Question 350962
First plot the point (-2,5)

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  circle(-2,5,.1),
  circle(-2,5,.12),
  circle(-2,5,.15)
)}}}


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

{{{slope=rise/run}}}


Also, because the slope is {{{-1}}}, this means:


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



So this means that you go down one unit and then go to the right one unit to plot the next point (-1,4)



{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  circle(-2,5,.1),
  circle(-2,5,.12),
  circle(-2,5,.15),
  circle(-1,4,.1),
  circle(-1,4,.12),
  circle(-1,4,.15),
  blue(arc(-2,5+(-1/2),2,-1,90,270)),
  blue(arc((-3/2),4,1,2, 0,180))
)}}}



Finally, draw a straight line through the points to graph the line



{{{drawing(500,500,-10,10,-10,10,
  graph(500,500,-10,10,-10,10,-x+3),
  grid(1),
  circle(-2,5,.1),
  circle(-2,5,.12),
  circle(-2,5,.15),
  circle(-1,4,.1),
  circle(-1,4,.12),
  circle(-1,4,.15),
  blue(arc(-2,5+(-1/2),2,-1,90,270)),
  blue(arc((-3/2),4,1,2, 0,180))
)}}}



So the graph of the line through (-2,5) with slope -1 is



{{{drawing(500,500,-10,10,-10,10,
  graph(500,500,-10,10,-10,10,-x+3),
  grid(1),
  circle(-2,5,.1),
  circle(-2,5,.12),
  circle(-2,5,.15),
  circle(-1,4,.1),
  circle(-1,4,.12),
  circle(-1,4,.15)
)}}}


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


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Jim