This is a really nice puzzle we looked at at the IB HL workshop:

**When x = 1, y = 1, when x = 2, y = 2, when x = 3, y = 3 but when x = 4, y does not equal 4. Find a sequence which describes these points.**

There are an infinite number of answers, though not necessarily easy to find!

If you are interested in the solutions, the answer is written below in white text, highlight to reveal!

Answer:

Two possible ways of tackling the problem –

1) as a polynomial – if y = (x-1)(x-2)(x-3) + x this satisfies the original question – as the brackets all cancel to zero for 1,2,3 but will remain for x = 4 onwards.

2) modelling as a function with absolute value. Notice that -(abs x ) will satisfy the correct shape – ie a linear increase and then divergence from this. By transformations therefore we can get -(abs(x-3) ) +3. This fits the graph for y=x for 1,2,3 before altering for y=4.

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