If the zeros are $0$ and $12$, write $f(x)=a\,x(x-12)$ for some constant $a$.

Use the point $(3,9)$ to solve for $a$ by substitution.

In Desmos, enter $y=a\,x(x-12)$ and create a slider for $a$.

Enter the point $(3,9)$.

Move the slider until the graph passes through $(3,9)$ while still crossing the $x$-axis at $x=0$ and $x=12$. Read the resulting equation from the expression list.

If a quadratic touches the floor at $x=0$ and $x=12$, then $f(0)=0$ and $f(12)=0$. So $f$ can be written as

Use $f(3)=9$:

So

Substitute $a=-\frac{1}{3}$ into $f(x)=ax(x-12)$:

Therefore, the correct choice is $f(x)=-\frac{1}{3}x(x-12)$.