Question 121255
#1


Looking at {{{y=3x+1}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3}}} and the y-intercept is {{{b=1}}} 



Since {{{b=1}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,1\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,1\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{3}}}, this means:


{{{rise/run=3/1}}}



which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3  and over 1




So starting at *[Tex \LARGE \left(0,1\right)], go up 3 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(arc(0,1+(3/2),2,3,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,4\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,1+(3/2),2,3,90,270)),
  blue(arc((1/2),4,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=3x+1}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,3x+1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(1,4,.15,1.5)),
  blue(circle(1,4,.1,1.5)),
  blue(arc(0,1+(3/2),2,3,90,270)),
  blue(arc((1/2),4,1,2, 180,360))
)}}} So this is the graph of {{{y=3x+1}}} through the points *[Tex \LARGE \left(0,1\right)] and *[Tex \LARGE \left(1,4\right)]




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#2




Looking at {{{y=(1/3)x+1}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1/3}}} and the y-intercept is {{{b=1}}} 



Since {{{b=1}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,1\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,1\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{1/3}}}, this means:


{{{rise/run=1/3}}}



which shows us that the rise is 1 and the run is 3. This means that to go from point to point, we can go up 1  and over 3




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

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


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,2\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(3,2,.15,1.5)),
  blue(circle(3,2,.1,1.5)),
  blue(arc(0,1+(1/2),2,1,90,270)),
  blue(arc((3/2),2,3,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/3)x+1}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(1/3)x+1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(3,2,.15,1.5)),
  blue(circle(3,2,.1,1.5)),
  blue(arc(0,1+(1/2),2,1,90,270)),
  blue(arc((3/2),2,3,2, 180,360))
)}}} So this is the graph of {{{y=(1/3)x+1}}} through the points *[Tex \LARGE \left(0,1\right)] and *[Tex \LARGE \left(3,2\right)]