**find a quadratic model (-4,8), (-1,5), (1,13)**

**Answer:**

You have three points you know the quadratic model (polynomial) passes through.

The general form of your model is:

y = ax^2 + bx + c

This gives you three unknowns you have to determine: (a,b,c).

But you also have three points:

(x1,y1) = (-4,8)

(x2,y2) = (-1,5)

(x3,y3) = (1,13)

so all you need to do is to first plug in

x=-4 and y=8, e.g.,:

8 = a*(-4)^2 + b*(-4) + c

Then solve for e.g. c.

Now you know c (but in terms of a and b still so we need more points to get rid of those; luckily we have two more points to use).

Then, plug in c in terms of a and b into your quadratic model and go to the next point:

5 = a*(-1)^2 + b*(-1) + c (where you have to insert c = whatever you found by inserting the first point).

You can then solve for, e.g., a in the above equation. If you plug a into your expression for c you see that the only parameter left which you dont know is b.

But by using the last point (1,13) in a similar way you can get that as well.