Hermite Interpolation extended to second derivative

[Haydo]

SOLVED

1. Homework Statement
Find polynomial of least degree satisfying:
p(1)=-1, p'(1)=2, p''(1)=0, p(2)=1, p'(2)=-2

Homework Equations

In general, a Hermite Polynomial is defined by the following:
∑[f(x_i)*h_i(x)+f'(x_i)*h²_i(x)]
where:
h_i(x_j)=1 if i=j and 0 otherwise. Similarly with h'². h'_i(x)=0 and h²(x)=0. i.e., they are zero if they are integrated or derived.

Here is a page from wolfram with general information: http://mathworld.wolfram.com/HermitesInterpolatingPolynomial.html

The Attempt at a Solution

First, I recognized x₀=1 and x₁=2. I tried to create some third coefficient term in order to satisfy p''(1)=0, but that seems to mean that I would have to make h_i(x_j) be zero for the second derivative, and I have no idea how to do this. I tried just setting this new h (call it h3) equal to (x)(x-2) when derived twice and integrating (so that h3''=1 for x₁ and zero otherwise) but that was a total flop.

I'm starting to think that, despite the question's section (it was in the section regarding Hermite Interpolating Polynomials), there is a better way to approach it. Any help would be greatly appreciated.

 

[Haydo]

For anyone reading this, I believe I have to use Newton's Formalism for the Hermite Interpolating polynomial, instead of Lagrange's.

EDIT- Yep, this is how to solve it. Answer is:
-1+2(x-1)-4(x-1)^3(x-2)

Here is a page describing Newton's Form for the Hermite Interpolant