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Folium of Descartes

Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.

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Witch of Agnesi

Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.

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Pitchfork

Plot the graph of x^y = y^x in the first quadrant and explain its properties.

Rational Request

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Steve said the following

I sketched four vertical asymptotes and a sketch showed that a function which decayed to zero from above at $x \rightarrow \pm \infty$ could have the right sorts of properties.

To get the right asymptotes and behaviour at $\pm \infty$ I guessed the following curve, choosing to make it symmetric about the origin for simplicity
$$
y = \frac{1}{(x-2)(x-1)(x+1)(x+2)}
$$
This worked: it has a turning point at $x$ between $-2$ and $-1$ another turning point at $x$ between $1$ and $2$ and a turning point at $x=0$.

The plot of this from graphmatica is as follows



It seems likely that many such curves, with differing constants, would also give the correct behaviour. To see why, upon differentiation, I get a cubic polynomial divided by another polynomial. For zeros the numerator would need to be zero and a cubic can have three real roots. I could choose the constants to have the correct number of real roots.

I then considered the second request. Initially, I thought that this seemed impossible, but then started to work through the possibilities for asymptotes. By turning the middle turning point into a point of inflection I would have a graph with the correct behaviour.


I wondered how to convert the behaviour of the central turning point and decided that the curve needed to be forced to pass through the origin and also to be antisymmetric. I therefore multiplied the expression by $x$, realising that this wouldn't affect the 'topological' behaviour at the other turning points. A plot of the curve
$$
y = \frac{x}{(x-2)(x-1)(x+1)(x+2)}
$$
gave graph



Which has the correct behaviour.