Question: 01 (20 points) Consider the following data points ($1791) :=(012)2 ($2,312) I: (212): ($3793) :2 (124)! ($4,194) I: (273)' a) Suppose that the graph of

01 (20 points) Consider the following data points
01 (20 points) Consider the following data points ($1791) :=(012)2 ($2,312) I: (212): ($3793) :2 (124)! ($4,194) I: (273)' a) Suppose that the graph of the real quadratic polynomial p(3:) = a0 + alx + 0,2172 passes through all four points. Derive the resulting linear system satisfied by the unknowns a0, a1, and a2, and show that it is inconsistent. Hint: For example, requiring the graph ofp to pass through (1,4) means that p(1) = 4, which simplifies to the linear equation a0 a1 + a2 = 4 in the three unknowns a0, a1, (22. b) Compute ATA by hand, showing every step. Then, use row reduction to show that ATA is invertible and to compute its inverse---you may use technology for this row reduction. c) Compute ATb by hand, and use your answer to part b) to solve the modified linear system (.10 ATA a1 =ATb. (12 You may use technology to help compute the resulting solution to this modified system. d) Plot the original data points together with the graph ofp(a:) = a0 + 0,153 l (1232, using the values of a0, a1, and a2 you solved for in part c). What do you notice? Remark: That the system from part a) is inconsistent means that you cannot find a perfect quadratic fit for the four data points provided. As we'll see later in the course, the modified system from part 0) provides the next best thing

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