Question

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.

x y
−3
54
−2
33
−1

0
1
2 9
(1 point)
The high or low point of this parabola is (
,
).
Remaining Attempts : 3

Answer 1

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:

- For \( x = -3 \):
\[
y = 3(-3)^2 - 6(-3) + 9 = 3(9) + 18 + 9 = 27 + 18 + 9 = 54
\]

- 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 \( 1 \):

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

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

- 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 \( (1, 6) \).