Question 773856
Looking at {{{y=7x}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=7}}} and the y-intercept is {{{b=0}}}  note: {{{y=7x}}} really looks like {{{y=7x+0}}} 



Since {{{b=0}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,0\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,0\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15))
)}}}



Now since the slope is comprised of the "rise" over the "run" this means

{{{slope=rise/run}}}


Also, because the slope is {{{7}}}, this means:


{{{rise/run=7/1}}}



which shows us that the rise is 7 and the run is 1. This means that to go from point to point, we can go up 7  and over 1




So starting at *[Tex \LARGE \left(0,0\right)], go up 7 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15)),
  blue(arc(0,0+(7/2),2,7,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,7\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15)),
  blue(circle(1,7,.15,1.5)),
  blue(circle(1,7,.1,1.5)),
  blue(arc(0,0+(7/2),2,7,90,270)),
  blue(arc((1/2),7,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=7x}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,0,7x),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15)),
  blue(circle(1,7,.15,1.5)),
  blue(circle(1,7,.1,1.5)),
  blue(arc(0,0+(7/2),2,7,90,270)),
  blue(arc((1/2),7,1,2, 180,360))
)}}} So this is the graph of {{{y=7x}}} through the points *[Tex \LARGE \left(0,0\right)] and *[Tex \LARGE \left(1,7\right)]