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)

Which is:

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.