1. The problem is to find the roots of the quadratic function $f(x) = -2x^2 - 8x - 3$ and describe its graph. 2. The roots of a quadratic function $ax^2 + bx + c = 0$ are found using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = -2$, $b = -8$, and $c = -3$. 3. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-8)^2 - 4(-2)(-3) = 64 - 24 = 40$$ Since $\Delta > 0$, there are two distinct real roots. 4. Find the roots: $$x = \frac{-(-8) \pm \sqrt{40}}{2(-2)} = \frac{8 \pm \sqrt{40}}{-4}$$ Simplify $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$: $$x = \frac{8 \pm 2\sqrt{10}}{-4} = \frac{\cancel{2} \times (4 \pm \sqrt{10})}{\cancel{2} \times (-2)} = \frac{4 \pm \sqrt{10}}{-2}$$ 5. Split into two roots: $$x_1 = \frac{4 + \sqrt{10}}{-2} = -2 - \frac{\sqrt{10}}{2}$$ $$x_2 = \frac{4 - \sqrt{10}}{-2} = -2 + \frac{\sqrt{10}}{2}$$ 6. The parabola opens downward because $a = -2 < 0$. 7. The vertex is at: $$x = -\frac{b}{2a} = -\frac{-8}{2 \times -2} = -\frac{8}{-4} = 2$$ Calculate $f(2)$: $$f(2) = -2(2)^2 - 8(2) - 3 = -8 - 16 - 3 = -27$$ So the vertex is at $(2, -27)$. 8. Summary: - Roots: $x = -2 - \frac{\sqrt{10}}{2}$ and $x = -2 + \frac{\sqrt{10}}{2}$ - Parabola opens downward - Vertex at $(2, -27)$ This matches the shape of a downward parabola crossing the x-axis near these roots.