Question 1066829
<pre>
{{{y=2x+1}}} which is in the slope intercept form

{{{y=mx+b}}}

Write it as

{{{y=expr(2/1)x+green(1)}}}

Start at <font color="green">1</font> on the y-axis:

{{{drawing(300,300,-3,3,-2,4, grid(1),green(circle(0,1,.09),circle(0,1,.08),circle(0,1,.07),circle(0,1,.06),circle(0,1,.05),circle(0,1,.04),circle(0,1,.03),circle(0,1,.02),circle(0,1,.01),circle(0,1,.005)) )}}}

The slope is the {{{2/1}}}.  From that green point draw a line
that goes up the numerator of the slope which is 2 units
('up' since it's positive), and then draw a line to the right 1 
unit, since the the denominator of the slope is 1, like the vertical
and horizontal green lines below:

{{{drawing(300,300,-3,3,-2,4, grid(1),green(line(0,1,0,3),line(0,3,1,3),circle(0,1,.09),circle(0,1,.08),circle(0,1,.07),circle(0,1,.06),circle(0,1,.05),circle(0,1,.04),circle(0,1,.03),circle(0,1,.02),circle(0,1,.01),circle(0,1,.005)) )}}}

Then get a ruler and draw a line from the point where you began 
to where you ended up:

{{{drawing(300,300,-3,3,-2,4, line(-7,-13,7,15),grid(1),green(line(0,1,0,3),line(0,3,1,3),circle(0,1,.09),circle(0,1,.08),circle(0,1,.07),circle(0,1,.06),circle(0,1,.05),circle(0,1,.04),circle(0,1,.03),circle(0,1,.02),circle(0,1,.01),circle(0,1,.005)) )}}}

Edwin</pre>