1. **Problem statement:** Given the function $f(x) = 2x^2 - 4x - 8$, determine if it has a minimum or maximum, find that value and where it occurs, and identify the domain and range. 2. **Formula and rules:** For a quadratic function $f(x) = ax^2 + bx + c$: - If $a > 0$, the parabola opens upward and has a minimum value. - If $a < 0$, it opens downward and has a maximum value. - The vertex $(h, k)$ gives the minimum or maximum point, where $h = -\frac{b}{2a}$ and $k = f(h)$. 3. **Determine minimum or maximum:** Here, $a = 2 > 0$, so the function has a minimum value. 4. **Find vertex:** Calculate $h = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1$. Calculate $k = f(1) = 2(1)^2 - 4(1) - 8 = 2 - 4 - 8 = -10$. 5. **Domain and range:** - Domain of any quadratic function is all real numbers: $(-\infty, \infty)$. - Since the parabola opens upward and minimum is $-10$, range is $[-10, \infty)$. **Final answers:** - The function has a minimum value. - Minimum value is $-10$ at $x = 1$. - Domain: $(-\infty, \infty)$. - Range: $[-10, \infty)$.