1. **Stating the problem:** We are given the quadratic function $$y = -\frac{1}{2}x^2 - x + 1$$ and need to analyze its properties. 2. **Formula and important rules:** The general form of a quadratic function is $$y = ax^2 + bx + c$$ where: - $$a$$ determines the direction of the parabola (upward if $$a > 0$$, downward if $$a < 0$$). - $$b$$ affects the position of the vertex. - $$c$$ is the y-intercept. 3. **Identify coefficients:** From the given equation, $$a = -\frac{1}{2}$$, $$b = -1$$, and $$c = 1$$. 4. **Direction of the parabola:** Since $$a = -\frac{1}{2} < 0$$, the parabola opens downward. 5. **Find the vertex:** The x-coordinate of the vertex is given by $$x = -\frac{b}{2a}$$. Calculate: $$x = -\frac{-1}{2 \times -\frac{1}{2}} = -\frac{-1}{-1} = -1$$. 6. **Find the y-coordinate of the vertex:** Substitute $$x = -1$$ into the equation: $$y = -\frac{1}{2}(-1)^2 - (-1) + 1 = -\frac{1}{2} + 1 + 1 = 1.5$$. 7. **Vertex coordinates:** The vertex is at $$(-1, 1.5)$$. 8. **Find y-intercept:** Set $$x=0$$: $$y = -\frac{1}{2}(0)^2 - 0 + 1 = 1$$. So the y-intercept is at $$(0,1)$$. 9. **Find x-intercepts:** Set $$y=0$$ and solve: $$0 = -\frac{1}{2}x^2 - x + 1$$ Multiply both sides by $$-2$$ to clear fraction: $$0 = x^2 + 2x - 2$$ Use quadratic formula: $$x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-2)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}$$ 10. **Final answer:** - Parabola opens downward. - Vertex at $$(-1, 1.5)$$. - Y-intercept at $$(0,1)$$. - X-intercepts at $$x = -1 + \sqrt{3}$$ and $$x = -1 - \sqrt{3}$$.