The given data is (-1, 0), (5, 5) and (6, 10).
That is,
{x} = [-1, 5, 6]
{y} = [0, 5, 10]
Let the curve of the best least squares fit be
y = ax² + b
The sum of the squares of errors between the data and the curve is
[tex]E = \sum_{i=1}^{3} (ax_{i}^{2}+b-y_{i})^{2}[/tex]
To minimize the error requires that
[tex] \frac{\partial E}{\partial a} =0\\ 2\sum x_{i}^{2} (ax_{i}^{} + b - y_{i} ) = 0[/tex]
Also,
[tex] \frac{\partial E}{\partial b} =0 \\ 2 \sum (ax_{i}^{2} + b - y_{i} ) = 0[/tex]
Therefore the normal equations are
(x₁⁴+x₂⁴+x₃⁴)a + (x₁²+x₂²+x₃²)b = x₁²y₁+x₂²y₂+x₃²y₃ (1)
(x₁²+x₂²+x₃²)a + 3b = y₁+y₂+y₃ (2)
From the given data, obtain
1922a + 62b = 485
62a + 3b = 15
The solution of these equations yields
a = 0.2732
b = -0.6452
The required curve is
y = 0.2732x² - 0.6452
A graph of the curve and the given data is shown below.
Answer: y = 0.2732x² - 0.6452
