Find the equation `y = β_0 + β_1 x` of the least-squares line that best fits the data points (-1, 0), (0, 1), (1, 2,),(2, 4).

Find the equation `y = β_0 +  β_1x` of the least-squares line that best fits the data points (-1, 0), (0, 1), (1, 2,),(2, 4).

Solution:

The least-squares line is a line that best fits the data points by minimizing the sum of the squares of the residuals (the differences between the observed values and the values predicted by the line).

To find the equation of the least-squares line, we first need to determine the slope `β_1` and y-intercept `β_0` of the line.

To find `β_1`, we use the following formula:

`β_1 = (nΣxy - ΣxΣy) / (nΣx^2 - (Σx)^2)`

Where:

n is the number of data points

`Σx` is the sum of all x-values

`Σy` is the sum of all y-values

`Σxy` is the sum of the product of x and y for each data point

`Σx^2` is the sum of the square of x for each data point

For our data points:

n = 4

`Σx = -1 + 0 + 1 + 2 = 2`

`Σy = 0 + 1 + 2 + 4 = 7`

`Σxy = (-1)(0) + (0)(1) + (1)(2) + (2)(4) = 10`

`Σx^2 = (-1)^2 + (0)^2 + (1)^2 + (2)^2 = 6`

Plugging these values into the formula for `β_1`, we get:

`β_1 = ((4)(10) - (2)(7)) / (4(6) - 2^2) = (40 - 14) / (24 - 4) = 26 / 20 = 13/10`

To find `β_0`, we use the following formula:

`β_0 = (Σy - β_1Σx) / n`

Plugging in the values we have for `β_1` and the data points:

`β_0 = (7 - (13/10) (2)) / 4 = (7 - 26/10) / 4 = (7 - 13/5) / 4 = (35 - 13) / 20 = 22/20`

So, the equation of the least-squares line is:

`y = β_0 + β_1x = (22/20) + (13/10)x = 11/10 + 13/10x`

Therefore, the equation of the least-squares line that best fits the data points (-1, 0), (0, 1), (1, 2), (2, 4) is `y = 11/10 + 13/10x`