8 Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfy $f(0)=1$, and for any $x, y \in \mathbf{R}$, there is $f(x y+1)=f(x) f(y)-f(y)-x+2$, then $f(x)=$ $\qquad$
For any $x, y \in \mathbf{R}$, we have
$$
f(x y+1)=f(x) f(y)-f(y)-x+2,
$$

so we also have $\quad f(y x+1)=f(y) f(x)-f(x)-y+2$.
Therefore, $\quad f(x) f(y)-f(y)-x+2=f(y) f(x)-f(x)-y+2$.
This simplifies to $f(x)+y=f(y)+x$.
Let $y=0$, we get $f(x)=x+1$.