Question 356931
The truth is that you never know what you're going to need in the future (otherwise, you'd probably just stick to making money as a psychic). So it helps to learn everything you can. You don't have to know everything, but it helps to get that exposure.



I'll do the first two to get you started. Please post these one at a time next time.



# 1


Looking at {{{y=2x-8}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=2}}} and the y-intercept is {{{b=-8}}} 



Since {{{b=-8}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-8\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,-8\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-8,.1)),
  blue(circle(0,-8,.12)),
  blue(circle(0,-8,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{2}}}, this means:


{{{rise/run=2/1}}}



which shows us that the rise is 2 and the run is 1. This means that to go from point to point, we can go up 2  and over 1




So starting at *[Tex \LARGE \left(0,-8\right)], go up 2 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-8,.1)),
  blue(circle(0,-8,.12)),
  blue(circle(0,-8,.15)),
  blue(arc(0,-8+(2/2),2,2,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-6\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-8,.1)),
  blue(circle(0,-8,.12)),
  blue(circle(0,-8,.15)),
  blue(circle(1,-6,.15,1.5)),
  blue(circle(1,-6,.1,1.5)),
  blue(arc(0,-8+(2/2),2,2,90,270)),
  blue(arc((1/2),-6,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=2x-8}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,2x-8),
  blue(circle(0,-8,.1)),
  blue(circle(0,-8,.12)),
  blue(circle(0,-8,.15)),
  blue(circle(1,-6,.15,1.5)),
  blue(circle(1,-6,.1,1.5)),
  blue(arc(0,-8+(2/2),2,2,90,270)),
  blue(arc((1/2),-6,1,2, 180,360))
)}}} So this is the graph of {{{y=2x-8}}} through the points *[Tex \LARGE \left(0,-8\right)] and *[Tex \LARGE \left(1,-6\right)]

================================================


# 2


{{{5x - 5 = 3x - y}}} Start with the given equation.



{{{5x - 5 - 3x = - y}}} Subtract 3x from both sides.



{{{2x - 5 = - y}}} Combine like terms.



{{{-2x + 5 =  y}}} Multiply every term by -1 to isolate y.



{{{y=-2x+5}}} Rearrange the equation.



Looking at {{{y=-2x+5}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-2}}} and the y-intercept is {{{b=5}}} 



Since {{{b=5}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,5\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,5\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{-2}}}, this means:


{{{rise/run=-2/1}}}



which shows us that the rise is -2 and the run is 1. This means that to go from point to point, we can go down 2  and over 1




So starting at *[Tex \LARGE \left(0,5\right)], go down 2 units 

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


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

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,5+(-2/2),2,-2,90,270)),
  blue(arc((1/2),3,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-2x+5}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x+5),
  blue(circle(0,5,.1)),
  blue(circle(0,5,.12)),
  blue(circle(0,5,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,5+(-2/2),2,-2,90,270)),
  blue(arc((1/2),3,1,2, 0,180))
)}}} So this is the graph of {{{y=-2x+5}}} through the points *[Tex \LARGE \left(0,5\right)] and *[Tex \LARGE \left(1,3\right)]