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Find the least-squares line $y=\beta_{0}+\beta_{1} x$ that best fits the data $(-2,0),(-1,0),(0,2),(1,4),$ and $(2,4),$ assuming that the first and last data points are less reliable. Weight them half as much as the three interior points.

$2+\left(\frac{3}{2}\right) x$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 8

Applications of Inner Product Spaces

Vectors

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finding that aggression line is essentially a least squares problem, and we know that to solve that, we need to form the metrics acts of the coefficients, what we have as many rose as the points we have. So we have four O's and the first column is all once and on the second column. We have the access off the given points, so we have 235 and six. And then for the constant term little why we have again for Rose as many as the points. And we put the wise of the given points or 3210 So says we need to compute the normal equation. We started by computing ex transfer, relax and x transpose y x transpose x Just a simple computation. Pistons out to be 4 16 16 74 and then X transpose y is 6 17 Now that we have everything, we can compute the equation. So we know that bottom is given by X transpose x inverse that multiplies extracts was why that we have just computed so in red we have X transpose x inverse, So one over determinant. They're multiplies, said the four months, 16 months 16 4 and then in green, we have X transpose y so 6 17 He's just an issue of multiplication. So one over 40 that multiplies the metrics 1 72 minus 28. So all in all, beta turns out to be 4.3, minus 0.7. That means that the confusions that we're looking for are bitter. Zero is 4.3 and bitter one is minus 0.7.

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