What is the vertex of the quadratic function f(x) (x 6)(x 2)?

To find the vertex of the quadratic function ( f(x) = (x - 6)(x - 2) ), we first expand it to get ( f(x) = x^2 - 8x + 12 ). The vertex form of a quadratic function ( ax^2 + bx + c ) has its vertex at ( x = -\frac{b}{2a} ). Here, ( a = 1 ) and ( b = -8 ), so the x-coordinate of the vertex is ( x = \frac{8}{2} = 4 ). Substituting ( x = 4 ) back into the function gives ( f(4) = (4 - 6)(4 - 2) = (-2)(2) = -4 ). Therefore, the vertex is at the point ( (4, -4) ).