The vertex of the parabola `y^(2) - 4y - 16 x - 12 = 0 ` is

Options

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

Answer

To find the vertex of the parabola given by the equation \( y^2 - 4y - 16x - 12 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ y^2 - 4y - 16x - 12 = 0 \]
We can rearrange this to isolate the terms involving \( y \):
\[ y^2 - 4y = 16x + 12 \]

**Hint:** Move all terms involving \( x \) to the right side of the equation.

### Step 2: Complete the square for \( y \)
To complete the square for the left side, we take the coefficient of \( y \) (which is -4), halve it to get -2, and then square it to get 4. We add and subtract 4 on the left side:
\[ y^2 - 4y + 4 - 4 = 16x + 12 \]
This simplifies to:
\[ (y - 2)^2 - 4 = 16x + 12 \]
Now, we can rewrite it as:
\[ (y - 2)^2 = 16x + 16 \]

**Hint:** Completing the square helps to rewrite the quadratic in a more manageable form.

### Step 3: Rewrite in standard form
Now we can simplify the equation:
\[ (y - 2)^2 = 16(x + 1) \]
This is now in the standard form of a parabola, which is \( (y - k)^2 = 4p(x - h) \), where \((h, k)\) is the vertex of the parabola.

**Hint:** The standard form helps identify the vertex and the direction of the parabola.

### Step 4: Identify the vertex
From the equation \( (y - 2)^2 = 16(x + 1) \), we can see that:
- \( h = -1 \)
- \( k = 2 \)

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

**Hint:** The vertex can be directly read from the standard form of the parabola.

### Final Answer
The vertex of the parabola \( y^2 - 4y - 16x - 12 = 0 \) is:
\[ \boxed{(-1, 2)} \]