1. **Problem:** Identify characteristics of the quadratic function $f(x) = -(x + 6)(x + 2)$. 2. **Step 1: Expand the function** $$f(x) = -(x + 6)(x + 2) = -(x^2 + 2x + 6x + 12) = -(x^2 + 8x + 12) = -x^2 - 8x - 12$$ 3. **Step 2: Find the vertex** The vertex form is $f(x) = a(x - h)^2 + k$ where $(h,k)$ is the vertex. Use vertex formula $h = -\frac{b}{2a}$ for $ax^2 + bx + c$. Here, $a = -1$, $b = -8$. $$h = -\frac{-8}{2(-1)} = \frac{8}{-2} = -4$$ Calculate $k = f(-4)$: $$k = -(-4)^2 - 8(-4) - 12 = -16 + 32 - 12 = 4$$ Vertex is $(-4, 4)$. 4. **Step 3: Axis of symmetry** Axis of symmetry is the vertical line through the vertex: $$x = -4$$ 5. **Step 4: Domain** Quadratic functions have domain all real numbers: $$\text{Domain}: (-\infty, \infty)$$ 6. **Step 5: Range** Since $a = -1 < 0$, parabola opens downward, so maximum at vertex $y=4$. $$\text{Range}: (-\infty, 4]$$ 7. **Step 6: Increasing and decreasing intervals** - Increasing on $(-\infty, -4)$ - Decreasing on $(-4, \infty)$ 8. **Step 7: End behavior** As $x \to -\infty$, $y \to -\infty$ (because of negative leading coefficient) As $x \to +\infty$, $y \to -\infty$ **Final answers:** - Vertex: $(-4, 4)$ - Axis of Symmetry: $x = -4$ - Domain: $(-\infty, \infty)$ - Range: $(-\infty, 4]$ - Increasing: $(-\infty, -4)$ - Decreasing: $(-4, \infty)$ - End Behavior: $y \to -\infty$ as $x \to -\infty$ and $y \to -\infty$ as $x \to +\infty$