Question 132892


Looking at {{{y=(9/5)x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=9/5}}} and the y-intercept is {{{b=3}}} 



Since {{{b=3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,3\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,3\right)]


{{{drawing(500,500,-13,13,-13,13,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{9/5}}}, this means:


{{{rise/run=9/5}}}



which shows us that the rise is 9 and the run is 5. This means that to go from point to point, we can go up 9  and over 5




So starting at *[Tex \LARGE \left(0,3\right)], go up 9 units 

{{{drawing(500,500,-13,13,-13,13,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(arc(0,3+(9/2),2,9,90,270))
)}}}


and to the right 5 units to get to the next point *[Tex \LARGE \left(5,12\right)]

{{{drawing(500,500,-13,13,-13,13,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(5,12,.15,1.5)),
  blue(circle(5,12,.1,1.5)),
  blue(arc(0,3+(9/2),2,9,90,270)),
  blue(arc((5/2),12,5,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(9/5)x+3}}}


{{{drawing(500,500,-13,13,-13,13,
  grid(1),
  graph(500,500,-13,13,-13,13,(9/5)x+3),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(5,12,.15,1.5)),
  blue(circle(5,12,.1,1.5)),
  blue(arc(0,3+(9/2),2,9,90,270)),
  blue(arc((5/2),12,5,2, 180,360))
)}}} So this is the graph of {{{y=(9/5)x+3}}} through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(5,12\right)]