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## 'Rational Request' printed from http://nrich.maths.org/

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.