1. **Stating the problem:** We are given a quadratic function $y = f(x)$ with a graph of a downward-opening parabola. 2. **Given information:** - Vertex at approximately $(0,12)$. - Roots (x-intercepts) near $-2$ and $3$. 3. **Goal:** Find the correct formula for $f(x)$ from the options. 4. **Recall the factored form of a quadratic:** $$f(x) = a(x - r_1)(x - r_2)$$ where $r_1$ and $r_2$ are roots. 5. **Using the roots:** Given roots $-2$ and $3$, the factored form is: $$f(x) = a(x + 2)(x - 3)$$ 6. **Use the vertex to find $a$:** The vertex is at $x=0$, so substitute $x=0$ and $f(0) = 12$: $$12 = a(0 + 2)(0 - 3) = a(2)(-3) = -6a$$ Solve for $a$: $$a = -2$$ 7. **Write the function:** $$f(x) = -2(x + 2)(x - 3)$$ 8. **Expand to verify:** $$f(x) = -2(x^2 - 3x + 2x - 6) = -2(x^2 - x - 6) = -2x^2 + 2x + 12$$ 9. **Check options:** Option C is $-2(x-2)(x+3)$ which is different roots. Option D is $-(x+2)(x-3)$ which has $a = -1$. Option A and B are vertex forms but do not match vertex or roots. 10. **Conclusion:** The correct formula is: $$f(x) = -2(x + 2)(x - 3)$$ **Final answer:** Option D matches the roots but with $a=-1$, so the correct formula is $-2(x+2)(x-3)$ which is not exactly listed but closest to the handwritten solution. Since the handwritten math shows $a=-2$, the correct function is: $$f(x) = -2(x + 2)(x - 3)$$