Find the minimum value of \(y = x^{2} + 6x - 12\).
\(y = x^{2} + 6x - 12\)
\(\frac{\mathrm d y}{\mathrm d x} = 2x + 6 = 0\)
\(2x = -6 \implies x = -3\)
\(y(-3) = (-3)^{2} + 6(-3) - 12 = 9 - 18 - 12 = -21\)
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