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# Curve Fitter

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Age 14 to 18

ShortChallenge Level

This problem looks at how the number of constraints can reduce the number of possible answers, by considering cubic graphs. It develops the idea that a general cubic $y=ax^3+bx^2+cx+d$ has 4 parameters (or 4 degrees of freedom) and so three pieces of information are not enough to fully determine all of the coefficients.

*This problem featured in an NRICH Secondary webinar in January 2021.*

Starting from the general form of the cubic, the information given by the points on the cubic can be used to generate equations. These can be used to find the values of some of the parameters, or to write some parameters in terms of another parameter and so reduce the number of "degrees of freedom".

As the problem progresses more pieces of information are given, so more equations can be generated which narrows down the options until the problem is "over-constrained" and it is impossible to satisfy all of the constraints at once.

Desmos can be used to help students investigate the graphs, or to check their solutions.

They could also look at cubic graphs which satisfy different sets of conditions, such as a cubic graph that has turning points at $(1,2)$ and $(2,1)$ which does not pass through the origin.

*The thinking that informed the creation of this problem is discussed in this video (after the introductory 3min 20sec):*

- What does the equation of a cubic curve look like?
- The curve passes through the origin. What does this tell us?
- How can we use the points that the curve passes through?
- If the curve has a turning point at $(1, 2)$, what information does that give us?

Students could be introduced to this problem by first asking them to find a quadratic which passes through the three points $(0,0)$, $(1, 2)$ and $(2,1)$. They could also be asked to show that it is impossible to find a straight line that passes through all three points.

For Part 1, students should find that there is more than one curve that passes through the three given points. This GeoGebra file shows how a family of cubics can pass through $(0,0)$, $(1, 2)$ and $(2,1)$. By varying $a$ it can be shown that either $(1,2)$ or $(2,1)$ can be a turning point, but not both at the same time.

Students could be asked to consider what questions might have been appropriate for parts 2(c), 2(d) and 2(e), and could find cubics which satisfy these. This GeoGebra file which shows a family of cubics which pass through $(0,0)$ and has turning points at $x=1$ and $x=2$ might help students visualise how four of the five conditions can be met at any one time. Students might also like to consider how this graph has been created.

This GeoGebra file allows you to change two parameters. Students could use it to find a cubic which has the required turning points at $(1,2)$ and $(2,1)$, and then see whether or not it passes through $(0,0)$.

Part 3 of the problem could be considered to be an extension, especially if students try and find a proof before being given the proof sorters: Proof Sorter 1 and Proof Sorter 2. Students could also use the given proof sorters to help them devise an alternative proof, or
even their own proof sorter to share with other members of the class.

You can find more short problems, arranged by curriculum topic, in our short problems collection.