1. **State the problem:** We have the quadratic function $f(x) = -2x^2 - 16x - 29$. We need to determine: (a) Whether the function has a minimum or maximum value. (b) The value of that minimum or maximum. (c) The $x$-value where this minimum or maximum occurs. 2. **Formula and rules:** A quadratic function $f(x) = ax^2 + bx + c$ has: - A maximum if $a < 0$. - A minimum if $a > 0$. The vertex (which gives the min or max) occurs at: $$x = -\frac{b}{2a}$$ The value of the function at the vertex is: $$f\left(-\frac{b}{2a}\right)$$ 3. **Identify coefficients:** Here, $a = -2$, $b = -16$, and $c = -29$. Since $a = -2 < 0$, the parabola opens downward, so the function has a **maximum** value. 4. **Find the $x$-value of the vertex:** $$x = -\frac{b}{2a} = -\frac{-16}{2 \times -2} = -\frac{-16}{-4} = -4$$ 5. **Find the maximum value by evaluating $f(-4)$:** $$f(-4) = -2(-4)^2 - 16(-4) - 29 = -2(16) + 64 - 29 = -32 + 64 - 29 = 3$$ **Final answers:** (a) The function has a **maximum** value. (b) The maximum value is $3$. (c) The maximum occurs at $x = -4$.