The function $f(x)$ is defined as:

$$f(x) = ax^2 - 7x + c$$

The graph of $y = f(x)$ is a parabola with vertex $(h, k)$, where $h$ and $k$ are constants. We are given that $f(-9) = f(-2)$ and $c > 0$. We need to determine whether:

• I. $a < -\frac{3}{4}$
• II. $k > 16$

Solution:

Since $f(-9) = f(-2)$, the line of symmetry (the vertex) lies midway between $-9$ and $-2$:

$$h = \frac{-9 + (-2)}{2} = -\frac{11}{2}$$

Substitute $h = -\frac{11}{2}$ into the vertex form of a parabola:

$$f(x) = a(x + \frac{11}{2})^2 + k$$ Expanding: $$f(x) = a\left(x^2 + 11x + \frac{121}{4}\right) + k$$ $$f(x) = ax^2 + 11ax + \frac{121a}{4} + k$$

Equating the coefficients with $f(x) = ax^2 - 7x + c$:

$$11a = -7 \implies a = -\frac{7}{11}$$

Now, calculate $k$:

At the vertex, $k = f(h)$, so substitute $h = -\frac{11}{2}$ into the original function:

$$f(h) = a\left(-\frac{11}{2}\right)^2 - 7\left(-\frac{11}{2}\right) + c$$ $$k = a\left(\frac{121}{4}\right) + \frac{77}{2} + c$$ $$k = \frac{121a}{4} + \frac{154}{4} + c$$ $$k = \frac{121a + 154}{4} + c$$ Substitute $a = -\frac{7}{11}$: $$k = \frac{121\left(-\frac{7}{11}\right) + 154}{4} + c$$ $$k = \frac{-77 + 154}{4} + c = \frac{77}{4} + c$$

Since $c > 0$, $k > \frac{77}{4} = 19.25$. Therefore, $k > 16$ is true.

Finally, check $a < -\frac{3}{4}$:

$-\frac{7}{11} > -\frac{3}{4}$ because $-\frac{7}{11}$ is closer to zero. Hence, $a < -\frac{3}{4}$ is false.

Conclusion: Only II is true.

Answer: B) II only