6. If the parabola $C_{m}: y=x^{2}-m x+m+1$ intersects the line segment $A B$ (where $A(0,4), B(4,0)$) at exactly two points, then the range of values for $m$ is $\qquad$
6. The equation of line $AB$ is $x+y=4(0 \leqslant x \leqslant 4)$. Substituting $y=4-x$ into $y=x^{2}-m x+m+1$ yields
$x^{2}-(m-1) x+(m-3)=0$.
Let $f(x)=x^{2}-(m-1) x+m-3$.
This problem is equivalent to the equation (1) having two distinct real roots $x_{1}, x_{2}$, and $0 \leqslant x_{1}0, \\
0 \leqslant \frac{m-1}{2} \leqslant 4, \\
f(0)=m-3 \geqslant 0, \\
f(4)=16-4(m-1)+m-3 \geqslant 0 .
\end{array}\right.
$$

Solving this, we get $3 \leqslant m \leqslant \frac{17}{3}$.