Question 537948
Looking at {{{y=-(2/3)x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-2/3}}} 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,-10,10,-10,10,
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 {{{-2/3}}}, this means:

{{{rise/run=-2/3}}}

which shows us that the rise is -2 and the run is 3. This means that to go from point to point, we can go down 2  and over 3

So starting at *[Tex \LARGE \left(0,3\right)], go down 2 units

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

and to the right 3 units to get to the next point *[Tex \LARGE \left(3,1\right)]

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

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

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

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Jim