Questions: The graph of the function has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. f(x)=-x^2-6x-9 The relative extreme point on the graph is (Type an ordered pair.)

The graph of the function has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. f(x)=-x^2-6x-9 The relative extreme point on the graph is (Type an ordered pair.)
Transcript text: The graph of the function has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. \[ f(x)=-x^{2}-6 x-9 \] The relative extreme point on the graph is $\square$ (Type an ordered pair.)
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Solution

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Solution Steps

Step 1: Find the derivative of the function

To find the relative extreme point, we first need to find the derivative of the function \( f(x) = -x^2 - 6x - 9 \).

\[ f'(x) = \frac{d}{dx}(-x^2 - 6x - 9) = -2x - 6 \]

Step 2: Set the derivative to zero to find critical points

Set the derivative equal to zero to find the critical points.

\[ -2x - 6 = 0 \]

Solving for \( x \):

\[ -2x = 6 \\ x = -3 \]

Step 3: Find the y-coordinate of the critical point

Substitute \( x = -3 \) back into the original function to find the y-coordinate.

\[ f(-3) = -(-3)^2 - 6(-3) - 9 = -9 + 18 - 9 = 0 \]

The relative extreme point is \((-3, 0)\).

Step 4: Check the concavity at the critical point

To check the concavity, we find the second derivative of the function.

\[ f''(x) = \frac{d}{dx}(-2x - 6) = -2 \]

Since \( f''(x) = -2 < 0 \), the function is concave down at \( x = -3 \), indicating a relative maximum.

Final Answer

The relative extreme point on the graph is \((-3, 0)\).

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