Question 150403
I'm assuming that you want to graph?


# 9


{{{2x - 4y - 14 = 0 }}} Start with the given equation.



{{{-4y-14=0-2x}}} Subtract {{{2x}}} from both sides.



{{{-4y=0-2x+14}}} Add {{{14}}} to both sides.



{{{-4y=-2x+14}}} Combine like terms on the right side.



{{{y=(-2x+14)/(-4)}}} Divide both sides by {{{-4}}} to isolate {{{y}}}.



{{{y=(1/2)x-7/2}}} Simplify.





Looking at {{{y=(1/2)x-7/2}}} 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=-7/2}}} 



Since {{{b=-7/2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-\frac{7}{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{7}{2}\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-7/2,.1)),
  blue(circle(0,-7/2,.12)),
  blue(circle(0,-7/2,.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 up 1  and over 2




So starting at *[Tex \LARGE \left(0,-\frac{7}{2}\right)], go up 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-7/2,.1)),
  blue(circle(0,-7/2,.12)),
  blue(circle(0,-7/2,.15)),
  blue(arc(0,-7/2+(1/2),2,1,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,-\frac{5}{2}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-7/2,.1)),
  blue(circle(0,-7/2,.12)),
  blue(circle(0,-7/2,.15)),
  blue(circle(2,-5/2,.15,1.5)),
  blue(circle(2,-5/2,.1,1.5)),
  blue(arc(0,-7/2+(1/2),2,1,90,270)),
  blue(arc((2/2),-5/2,2,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/2)x-7/2}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(1/2)x-7/2),
  blue(circle(0,-7/2,.1)),
  blue(circle(0,-7/2,.12)),
  blue(circle(0,-7/2,.15)),
  blue(circle(2,-5/2,.15,1.5)),
  blue(circle(2,-5/2,.1,1.5)),
  blue(arc(0,-7/2+(1/2),2,1,90,270)),
  blue(arc((2/2),-5/2,2,2, 180,360))
)}}} So this is the graph of {{{y=(1/2)x-7/2}}} through the points *[Tex \LARGE \left(0,-\frac{7}{2}\right)] and *[Tex \LARGE \left(2,-\frac{5}{2}\right)]



<hr>



# 15






Looking at {{{y=-x-15}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1}}} and the y-intercept is {{{b=-15}}} 



Since {{{b=-15}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-15\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,-15\right)]


{{{drawing(500,500,-10,10,-18,2,
  grid(1),
  blue(circle(0,-15,.1)),
  blue(circle(0,-15,.12)),
  blue(circle(0,-15,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=-1/1}}}



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




So starting at *[Tex \LARGE \left(0,-15\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-18,2,
  grid(1),
  blue(circle(0,-15,.1)),
  blue(circle(0,-15,.12)),
  blue(circle(0,-15,.15)),
  blue(arc(0,-15+(-1/2),2,-1,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-16\right)]

{{{drawing(500,500,-10,10,-18,2,
  grid(1),
  blue(circle(0,-15,.1)),
  blue(circle(0,-15,.12)),
  blue(circle(0,-15,.15)),
  blue(circle(1,-16,.15,1.5)),
  blue(circle(1,-16,.1,1.5)),
  blue(arc(0,-15+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-16,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-x-15}}}


{{{drawing(500,500,-10,10,-18,2,
  grid(1),
  graph(500,500,-10,10,-18,2,-x-15),
  blue(circle(0,-15,.1)),
  blue(circle(0,-15,.12)),
  blue(circle(0,-15,.15)),
  blue(circle(1,-16,.15,1.5)),
  blue(circle(1,-16,.1,1.5)),
  blue(arc(0,-15+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-16,1,2, 0,180))
)}}} So this is the graph of {{{y=-x-15}}} through the points *[Tex \LARGE \left(0,-15\right)] and *[Tex \LARGE \left(1,-16\right)]