Questions on Algebra: Graphs, graphing equations and inequalities answered by real tutors!

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Looking at y=(1/3)x

we can see that the equation is in slope-intercept form y=mx+b

where the slope is m=1/3

and the y-intercept is b=0

note:

really looks like y=(1/3)x+0

Since

this tells us that the y-intercept is

.Remember the y-intercept is the point where the graph intersects with the y-axis

So we have one point

drawing(500,500,-10,10,-10,10,<BR>
grid(1),<BR>
blue(circle(0,0,.1)),<BR>
blue(circle(0,0,.12)),<BR>
blue(circle(0,0,.15))<BR>
)

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

Also, because the slope is 1/3

, this means:

rise/run=1/3

which shows us that the rise is 1 and the run is 3. This means that to go from point to point, we can go up 1 and over 3

So starting at

, go up 1 unit
drawing(500,500,-10,10,-10,10, 
grid(1), 
blue(circle(0,0,.1)), 
blue(circle(0,0,.12)), 
blue(circle(0,0,.15)), 
blue(arc(0,0+(1/2),2,1,90,270)) 
)

and to the right 3 units to get to the next point

drawing(500,500,-10,10,-10,10,<BR>
grid(1),<BR>
blue(circle(0,0,.1)),<BR>
blue(circle(0,0,.12)),<BR>
blue(circle(0,0,.15)),<BR>
blue(circle(3,1,.15,1.5)),<BR>
blue(circle(3,1,.1,1.5)),<BR>
blue(arc(0,0+(1/2),2,1,90,270)),<BR>
blue(arc((3/2),1,3,2, 180,360))<BR>
)

Now draw a line through these points to graph y=(1/3)x

drawing(500,500,-10,10,-10,10,<BR>
grid(1),<BR>
graph(500,500,-10,10,-10,10,(1/3)x),<BR>
blue(circle(0,0,.1)),<BR>
blue(circle(0,0,.12)),<BR>
blue(circle(0,0,.15)),<BR>
blue(circle(3,1,.15,1.5)),<BR>
blue(circle(3,1,.1,1.5)),<BR>
blue(arc(0,0+(1/2),2,1,90,270)),<BR>
blue(arc((3/2),1,3,2, 180,360))<BR>
)

So this is the graph of y=(1/3)x

through the points

and