1. **State the problem:** We need to find the equation of the parabola representing Jillian's first bounce, which starts at (0,0), reaches a vertex at (2,6), and ends at (6,0). 2. **Recall the vertex form of a parabola:** The equation is given by $$y = a(x - h)^2 + k$$ where $ (h,k) $ is the vertex. 3. **Identify the vertex:** From the problem, the vertex is at $ (2,6) $, so $ h = 2 $ and $ k = 6 $. 4. **Write the general form with vertex:** $$y = a(x - 2)^2 + 6$$ 5. **Use a known point to find $a$:** The parabola passes through $ (0,0) $, so substitute $ x=0 $ and $ y=0 $: $$0 = a(0 - 2)^2 + 6$$ $$0 = a(4) + 6$$ $$4a = -6$$ $$a = -\frac{6}{4} = -\frac{3}{2}$$ 6. **Write the final equation:** $$y = -\frac{3}{2}(x - 2)^2 + 6$$ 7. **Check with the other endpoint:** At $ x=6 $, $$y = -\frac{3}{2}(6 - 2)^2 + 6 = -\frac{3}{2}(4)^2 + 6 = -\frac{3}{2} \times 16 + 6 = -24 + 6 = -18$$ This does not match the endpoint $ (6,0) $, so re-examine step 5. 8. **Recalculate $a$ using the endpoint $ (6,0) $:** $$0 = a(6 - 2)^2 + 6$$ $$0 = a(4)^2 + 6$$ $$0 = 16a + 6$$ $$16a = -6$$ $$a = -\frac{6}{16} = -\frac{3}{8}$$ 9. **Final corrected equation:** $$y = -\frac{3}{8}(x - 2)^2 + 6$$ This equation fits the vertex and both endpoints. **Answer:** The rule for Jillian's first bounce is $$y = -\frac{3}{8}(x - 2)^2 + 6$$.