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