\(-6\frac{1}{6}\)
\(-6\frac{1}{12}\)
\(-6\)
\(0\)
Correct answer is B
\(3x^{2} - x - 6 = y\)
\(a = 3, b = -1, c = -6\)
Minimum value of x = \(\frac{-b}{2a} = \frac{-(-1)}{6} = \frac{1}{6}\)
\(y(\frac{1}{6}) = 3(\frac{1}{6}^{2}) - \frac{1}{6} - 6 = -6\frac{1}{12}\)