Number of different quadratic functions mod 12. How many different quadratic functions are there of form:

$x \mapsto ax^2+bx+c \pmod{12}$

All I could come up with is an upper bound of $11*12*12$. This is a puzzle from a YouTube video by Socratica.
 A: The map $$\varphi: (a,b,c)\mapsto \bigl[ x \mapsto ax^2+bx+c \bigr]$$ is a homomorphism from $(\mathbb Z_{12})^3$ to $(\mathbb Z_{12})^{\mathbb Z_{12}}$ as groups under pointwise addition. We want to find the number of elements in the image
$$ \{ \varphi(a,b,c) \mid a\ne 0 \} $$
Fortunately this is the same as the image of $\varphi$ itself. The requirement that $a\ne 0$ doesn't change the image because we always have $\varphi(0,b,c)=\varphi(6,b+6,c)$ since $6x^2+6x\equiv 0\pmod{12}$ for all $x$. (To see this note that $6x^2+6x=6x(x+1)$ and either $x$ or $x+1$ will be even).
The size of the image of a homomorphism between finite groups is the size of the domain divided by the size of the kernel, so we need to count the number of triples $(a,b,c)$ such that
$$ ax^2 + bx + c $$
is the zero function.
First, of course $c$ must be $0$, because $\varphi(a,b,c)(0)=c$.
Next, calculating modulo $3$ we find that $a\equiv b\equiv c\equiv 0\pmod 3$ because $\mathbb Z_3$ is a field and a nonzero (formal) polynomial over a field can only have as many roots as its degree.
So $a,b\in\{0,3,6,9\}$.
Furthermore, plugging in $x=-1$ gives us $a-b\equiv 0$, so $a=b$. So the only triples that could possibly lead to the zero function are
$$ (0,0,0) \qquad (3,3,0) \qquad (6,6,0) \qquad (9,9,0) $$
We already know that $(6,6,0)$ (and of course $(0,0,0)$) is in the kernel, and plugging in $x=1$ elimintates $(3,3,0)$ and $(9,9,0)$.
In total, the kernel of $\varphi$ is exactly $\{(0,0,0),(6,6,0)\}$, so the size of the image of $\varphi$ is
$$ \frac{12^3}{2} = 864 $$
