Question 627133
Looking at {{{y=-4x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-4}}} and the y-intercept is {{{b=3}}} 



Since {{{b=3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,3\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,3\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=-4/1}}}



which shows us that the rise is -4 and the run is 1. This means that to go from point to point, we can go down 4  and over 1




So starting at *[Tex \LARGE \left(0,3\right)], go down 4 units 

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


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

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



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


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


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