Complete the square to find the vertex of this parabola. \( y^{2}+12 x-10 y+1=0 \) \( ([?],[]) \)

To find the vertex of the parabola defined by the equation: \[ y^{2} + 12x - 10y + 1 = 0 \] we will **complete the square** for the \( y \)-terms. ### Step 1: Rearrange the Equation First, let's group the \( y \)-terms and move the other terms to the opposite side: \[ y^{2} - 10y = -12x - 1 \] ### Step 2: Complete the Square for \( y \) To complete the square for the expression \( y^{2} - 10y \): 1. **Find the coefficient of \( y \)**: Here, it's \(-10\). 2. **Take half of this coefficient**: \(-10 / 2 = -5\). 3. **Square this value**: \((-5)^2 = 25\). Add and subtract this square inside the equation: \[ y^{2} - 10y + 25 = -12x - 1 + 25 \] This simplifies to: \[ (y - 5)^2 = -12x + 24 \] ### Step 3: Rewrite in Vertex Form The equation now resembles the standard vertex form of a horizontal parabola: \[ (y - k)^2 = 4p(x - h) \] Comparing both equations: \[ (y - 5)^2 = -12(x - 2) \] Here: - \( h = 2 \) - \( k = 5 \) - \( 4p = -12 \) (which indicates the parabola opens to the left) ### Step 4: Identify the Vertex From the vertex form, the vertex \((h, k)\) is: \[ \boxed{(2,\ 5)} \]