To find the minimum value of the quadratic function, we first need to find the vertex of the parabola. We can do this by using the formula:
\(-\frac{b}{2a}\)
where \(a\) is the coefficient of the squared term, and \(b\) is the coefficient of the linear term.
In this case, \(a = 3\) and \(b = -1\), so:
\(-\frac{b}{2a} = -\frac{-1}{2(3)} = \frac{1}{6}\)
This tells us that the vertex of the parabola occurs at \(x = \frac{1}{6}\). To find the minimum value of the function, we substitute this value of \(x\) into the function:
\(y = 3\left(\frac{1}{6}\right)^{2} - \frac{1}{6} - 6 = -6\frac{1}{12} = -\frac{25}{4}\)
Therefore, the minimum value of the function is \(-6\frac{1}{12}\), which is equivalent to \(-\frac{25}{4}\). So the answer is option B.