1. The problem is to simplify the expression $-3x^2 - 2x + 1$. 2. This is a quadratic expression in standard form $ax^2 + bx + c$ where $a = -3$, $b = -2$, and $c = 1$. 3. Since the expression is already simplified, we can analyze its properties such as vertex, intercepts, or factorization if needed. 4. To find the roots, use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. Substitute $a = -3$, $b = -2$, and $c = 1$: $$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-3)(1)}}{2(-3)} = \frac{2 \pm \sqrt{4 + 12}}{-6} = \frac{2 \pm \sqrt{16}}{-6}$$ 6. Simplify the square root: $$x = \frac{2 \pm 4}{-6}$$ 7. Calculate the two roots: - For $+$: $$x = \frac{2 + 4}{-6} = \frac{6}{-6} = -1$$ - For $-$: $$x = \frac{2 - 4}{-6} = \frac{-2}{-6} = \frac{1}{3}$$ 8. Therefore, the roots of the quadratic are $x = -1$ and $x = \frac{1}{3}$. 9. The vertex can be found using $x = -\frac{b}{2a}$: $$x = -\frac{-2}{2(-3)} = \frac{2}{-6} = -\frac{1}{3}$$ 10. Substitute $x = -\frac{1}{3}$ back into the expression to find the vertex $y$-value: $$y = -3\left(-\frac{1}{3}\right)^2 - 2\left(-\frac{1}{3}\right) + 1 = -3\left(\frac{1}{9}\right) + \frac{2}{3} + 1 = -\frac{1}{3} + \frac{2}{3} + 1 = \frac{4}{3}$$ 11. The vertex is at $\left(-\frac{1}{3}, \frac{4}{3}\right)$. Final answer: The quadratic $-3x^2 - 2x + 1$ has roots $x = -1$ and $x = \frac{1}{3}$ and vertex at $\left(-\frac{1}{3}, \frac{4}{3}\right)$.