Find the vertex and focus of
`x^(2)-6x-6y+6=0`

Answer

To find the vertex and focus of the parabola given by the equation \( x^2 - 6x - 6y + 6 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start by rearranging the equation to isolate the \( y \) term:
\[ x^2 - 6x + 6 = 6y \]

### Step 2: Complete the square for the \( x \) terms
To complete the square for the expression \( x^2 - 6x \), we take half of the coefficient of \( x \) (which is -6), square it, and add it inside the equation:
\[ x^2 - 6x = (x - 3)^2 - 9 \]
Now substitute this back into the equation:
\[ (x - 3)^2 - 9 + 6 = 6y \]
This simplifies to:
\[ (x - 3)^2 - 3 = 6y \]

### Step 3: Rearrange to standard form
Now, rearranging gives:
\[ 6y = (x - 3)^2 - 3 \]
\[ y = \frac{1}{6}(x - 3)^2 + \frac{1}{2} \]

### Step 4: Identify vertex and focus
From the standard form of the parabola \( y = a(x - h)^2 + k \), we can identify:
- Vertex \((h, k) = (3, \frac{1}{2})\)
- The value of \( a = \frac{1}{6} \)

The focus of a parabola in this form is given by the point:
\[ (h, k + \frac{1}{4a}) \]
Calculating \( \frac{1}{4a} \):
\[ \frac{1}{4a} = \frac{1}{4 \cdot \frac{1}{6}} = \frac{6}{4} = \frac{3}{2} \]
Thus, the coordinates of the focus are:
\[ (3, \frac{1}{2} + \frac{3}{2}) = (3, 2) \]

### Final Answer
- **Vertex**: \( (3, \frac{1}{2}) \)
- **Focus**: \( (3, 2) \)