Polynomial of Degree 2 through Given Points

frasifrasi
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Linear Algebra Question...

So, the professor assigned this problem even though the book never touches on it whatsoever. I have no clud how to do it, so I would appreciate if anyone can describe the process of demonstrate.

Q. Find the polynomial of degree 2 whose graph goes through points (1,-1), (2,3), and (3,13).

Thank you for the help!
 
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A polynomial of degree 2 is a*x^2+b*x+c=y. Substitute your three points and you get three linear equation in the variables a,b,c. Solve the system.
 
See the section where it talks about AX = 0, and about homogeneous trivial solutions (or non-trivial).
I strongly think that it would have something about this question in that section.
 
Dick, I think you got it...I will try it.

Should I solve for it by making it in rref ?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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