clarification or if you have any questions!

Here are 5 related questions for further practice:
1. How would the vertex change if the coefficient of \( x^2 \) was different?
2. How do you determine if a parabola opens upwards or downwards?
3. How would you solve this if the quadratic was not tangential to a line?
4. What is the significance of the derivative being 0 at the vertex?
5. How would you find the vertex for a non-quadratic polynomial?

**Tip:** For any quadratic function, the vertex formula \( x = -\frac{b}{2a} \) is essential to quickly finding key properties!

For the quadratic function f(x) = -2x^2 + ax + b, a and b are constants such that the x-coordinate of the vertex is 3. If the graph of the function is tangential to the line y = 10, then find the value of f(2).

Solve the quadratic function f(x) = -2x^2 + ax + b with given conditions: vertex at x = 3 and tangency to y = 10. Learn how to apply the vertex formula and derivative for tangency to find f(2). Suitable for Grades 10-12.

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