Step 1 :The given function is a quadratic function in the form of \(f(x) = ax^2 + bx + c\), where \(a = 1\), \(b = -8\), and \(c = 12\).
Step 2 :The vertex of a parabola \(y = ax^2 + bx + c\) is given by the point \((-b/2a, f(-b/2a))\).
Step 3 :Since the coefficient of \(x^2\) is positive, the parabola opens upward and has a minimum value at the vertex.
Step 4 :The range of the function is all the y-values that the function can take. Since the parabola opens upward and has a minimum value, the range is \([f(-b/2a), +\infty)\).
Step 5 :The function is increasing on the interval \((-b/2a, +\infty)\) and decreasing on the interval \((-\infty, -b/2a)\).
Step 6 :Calculating these values, we find that the vertex is \((4, -4)\).
Step 7 :The parabola opens upward and has a minimum value of \(-4\).
Step 8 :The range of \(f(x)\) is \([-4, +\infty)\).
Step 9 :The function is increasing on the interval \((4, +\infty)\) and decreasing on the interval \((-\infty, 4)\).
Step 10 :\(\boxed{\text{Final Answer:}}\)
Step 11 :\(\boxed{\text{(a) The vertex is (4, -4).}}\)
Step 12 :\(\boxed{\text{(b) The parabola opens upward and has a minimum value of -4.}}\)
Step 13 :\(\boxed{\text{(c) The range of } f(x) \text{ is } [-4, +\infty).}\)
Step 14 :\(\boxed{\text{(d) The function is increasing on the interval } (4, +\infty) \text{ and decreasing on the interval } (-\infty, 4).}\)