One, (20 points) Given the function
$$
y=k^{2} x^{2}+k\left(2 x-3 x^{2}\right)+2 x^{2}-2 x+1
$$

the graph does not pass through the fourth quadrant. Find the range of values for the constant $k$.
The original function can be transformed into
$$
y=(k-1)(k-2) x^{2}+2(k-1) x+1 \text {. }
$$
(1) When the function is a quadratic function, it can be known from the graph not passing through the fourth quadrant that the parabola must open upwards. There are two possibilities: one is that the parabola has no intersection points or only one intersection point with the $x$-axis; the other is that the parabola has two intersection points with the $x$-axis and the axis of symmetry is to the left of the $y$-axis. Then
$$
\left\{\begin{array}{l}
(k-1)(k-2)>0, \\
\Delta=[2(k-1)]^{2}-4(k-1)(k-2) \leqslant 0
\end{array}\right.
$$

or $\left\{\begin{array}{l}(k-1)(k-2)>0, \\ \Delta=[2(k-1)]^{2}-4(k-1)(k-2)>0, \\ -\frac{2(k-1)}{2(k-1)(k-2)}2\right.$.
(2) When the function is a linear function, it can be derived from the graph not passing through the fourth quadrant that
$$
\left\{\begin{array}{l}
(k-1)(k-2)=0, \\
2(k-1)>0
\end{array} \Rightarrow k=2 .\right.
$$
(3) When the function is a constant function,
$$
\left\{\begin{array}{l}
(k-1)(k-2)=0, \\
2(k-1)=0
\end{array} \Rightarrow k=1\right. \text {. }
$$

At this point, the function is $y=1$. Its graph does not pass through the fourth quadrant, which meets the requirement.
In summary, the range of the constant $k$ is $k \leqslant 1$ or $k \geqslant 2$.