Monday, May 11 2009
6:35 am
I want to see the graphs for a variety of reciprocal polynomial functions.

In[2]:=

Plot[1/x, {x, -10, 10}, PlotStyleRed] ;

[Graphics:HTMLFiles/index_2.gif]

In[4]:=

Plot[{1/x^2, 1/(x^2 + x)}, {x, -10, 10}, PlotStyle {Red, Blue}] ;

[Graphics:HTMLFiles/index_4.gif]

Hmmm. That surprised me. Let's have a closer look at just the second function.

In[5]:=

Plot[1/(x^2 + x), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_6.gif]

This curve is much more complicated than I expected.
The curve is symmetric about the line x = -1/2. The function also approaches + or - infinity as x approaches -1.

In[6]:=

Plot[1/(x^2 - x), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_8.gif]

Changing the sign of the x-coefficient does not alter the general shape of the curve, but it shifts the entire curve 1 unit to the right

In[7]:=

Plot[1/(x^2 + x + 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_10.gif]

Another surprise! Adding an insignificant term to the denominator totally changes the shape of the function.

In[8]:=

Plot[1/(x^2 + x - 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_12.gif]

Fascinating! Changing the sign of the coefficient again changes the shape of the function. It now looks much like the earlier graphs.

In[9]:=

Plot[1/(x^2 - x + 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_14.gif]

In[10]:=

Plot[1/(x^2 - x - 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_16.gif]

In[11]:=

Plot[1/(x^2 + 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_18.gif]

In[12]:=

Plot[1/(x^2 - 5), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_20.gif]

In[14]:=

Plot[1/(x^2 + 1), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_22.gif]

In[15]:=

Plot[1/(x^2 + .1), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_24.gif]

In[16]:=

Plot[1/(x^2 + 10), {x, -5, 5}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_26.gif]

In[18]:=

Plot[1/(x^2 + 10), {x, -5, 5}, AxesOrigin {0, 0}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_28.gif]

In[19]:=

Plot[1/(x^2 + 10), {x, -50, 50}, AxesOrigin {0, 0}, PlotStyleBlue] ;

[Graphics:HTMLFiles/index_30.gif]

This is turning out to be much more complicated, and interesting, than I suspected. The reciprocal function has many nuances.
7:20 am

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