1. **State the problem:** We are given the function $f(x) = 3x^2 - 8x$ and want to analyze it. 2. **Identify the function type:** This is a quadratic function of the form $ax^2 + bx + c$ where $a=3$, $b=-8$, and $c=0$. 3. **Find the roots (x-intercepts):** Set $f(x) = 0$: $$3x^2 - 8x = 0$$ Factor out $x$: $$x(3x - 8) = 0$$ So the roots are: $$x = 0 \quad \text{or} \quad 3x - 8 = 0 \Rightarrow x = \frac{8}{3}$$ 4. **Find the vertex (extremum):** The vertex $x$-coordinate is given by: $$x = -\frac{b}{2a} = -\frac{-8}{2 \times 3} = \frac{8}{6} = \frac{4}{3}$$ Calculate $f\left(\frac{4}{3}\right)$: $$f\left(\frac{4}{3}\right) = 3\left(\frac{4}{3}\right)^2 - 8\left(\frac{4}{3}\right) = 3 \times \frac{16}{9} - \frac{32}{3} = \frac{48}{9} - \frac{96}{9} = -\frac{48}{9} = -\frac{16}{3}$$ 5. **Interpretation:** The parabola opens upwards (since $a=3 > 0$), so the vertex at $\left(\frac{4}{3}, -\frac{16}{3}\right)$ is a minimum point. **Final answer:** - Roots: $x=0$ and $x=\frac{8}{3}$ - Vertex (minimum): $\left(\frac{4}{3}, -\frac{16}{3}\right)$