Question 342076
Looking at {{{y=-(5/2)x}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-5/2}}} and the y-intercept is {{{b=0}}}  note: {{{y=-(5/2)x}}} really looks like {{{y=-(5/2)x+0}}} 



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


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


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{-5/2}}}, this means:


{{{rise/run=-5/2}}}



which shows us that the rise is -5 and the run is 2. This means that to go from point to point, we can go down 5  and over 2




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

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


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,-5\right)]

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



Now draw a line through these points to graph {{{y=-(5/2)x}}}


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