Quadratic Interpolation question

doggitz
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I have to find a system of linear equations to determine a quadratic polynomial

p(x) = ax2+bx+c

Given the conditions:

p(1) = f(1)
p'(1) = f'(1)
p''(1) = f''(1)

where f(x) = xex-1

I know:

p(x) = ax2+bx+c
p'(x) = 2ax +b
p''(x) = 2x

f(x) = xex-1
f'(x) = ex-1 + xex-1
f''(x) = 2ex-1 + xex-1

I am just learning Linear Algebra, and I am in the early Chapters of my book, so I can only use the Gauss method or the Gauss-Jordan method for this matrix. Just to clarify, I am hoping to return to school, so I am trying to learn this on my own.

I understand that the x values are known and I am finding the coefficients values in the matrix operations, but I am a little confused about the column order of the 'a' and 'b' coefficients.

If

Code:
A = | x[SUP]2[/SUP]  x  1  | 
    | x[SUP]2[/SUP]  x  1  | 
    | x[SUP]2[/SUP]  x  1  |

is my matrix "formula" how do I account for the 'a' coefficient "changing order" to the lower order coefficient of x, and what does my matrix look like to solve this problem?

Without the derivatives I am not confused, I would substitute for the 'x' values and set each equation to the corresponding 'y' value and solve the matrix to get the values of the coefficients for 'a' 'b' and 'c'. With the derivative involved I am in need of help.
 
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This isn't as complicated as you think. f(1)=1 and p(1)=a+b+c. So your first equation is 1=a+b+c. What are the second and third? It's not really about the x's.
 
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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