For $y=ax^2+bx+c$, the axis of symmetry is at $x=-\frac{b}{2a}$.

Point $C$ is on the y-axis. Plug in $x=0$ to determine $c$ immediately.

Enter the three points: $(-1,6)$, $(5,6)$, and $(0,2)$.

Since $(0,2)$ is on the parabola, set $c=2$ (type c=2).

Substitute $(-1,6)$ and $(5,6)$ into $y=ax^2+bx+c$ with $c=2$ to get two linear equations in $a$ and $b$:

$6=a(-1)^2+b(-1)+2$ and $6=a(5)^2+b(5)+2$.

Rewrite each equation in the form $b=\text{(expression in }a)$, then type both into Desmos. Click their intersection to read the corresponding values of $a$ and $b$.

From the graph, $A$ is at $(-1,6)$ and $B$ is at $(5,6)$. Since these points have the same $y$-value, they are the same horizontal distance from the axis of symmetry.

So the axis of symmetry is halfway between their $x$-coordinates:

Thus the axis of symmetry is $x=2$.

For $y=ax^2+bx+c$, the $x$-coordinate of the vertex (the axis of symmetry) is

So

Point $C$ is the y-intercept at $(0,2)$, so

Use point $A(-1,6)$ in $y=ax^2+bx+c$ with $c=2$:

Simplify:

Substitute $b=-4a$:

Since $b=-4a$,

Therefore, the correct choice is $-\frac{16}{5}$.