Question 131152


Looking at {{{y=(9/5)x+2}}} 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=2}}} 



Since {{{b=2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,2\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,2\right)]


{{{drawing(500,500,-12,12,-12,12,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.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(0,2\right)], go up 9 units 

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


and to the right 5 units to get to the next point *[Tex \LARGE \left(5,11\right)]

{{{drawing(500,500,-12,12,-12,12,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(5,11,.15,1.5)),
  blue(circle(5,11,.1,1.5)),
  blue(arc(0,2+(9/2),2,9,90,270)),
  blue(arc((5/2),11,5,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(9/5)x+2}}}


{{{drawing(500,500,-12,12,-12,12,
  grid(1),
  graph(500,500,-12,12,-12,12,(9/5)x+2),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(5,11,.15,1.5)),
  blue(circle(5,11,.1,1.5)),
  blue(arc(0,2+(9/2),2,9,90,270)),
  blue(arc((5/2),11,5,2, 180,360))
)}}} So this is the graph of {{{y=(9/5)x+2}}} through the points *[Tex \LARGE \left(0,2\right)] and *[Tex \LARGE \left(5,11\right)]