Question 146779


Looking at {{{y=(9/4)x+6}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=9/4}}} and the y-intercept is {{{b=6}}} 



Since {{{b=6}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,6\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,6\right)]


{{{drawing(500,500,-10,10,-5,17,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{9/4}}}, this means:


{{{rise/run=9/4}}}



which shows us that the rise is 9 and the run is 4. This means that to go from point to point, we can go up 9  and over 4




So starting at *[Tex \LARGE \left(0,6\right)], go up 9 units 

{{{drawing(500,500,-10,10,-5,17,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(arc(0,6+(9/2),2,9,90,270))
)}}}


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,15\right)]

{{{drawing(500,500,-10,10,-5,17,
  grid(1),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(circle(4,15,.15,1.5)),
  blue(circle(4,15,.1,1.5)),
  blue(arc(0,6+(9/2),2,9,90,270)),
  blue(arc((4/2),15,4,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(9/4)x+6}}}


{{{drawing(500,500,-10,10,-5,17,
  grid(1),
  graph(500,500,-10,10,-5,17,(9/4)x+6),
  blue(circle(0,6,.1)),
  blue(circle(0,6,.12)),
  blue(circle(0,6,.15)),
  blue(circle(4,15,.15,1.5)),
  blue(circle(4,15,.1,1.5)),
  blue(arc(0,6+(9/2),2,9,90,270)),
  blue(arc((4/2),15,4,2, 180,360))
)}}} So this is the graph of {{{y=(9/4)x+6}}} through the points *[Tex \LARGE \left(0,6\right)] and *[Tex \LARGE \left(4,15\right)]