Given this table of values, complete the table for the quadratic equation 3x^2-6x + 9 to find the values of the high or low point of the parabola
x y
-3 54
-2 33
-1
0
1
2 9
The high or low of this parabola is (__, __)
To complete the table, we need to substitute the values of x into the quadratic equation \( y = 3x^2 - 6x + 9 \) and calculate the corresponding y values. However, before doing that, we can find the vertex of the parabola—which represents the high or low point—by using the vertex formula for a quadratic equation in the form of \( y = ax^2 + bx + c \).
The x-coordinate of the vertex (h) is given by \( h = -\frac{b}{2a} \). For the given equation:
\( a = 3 \)
\( b = -6 \)
So,
\( h = -\frac{-6}{2 \times 3} \)
\( h = -\frac{-6}{6} \)
\( h = 1 \)
Now that we have the x-coordinate of the vertex, let's find the y-coordinate by substituting x = 1 into the equation:
\( y = 3(1)^2 - 6 \times 1 + 9 \)
\( y = 3 - 6 + 9 \)
\( y = 6 \)
Therefore, the vertex (high or low point) of the parabola represented by the quadratic equation \( y = 3x^2 - 6x + 9 \) is at (1, 6).
To complete the table for the missing y-values:
For x = -1:
\( y = 3(-1)^2 - 6 \times -1 + 9 \)
\( y = 3 + 6 + 9 \)
\( y = 18 \)
For x = 0:
\( y = 3(0)^2 - 6 \times 0 + 9 \)
\( y = 0 + 0 + 9 \)
\( y = 9 \)
For x = 1 (which we already calculated as the vertex):
\( y = 6 \)
So the completed table of values is:
x y
-3 54
-2 33
-1 18
0 9
1 6
2 9
Therefore, the high or low point of this parabola is at the vertex (1, 6). Note that because the coefficient of \( x^2 \) is positive (3), the parabola opens upwards, and therefore, the vertex is the low point of the parabola.