# Rational request

Can you make a curve to match my friend's requirements?

For a secret reason, my friend wants a curve which has 4 vertical asymptotes and 3 turning points.

Could you sketch him such a curve? Could you find an algebraic form for such a curve? Could you find many different curves with such properties?

My other friend wants a curve which also has 4 vertical asymptotes, but only 2 turning points. Can her needs be met algebraically?

As you consider this problem, many questions might emerge in your mind such as: "what makes one type of curve 'the same' as or 'different' from another?" or "can I satisfy requests for other numbers of asymptotes and turning points?". Why not make a note of these questions and ask your teacher, yourself or your friends to try to solve them?

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.

Image

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

Image

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.

Image

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

Image

Which has the correct behaviour.