Question 166523

Looking at {{{y=4x-3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=4}}} 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 {{{4}}}, this means:

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

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

So starting at *[Tex \LARGE \left(0,-3\right)], go up 4 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+(4/2),2,4,90,270))
)}}}

and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,1\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(1,1,.15,1.5)),
blue(circle(1,1,.1,1.5)),
blue(arc(0,-3+(4/2),2,4,90,270)),
blue(arc((1/2),1,1,2, 180,360))
)}}}

Now draw a line through these points to graph {{{y=4x-3}}}

{{{drawing(500,500,-10,10,-10,10,
grid(1),
graph(500,500,-10,10,-10,10,4x-3),
blue(circle(0,-3,.1)),
blue(circle(0,-3,.12)),
blue(circle(0,-3,.15)),
blue(circle(1,1,.15,1.5)),
blue(circle(1,1,.1,1.5)),
blue(arc(0,-3+(4/2),2,4,90,270)),
blue(arc((1/2),1,1,2, 180,360))
)}}} So this is the graph of {{{y=4x-3}}} through the points *[Tex \LARGE \left(0,-3\right)] and *[Tex \LARGE \left(1,1\right)]