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).