Question 319205


Looking at {{{y=(2/3)x-12}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=2/3}}} and the y-intercept is {{{b=-12}}} 



Since {{{b=-12}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-12\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,-12\right)]


{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,-12,.1)),
  blue(circle(0,-12,.12)),
  blue(circle(0,-12,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{2/3}}}, this means:


{{{rise/run=2/3}}}



which shows us that the rise is 2 and the run is 3. This means that to go from point to point, we can go up 2  and over 3




So starting at *[Tex \LARGE \left(0,-12\right)], go up 2 units 

{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,-12,.1)),
  blue(circle(0,-12,.12)),
  blue(circle(0,-12,.15)),
  blue(arc(0,-12+(2/2),2,2,90,270))
)}}}


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,-10\right)]

{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,-12,.1)),
  blue(circle(0,-12,.12)),
  blue(circle(0,-12,.15)),
  blue(circle(3,-10,.15,1.5)),
  blue(circle(3,-10,.1,1.5)),
  blue(arc(0,-12+(2/2),2,2,90,270)),
  blue(arc((3/2),-10,3,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(2/3)x-12}}}


{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  graph(500,500,-10,10,-15,5,(2/3)x-12),
  blue(circle(0,-12,.1)),
  blue(circle(0,-12,.12)),
  blue(circle(0,-12,.15)),
  blue(circle(3,-10,.15,1.5)),
  blue(circle(3,-10,.1,1.5)),
  blue(arc(0,-12+(2/2),2,2,90,270)),
  blue(arc((3/2),-10,3,2, 180,360))
)}}} So this is the graph of {{{y=(2/3)x-12}}} through the points *[Tex \LARGE \left(0,-12\right)] and *[Tex \LARGE \left(3,-10\right)]