Find the vertex and axis of the parabola `x^(2)-4x-3y+7=0`.

Answer

To find the vertex and axis of the parabola given by the equation \( x^2 - 4x - 3y + 7 = 0 \), we can follow these steps:

### Step 1: Rearrange the Equation
Start by rearranging the equation to isolate the \( y \) term:
\[ x^2 - 4x + 7 = 3y \]

### Step 2: Complete the Square
Next, we will complete the square for the \( x \) terms. The expression \( x^2 - 4x \) can be rewritten as follows:
\[ x^2 - 4x = (x - 2)^2 - 4 \]
So, substituting this back into the equation gives:
\[ (x - 2)^2 - 4 + 7 = 3y \]
This simplifies to:
\[ (x - 2)^2 + 3 = 3y \]

### Step 3: Rearrange to Standard Form
Now, rearranging this equation gives:
\[ (x - 2)^2 = 3y - 3 \]
or
\[ (x - 2)^2 = 3(y - 1) \]

### Step 4: Identify the Vertex
From the standard form of the parabola \( (x - x_1)^2 = 4a(y - y_1) \), we can identify the vertex \((x_1, y_1)\):
- Here, \( x_1 = 2 \) and \( y_1 = 1 \).
Thus, the vertex of the parabola is:
\[ \text{Vertex} = (2, 1) \]

### Step 5: Find the Axis of Symmetry
The axis of symmetry for a parabola in this form is a vertical line given by:
\[ x = x_1 \]
From our vertex, we have:
\[ \text{Axis of symmetry} = x = 2 \]

### Final Answer
- **Vertex**: \( (2, 1) \)
- **Axis of Symmetry**: \( x = 2 \)

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