Identify any extrema of the function by recognizing its given form or its form after completing the square. extrema. \[ f(x, y)=x^{2}+y^{2}+12 x-8 y+8 \]

We start with the function \[ f(x, y) = x^2 + y^2 + 12x - 8y + 8. \] We complete the square for both \(x\) and \(y\). 1. For the \(x\)-terms: \[ x^2 + 12x = \left(x^2 + 12x + 36\right) - 36 = (x+6)^2 - 36. \] 2. For the \(y\)-terms: \[ y^2 - 8y = \left(y^2 - 8y + 16\right) - 16 = (y-4)^2 - 16. \] 3. Substituting these back into the function gives: \[ f(x, y) = (x+6)^2 - 36 + (y-4)^2 - 16 + 8. \] 4. Combine the constant terms: \[ -36 - 16 + 8 = -44. \] Thus, we have: \[ f(x, y) = (x+6)^2 + (y-4)^2 - 44. \] Since \((x+6)^2\) and \((y-4)^2\) are always non-negative, the minimum value of \(f(x,y)\) occurs when both are zero, i.e., when \[ x+6=0 \quad \text{and} \quad y-4=0, \] which implies \[ x=-6 \quad \text{and} \quad y=4. \] At \((x, y) = (-6, 4)\), the function attains its minimum value: \[ f(-6, 4) = -44. \] Since the squared terms can grow arbitrarily large, there is no maximum value. In summary, the function has a unique minimum at \((-6, 4)\) with a value of \(-44\), and no maximum exists.