Consider the function \( f(x)=2 x^{2}-4 x-8 \). a. Determine, without graphing, whether the function has a minimum value or a max 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.

1. Since \( f(x) = 2x^2 - 4x - 8 \) is a quadratic function with \( a = 2 \) (which is positive), the parabola opens upward. Thus, the function has a minimum value (and no maximum value). 2. The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) occurs at \[ x = -\frac{b}{2a}. \] For this function, \( a = 2 \) and \( b = -4 \), so \[ x = -\frac{-4}{2 \cdot 2} = \frac{4}{4} = 1. \] Substitute \( x = 1 \) into the function to find the minimum value: \[ f(1) = 2(1)^2 - 4(1) - 8 = 2 - 4 - 8 = -10. \] Therefore, the minimum value of \(-10\) occurs at \( x = 1 \). 3. The domain of a quadratic function is all real numbers, so \[ \text{Domain: } (-\infty, \infty). \] Since the function has a minimum value of \(-10\) and the parabola opens upward, the range is: \[ \text{Range: } [-10, \infty). \]