Question 373051
Looking at {{{y=-(1/4)x-3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1/4}}} 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 {{{-1/4}}}, this means:


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



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




So starting at *[Tex \LARGE \left(0,-3\right)], go down 1 unit 

{{{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+(-1/2),2,-1,90,270))
)}}}


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,-4\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(4,-4,.15,1.5)),
  blue(circle(4,-4,.1,1.5)),
  blue(arc(0,-3+(-1/2),2,-1,90,270)),
  blue(arc((4/2),-4,4,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(1/4)x-3}}}


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



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim