1. **State the problem:** We have a parabola with vertex at $(9, -14)$ and two x-intercepts. The parabola's equation is $y = ax^2 + bx + c$. We need to find which value among $a+b+c$ could be true. 2. **Recall vertex form:** The parabola can be written as $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. Here, $h=9$ and $k=-14$, so $$y = a(x - 9)^2 - 14.$$ 3. **Use the fact that the parabola crosses the x-axis twice:** This means the quadratic has two distinct real roots. The roots come from solving $$a(x - 9)^2 - 14 = 0 \\ (x - 9)^2 = \frac{14}{a}.$$ For two distinct real roots, we need $\frac{14}{a} > 0$, so $a > 0$. 4. **Convert vertex form to standard form:** $$y = a(x^2 - 18x + 81) - 14 = a x^2 - 18 a x + 81 a - 14.$$ So, $$a = a, \quad b = -18 a, \quad c = 81 a - 14.$$ 5. **Calculate $a + b + c$:** $$a + b + c = a + (-18 a) + (81 a - 14) = (a - 18 a + 81 a) - 14 = 64 a - 14.$$ 6. **Check which given options can be written as $64 a - 14$ with $a > 0$:** Set $64 a - 14 = ext{option}$ and solve for $a$: - For $-23$: $64 a - 14 = -23 \Rightarrow 64 a = -9 \Rightarrow a = -\frac{9}{64} < 0$ (not valid) - For $-19$: $64 a - 14 = -19 \Rightarrow 64 a = -5 \Rightarrow a = -\frac{5}{64} < 0$ (not valid) - For $-14$: $64 a - 14 = -14 \Rightarrow 64 a = 0 \Rightarrow a = 0$ (not valid, $a$ cannot be zero) - For $-12$: $64 a - 14 = -12 \Rightarrow 64 a = 2 \Rightarrow a = \frac{1}{32} > 0$ (valid) 7. **Conclusion:** The only valid value for $a + b + c$ is $-12$. **Final answer:** $\boxed{-12}$