1. **State the problem:** We are given the quadratic function $y = -2(x+4)^2 - 1$ and want to analyze it. 2. **Formula and rules:** This is a quadratic function in vertex form $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex. 3. **Identify vertex:** Here, $a = -2$, $h = -4$, and $k = -1$, so the vertex is at $(-4, -1)$. 4. **Direction of parabola:** Since $a = -2 < 0$, the parabola opens downward. 5. **Axis of symmetry:** The axis of symmetry is the vertical line $x = -4$. 6. **Find y-intercept:** Set $x=0$: $$y = -2(0+4)^2 - 1 = -2(16) - 1 = -32 - 1 = -33$$ So the y-intercept is $(0, -33)$. 7. **Find x-intercepts:** Set $y=0$: $$0 = -2(x+4)^2 - 1$$ $$-2(x+4)^2 = 1$$ $$(x+4)^2 = \frac{1}{-2}$$ Since the right side is negative, there are no real x-intercepts. 8. **Summary:** The parabola has vertex $(-4,-1)$, opens downward, y-intercept at $(0,-33)$, and no real x-intercepts. **Final answer:** Vertex at $(-4,-1)$, opens downward, y-intercept $(0,-33)$, no real roots.