Question

Graphs of Quadratic Equations Practice
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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
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Answer 1

To fill in the table for the quadratic equation \( y = 3x^2 - 6x + 9 \), we can calculate the values of \( y \) for each given \( x \).

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

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

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 fill in the missing values in the table:

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

Next, to find the high or low point of the parabola (the vertex), we can use the vertex formula for the vertex of a quadratic equation given in the standard form \( y = ax^2 + bx + c \).

The x-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 3 \) and \( b = -6 \):

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

Now we substitute \( x = 1 \) back into the original equation to find the corresponding \( y \):
\[
y = 3(1)^2 - 6(1) + 9 = 3 - 6 + 9 = 6
\]

Therefore, the high or low point (the vertex) of the parabola is \( (1, 6) \).

The final answer is:
The high or low point of this parabola is \( (1, 6) \).