8. Given points $A(1,0)$ and $B(2,0)$. If the graph of the quadratic function $y=x^{2}+(a-3)x+3$ intersects the line segment $AB$ at only one point, then the range of values for $a$ is $\qquad$.
8. $-1 \leqslant a<-\frac{1}{2}$ or $a=3-2 \sqrt{3}$.

In two cases:
(1) Since the graph of the quadratic function $y=x^{2}+(a-3) x+3$ intersects the line segment $A B$ at only one point, and $A(1,0), B(2,0)$, then
$$
\left[1^{2}+(a-3) \times 1+3\right] \times\left[2^{2}+(a-3) \times 2+3\right]<0 \text {. }
$$

Solving this, we get $-1<a<-\frac{1}{2}$.
By $1^{2}+(a-3) \times 1+3=0$, we get $a=-1$, at this time, $x_{1}=1, x_{2}=3$, which meets the requirement;

By $2^{2}+(a-3) \times 2+3=0$, we get $a=-\frac{1}{2}$, at this time, $x_{1}=2, x_{2}=\frac{3}{2}$, which does not meet the requirement.
(2) Let $x^{2}+(a-3) x+3=0$, by the discriminant $\Delta=0$, we get $a=3 \pm 2 \sqrt{3}$.
When $a=3+2 \sqrt{3}$, $x_{1}=x_{2}=-\sqrt{3}$, which does not meet the requirement;
When $a=3-2 \sqrt{3}$, $x_{1}=x_{2}=\sqrt{3}$, which meets the requirement.
In summary, the range of $a$ is $-1 \leqslant a<-\frac{1}{2}$ or $a=3-2 \sqrt{3}$.