Thank you for sharing the problem! Since derivatives and asymptotes are not part of your studies yet, we will approach this question using domain analysis and elementary methods to describe the behavior of the functions. Here's the breakdown of the tasks:

1. Determining the domains ($D_f$, $D_g$, $D_h$)

We analyze the domains of each function:

$f(x) = -x^2 + 2x - 2$

This is a polynomial, so it is defined for all real numbers:
$D_f = \mathbb{R}.$

$g(x) = \frac{-2x - 6}{2x + 1}$

For $g(x)$, the denominator $2x + 1 \neq 0$. Solving $2x + 1 = 0$:
$x = -\frac{1}{2}.$
Thus, the domain is:
$D_g = \mathbb{R} \setminus \{-\frac{1}{2}\}.$

$h(x) = \sqrt{x - 2}$

For $h(x)$, the expression under the square root must be non-negative:
$x - 2 \geq 0 \implies x \geq 2.$
Thus, the domain is:
$D_h = [2, \infty[.$

2. (a) Variation Tables for $f(x)$, $g(x)$, and $h(x)$

$f(x) = -x^2 + 2x - 2$

This is a quadratic function with a negative leading coefficient ($-x^2$), so it opens downward. The vertex is found using:
$x_{\text{vertex}} = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1.$
Substituting $x = 1$ into $f(x)$:
$f(1) = -(1)^2 + 2(1) - 2 = -1 + 2 - 2 = -1.$
So the vertex is $(1, -1)$. The function decreases for $x > 1$ and increases for $x < 1$.

$g(x) = \frac{-2x - 6}{2x + 1}$

We analyze the behavior of the function around $x = -\frac{1}{2}$, where the function is undefined. As $x \to -\frac{1}{2}^+$, the denominator approaches $0^+$, and the numerator is negative, so $g(x) \to -\infty$. For $x \to -\frac{1}{2}^-$, $g(x) \to +\infty$.

We will also observe its behavior at other points. The exact variation table requires calculus (not applicable now), so we rely on intuition and computation of key points.

$h(x) = \sqrt{x - 2}$

This function is increasing because as $x$ increases, $h(x)$ increases. Its variation is straightforward:

2. (b) Evaluation of Specific Intervals

$h([3, 6])$:

Substituting $x = 3$ and $x = 6$ into $h(x)$:
$h(3) = \sqrt{3 - 2} = \sqrt{1} = 1, \quad h(6) = \sqrt{6 - 2} = \sqrt{4} = 2.$
So, $h([3, 6]) = [1, 2]$.

$g([-2, -1])$:

For $g(x)$:
$g(-2) = \frac{-2(-2) - 6}{2(-2) + 1} = \frac{4 - 6}{-4 + 1} = \frac{-2}{-3} = \frac{2}{3}.$ $g(-1) = \frac{-2(-1) - 6}{2(-1) + 1} = \frac{2 - 6}{-2 + 1} = \frac{-4}{-1} = 4.$
So, $g([-2, -1]) = \left[\frac{2}{3}, 4\right]$.

$f([-1, 3])$:

Substitute $x = -1$ and $x = 3$ into $f(x)$:
$f(-1) = -(-1)^2 + 2(-1) - 2 = -1 - 2 - 2 = -5,$ $f(3) = -(3)^2 + 2(3) - 2 = -9 + 6 - 2 = -5.$
So, $f([-1, 3]) = [-5, -5]$.

Would you like me to clarify or expand on any of these solutions?

Related Questions:

1. What are the properties of quadratic functions like $f(x)$?
2. How do you determine whether a rational function has vertical asymptotes?
3. What is the general process for finding the domain of a square-root function?
4. How can we compare behaviors of $f(x)$, $g(x)$, and $h(x)$ on the same graph?
5. How does symmetry influence the analysis of functions?

Tip:

Always test critical points (roots, vertices, or undefined points) to understand the shape and domain of a function better.