[quote]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).There is one point given at: (-1,-2)
This is impossible. If this is the graph of any function, it [b]cannot go through both (-1, 0) and (-1, -2).The graph starts from the bottom goes up towards -1(x-axis) then curves and goes through zero, down to the point (-1,-2), curves at that point and goes up through the point (2,0). The curves are consistent in size.
So what I was thinking was something like:
y = a (x+1)(x-2)^2
y = 1 (x+1)(x-2)^2
...This get's me somewhat close but is not correct.
I know you need +1 and -2 in brackets, since these will be opposite of the zero's which are -1 and +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 .__________________________________________________ _________________
Sort of related so I though I'd ask in the same thread. This is a different question.
Determine an equation in factored form for the polynomial function with zeros 2 (order2), 2/3, and 3 that passes through the point (1,6).
What does "(Order 2)" mean? What are orders? I know zero's are where the graph hit's 0-on-the-x-axis. How would I start this question?
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.