### Poly Fibs

A sequence of polynomials starts 0, 1 and each poly is given by combining the two polys in the sequence just before it. Investigate and prove results about the roots of the polys.

### Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

### Janusz Asked

In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?

# Curve Fitter

##### Age 14 to 18Challenge Level

This problem challenges you to find cubic equations which satisfy different conditions. You may like to use Desmos to help you investigate possible cubics.

Part 1
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$?
Can you find more than one possible cubic?

Can you write down the general algebraic form of a cubic equation?
How can you use this together with the information provided in the question?
Try using this GeoGebra page to investigate possible cubics.

Part 2 (a)
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$, and where the point $(1,2)$ is a turning point of the cubic?
Can you find more than one cubic satisfying all the conditions?

What extra information do you now have?

Part 2 (b)
Can you find a cubic which passes through $(0,0)$ and the points $(1, 2)$ and $(2,1)$, and where the point $(2,1)$ is a turning point of the cubic?
Can you find more than one cubic satisfying all the conditions?

Part 3
Can you find a cubic which passes through $(0,0)$ and where the points $(1, 2)$ and $(2,1)$ are both turning points?

Try using this GeoGebra page to help you investigate the problem.

Your answers to Part 2 may make you suspect that it is impossible to find a cubic which satisfies the last set of conditions. Can you prove that it is impossible?

You can use this proof sorter to show one way of proving that there are no cubics that satisfy the last set of conditions. Alternatively, this second proof sorter shows another possible way. Can you work out a third proof?