Nonlinear functions SAT question

Choice D is correct. It's given that the graph of $y=f(x)$ in the xy-plane is a parabola with vertex $(h,k)$. If $f(-9)=f(3)$, then for the graph of $y=f(x)$, the point with an x-coordinate of $-9$ and the point with an x-coordinate of $3$ have the same y-coordinate. In the xy-plane, a parabola is a symmetric graph such that when two points have the same y-coordinate, these points are equidistant from the vertex, and the x-coordinate of the vertex is halfway between the x-coordinates of these two points. Therefore, for the graph of $y=f(x)$, the points with x-coordinates $-9$ and $3$ are equidistant from the vertex, $(h,k)$, and $h$ is halfway between $-9$ and $3$. The value that is halfway between $-9$ and $3$ is $\frac{-9+3}{2}$, or $-3$. Therefore, $h=-3$. The equation defining $f$ can also be written in vertex form, $f(x)=a{(x-h)}^2+k$. Substituting $-3$ for $h$ in this equation yields $f(x)=a{(x-(-3))}^2+k$, or $f(x)=a{(x+3)}^2+k$. This equation is equivalent to $f(x)=a(x^2+6x+9)+k$, or $f(x)=a{x}^2+6ax+9a+k$. Since $f(x)=a{x}^2+4x+c$, it follows that $6a=4$ and $9a+k=c$. Dividing both sides of the equation $6a=4$ by $6$ yields $a=\frac{4}{6}$, or $a=\frac{2}{3}$. Since $\frac{2}{3}<1$, it's not true that $a\ge 1$. Therefore, statement II isn't true. Substituting $\frac{2}{3}$ for $a$ in the equation $9a+k=c$ yields $9(\frac{2}{3})+k=c$, or $6+k=c$. Subtracting $6$ from both sides of this equation yields $k=c-6$. If $k<0$, then $c-6<0$, or $c<6$. Since $c$ could be any value less than $6$, it's not necessarily true that $c<0$. Therefore, statement I isn't necessarily true. Thus, neither I nor II must be true.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.