Complete the square to find the vertex of this parabola. \( y^{2}-4 x-8 y-12=0 \) ([?], [ ])

We start with the equation \[ y^{2} - 4x - 8y - 12 = 0. \] **Step 1. Isolate the terms involving \( y \):** \[ y^{2} - 8y = 4x + 12. \] **Step 2. Complete the square for the \( y \)-terms.** The expression \( y^{2} - 8y \) can be rewritten by completing the square. Recall that \[ y^{2} - 8y = (y - 4)^{2} - 16. \] So, substitute back: \[ (y - 4)^{2} - 16 = 4x + 12. \] **Step 3. Solve for \( x \):** Add \( 16 \) to both sides: \[ (y - 4)^{2} = 4x + 28. \] Now, solve for \( x \) by isolating it: \[ 4x = (y - 4)^{2} - 28, \] \[ x = \frac{1}{4}(y - 4)^{2} - 7. \] **Step 4. Identify the vertex.** The equation is now in the form \[ x = \frac{1}{4}(y - 4)^{2} - 7, \] which is the standard form of a sideways parabola \( x = a(y-k)^2 + h \) with vertex at \((h, k)\). Here, we have \( h = -7 \) and \( k = 4 \). Thus, the vertex of the parabola is \[ \boxed{(-7, \, 4)}. \]