1. **State the problem:** We are given the function $f(x) = 3x^{2} - 6x$ and want to analyze it. 2. **Formula and rules:** This is a quadratic function of the form $ax^{2} + bx + c$ where $a=3$, $b=-6$, and $c=0$. 3. **Find the vertex:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a} = -\frac{-6}{2 \times 3} = 1$. 4. **Calculate the vertex $y$-value:** Substitute $x=1$ into $f(x)$: $$f(1) = 3(1)^2 - 6(1) = 3 - 6 = -3$$ So the vertex is at $(1, -3)$. 5. **Find the roots (x-intercepts):** Set $f(x) = 0$: $$3x^{2} - 6x = 0$$ Factor out $3x$: $$3x(x - 2) = 0$$ So $x=0$ or $x=2$. 6. **Find the y-intercept:** Set $x=0$: $$f(0) = 3(0)^2 - 6(0) = 0$$ So the y-intercept is at $(0,0)$. 7. **Summary:** The parabola opens upwards (since $a=3>0$), has vertex at $(1,-3)$, roots at $x=0$ and $x=2$, and y-intercept at $(0,0)$. **Final answer:** The function $f(x) = 3x^{2} - 6x$ has vertex $(1,-3)$, roots $0$ and $2$, and y-intercept $0$.