Consider the function $f(x)=3 x^{2}-6 x-1$ a. Determine, without graphing, whether the function has a minimum value or a maximun value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. a. The function has a minimum value. b. The minimum/maximum value is -4 . It occurs at $x=1$. c. The domain of $\mathrm{f}$ is $(-\infty, \infty)$. (Type your answer in interval notation.) The range of $f$ is $[-4, \infty)$. (Type your answer in interval notation.)

Step 1 ：The given function is a quadratic function of the form \(f(x) = ax^2 + bx + c\). For such functions, the vertex form is given by \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. The x-coordinate of the vertex is given by \(h = -\frac{b}{2a}\). Since the coefficient of \(x^2\) is positive, the parabola opens upwards and hence, the function has a minimum value.

Step 2 ：The minimum value is the y-coordinate of the vertex, which is \(f(h)\).

Step 3 ：The domain of the function is all real numbers, i.e., \((-\infty, \infty)\).

Step 4 ：The range of the function is all values greater than or equal to the minimum value, i.e., \([f(h), \infty)\).

Step 5 ：Final Answer: a. The function has a minimum value. b. The minimum value is \(\boxed{-4}\) and it occurs at \(x=\boxed{1}\). c. The domain of the function is \(\boxed{(-\infty, \infty)}\) and the range of the function is \(\boxed{[-4, \infty)}\).