Question 923486
Looking at {{{y=(1/2)x+10}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1/2}}} and the y-intercept is {{{b=10}}} 



Since {{{b=10}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,10\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,10\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{1/2}}}, this means:


{{{rise/run=1/2}}}



which shows us that the rise is 1 and the run is 2. This means that to go from point to point, we can go up 1  and over 2




So starting at *[Tex \LARGE \left(0,10\right)], go up 1 unit 

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


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,11\right)]

{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,10,.1)),
  blue(circle(0,10,.12)),
  blue(circle(0,10,.15)),
  blue(circle(2,11,.15,1.5)),
  blue(circle(2,11,.1,1.5)),
  blue(arc(0,10+(1/2),2,1,90,270)),
  blue(arc((2/2),11,2,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/2)x+10}}}


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  graph(500,500,-10,10,-5,15,0,(1/2)x+10),
  blue(circle(0,10,.1)),
  blue(circle(0,10,.12)),
  blue(circle(0,10,.15)),
  blue(circle(2,11,.15,1.5)),
  blue(circle(2,11,.1,1.5)),
  blue(arc(0,10+(1/2),2,1,90,270)),
  blue(arc((2/2),11,2,2, 180,360))
)}}} So this is the graph of {{{y=(1/2)x+10}}} through the points *[Tex \LARGE \left(0,10\right)] and *[Tex \LARGE \left(2,11\right)]


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Let me know if that helps or not. Thanks.


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