Question 283588
{{{4x-3y-12=0}}} Start with the given equation.



{{{4x-3y=12}}} Add 12 to both sides.



{{{-3y=12-4x}}} Subtract {{{4x}}} from both sides.



{{{-3y=-4x+12}}} Rearrange the terms.



{{{y=(-4x+12)/(-3)}}} Divide both sides by {{{-3}}} to isolate y.



{{{y=((-4)/(-3))x+(12)/(-3)}}} Break up the fraction.



{{{y=(4/3)x-4}}} Reduce.




Looking at {{{y=(4/3)x-4}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=4/3}}} and the y-intercept is {{{b=-4}}} 



Since {{{b=-4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-4\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,-4\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{4/3}}}, this means:


{{{rise/run=4/3}}}



which shows us that the rise is 4 and the run is 3. This means that to go from point to point, we can go up 4  and over 3




So starting at *[Tex \LARGE \left(0,-4\right)], go up 4 units 

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


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,0\right)]

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



Now draw a line through these points to graph {{{y=(4/3)x-4}}}


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