If $f(u)=2$ at two x-values, then $f(f(x))=2$ becomes $f(x)=u_1$ or $f(x)=u_2$.

Check the smallest y-value on the graph to see whether an equation like $f(x)=-6$ can have any solutions.

Enter a vertex-form function with a slider:

This matches the vertex at $(-1,-4)$.

Adjust $a$ so that the graph passes through $(-6,2)$ (it will then also pass through $(4,2)$, as in the figure).

In a new line, enter $y=f(f(x))$, and also enter $y=2$.

Use the graph to count how many intersection points $y=f(f(x))$ has with the line $y=2$. That count is the answer.

The equation $f(f(x))=2$ means: the outer function outputs 2 when its input is $f(x)$.

So first, look for where the graph of $f$ has $y=2$.

From the graph, the curve passes through the points $(-6,2)$ and $(4,2)$, so:

If $f(f(x))=2$ and the only inputs that produce 2 are $-6$ and $4$, then $f(x)$ must be one of those values:

• For $f(x)=-6$: the graph’s lowest point is the vertex at $(-1,-4)$, so $f(x)$ never reaches $-6$. That gives 0 solutions.

• For $f(x)=4$: the horizontal line $y=4$ crosses the parabola in 2 points (one on each side of the axis of symmetry). That gives 2 solutions.

Total solutions $=0+2=2$.

So the correct choice is $2$.