1. **Problem:** Determine the rule, in general form, of a quadratic function with vertex $V(1, -8)$ and one zero at $x=3$. 2. **Recall the vertex form of a quadratic function:** $$f(x) = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. **Substitute the vertex:** $$f(x) = a(x - 1)^2 - 8$$ 4. **Use the zero at $x=3$ to find $a$:** Since $f(3) = 0$, substitute $x=3$: $$0 = a(3 - 1)^2 - 8$$ $$0 = a(2)^2 - 8$$ $$0 = 4a - 8$$ 5. **Solve for $a$:** $$4a = 8$$ $$a = \frac{8}{4} = 2$$ 6. **Write the function with $a=2$:** $$f(x) = 2(x - 1)^2 - 8$$ 7. **Expand to general form:** $$f(x) = 2(x^2 - 2x + 1) - 8$$ $$= 2x^2 - 4x + 2 - 8$$ $$= 2x^2 - 4x - 6$$ **Final answer:** $$\boxed{f(x) = 2x^2 - 4x - 6}$$