The problem is to consider a fixed-step algorithm applied to the functions and find the largest value of the step size $\alpha$ for which the algorithm is globally convergent.
I figured out part a) of this problem, but part b) is $f(x)=x^T\begin{bmatrix}3 & 3 \\ 1 & 3\end{bmatrix}x+[16,23]x+\pi^2$.
My attempt:
It must be in the form $f(x)=\frac{1}{2}x^TQx-b^Tx$. So a quick adjustment yields $f(x)=\frac{1}{2}x^T\begin{bmatrix}6 & 6 \\ 2 & 6\end{bmatrix}x-(-[16,23])x+\pi^2$. Thus $Q=\begin{bmatrix} 6 & 6 \\ 2 & 6 \end{bmatrix}$. Finding the eigenvalues yields $(6-\lambda)^2-12=0 \implies \lambda=6\pm2\sqrt{3}$. So then $0<\alpha<\frac{2}{6+2\sqrt{3}}$.
I just want confirmation if this looks right.