Bivariate Case (OLS)

Given
$$x = x_1,x_2,\cdots, x_n   \,\,\ \text{and}  \,\,\,  y = y_1,y_2,\cdots, y_n$$

We define the optimal minimum or the optimized mininum as:
$$\text{Opt Min F} = \sum{[y_\text{actual}-f(x)]^2}$$
where
$$\text{line  } y = f(x)= a +bx$$
Substituting the line $f(x)$ to the optimized minimum $F$, we have
$$F = \sum{[y-a -bx)]^2}$$
Taking the first derivative with respect to $a$ and $b$,
$$\begin{split} \frac{\partial F}{\partial a} &= 2 \sum{(y-a-bx)(-1)} = 0\\ \frac{\partial F}{\partial b} &= 2 \sum{(y-a-bx)(-x)} = 0 \end{split}$$
We can now derive the normal equation for the OLS as:

$$\begin{split} \sum{y} &= an + b \sum{x}\\ \sum{xy} &= a \sum{x} + b \sum{x^2} \end{split}$$

Solve for $a$ and $b$ to get critical points and find out if they really produced the optimum.