use the drop - down menus to describe the key aspects of the function $f(x)=-x^{2}-2x - 1$. the vertex is…

Explanation:

Step1: Rewrite function in vertex - form

The general form of a quadratic function is $y = ax^{2}+bx + c$. For $f(x)=-x^{2}-2x - 1$, where $a=-1$, $b = - 2$, $c=-1$. We use the formula $y=a(x - h)^{2}+k$, and $h=-\frac{b}{2a}$, $k = f(h)$. First, $h=-\frac{-2}{2\times(-1)}=-1$. Then $k=f(-1)=-(-1)^{2}-2\times(-1)-1=-1 + 2-1=0$. So $f(x)=-(x + 1)^{2}$.

Step2: Analyze vertex

The vertex - form of the function is $f(x)=-(x + 1)^{2}+0$, so the vertex is $(-1,0)$. Since $a=-1\lt0$, the parabola opens downwards, and the vertex is the maximum point.

Step3: Analyze increasing and decreasing intervals

The function $y = f(x)$ is increasing when $x\lt - 1$ (because the slope of the tangent line to the parabola is positive for $x\lt - 1$) and decreasing when $x\gt - 1$ (because the slope of the tangent line to the parabola is negative for $x\gt - 1$).

Step4: Analyze domain

For a quadratic function $y = ax^{2}+bx + c$, the domain is all real numbers, i.e., $(-\infty,\infty)$.

Step5: Analyze range

Since the parabola opens downwards and the vertex is at $( - 1,0)$, the maximum value of the function is $y = 0$. So the range is $(-\infty,0]$.

Answer:

The vertex is the maximum point. The function is increasing for $x\lt - 1$. The function is decreasing for $x\gt - 1$. The domain of the function is all real numbers ($(-\infty,\infty)$). The range of the function is $y\leqslant0$ ($(-\infty,0]$).