1. **State the problem:** Find the vertex and x-intercepts of the quadratic function $f(x) = 3x^2 - 2x - 2$. 2. **Formula for vertex:** The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. 3. **Calculate the vertex:** Here, $a=3$, $b=-2$, so $$x = -\frac{-2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}.$$ Find $f\left(\frac{1}{3}\right)$: $$f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) - 2 = 3\times \frac{1}{9} - \frac{2}{3} - 2 = \frac{1}{3} - \frac{2}{3} - 2 = -\frac{1}{3} - 2 = -\frac{7}{3} \approx -2.33.$$ So the vertex is at $\left(\frac{1}{3}, -\frac{7}{3}\right)$ or approximately $(0.33, -2.33)$. 4. **Find x-intercepts:** Solve $3x^2 - 2x - 2 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 3 \times (-2)}}{2 \times 3} = \frac{2 \pm \sqrt{4 + 24}}{6} = \frac{2 \pm \sqrt{28}}{6}.$$ Simplify $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$: $$x = \frac{2 \pm 2\sqrt{7}}{6} = \frac{\cancel{2}(1 \pm \sqrt{7})}{\cancel{6}3} = \frac{1 \pm \sqrt{7}}{3}.$$ 5. **Calculate approximate values:** $$\sqrt{7} \approx 2.65,$$ so $$x_1 = \frac{1 + 2.65}{3} = \frac{3.65}{3} \approx 1.22,$$ $$x_2 = \frac{1 - 2.65}{3} = \frac{-1.65}{3} \approx -0.55.$$ 6. **Final answers:** - Vertex: $(0.33, -2.33)$ - x-intercepts: $1.22, -0.55$