Given this table of values, complete the table for the quadratic equation y=3x2−6x+9 to find the values of the high or low point of the parabola.

To find the values of \( y \) for \( x = -1 \) and \( x = 0 \) in the quadratic equation \( y = 3x^2 - 6x + 9 \), we will substitute these values of \( x \) into the equation.

### For \( x = -1 \):
\[
y = 3(-1)^2 - 6(-1) + 9
\]
\[
y = 3(1) + 6 + 9
\]
\[
y = 3 + 6 + 9 = 18
\]

### For \( x = 0 \):
\[
y = 3(0)^2 - 6(0) + 9
\]
\[
y = 0 - 0 + 9 = 9
\]

Now we can fill in the table:

\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-3 & 54 \\
-2 & 33 \\
-1 & 18 \\
0 & 9 \\
1 & 6 \\
2 & 9 \\
\hline
\end{array}
\]

Next, we need to determine the high or low point of the parabola. The vertex \( x \) coordinate of a parabola given by the equation \( y = ax^2 + bx + c \) can be calculated with the formula \( x = -\frac{b}{2a} \).

In this case:
- \( a = 3 \)
- \( b = -6 \)

Calculating the vertex:

\[
x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1
\]

Now substitute \( x = 1 \) back into the equation to find the corresponding \( y \):

\[
y = 3(1)^2 - 6(1) + 9 = 3 - 6 + 9 = 6
\]

Therefore, the high or low point of the parabola is:

\[
(1, 6)
\]

Thus, the final answer is:
The high or low point of this parabola is **(1, 6)**.