Question 915724
Looking at {{{y=8x}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=8}}} and the y-intercept is {{{b=0}}}  note: {{{y=8x}}} really looks like {{{y=8x+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 {{{8}}}, this means:


{{{rise/run=8/1}}}



which shows us that the rise is 8 and the run is 1. This means that to go from point to point, we can go up 8  and over 1




So starting at *[Tex \LARGE \left(0,0\right)], go up 8 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+(8/2),2,8,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,8\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(1,8,.15,1.5)),
  blue(circle(1,8,.1,1.5)),
  blue(arc(0,0+(8/2),2,8,90,270)),
  blue(arc((1/2),8,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=8x}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,0,8x),
  blue(circle(0,0,.1)),
  blue(circle(0,0,.12)),
  blue(circle(0,0,.15)),
  blue(circle(1,8,.15,1.5)),
  blue(circle(1,8,.1,1.5)),
  blue(arc(0,0+(8/2),2,8,90,270)),
  blue(arc((1/2),8,1,2, 180,360))
)}}} So this is the graph of {{{y=8x}}} through the points *[Tex \LARGE \left(0,0\right)] and *[Tex \LARGE \left(1,8\right)]



Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
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Thanks,


Jim