1. **State 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 rules:** A quadratic function is generally written as $$y = ax^{2} + bx + c$$ where $a$, $b$, and $c$ are constants. - The parabola opens downward if $a < 0$ and upward if $a > 0$. - The vertex (maximum or minimum point) is at $$x = -\frac{b}{2a}$$. - The y-intercept is at $y = c$. 3. **Identify coefficients:** Here, $a = -\frac{1}{2}$, $b = -1$, and $c = 1$. 4. **Find the vertex:** $$x = -\frac{b}{2a} = -\frac{-1}{2 \times -\frac{1}{2}} = -\frac{-1}{-1} = -1$$ 5. **Calculate the y-coordinate of the vertex:** $$y = -\frac{1}{2}(-1)^{2} - (-1) + 1 = -\frac{1}{2} \times 1 + 1 + 1 = -\frac{1}{2} + 2 = \frac{3}{2}$$ So the vertex is at $$(-1, \frac{3}{2})$$. 6. **Find the y-intercept:** Set $x=0$: $$y = -\frac{1}{2} \times 0^{2} - 0 + 1 = 1$$ So the y-intercept is at $(0,1)$. 7. **Find the x-intercepts:** Set $y=0$: $$0 = -\frac{1}{2}x^{2} - x + 1$$ Multiply both sides by $-2$ to clear the fraction: $$0 = x^{2} + 2x - 2$$ Use the quadratic formula: $$x = \frac{-2 \pm \sqrt{(2)^{2} - 4 \times 1 \times (-2)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}$$ So the x-intercepts are at $$x = -1 + \sqrt{3}$$ and $$x = -1 - \sqrt{3}$$. **Final answer:** The parabola opens downward with vertex at $$(-1, \frac{3}{2})$$, y-intercept at $(0,1)$, and x-intercepts at $$x = -1 \pm \sqrt{3}$$.