Let f(x) = x^{2} + bx + c. Let C be the curve y = f(x) and let P_{i} be the point (i, f(i) ) on C. Let A_{i} be the point of intersection of the tangents at P_{i} and P_{i+1}. Find the polynomial of smallest degree passing through A_{1}, A_{2}, ... , A_{9}.

**Solution**

y = f(x) - 1/4.

*Easy*

The answer suggests there ought to be a one-line solution, but I cannot see it.

The equation of the tangent at x = i is y - (i^{2} + ib + c) = (2i + b)(x - i), or y - (2i + b)x = c - i^{2}. Solving, we find that A_{i} is (i + 1/2, i^{2} + i + bi + b/2 + c). Clearly these do not lie on a straight line, so the degree is at least 2. But we see at once that *any* A_{i} lies on the second degree y = x^{2} + bx + c - 1/4.

© John Scholes

jscholes@kalva.demon.co.uk

3 Nov 1999

Last corrected/updated 21 Nov 02