Question 725574


{{{2x+4y=6}}} Start with the given equation.



{{{4y=6-2x}}} Subtract {{{2x}}} from both sides.



{{{4y=-2x+6}}} Rearrange the terms.



{{{y=(-2x+6)/(4)}}} Divide both sides by {{{4}}} to isolate y.



{{{y=((-2)/(4))x+(6)/(4)}}} Break up the fraction.



{{{y=-(1/2)x+3/2}}} Reduce.


Looking at {{{y=-(1/2)x+3/2}}} 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=3/2}}} 



Since {{{b=3/2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,\frac{3}{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,\frac{3}{2}\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3/2,.1)),
  blue(circle(0,3/2,.12)),
  blue(circle(0,3/2,.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 down 1  and over 2




So starting at *[Tex \LARGE \left(0,\frac{3}{2}\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3/2,.1)),
  blue(circle(0,3/2,.12)),
  blue(circle(0,3/2,.15)),
  blue(arc(0,3/2+(-1/2),2,-1,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,\frac{1}{2}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3/2,.1)),
  blue(circle(0,3/2,.12)),
  blue(circle(0,3/2,.15)),
  blue(circle(2,1/2,.15,1.5)),
  blue(circle(2,1/2,.1,1.5)),
  blue(arc(0,3/2+(-1/2),2,-1,90,270)),
  blue(arc((2/2),1/2,2,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(1/2)x+3/2}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,0,-(1/2)x+3/2),
  blue(circle(0,3/2,.1)),
  blue(circle(0,3/2,.12)),
  blue(circle(0,3/2,.15)),
  blue(circle(2,1/2,.15,1.5)),
  blue(circle(2,1/2,.1,1.5)),
  blue(arc(0,3/2+(-1/2),2,-1,90,270)),
  blue(arc((2/2),1/2,2,2, 0,180))
)}}} So this is the graph of {{{y=-(1/2)x+3/2}}} through the points *[Tex \LARGE \left(0,\frac{3}{2}\right)] and *[Tex \LARGE \left(2,\frac{1}{2}\right)]