Consider the cubic equation for Euler's Elegant Example: \[ t(y^3-y) = s(x^3-x).\]
As long as \(t \ne 0,\) we can divide through by \(t.\) If we then let \(p={s}/{t},\) we have the parameterized cubic equation \[ y^3-y = p(x^3-x).\]
Now for any real number that we assign to the parameter \(p,\) we have an equation whose graph passes through the nine points of the \(3 \times 3\) grid. You can explore these curves using the applet in Figure 8. You can set particular values of \(p\) using the slider control, or put the applet into play mode and watch as \(p\) cycles through values between \(-4\) and \(+4.\)
Figure 8. Euler's Elegant Example. Set values of \(p\) using the slider control, or click the arrow at lower left and watch as \(p\) cycles through values between \(-4\) and \(+4.\) (Interactive applet created using GeoGebra.)
For the particular values \(p=0,\, p=1,\) and \(p=−1,\) we can factor the parameterized cubic equation \[ y^3-y = p(x^3-x)\] into factors of lower order.
Value of \(p\) |
Factorization |
\(−1\) |
\((y+x)(y^2−xy+x^2−1)=0\) |
\(0\) |
\(y(y−1)(y+1)=0\) |
\(1\) |
\((y−x)(y^2+xy+x^2−1)=0\) |
There is one more important special case, when \(t=0\) in the original equation \[t(y^3-y) = s(x^3-x)\] for Euler's Elegant Example. In this case, we have \(x(x-1)(x+1) = 0,\) so that the graph consists of three vertical lines: \(x=0,\) \(x=1\) and \(x=-1.\) This is the asymptotic case to which the parameterized cubic equation \[ y^3-y = p(x^3-x)\] tends when either \(p \rightarrow +\infty\) or \(p \rightarrow -\infty.\)