1. **State the problem:** We have the quadratic equation $y = ax^2 + bx + c$ and the conditions $-\frac{b}{2a} = 2$, $4a - 8a + c = -3$, and we need to find $\frac{ac}{b}$. 2. **Use the vertex formula:** The vertex $x$-coordinate of a quadratic $ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. Given $-\frac{b}{2a} = 2$, we have $$-\frac{b}{2a} = 2 \implies b = -4a.$$ 3. **Simplify the second condition:** $$4a - 8a + c = -3 \implies -4a + c = -3 \implies c = -3 + 4a.$$ 4. **Calculate $\frac{ac}{b}$:** Substitute $b = -4a$ and $c = -3 + 4a$: $$\frac{ac}{b} = \frac{a(-3 + 4a)}{-4a} = \frac{a(-3 + 4a)}{\cancel{-4a}} \times \frac{\cancel{a}}{1} = \frac{-3 + 4a}{-4} = \frac{3 - 4a}{4}.$$ 5. **Final answer:** $$\boxed{\frac{ac}{b} = \frac{3 - 4a}{4}}.$$