Finding a super-attractive orbit

Let $f(x)=x^2-4x+5$. Show that $f$ has a super-attractive orbit of period 2.

Note: This is under Test Prep, you should be able to do this easily without a calculator.

If $f(x)=x^2−4x+5$ then $f'(x)=2x-4$.

To be a super attractive orbit, $f'(x)=0$. So, $x=2$ and it's orbit are super attractive.

If it is a period 2 orbit, then $f(f(2))=2$
$$f(f(x))=f(2^2−4(2)+5)=f(1)=f(x)=(1)^2−4(1)+5=2$$
and $f$ has a super-attractive orbit , namely $2\to1\to2$, of period 2.

@Rick Looks good! I did make one little edit to clarify the fact that the orbit the specific sequence of two points $2\to1\to2$.