Question 923946
This is the equation y = -5x + 8


Looking at {{{y=-5x+8}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-5}}} and the y-intercept is {{{b=8}}} 



Since {{{b=8}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,8\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,8\right)]


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,8,.1)),
  blue(circle(0,8,.12)),
  blue(circle(0,8,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=-5/1}}}



which shows us that the rise is -5 and the run is 1. This means that to go from point to point, we can go down 5  and over 1




So starting at *[Tex \LARGE \left(0,8\right)], go down 5 units 

{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,8,.1)),
  blue(circle(0,8,.12)),
  blue(circle(0,8,.15)),
  blue(arc(0,8+(-5/2),2,-5,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,3\right)]

{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,8,.1)),
  blue(circle(0,8,.12)),
  blue(circle(0,8,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,8+(-5/2),2,-5,90,270)),
  blue(arc((1/2),3,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-5x+8}}}


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  graph(500,500,-10,10,-5,15,0,-5x+8),
  blue(circle(0,8,.1)),
  blue(circle(0,8,.12)),
  blue(circle(0,8,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5))
)}}} So this is the graph of {{{y=-5x+8}}} through the points *[Tex \LARGE \left(0,8\right)] and *[Tex \LARGE \left(1,3\right)]


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Let me know if that helps or not. Thanks.


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