Consider the function \( f(x)=3 x^{2}-12 x-3 \). a. Determine, without graphing, whether the function has a minimum value or a maximum 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 value.

**a. Determining if the function has a minimum or maximum value** The function is given by \[ f(x)=3x^2-12x-3. \] For a quadratic function in the form \[ f(x)=ax^2+bx+c, \] if \( a > 0 \), the parabola opens upward and the function attains a minimum value. Here, \( a=3>0 \), so the function has a minimum value. --- **b. Finding the minimum value and where it occurs** The vertex of a quadratic function \( f(x)=ax^2+bx+c \) occurs at \[ x=-\frac{b}{2a}. \] For our function, \( a=3 \) and \( b=-12 \), hence: \[ x=-\frac{-12}{2\cdot3}=\frac{12}{6}=2. \] Now, substitute \( x=2 \) into \( f(x) \) to find the minimum value: \[ f(2)=3(2)^2-12(2)-3. \] Calculate step by step: \[ 3(2)^2=3\cdot4=12, \] \[ 12-24-3=-15. \] Thus, the minimum value is \(-15\) and it occurs at \( x=2 \). --- **c. Determining the domain and range** - **Domain:** The function \( f(x)=3x^2-12x-3 \) is a polynomial, hence its domain is all real numbers: \[ (-\infty, \infty). \] - **Range:** Since the function has a minimum value of \(-15\) (at \( x=2 \)) and opens upward, the range is all values greater than or equal to \(-15\): \[ [-15, \infty). \]