Question 193200


Looking at {{{y=(1/5)x}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1/5}}} and the y-intercept is {{{b=0}}}  note: {{{y=(1/5)x}}} really looks like {{{y=(1/5)x+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 {{{1/5}}}, this means:


{{{rise/run=1/5}}}



which shows us that the rise is 1 and the run is 5. This means that to go from point to point, we can go up 1  and over 5




So starting at *[Tex \LARGE \left(0,0\right)], go up 1 unit 

{{{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+(1/2),2,1,90,270))
)}}}


and to the right 5 units to get to the next point *[Tex \LARGE \left(5,1\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(5,1,.15,1.5)),
  blue(circle(5,1,.1,1.5)),
  blue(arc(0,0+(1/2),2,1,90,270)),
  blue(arc((5/2),1,5,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/5)x}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(1/5)x),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15)),
  blue(circle(5,1,.15,1.5)),
  blue(circle(5,1,.1,1.5)),
  blue(arc(0,0+(1/2),2,1,90,270)),
  blue(arc((5/2),1,5,2, 180,360))
)}}} So this is the graph of {{{y=(1/5)x}}} through the points *[Tex \LARGE \left(0,0\right)] and *[Tex \LARGE \left(5,1\right)]