Answer: D. Neither I nor II

Explanation:1. Analyze the given information: - $f(x)=ax^{2}+4x+c$. - Parabola opens upward $\implies a > 0$. - Vertex at $(h, k)$ with $k<0$. This means the minimum value of the function is negative. Since the parabola opens upward and its minimum value is negative, it must have two distinct real roots (it crosses the x-axis). - $f(-8)=f(2)$. Due to the symmetry of a parabola, the x-coordinate of the vertex (h) is the midpoint of any two x-values that have the same f(x) value. So, $h = \dfrac{-8+2}{2} = \dfrac{-6}{2} = -3$.

2. Relate h to the coefficients of the quadratic: The x-coordinate of the vertex of $f(x)=ax^2+bx'+c'$ (using $b'$ to avoid confusion with the constant $c$ in the problem) is given by $h = -\dfrac{b'}{2a}$. In our function, $f(x)=ax^{2}+4x+c$, the coefficient of x is $4$. So, $h = -\dfrac{4}{2a} = -\dfrac{2}{a}$.

3. Use the value of h: We found $h=-3$ and $h = -\dfrac{2}{a}$. Therefore, $-3 = -\dfrac{2}{a}$. $3 = \dfrac{2}{a}$ $3a = 2$ $a = \dfrac{2}{3}$. Since $a = \dfrac{2}{3}$, we know $a > 0$, which is consistent with the parabola opening upward.

4. Analyze statement II: $a \ge 1$. We found $a = \dfrac{2}{3}$. Is $\dfrac{2}{3} \ge 1$? No, $\dfrac{2}{3} < 1$. So, statement II is false.

5. Analyze statement I: $c<0$. The y-coordinate of the vertex is $k = f(h)$. We know $k<0$. $k = f(-3) = a(-3)^2 + 4(-3) + c$ $k = 9a - 12 + c$ Substitute $a = \dfrac{2}{3}$: $k = 9(\dfrac{2}{3}) - 12 + c$ $k = 3 \times 2 - 12 + c$ $k = 6 - 12 + c$ $k = -6 + c$ We are given $k<0$. So, $-6 + c < 0$. $c < 6$. Statement I says $c<0$. Does $c<6$ mean that $c$ *must* be less than $0$? No. For example, if $c=1$, then $1<6$. In this case, $k = -6+1 = -5$, which is $<0$. Here $c=1$ is not $<0$. So, $c<0$ is not necessarily true. For instance, $c$ could be $1, 2, 3, 4, 5$. For $c=5$, $k = -6+5 = -1 < 0$. Here $c=5$ is not $<0$. For $c=-1$, $k = -6-1 = -7 < 0$. Here $c=-1$ is $<0$. Since we can find a case where $k<0$ but $c$ is not $<0$ (e.g., $c=1$), statement I is not *must be true*.

Conclusion:Neither statement I nor statement II must be true.- Statement I ($c<0$): We found $c<6$. This does not guarantee $c<0$.- Statement II ($a \ge 1$): We found $a = \dfrac{2}{3}$, which is not $\ge 1$.Therefore, the answer is Neither I nor II.