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Let's solve the problem step by step. The function given is:

\[
f(x) = -2x^2 + 4x + 9
\]

### (a) Find the critical numbers of \( f(x) \).
To find the critical numbers, we first take the derivative of \( f(x) \), set it equal to zero, and solve for \( x \).

\[
f'(x) = \frac{d}{dx} \left( -2x^2 + 4x + 9 \right)
\]

The derivative is:

\[
f'(x) = -4x + 4
\]

Now, set \( f'(x) = 0 \) and solve for \( x \):

\[
-4x + 4 = 0
\]
\[
4x = 4
\]
\[
x = 1
\]

So, the critical number is \( x = 1 \).

### (b) Find the open intervals where the function is increasing or decreasing.
The function is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \).

We know \( f'(x) = -4x + 4 \), so let's analyze its sign by testing values of \( x \) around the critical number \( x = 1 \).

1. For \( x < 1 \) (e.g., \( x = 0 \)):
   \[
   f'(0) = -4(0) + 4 = 4 > 0
   \]
   So, the function is increasing on \( (-\infty, 1) \).

2. For \( x > 1 \) (e.g., \( x = 2 \)):
   \[
   f'(2) = -4(2) + 4 = -8 + 4 = -4 < 0
   \]
   So, the function is decreasing on \( (1, \infty) \).

Thus, the function is:
- Increasing on \( (-\infty, 1) \)
- Decreasing on \( (1, \infty) \)

### (c) Apply the First Derivative Test to identify the relative extrema.
The First Derivative Test tells us that if the derivative changes from positive to negative at a critical point, there is a relative maximum, and if it changes from negative to positive, there is a relative minimum.

Since \( f'(x) \) changes from positive to negative at \( x = 1 \), there is a relative maximum at \( x = 1 \).

To find the \( y \)-coordinate of the relative maximum, we plug \( x = 1 \) into the original function \( f(x) \):

\[
f(1) = -2(1)^2 + 4(1) + 9 = -2 + 4 + 9 = 11
\]

So, the relative maximum is at \( (1, 11) \), and there is no relative minimum (DNE).

### Final answers:
- **(a)** Critical numbers: \( x = 1 \)
- **(b)** Increasing on \( (-\infty, 1) \), Decreasing on \( (1, \infty) \)
- **(c)** Relative minimum: DNE, Relative maximum: \( (1, 11) \)

Would you like more details or have any questions?

Here are 5 related questions:
1. What are the conditions for identifying critical points?
2. How can we apply the second derivative test for concavity?
3. How do we determine the global maximum or minimum on a closed interval?
4. What is the significance of the sign of the first derivative?
5. How does the behavior of a quadratic function relate to its derivative?

**Tip:** The derivative gives you the slope of the function at any point. Positive slope means increasing, and negative slope means decreasing!

Consider the function below.
 f(x) = -2x^2 + 4x + 9 
(a) Find the critical numbers of f. (Enter your answers as a comma-separated list.)
(b) Find the open intervals on which the function is increasing or decreasing.
(c) Apply the First Derivative Test to identify the relative extrema.

Solve the quadratic function f(x) = -2x^2 + 4x + 9 by finding the critical points, determining where the function is increasing or decreasing, and applying the First Derivative Test to find the relative extrema. Perfect for high school calculus students.

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