Question 916079
I'll show you how to graph something like {{{y = (3/2)x-1}}}


Looking at {{{y=(3/2)x-1}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3/2}}} and the y-intercept is {{{b=-1}}} 



Since {{{b=-1}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-1\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,-1\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{3/2}}}, this means:


{{{rise/run=3/2}}}



which shows us that the rise is 3 and the run is 2. This means that to go from point to point, we can go up 3  and over 2




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

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


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

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-1,.1)),
  blue(circle(0,-1,.12)),
  blue(circle(0,-1,.15)),
  blue(circle(2,2,.15,1.5)),
  blue(circle(2,2,.1,1.5)),
  blue(arc(0,-1+(3/2),2,3,90,270)),
  blue(arc((2/2),2,2,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(3/2)x-1}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,0,(3/2)x-1),
  blue(circle(0,-1,.1)),
  blue(circle(0,-1,.12)),
  blue(circle(0,-1,.15)),
  blue(circle(2,2,.15,1.5)),
  blue(circle(2,2,.1,1.5)),
  blue(arc(0,-1+(3/2),2,3,90,270)),
  blue(arc((2/2),2,2,2, 180,360))
)}}} So this is the graph of {{{y=(3/2)x-1}}} through the points *[Tex \LARGE \left(0,-1\right)] and *[Tex \LARGE \left(2,2\right)]


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>
or you can visit my website here: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a>


Thanks,


Jim