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