Question: User: f(x)=3x^2-2x-8
1. **State the problem:** We are given the quadratic function $$f(x)=3x^2-2x-8$$ and we want to analyze it.
2. **Recall the general form:** A quadratic function is generally written as $$f(x)=ax^2+bx+c$$ where $$a=3$$, $$b=-2$$, and $$c=-8$$.
3. **Find the roots (zeros) of the function:** To find the roots, solve $$3x^2-2x-8=0$$ using the quadratic formula:
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
4. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = (-2)^2 - 4 \times 3 \times (-8) = 4 + 96 = 100$$
5. **Calculate the roots:**
$$x=\frac{-(-2) \pm \sqrt{100}}{2 \times 3} = \frac{2 \pm 10}{6}$$
6. **Evaluate each root:**
- $$x_1 = \frac{2 + 10}{6} = \frac{12}{6} = 2$$
- $$x_2 = \frac{2 - 10}{6} = \frac{-8}{6} = -\frac{4}{3}$$
7. **Find the vertex:** The vertex $$x$$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{-2}{2 \times 3} = \frac{2}{6} = \frac{1}{3}$$.
8. **Calculate the vertex $$y$$-coordinate:**
$$f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) - 8 = 3 \times \frac{1}{9} - \frac{2}{3} - 8 = \frac{1}{3} - \frac{2}{3} - 8 = -\frac{1}{3} - 8 = -\frac{25}{3}$$
9. **Summary:**
- Roots: $$x=2$$ and $$x=-\frac{4}{3}$$
- Vertex: $$\left(\frac{1}{3}, -\frac{25}{3}\right)$$
- Since $$a=3>0$$, the parabola opens upwards.
**Final answer:** The function $$f(x)=3x^2-2x-8$$ has roots at $$x=2$$ and $$x=-\frac{4}{3}$$ and a vertex at $$\left(\frac{1}{3}, -\frac{25}{3}\right)$$.