Axis of the parabola `x^(2)-3y-6x+6 = 0` is

Options

A. x=-3
B. y=-1
C. x=3
D. y=1

Answer

To find the axis of the parabola given by the equation \(x^2 - 3y - 6x + 6 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x^2 - 3y - 6x + 6 = 0 \]
Rearranging it to isolate \(y\):
\[ -3y = -x^2 + 6x - 6 \]
Dividing the entire equation by -3 gives:
\[ y = \frac{1}{3}x^2 - 2x + 2 \]

### Step 2: Complete the square for the \(x\) terms
Next, we need to complete the square for the quadratic expression in \(x\):
\[ y = \frac{1}{3}(x^2 - 6x) + 2 \]
To complete the square, we take half of the coefficient of \(x\) (which is -6), square it, and add/subtract it inside the parentheses:
\[ \left(-\frac{6}{2}\right)^2 = 9 \]
So we rewrite the equation as:
\[ y = \frac{1}{3}((x^2 - 6x + 9) - 9) + 2 \]
This simplifies to:
\[ y = \frac{1}{3}(x - 3)^2 - 3 + 2 \]
\[ y = \frac{1}{3}(x - 3)^2 - 1 \]

### Step 3: Identify the vertex and axis of the parabola
The equation is now in the vertex form \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Here, \(h = 3\) and \(k = -1\), so the vertex is at the point \((3, -1)\).

Since the parabola opens upwards (the coefficient of \((x - 3)^2\) is positive), the axis of symmetry is a vertical line that passes through the vertex. Therefore, the axis of the parabola is:
\[ x = 3 \]

### Final Answer
The axis of the parabola is:
\[ \text{Axis: } x = 3 \]

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