1. **State the problem:** Simplify and analyze the quadratic expression $-x^{2} + 2x - \frac{3}{4}$. 2. **Recall the standard form:** A quadratic expression is generally written as $ax^{2} + bx + c$. Here, $a = -1$, $b = 2$, and $c = -\frac{3}{4}$. 3. **Find the vertex:** The vertex of a parabola $y = ax^{2} + bx + c$ is at $x = -\frac{b}{2a}$. Substitute $a$ and $b$: $$x = -\frac{2}{2 \times (-1)} = -\frac{2}{-2} = 1$$ 4. **Calculate the vertex's y-coordinate:** Substitute $x=1$ into the expression: $$y = -(1)^{2} + 2(1) - \frac{3}{4} = -1 + 2 - \frac{3}{4} = 1 - \frac{3}{4} = \frac{1}{4}$$ 5. **Determine the direction of the parabola:** Since $a = -1 < 0$, the parabola opens downward. 6. **Find the roots (x-intercepts):** Solve $-x^{2} + 2x - \frac{3}{4} = 0$. Multiply both sides by $-1$ to simplify: $$\cancel{-}x^{2} + \cancel{2}x - \cancel{\frac{3}{4}} = 0 \Rightarrow x^{2} - 2x + \frac{3}{4} = 0$$ 7. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(\frac{3}{4})}}{2(1)} = \frac{2 \pm \sqrt{4 - 3}}{2} = \frac{2 \pm 1}{2}$$ 8. **Calculate the roots:** $$x_1 = \frac{2 + 1}{2} = \frac{3}{2} = 1.5$$ $$x_2 = \frac{2 - 1}{2} = \frac{1}{2} = 0.5$$ 9. **Summary:** The parabola opens downward, has vertex at $(1, \frac{1}{4})$, and roots at $x=0.5$ and $x=1.5$.