To complete the table for the quadratic equation \( y = 3x^2 - 6x + 9 \), we can substitute different values of \( x \) into the equation to calculate the corresponding \( y \) values.
Given values already in the table are:
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For \( x = -3 \): \[ y = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54 \]
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For \( x = -2 \): \[ y = 3(-2)^2 - 6(-2) + 9 = 3(4) + 12 + 9 = 12 + 12 + 9 = 33 \]
Now, let's compute the missing \( y \) values for \( x = -1 \), \( 0 \), and \( 1 \):
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For \( x = -1 \): \[ y = 3(-1)^2 - 6(-1) + 9 = 3(1) + 6 + 9 = 3 + 6 + 9 = 18 \]
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For \( x = 0 \): \[ y = 3(0)^2 - 6(0) + 9 = 0 + 0 + 9 = 9 \]
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For \( x = 1 \): \[ y = 3(1)^2 - 6(1) + 9 = 3(1) - 6 + 9 = 3 - 6 + 9 = 6 \]
Now, we can complete the table with the following values:
\[ \begin{array}{|c|c|} \hline x & y \ \hline -3 & 54 \ -2 & 33 \ -1 & 18 \ 0 & 9 \ 1 & 6 \ 2 & 9 \ \hline \end{array} \]
Next, to find the high or low point of the parabola, we can use the vertex formula for a quadratic equation of the form \( y = ax^2 + bx + c \). The \( x \)-coordinate of the vertex is given by: \[ x = -\frac{b}{2a} \]
In our case, \( a = 3 \) and \( b = -6 \): \[ x = -\frac{-6}{2(3)} = \frac{6}{6} = 1 \]
Now, we can find the \( y \)-coordinate of the vertex by substituting \( x = 1 \) into the original equation: \[ y = 3(1)^2 - 6(1) + 9 = 3(1) - 6 + 9 = 3 - 6 + 9 = 6 \]
Therefore, the vertex (high or low point) of the parabola is: \[ (1, 6) \]
The high or low point of this parabola is \( (1, 6) \).