The score boundary can be solved directly in input space. With w=1 and b=-2, setting z=0 gives the exact boundary x=2.

highlighted = computed this step

Start with the score rule

The displayed rule is z = w*x + b. Here w=1 and b=-2, so the score rule is z = 1*x - 2.

z=1x2z=1\cdot x - 2
Solve the boundaryExact score-zero boundary in input space.Classifier score tablexzdecision0-201-10201311Decision boundaryThe exact point where the score is zero.ABCDz=0 at x=2

Set the boundary score to zero

The decision boundary is the input where the score z equals 0. Put z=0 into the exact score rule.

0=1x20=1\cdot x - 2
Solve the boundaryExact score-zero boundary in input space.Classifier score tablexzdecision0-201-10201311Decision boundaryThe exact point where the score is zero.ABCDz=0 at x=2

Solve for x

Adding 2 to both sides gives x=2. That is why the plot marks the boundary at x=2.

x=2x=2
Solve the boundaryExact score-zero boundary in input space.Classifier score tablexzdecision0-201-10201311Decision boundaryThe exact point where the score is zero.ABCDz=0 at x=2

Check the displayed rows

The table recomputes scores -2, -1, 0, and 1. At x=2, z=0. This is exact boundary bookkeeping for the displayed score rule only, not a probability or a claim about later rows.

x=2z=0x=2\Rightarrow z=0
Solve the boundaryExact score-zero boundary in input space.Classifier score tablexzdecision0-201-10201311Decision boundaryThe exact point where the score is zero.ABCDz=0 at x=2