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!