9. Given the quadratic function $y=a x^{2}+b x+c$, and the linear function $y=$ $k(x-1)-\frac{k^{2}}{4}$. If their graphs have only one common point for any real number $k$, then the analytical expression of the quadratic function is $\qquad$ .
9. $y=x^{2}-2 x+1$.

From the problem, we get $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4} .\end{array}\right.$
Rearranging gives $a x^{2}+(b-k) x+\left(c+k+\frac{k^{2}}{4}\right)=0$.
Also, $\Delta=(b-k)^{2}-4 a\left(c+k+\frac{k^{2}}{4}\right)=0$,
which simplifies to $(1-a) k^{2}-2(b+2 a) k+\left(b^{2}-4 a c\right)=0$.
Since the above equation holds for any real number $k$, we solve to get $a=1, b=-2, c=1$.