Question 1162807
f(x)=x(x-3)^2 roots at 0 and 3, with a bounce at 3.  Crosses and touches
{{{graph(300,300,-10,10,-10,10,x*(x-3)^2)}}}

g(x)=x(x+2)(x-3), roots at 0,-2, 3 crosses
{{{graph(300,300,-10,10,-10,10,x*(x+2)(x-3))}}}

H(x)=(x-3)^3 crosses at x=3
{{{graph(300,300,-10,10,-10,10,(x-3)^3)}}}

4. -(x+3)(x-1)^2. This is a cubic, so making it negative will have the beginning positive and the end negative. Put in a bounce as the last pair of roots, and the function crosses once and touches.
{{{graph(300,300,-10,10,-10,10,-(x+3)(x-1)^2)}}}