Question 192961
Do you want to graph this? If so, then...

Looking at {{{y=(1/4)x}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1/4}}} and the y-intercept is {{{b=0}}}  note: {{{y=(1/4)x}}} really looks like {{{y=(1/4)x+0}}}

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

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

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

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

{{{slope=rise/run}}}

Also, because the slope is {{{1/4}}}, this means:

{{{rise/run=1/4}}}

which shows us that the rise is 1 and the run is 4. This means that to go from point to point, we can go up 1  and over 4

So starting at *[Tex \LARGE \left(0,0\right)], 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 4 units to get to the next point *[Tex \LARGE \left(4,1\right)]

{{{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(circle(4,1,.15,1.5)),
blue(circle(4,1,.1,1.5)),
blue(arc(0,0+(1/2),2,1,90,270)),
blue(arc((4/2),1,4,2, 180,360))
)}}}

Now draw a line through these points to graph {{{y=(1/4)x}}}

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