The vertex of the parabola `y^2+6x-2y+13=0` is

Options

A. (-2,1)
B. (2,-1)
C. (1,1)
D. (1,-1)

Answer

To find the vertex of the parabola given by the equation \( y^2 + 6x - 2y + 13 = 0 \), we will convert it into standard form by completing the square.

### Step 1: Rearranging the equation
Start with the given equation:
\[ y^2 + 6x - 2y + 13 = 0 \]
Rearranging gives:
\[ y^2 - 2y + 6x + 13 = 0 \]

### Step 2: Completing the square for the \( y \) terms
Focus on the \( y \) terms \( y^2 - 2y \). To complete the square, take half of the coefficient of \( y \) (which is -2), square it, and add/subtract it:
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Now substitute this back into the equation:
\[ (y - 1)^2 - 1 + 6x + 13 = 0 \]
This simplifies to:
\[ (y - 1)^2 + 6x + 12 = 0 \]

### Step 3: Isolating \( x \)
Now, isolate \( x \):
\[ (y - 1)^2 = -6x - 12 \]
Rearranging gives:
\[ (y - 1)^2 = -6(x + 2) \]

### Step 4: Identifying the vertex
The equation is now in the standard form of a parabola:
\[ (y - k)^2 = 4a(x - h) \]
where \( (h, k) \) is the vertex. From our equation:
- \( k = 1 \)
- \( h = -2 \)

Thus, the vertex of the parabola is:
\[ (-2, 1) \]

### Final Answer
The vertex of the parabola \( y^2 + 6x - 2y + 13 = 0 \) is:
\[ \boxed{(-2, 1)} \]