Question 122814
# 1




Looking at {{{y=x+1}}} 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=1}}} 



Since {{{b=1}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,1\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,1\right)]


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




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

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,2\right)]

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



Now draw a line through these points to graph {{{y=x+1}}}


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





<hr>




# 2


In order to graph {{{y=2x^2}}}, we need to plot some points.


We can start at any x value, so lets start at x=-2





Lets evaluate {{{f(-2)}}}


{{{f(x)=2x^2}}} Start with the given polynomial



{{{f(-2)=2(-2)^2}}} Plug in {{{x=-2}}}



{{{f(-2)=2(4)}}} Evaluate {{{(-2)^2}}} to get 4

 

{{{f(-2)=+8}}} Multiply 2 and  4  to get 8





So when {{{x=-2}}}, {{{f(x)=8}}} which means our 1st point is (-2,8)




-------Now lets find another point-------




Lets evaluate {{{f(-1)}}}


{{{f(x)=2x^2}}} Start with the given polynomial



{{{f(-1)=2(-1)^2}}} Plug in {{{x=-1}}}



{{{f(-1)=2(1)}}} Evaluate {{{2(-1)^2}}} to get 21

 

{{{f(-1)=+2}}} Multiply 2 and  1  to get 2





So when {{{x=-1}}}, {{{f(x)=2}}}  which means our 2nd point is (-1,2)




-------Now lets find another point-------




Lets evaluate {{{f(0)}}}


{{{f(x)=2x^2}}} Start with the given polynomial



{{{f(0)=2(0)^2}}} Plug in {{{x=0}}}



{{{f(0)=2(0)}}} Evaluate {{{2(0)^2}}} to get 20

 

{{{f(0)=0}}} Multiply 2 and  0  to get 0

 


So when {{{x=0}}}, {{{f(x)=0}}}  which means our 3rd point is (0,0)




-------Now lets find another point-------




Lets evaluate {{{f(1)}}}


{{{f(x)=2x^2}}} Start with the given polynomial



{{{f(1)=2(1)^2}}} Plug in {{{x=1}}}



{{{f(1)=2(1)}}} Evaluate {{{2(1)^2}}} to get 21

 

{{{f(1)=+2}}} Multiply 2 and  1  to get 2

   




So when {{{x=1}}}, {{{f(x)=2}}}  which means our 4th point is (1,2)




-------Now lets find another point-------




Lets evaluate {{{f(2)}}}


{{{f(x)=2x^2}}} Start with the given polynomial



{{{f(2)=2(2)^2}}} Plug in {{{x=2}}}



{{{f(2)=2(4)}}} Evaluate {{{2(2)^2}}} to get 24

 

{{{f(2)=+8}}} Multiply 2 and  4  to get 8

   

{{{f(2)=+8}}} Multiply 0 and    to get 0

  

{{{f(2)=8}}} Now combine like terms



So when {{{x=2}}}, {{{f(x)=8}}}  which means our 5th point is (2,8)







Now lets make a table of the values we have calculated

<pre>
<TABLE width=500>

<TR><TD> x</TD><TD>y</TD></TR>

<TR><TD> -2</TD><TD>8</TD></TR> 
<TR><TD> -1</TD><TD>2</TD></TR> 
<TR><TD> 0</TD><TD>0</TD></TR> 
<TR><TD> 1</TD><TD>2</TD></TR> 
<TR><TD> 2</TD><TD>8</TD></TR> 
</TABLE>
</pre>Now plot the points

{{{drawing(900,900,-10,10,-10,10,
  grid( 1 ),
circle(-2,8,0.05),
circle(-2,8,0.08),
circle(-1,2,0.05),
circle(-1,2,0.08),
circle(0,0,0.05),
circle(0,0,0.08),
circle(1,2,0.05),
circle(1,2,0.08),
circle(2,8,0.05),
circle(2,8,0.08)
)}}}



Now connect the points to graph {{{y=2x^2}}}  (note: the more points you plot, the easier it is to draw the graph)

{{{drawing(900,900,-10,10,-10,10,
grid( 1 ),
graph(900,900,-10,10,-10,10, 2x^2),
circle(-2,8,0.05),
circle(-2,8,0.08),
circle(-1,2,0.05),
circle(-1,2,0.08),
circle(0,0,0.05),
circle(0,0,0.08),
circle(1,2,0.05),
circle(1,2,0.08),
circle(2,8,0.05),
circle(2,8,0.08)
)}}}