Question 1194729
<pre>
Might as well let the line be the x-axis.
To have a point of tangency to the x-axis at the origin, 
(i.e., "bounce" off the x-axis there), 
it must have a root or zero at 0 with an even multiplicity, say 2.
So it must have a factor of (x-0)<sup>2</sup> or x<sup>2</sup>

In order for it to intersect (cut through the x-axis) at 1, 2, and 3,
it must have factors (x-1), (x-2), (x-3) of odd multiplicity, say 
multiplicity 1 each. So it must have factors (x-1)<sup>1</sup>, (x-2)<sup>1</sup>, and 
(x-3)<sup>1</sup>, or (x-1), (x-2), and (x-3).

So one such polynomial curve would be

{{{f(x)}}}{{{""=""}}}{{{x^2(x-1)(x-2)(x-3)}}}

{{{graph(400,400,-2,4,-4,2,x^2(x-1)(x-2)(x-3))}}}

Edwin</pre>