I think that there are many many possible equations. For me, I would start by using the maximum point and 'guess' the minimum point.
The derivative of the cubic gives a quadratic, with one solution at (x-2).
Let the other solution be (x+3). The quadratic then becomes:
y' = a(x-2)(x+3)
y' = ax^2 + ax - 6a
I put the a since we don't know 'how fast the quadratic is' and also allows for a negative value.
Integrating this should now give me the cubic.
Now I take (0, 1)
A suitable value of a is -1 since we know the cubic is descending.
So.. we get:
And making it easier on the eyes...
This is one of the equations. I could have taken another minimum and another value of a to get different equations.