Find the vertex, focus and directrix of the parabola
`x^(2) + 4x + 2y -7 = 0 ` .

Answer

To find the vertex, focus, and directrix of the parabola given by the equation \( x^2 + 4x + 2y - 7 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^2 + 4x + 2y - 7 = 0 \]
Rearranging it gives:
\[ x^2 + 4x = 7 - 2y \]
Now we will complete the square for the \( x \) terms.

### Step 2: Completing the Square
To complete the square for \( x^2 + 4x \):
\[ x^2 + 4x = (x + 2)^2 - 4 \]
Substituting this back into the equation gives:
\[ (x + 2)^2 - 4 = 7 - 2y \]
This simplifies to:
\[ (x + 2)^2 = 2y + 11 \]

### Step 3: Rearranging to Standard Form
Now, rearranging gives:
\[ (x + 2)^2 = 2(y + \frac{11}{2}) \]
This is in the standard form of a parabola \( (x - h)^2 = 4p(y - k) \), where \( (h, k) \) is the vertex.

### Step 4: Identifying Vertex
From the equation \( (x + 2)^2 = 2(y + \frac{11}{2}) \):
- The vertex \( (h, k) \) is \( (-2, -\frac{11}{2}) \).

### Step 5: Finding Focus and Directrix
In the standard form \( (x - h)^2 = 4p(y - k) \):
- Here, \( 4p = 2 \) implies \( p = \frac{1}{2} \).
- The focus is located at \( (h, k + p) = (-2, -\frac{11}{2} + \frac{1}{2}) = (-2, -5) \).
- The directrix is given by the line \( y = k - p = -\frac{11}{2} - \frac{1}{2} = -6 \).

### Summary of Results
- **Vertex**: \( (-2, -\frac{11}{2}) \)
- **Focus**: \( (-2, -5) \)
- **Directrix**: \( y = -6 \)