Let’s solve the problem step by step based on the uploaded image.

Given Functions:

1. $f(x) = -x^2 + 2x - 2$
2. $g(x) = \frac{-2x - 6}{2x + 1}$
3. $h(x) = \sqrt{x - 2}$

Questions:

1. Determine $D_f$, $D_g$, $D_h$ (domains of $f$, $g$, and $h$).
2. (a) Create the variation tables for $f$, $g$, and $h$.
   (b) Calculate specific values:
   • $h([3,6])$
   • $g([-2,-1])$
   • $f([-1,3])$.

Step 1: Domains

1. Domain of $f(x) = -x^2 + 2x - 2$: $f(x)$ is a quadratic polynomial, so it is defined for all $x \in \mathbb{R}$.
   $\boxed{D_f = \mathbb{R}}$.

2. Domain of $g(x) = \frac{-2x - 6}{2x + 1}$: The denominator must not be zero: $2x + 1 \neq 0 \implies x \neq -\frac{1}{2}.$ Hence,
   $\boxed{D_g = \mathbb{R} \setminus \{-\frac{1}{2}\}}$.

3. Domain of $h(x) = \sqrt{x - 2}$: The expression inside the square root must be non-negative: $x - 2 \geq 0 \implies x \geq 2.$ Hence,
   $\boxed{D_h = [2, \infty)}$.

Step 2: Variation Tables

(a) Behavior of the Functions

1. Variation Table for $f(x)$:

   • $f(x) = -x^2 + 2x - 2$ is a quadratic function.
   • Rewrite in vertex form: $f(x) = -(x^2 - 2x) - 2 = -(x - 1)^2 + 1 - 2 = -(x - 1)^2 - 1.$
   • Vertex: $(1, -1)$. Since the coefficient of $x^2$ is negative, the parabola opens downward.
   • Critical points:
     • $f(x)$ increases on $(-\infty, 1]$ and decreases on $[1, \infty)$.

2. Variation Table for $g(x)$:

   • $g(x) = \frac{-2x - 6}{2x + 1}$.
   • Vertical asymptote: $x = -\frac{1}{2}$.
   • Horizontal asymptote: As $x \to \pm\infty$, $g(x) \to -1$ (degree of numerator = degree of denominator).

3. Variation Table for $h(x)$:

   • $h(x) = \sqrt{x - 2}$ is defined for $x \geq 2$.