To determine whether the quadratic function has a maximum or minimum value, we can examine the coefficient of the $x^2$ term. The function given is:

$f(x) = -3x^2 - 12x$

Step 1: Check the sign of the quadratic term

• The coefficient of $x^2$ is $-3$, which is negative.
• When the coefficient of $x^2$ is negative, the parabola opens downward, meaning the function has a maximum value.

Step 2: Find the vertex

The vertex of a quadratic function in the form $ax^2 + bx + c$ occurs at $x = \frac{-b}{2a}$, where $a$ and $b$ are the coefficients of $x^2$ and $x$, respectively.

For $f(x) = -3x^2 - 12x$:

• $a = -3$
• $b = -12$

Using the vertex formula: $x = \frac{-(-12)}{2(-3)} = \frac{12}{-6} = -2$

Step 3: Find the value of the function at the vertex

Now, substitute $x = -2$ into the function to find the maximum value: $f(-2) = -3(-2)^2 - 12(-2)$ $f(-2) = -3(4) + 24 = -12 + 24 = 12$

Conclusion:

• The quadratic function has a maximum value.
• The maximum value is $12$.

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Would you like more details or have any further questions? Here are some related questions:

1. How do you find the vertex of a quadratic function in general form?
2. Why does a negative coefficient of $x^2$ indicate a maximum?
3. How would the graph of $f(x) = -3x^2 - 12x$ look like?
4. What happens if the coefficient of $x^2$ is positive instead?
5. How can completing the square also help find the maximum or minimum?

Tip: The vertex formula $x = \frac{-b}{2a}$ is key to quickly finding the maximum or minimum of any quadratic function!