Find the minimum value of \(y = x^{2} + 6x - 12\).

To find the minimum value of the quadratic function, we need to determine its vertex. We can do this by completing the square or using the formula for the x-coordinate of the vertex, which is given by: $$x = -\frac{b}{2a}$$ where a and b are the coefficients of the quadratic function in standard form: \(y = ax^{2} + bx + c\). In this case, we have: $$y = x^{2} + 6x - 12$$ where a = 1 and b = 6. Therefore, the x-coordinate of the vertex is: $$x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3$$ To find the minimum value of y, we substitute x = -3 into the quadratic function: $$y = (-3)^{2} + 6(-3) - 12 = -21$$ Therefore, the minimum value of y is -21. So, the correct option is (a) -21.