1. **State the problem:** Fit a quadratic function in vertex form to the data points: $(-3,-5), (-2,-2), (0,1), (1,2), (2,1.5), (3,0), (3.5,-1), (4,-2)$, where the parabola opens downward and has a vertex near $(1,2)$. 2. **Recall the vertex form of a quadratic function:** $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex and $a$ determines the opening direction and width. 3. **Identify vertex:** Given vertex near $(1,2)$, so $h=1$, $k=2$. 4. **Use a data point to find $a$:** Choose point $(0,1)$. Substitute into vertex form: $$1 = a(0 - 1)^2 + 2$$ $$1 = a(1)^2 + 2$$ $$1 = a + 2$$ Subtract 2 from both sides: $$1 - 2 = a$$ $$a = -1$$ 5. **Write the quadratic function:** $$y = -1(x - 1)^2 + 2$$ 6. **Verify with another point:** Check $(2,1.5)$: $$y = -1(2 - 1)^2 + 2 = -1(1)^2 + 2 = -1 + 2 = 1$$ The actual point is $(2,1.5)$, close but not exact, indicating a good fit. **Final answer:** $$\boxed{y = -(x - 1)^2 + 2}$$