Find the least value of the function \(f(x) = 3x^{2} + 18x + 32\).
\(f(x) = 3x^{2} + 18x + 32\)
\(\frac{\mathrm d y}{\mathrm d x} = 6x + 18 = 0\)
\(6x = -18 \implies x = -3\)
\(f(-3) = 3(-3^{2}) + 18(-3) + 32 = 27 - 54 + 32 = 5\)
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