The vertex of the parabola `y^(2)+4x-2y+3=0 ` is

Options

A. `(-3,1)`
B. `(-3,2)`
C. `(-(3)/(2),3)`
D. `(-(1)/(2),1)`

Answer

To find the vertex of the parabola given by the equation \( y^2 + 4x - 2y + 3 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation to isolate the terms involving \( y \):
\[ y^2 - 2y + 4x + 3 = 0 \]

### Step 2: Completing the Square for \( y \)
To complete the square for the \( y \) terms, we take the coefficient of \( y \) (which is -2), halve it to get -1, and then square it to get 1. We can rewrite the equation as:
\[ (y^2 - 2y + 1) + 4x + 3 - 1 = 0 \]
This simplifies to:
\[ (y - 1)^2 + 4x + 2 = 0 \]

### Step 3: Isolate the \( x \) term
Now, isolate \( x \):
\[ (y - 1)^2 = -4x - 2 \]
This can be rearranged to:
\[ (y - 1)^2 = -4(x + \frac{1}{2}) \]

### Step 4: Identify the Vertex
The equation is now in the standard form of a parabola:
\[ (y - k)^2 = 4p(x - h) \]
where \((h, k)\) is the vertex. From our equation, we can identify:
- \( h = -\frac{1}{2} \)
- \( k = 1 \)

Thus, the vertex of the parabola is:
\[ \left(-\frac{1}{2}, 1\right) \]

### Final Answer
The vertex of the parabola \( y^2 + 4x - 2y + 3 = 0 \) is:
\[ \boxed{\left(-\frac{1}{2}, 1\right)} \]

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