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

Answer 1

To complete the table using the quadratic equation \( y = 3x^2 - 6x + 9 \), we need to calculate the values of \( y \) for each \( x \) value provided.

Let's compute \( y \) for \( x = -3, -2, -1, 0, 1 \), and \( 2 \):

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

2. **For \( x = -2 \)**:
\[
y = 3(-2)^2 - 6(-2) + 9 = 3(4) + 12 + 9 = 12 + 12 + 9 = 33
\]

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

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

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

6. **For \( x = 2 \)**:
\[
y = 3(2)^2 - 6(2) + 9 = 3(4) - 12 + 9 = 12 - 12 + 9 = 9
\]

Now, we can complete 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, to find the high or low point of the parabola, we should identify the vertex of the parabola given by the formula for the vertex of a quadratic \( y = ax^2 + bx + c \), which occurs at \( x = -\frac{b}{2a} \).

For the equation \( y = 3x^2 - 6x + 9 \):
- \( a = 3 \)
- \( b = -6 \)

Calculating the x-coordinate of the vertex:

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

Now substituting \( x = 1 \) back into the equation to find the y-coordinate:

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

Thus, the high or low point of the parabola (the vertex) is:

\[
\text{The high or low point of this parabola is } (1, 6).
\]

Since the coefficient of \( x^2 \) (which is 3) is positive, the parabola opens upwards, meaning this point is a low point.