1. **State the problem:** We need to analyze and solve the quadratic function $$y = -x^2 - 6x - 10$$. 2. **Formula and rules:** The quadratic function is in the form $$y = ax^2 + bx + c$$ where $$a = -1$$, $$b = -6$$, and $$c = -10$$. 3. **Find the vertex:** The vertex of a parabola $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. Calculate: $$x = -\frac{-6}{2 \times -1} = -\frac{-6}{-2} = -3$$. 4. **Find the y-coordinate of the vertex:** Substitute $$x = -3$$ into the equation: $$y = -(-3)^2 - 6(-3) - 10 = -9 + 18 - 10 = -1$$. So, the vertex is $$(-3, -1)$$. 5. **Find the y-intercept:** Set $$x = 0$$: $$y = -0 - 0 - 10 = -10$$. 6. **Find the x-intercepts:** Set $$y = 0$$ and solve for $$x$$: $$0 = -x^2 - 6x - 10$$ Multiply both sides by $$-1$$: $$0 = x^2 + 6x + 10$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2}$$ Since the discriminant is negative, there are no real x-intercepts. 7. **Summary:** The parabola opens downward (since $$a = -1 < 0$$), has vertex at $$(-3, -1)$$, y-intercept at $$(0, -10)$$, and no real x-intercepts. **Final answer:** Vertex: $$(-3, -1)$$, y-intercept: $$(0, -10)$$, no real x-intercepts.