SAT Question 4dd4efcf: Nonlinear Functions answer and explanation with Desmos solution

Choice D is correct. It's given that the graph of $y=f\left(x\right)$ in the xy-plane is a parabola with vertex $\left(h,k\right)$. If $f\left(-9\right)=f\left(3\right)$, then for the graph of $y=f\left(x\right)$, 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 $y=f\left(x\right)$, the points with x-coordinates $-9$ and $3$ are equidistant from the vertex, $\left(h,k\right)$, 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, $f\left(x\right)=a{\left(x-h\right)}^2+k$. Substituting $-3$ for $h$ in this equation yields $f\left(x\right)=a{\left(x-\left(-3\right)\right)}^2+k$, or $f\left(x\right)=a{\left(x+3\right)}^2+k$. This equation is equivalent to $f\left(x\right)=a\left(x^2+6x+9\right)+k$, or $f\left(x\right)=a{x}^2+6ax+9a+k$. Since $f\left(x\right)=a{x}^2+4x+c$, 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 $9\left(\frac{2}{3}\right)+k=c$, 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.