Question 334566


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



Since {{{b=9}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,9\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,9\right)]


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,9,.1)),
  blue(circle(0,9,.12)),
  blue(circle(0,9,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=5/1}}}



which shows us that the rise is 5 and the run is 1. This means that to go from point to point, we can go up 5  and over 1




So starting at *[Tex \LARGE \left(0,9\right)], go up 5 units 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,14\right)]

{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,9,.1)),
  blue(circle(0,9,.12)),
  blue(circle(0,9,.15)),
  blue(circle(1,14,.15,1.5)),
  blue(circle(1,14,.1,1.5)),
  blue(arc(0,9+(5/2),2,5,90,270)),
  blue(arc((1/2),14,1,2, 180,360))
)}}}



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


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  graph(500,500,-10,10,-5,15,5x+9),
  blue(circle(0,9,.1)),
  blue(circle(0,9,.12)),
  blue(circle(0,9,.15)),
  blue(circle(1,14,.15,1.5)),
  blue(circle(1,14,.1,1.5)),
  blue(arc(0,9+(5/2),2,5,90,270)),
  blue(arc((1/2),14,1,2, 180,360))
)}}} So this is the graph of {{{y=5x+9}}} through the points *[Tex \LARGE \left(0,9\right)] and *[Tex \LARGE \left(1,14\right)]