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 will calculate the value of \( y \) for each value of \( x \).

Let's first fill in the missing \( y \) values from the given \( x \) values in the table.

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

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

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

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

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

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

Now, let's summarize the completed table:

| x | y |
|-----|-------|
| -3 | 54 |
| -2 | 33 |
| -1 | 18 |
| 0 | 9 |
| 1 | 6 |
| 2 | 9 |

### Finding the Vertex (High or Low Point)
Since this is a quadratic function of the form \( y = ax^2 + bx + c \) with \( a = 3 > 0 \), it opens upwards. Therefore, it has a low point (vertex).

The x-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
where \( a = 3 \) and \( b = -6 \):
\[
x = -\frac{-6}{2 \cdot 3} = \frac{6}{6} = 1
\]

Now, we can substitute \( x = 1 \) back into the equation to find the y-coordinate of the vertex:
\[
y = 3(1)^2 - 6(1) + 9 = 3 - 6 + 9 = 6
\]

Thus, the low point of the parabola is at:

**The low point of this parabola is (1, 6).**