Question 158031
First let's find the equation of the line 


If you want to find the equation of line with a given a slope of {{{-3}}} which goes through the point (2,-1), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---


{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point



So lets use the Point-Slope Formula to find the equation of the line



{{{y--1=(-3)(x-2)}}} Plug in {{{m=-3}}}, {{{x[1]=2}}}, and {{{y[1]=-1}}} (these values are given)



{{{y+1=(-3)(x-2)}}} Rewrite {{{y--1}}} as {{{y+1}}}



{{{y+1=-3x+(-3)(-2)}}} Distribute {{{-3}}}



{{{y+1=-3x+6}}} Multiply {{{-3}}} and {{{-2}}} to get {{{6}}}


{{{y=-3x+6-1}}} Subtract 1 from  both sides to isolate y



{{{y=-3x+5}}} Combine like terms {{{6}}} and {{{-1}}} to get {{{5}}} 



------------------------------------------------------------------------------------------------------------

Answer:



So the equation of the line with a slope of {{{-3}}} which goes through the point (2,-1) is:



{{{y=-3x+5}}} which is now in {{{y=mx+b}}} form where the slope is {{{m=-3}}} and the y-intercept is {{{b=5}}}




Since {{{b=5}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,5\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,5\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.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 down 3  and over 1




So starting at *[Tex \LARGE \left(0,5\right)], go down 3 units 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,2\right)]

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



Now draw a line through these points to graph {{{y=-3x+5}}}


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



Also, notice how the graph has a slope of -3 (since {{{m=-3}}}) and it goes through the point (2,-1).