Solution
= Solution
Put $h=x-1$ and $k=y-1$. At $(1,1)$, the value is $5+4\alpha$, the <gradient> is $(5+4\alpha,4+2\alpha)$, and the <Hessian matrix> is $\begin{pmatrix}8&2\alpha\\2\alpha&2\end{pmatrix}$. Thus the quadratic <Taylor expansion> is
$$
\boxed{f(1+h,1+k)
=5+4\alpha+(5+4\alpha)h+(4+2\alpha)k
+4h^2+2\alpha hk+k^2+O((|h|+|k|)^3).}
$$
The exact remainder is $h^3$, since the original <polynomial> has only the cubic term $x^3$ beyond degree two.