Sure, let's analyze the quadratic function [tex]\( f(x) = -x^2 - 2x - 1 \)[/tex].
1. The vertex:
The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] is given by [tex]\( x = -\frac{b}{2a} \)[/tex]. For the function [tex]\( f(x) = -x^2 - 2x - 1 \)[/tex], we have [tex]\( a = -1 \)[/tex] and [tex]\( b = -2 \)[/tex].
[tex]\[
x = -\frac{-2}{2 \cdot -1} = -\frac{2}{-2} = 1
\][/tex]
Substituting [tex]\( x = -1 \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[
f(-1) = -(-1)^2 - 2(-1) - 1 = -1 + 2 - 1 = 0
\][/tex]
So, the vertex is [tex]\((-1, 0)\)[/tex].
2. Increasing interval:
The function is increasing where its derivative is positive. The derivative of [tex]\( f(x) = -x^2 - 2x - 1 \)[/tex] is [tex]\( f'(x) = -2x - 2 \)[/tex]. Set [tex]\( f'(x) > 0 \)[/tex]:
[tex]\[
-2x - 2 > 0 \implies -2x > 2 \implies x < -1
\][/tex]
So, the function is increasing on the interval [tex]\((-∞, -1)\)[/tex].
3. Decreasing interval:
The function is decreasing where its derivative is negative. From the earlier derivative [tex]\( f'(x) = -2x - 2 \)[/tex], set [tex]\( f'(x) < 0 \)[/tex]:
[tex]\[
-2x - 2 < 0 \implies -2x < 2 \implies x > -1
\][/tex]
So, the function is decreasing on the interval [tex]\((-1, ∞)\)[/tex].
4. Domain:
The domain of a quadratic function is all real numbers, so the domain is [tex]\((-∞, ∞)\)[/tex].
5. Range:
Since the quadratic function [tex]\( f(x) = -x^2 - 2x - 1 \)[/tex] opens downwards (the leading coefficient [tex]\( a = -1 \)[/tex] is negative), the range is from the y-coordinate of the vertex down to negative infinity.
The y-coordinate of the vertex is [tex]\( 0 \)[/tex]. So, the range is [tex]\((-∞, 0]\)[/tex].
To summarize:
- The vertex is the [tex]\(\boxed{(-1, 0)}\)[/tex]
- The function is increasing [tex]\(\boxed{(-∞, -1)}\)[/tex]
- The function is decreasing [tex]\(\boxed{(-1, ∞)}\)[/tex]
- The domain of the function is [tex]\(\boxed{(-∞, ∞)}\)[/tex]
- The range of the function is [tex]\(\boxed{(-∞, 0]}\)[/tex]
