Let's complete the table by evaluating the function $h(x) = (-2x + 6)^2$ for each given value of $x$.

1. When $x = -3$:

$h(-3) = (-2(-3) + 6)^2 = (6 + 6)^2 = (12)^2 = 144$

2. When $x = -2$:

$h(-2) = (-2(-2) + 6)^2 = (4 + 6)^2 = (10)^2 = 100$

3. When $x = -1$:

$h(-1) = (-2(-1) + 6)^2 = (2 + 6)^2 = (8)^2 = 64$

4. When $x = 0$:

$h(0) = (-2(0) + 6)^2 = (0 + 6)^2 = (6)^2 = 36$

5. When $x = 1$:

$h(1) = (-2(1) + 6)^2 = (-2 + 6)^2 = (4)^2 = 16$

6. When $x = 2$:

$h(2) = (-2(2) + 6)^2 = (-4 + 6)^2 = (2)^2 = 4$

7. When $x = 3$:

$h(3) = (-2(3) + 6)^2 = (-6 + 6)^2 = (0)^2 = 0$

Completed Table:

$x$	$h(x) = (-2x + 6)^2$
-3	144
-2	100
-1	64
0	36
1	16
2	4
3	0

Would you like further details or have any questions about this solution?

Here are some related questions to explore:

1. How would the graph of the function $h(x) = (-2x + 6)^2$ look like?
2. What is the vertex of the function $h(x) = (-2x + 6)^2$?
3. How does changing the coefficient of $x$ affect the graph of the function?
4. What is the effect of adding a constant to the function $h(x) = (-2x + 6)^2 + k$?
5. How would this function behave for very large or very small values of $x$?

Tip: When squaring a binomial like $(-2x + 6)$, always remember to apply the distributive property, and then square each term.