Answer :
To determine the interval over which the function [tex]\( f(x)=-(x+8)^2-1 \)[/tex] is decreasing, we need to analyze its behavior by looking at certain characteristics of the function.
1. Understand the Function Form:
The function [tex]\( f(x)=-(x+8)^2-1 \)[/tex] is a quadratic function in the form of [tex]\( y = a(x-h)^2 + k \)[/tex], which is a standard form for parabolas. Here, [tex]\( a = -1 \)[/tex], [tex]\( h = -8 \)[/tex], and [tex]\( k = -1 \)[/tex].
2. Direction of the Parabola:
Since [tex]\( a = -1 \)[/tex] is negative, the parabola opens downward. For downward-opening parabolas, the function value decreases as we move from the right towards the vertex and increases as we move from the left of the vertex to the right.
3. Finding the Vertex:
The vertex of a parabola [tex]\( y = a(x-h)^2 + k \)[/tex] is at the point [tex]\( (h, k) \)[/tex].
- In this case, [tex]\( h = -8 \)[/tex] and [tex]\( k = -1 \)[/tex].
- Therefore, the vertex is at the point [tex]\((-8, -1)\)[/tex].
4. Intervals of Increase and Decrease:
For a downward-opening parabola, it decreases to the left of the vertex and increases to the right of the vertex.
- Since the vertex is at [tex]\( x = -8 \)[/tex], the function [tex]\( f(x) \)[/tex] is decreasing for all [tex]\( x < -8 \)[/tex].
5. Conclusion:
The interval over which [tex]\( f(x) \)[/tex] is decreasing is [tex]\( (-\infty, -8) \)[/tex].
Therefore, the correct answer is:
[tex]\[ (-\infty, -8) \][/tex]
1. Understand the Function Form:
The function [tex]\( f(x)=-(x+8)^2-1 \)[/tex] is a quadratic function in the form of [tex]\( y = a(x-h)^2 + k \)[/tex], which is a standard form for parabolas. Here, [tex]\( a = -1 \)[/tex], [tex]\( h = -8 \)[/tex], and [tex]\( k = -1 \)[/tex].
2. Direction of the Parabola:
Since [tex]\( a = -1 \)[/tex] is negative, the parabola opens downward. For downward-opening parabolas, the function value decreases as we move from the right towards the vertex and increases as we move from the left of the vertex to the right.
3. Finding the Vertex:
The vertex of a parabola [tex]\( y = a(x-h)^2 + k \)[/tex] is at the point [tex]\( (h, k) \)[/tex].
- In this case, [tex]\( h = -8 \)[/tex] and [tex]\( k = -1 \)[/tex].
- Therefore, the vertex is at the point [tex]\((-8, -1)\)[/tex].
4. Intervals of Increase and Decrease:
For a downward-opening parabola, it decreases to the left of the vertex and increases to the right of the vertex.
- Since the vertex is at [tex]\( x = -8 \)[/tex], the function [tex]\( f(x) \)[/tex] is decreasing for all [tex]\( x < -8 \)[/tex].
5. Conclusion:
The interval over which [tex]\( f(x) \)[/tex] is decreasing is [tex]\( (-\infty, -8) \)[/tex].
Therefore, the correct answer is:
[tex]\[ (-\infty, -8) \][/tex]