To find the range of the function \( f(x) = -x^2 - 4x - 4 \), we first note that this is a quadratic function, which opens downwards since the coefficient of \( x^2 \) is negative.
To determine the maximum value of this function, we can complete the square or use the vertex formula for a quadratic function.
The vertex form of a quadratic function is given by:
\[ f(x) = a(x-h)^2 + k \]
where \((h, k)\) is the vertex of the parabola. The \(x\)-coordinate of the vertex can be found using:
\[ h = -\frac{b}{2a} \]
For the quadratic \(f(x) = -x^2 - 4x - 4\), we have:
- \(a = -1\)
- \(b = -4\)
Now, calculate \(h\):
\[ h = -\frac{-4}{2 \cdot -1} = -\frac{4}{-2} = 2 \]
Now we can calculate the value of \(f\) at \(x = -2\):
\[ f(-2) = -(-2)^2 - 4(-2) - 4 = -4 + 8 - 4 = 0 \]
The vertex of the parabola is at the point \((-2, 0)\). Since the parabola opens downward, the highest value (maximum value) of \(f(x)\) is \(0\) which occurs at \(x = -2\).
The function goes to \(-\infty\) as \(x\) moves away from \(-2\) in either direction.
Thus, the range of the function is all values less than or equal to \(0\):
\[ \text{Range} = (-\infty, 0] \]
So, the correct response is:
(−∞, 0]
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