Here is a plot of the data points $(0, 8.5), \ \ (1,10), \ \ (2,10), \ \ (3,11):$
The observed values of $y$ are $8.5,\ 10,\ 10,$ and $11$.

If we approximate the data by the linear equation $\hat{y} = %0x + %1$, then the predicted values are
$\hat{y}_1 = %0(0) + %1 = %1$
$\hat{y}_2 = $BOX
$\hat{y}_3 = $BOX
$\hat{y}_4 = $BOX

Residuals and sum of squares error

Using the above data and linear equation, we get the following residuals:
Residual at $x = x_1: \ \ y_1-\hat{y}_1 = 8.5 - %1 = %3$
Residual at $x = x_2:$ BOX
Residual at $x = x_3:$ BOX
Residual at $x = x_4:$ BOX
Squaring and adding the residuals above gives
SSE = BOX