Question 662369
{{{drawing(300,300,-5,5,-5,5,
grid(0),
blue(line(-5,1,2,1)),blue(line(2.1,-2,5,-2)),
red(circle(2,-2,0.1)),
blue(circle(2,1,0.1)), blue(circle(2,1,0.05)),
green(line(2,-5,2,-4.25)),green(line(2,-5,2,-4.25)),
green(line(2,-5,2,-4.25)),green(line(2,-3.75,2,-3.25)),
green(line(2,-2.75,2,-2.25)),green(line(2,-1.75,2,-1.25)),
green(line(2,-0.75,2,-0.25)),green(line(2,0.25,2,0.75)),
green(line(2,1.25,2,1.75)),green(line(2,2.255,2,2.75)),
green(line(2,3.25,2,3.75)),green(line(2,4.255,2,5))
)}}} The solid blue (2,1) point, and the blue line to its left
represent the function for {{{x=2}}} ({{{f(2)=1}}}),
and for {{{x<2}}} ({{{y=f(x)=1}}}.
The open red circle at (2,-2) shows that
point (2,-2) is not part of the function.
All the points with {{{y=-2}}} to the right of (2,-2)
are part of the function, with {{{x>2}}}, and {{{y=f(x)=-2}}}.
I drew the lower blue line from (2.1,-2) to {5,-2)
to make it clear that the line does not reach (2,-1),
but of course there are infinity of points with {{{y=-2}}}
and {{{x}}} just a little larger than {{{2}}} that belong to the function.