Question 316367


In order to graph {{{y=(x-1)^3}}}, we need to plot a few points.



To get points in the form of (x,y), we need to find corresponding 'y' values to given 'x' values.



Let's find the y value when {{{x=-1}}} note: you can start at any x value. 



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(-1-1)^3}}} Plug in {{{x=-1}}}.



{{{y=(-2)^3}}} Subtract.



{{{y=-8}}} Cube {{{-2}}} to get {{{-8}}}.



So if {{{x=-1}}}, then {{{y=-8}}} which gives us the point (-1,-8).



----------------------------



Let's find the y value when {{{x=-0.5}}}  



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(-0.5-1)^3}}} Plug in {{{x=-0.5}}}.



{{{y=(-1.5)^3}}} Subtract.



{{{y=-3.375}}} Cube {{{-1.5}}} to get {{{-3.375}}}.



So if {{{x=-0.5}}}, then {{{y=-3.375}}} which gives us the point (-0.5,-3.375).



----------------------------



Let's find the y value when {{{x=0.5}}}  



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(0.5-1)^3}}} Plug in {{{x=0.5}}}.



{{{y=(-0.5)^3}}} Subtract.



{{{y=-0.125}}} Cube {{{-0.5}}} to get {{{-0.125}}}.



So if {{{x=0.5}}}, then {{{y=-0.125}}} which gives us the point (0.5,-0.125).



----------------------------



Let's find the y value when {{{x=1}}}  



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(1-1)^3}}} Plug in {{{x=1}}}.



{{{y=(0)^3}}} Subtract.



{{{y=0}}} Cube {{{0}}} to get {{{0}}}.



So if {{{x=1}}}, then {{{y=0}}} which gives us the point (1,0).



----------------------------



Let's find the y value when {{{x=2}}}  



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(2-1)^3}}} Plug in {{{x=2}}}.



{{{y=(1)^3}}} Subtract.



{{{y=1}}} Cube {{{1}}} to get {{{1}}}.



So if {{{x=2}}}, then {{{y=1}}} which gives us the point (2,1).



----------------------------



Let's find the y value when {{{x=3}}}  



{{{y=(x-1)^3}}} Start with the given equation.



{{{y=(3-1)^3}}} Plug in {{{x=3}}}.



{{{y=(2)^3}}} Subtract.



{{{y=8}}} Cube {{{2}}} to get {{{8}}}.



So if {{{x=3}}}, then {{{y=8}}} which gives us the point (3,8).



----------------------------



Now let's make a table of the values we just found.



<h4>Table of Values:</h4><pre>

<TABLE border="1" width="100">
<TR><TD>x</TD><TD>y</TD></TR><tr><td>-1</td><td>-8</td></tr>
<tr><td>-0.5</td><td>-3.375</td></tr>
<tr><td>0.5</td><td>-0.125</td></tr>
<tr><td>1</td><td>0</td></tr>
<tr><td>2</td><td>1</td></tr>
<tr><td>3</td><td>8</td></tr>
</TABLE>

</pre>

Now let's plot the points:



{{{ drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10, 0),
circle(-1,-8,0.08),circle(-1,-8,0.10),circle(-1,-8,0.12),circle(-1,-8,0.14),
circle(-0.5,-3.375,0.08),circle(-0.5,-3.375,0.10),circle(-0.5,-3.375,0.12),circle(-0.5,-3.375,0.14),
circle(0.5,-0.125,0.08),circle(0.5,-0.125,0.10),circle(0.5,-0.125,0.12),circle(0.5,-0.125,0.14),
circle(1,0,0.08),circle(1,0,0.10),circle(1,0,0.12),circle(1,0,0.14),
circle(2,1,0.08),circle(2,1,0.10),circle(2,1,0.12),circle(2,1,0.14),
circle(3,8,0.08),circle(3,8,0.10),circle(3,8,0.12),circle(3,8,0.14)

)}}}


<h4>Graph:</h4>

Now draw a curve through all of the points to graph {{{y=(x-1)^3}}}:



{{{ drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10, (x-1)^3),
circle(-1,-8,0.08),circle(-1,-8,0.10),circle(-1,-8,0.12),circle(-1,-8,0.14),
circle(-0.5,-3.375,0.08),circle(-0.5,-3.375,0.10),circle(-0.5,-3.375,0.12),circle(-0.5,-3.375,0.14),
circle(0.5,-0.125,0.08),circle(0.5,-0.125,0.10),circle(0.5,-0.125,0.12),circle(0.5,-0.125,0.14),
circle(1,0,0.08),circle(1,0,0.10),circle(1,0,0.12),circle(1,0,0.14),
circle(2,1,0.08),circle(2,1,0.10),circle(2,1,0.12),circle(2,1,0.14),
circle(3,8,0.08),circle(3,8,0.10),circle(3,8,0.12),circle(3,8,0.14)

)}}} Graph of {{{y=(x-1)^3}}}