1. **Problem Statement:** We have a parabola with vertex at $(9, -14)$ and it intersects the x-axis at two points. The parabola's equation is $y = ax^2 + bx + c$. We need to find which value among $a+b+c$ could be correct. 2. **Recall the vertex form of a parabola:** $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. Here, $h=9$ and $k=-14$, so: $$y = a(x - 9)^2 - 14$$ 3. **Convert vertex form to standard form:** $$y = a(x^2 - 18x + 81) - 14 = a x^2 - 18 a x + 81 a - 14$$ So, $$a = a, \quad b = -18 a, \quad c = 81 a - 14$$ 4. **Sum $a + b + c$:** $$a + b + c = a + (-18 a) + (81 a - 14) = (a - 18 a + 81 a) - 14 = 64 a - 14$$ 5. **Use the fact that the parabola intersects the x-axis at two points:** This means the quadratic has two distinct real roots, so the discriminant $\Delta > 0$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4 a c = (-18 a)^2 - 4 a (81 a - 14) = 324 a^2 - 4 a (81 a - 14) = 324 a^2 - 324 a^2 + 56 a = 56 a$$ 7. **For two distinct roots, $\Delta > 0$ implies:** $$56 a > 0 \implies a > 0$$ 8. **Check which given value of $a + b + c = 64 a - 14$ can be obtained with $a > 0$:** Add 14 to each option: - A) $-23 + 14 = -9$ (would require $64 a = -9 \Rightarrow a = -9/64 < 0$ no) - B) $-19 + 14 = -5$ (would require $a = -5/64 < 0$ no) - C) $-14 + 14 = 0$ (would require $a = 0$ no, parabola would be linear) - D) $-12 + 14 = 2$ (would require $a = 2/64 = 1/32 > 0$ yes) 9. **Conclusion:** The only possible value for $a + b + c$ given the conditions is **-12**. **Final answer:** $\boxed{-12}$