There are three zero's marked at: -1, 0, +2

So we have (-1, 0), (0, 0), and (2, 0) which means,

**assuming** this is a polynomial, we have factors of x+ 1, x, and x- 2.

That's contradictory- you said above that was a 0 at -1 and if the graph of a function goes through (-1, 0) it

**cannot** also go through (-1, -2).

This is impossible. If this is the graph of any function, it [b]cannot go through both (-1, 0) and (-1, -2).

I suspect that "order of a zero" is defined in the text where you got this problem. A zero, a, of a polynomial is of "order n" if the polynomial has a factor of $\displaystyle (x- a)^n$.

Saying that this polynomial function has "zeros 2 (order2), 2/3, and 3" means it is (at least) of the form [ex]y= a(x- 2)^2(x- 2/3)(x- 3)[tex]. Assuming there are no other factors you can determine a so that the graph "passes through the point (1,6)" by setting x= 1, y= 6 in that and solving for a.