To determine the interval over which the graph of the function [tex]\( f(x) = -(x+8)^2 - 1 \)[/tex] is decreasing, we need to analyze its properties and behavior.
1. Understand the Function's Shape:
The function [tex]\( f(x) = -(x+8)^2 - 1 \)[/tex] is a quadratic function with the general form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], where [tex]\( a < 0 \)[/tex] indicates a downward-opening parabola. Here, [tex]\( a = -1 \)[/tex], [tex]\( h = -8 \)[/tex], and [tex]\( k = -1 \)[/tex].
2. Identify the Vertex:
For the quadratic function [tex]\( f(x) = a(x-h)^2 + k \)[/tex], the vertex is at the point [tex]\((h, k)\)[/tex]. So, the vertex of [tex]\( f(x) = -(x+8)^2 - 1 \)[/tex] is at [tex]\((h, k) = (-8, -1)\)[/tex].
3. Determine the Intervals of Increase and Decrease:
Since the parabola opens downward ( [tex]\( a < 0 \)[/tex] ), it reaches its maximum value at the vertex and decreases on both sides of the vertex.
Thus, for [tex]\( x < -8 \)[/tex], the function is increasing as it approaches the vertex.
For [tex]\( x > -8 \)[/tex], the function is decreasing as it moves away from the vertex.
4. Conclusion:
The function [tex]\( f(x) = -(x+8)^2 - 1 \)[/tex] is decreasing for all [tex]\( x \)[/tex] values greater than the x-coordinate of the vertex, which is [tex]\( x = -8 \)[/tex].
Therefore, the interval over which the function [tex]\( f(x) = -(x+8)^2 - 1 \)[/tex] is decreasing is [tex]\( (-8, \infty) \)[/tex].
Thus, the correct answer is:
[tex]\[ (-8, \infty) \][/tex]
