Optimal points of $f(x,y)=x^2 + y^2 + beta xy + x + 2y$

I am self-learning basic optimization theory and algorithms from "An Introduction to Optimization" by Chong and Zak. I would like someone to verify my solution to this problem, on finding the minimizer/maximizer of a function of two variables, or any tips/hint to proceed ahead.

For each value of the scalar $beta$, find the set of all stationary points of the following two variables $x$ and $y$

$$f(x,y) = x^2 + y^2 + beta xy + x + 2y$$

Which of those stationary points are local minima? Which are global minima and why? Does this function have a global maximum for some value of $beta$.

Solution.

We have:

begin{align*}
f(x,y) &= x^2 + y^2 + beta xy + x +2y
f_x(x,y) &= 2x+beta y + 1
f_y(x,y) &= beta x + 2y + 2
f_{xx}(x,y) &= 2
f_{xy}(x,y) &= beta
f_{yy}(x,y) &= 2
end{align*}

By the first order necessary condition(FONC) for optimality, we know that if $nabla f(mathbf{x})=0$, then $mathbf{x}$ is a critical point.

Thus,

begin{align*}
f_x(x,y) &= 2x+beta y + 1 = 0
f_y(x,y) &= beta x + 2y + 2 = 0
end{align*}

Solving for $x$ and $y$, we find that:

begin{align*}
x = frac{begin{array}{|cc|}
-1 & beta
-2 & 2
end{array}}{begin{array}{|cc|}
2 & beta
beta & 2
end{array}}=frac{-2+2beta}{4-beta^2}=frac{2beta-2}{4 -beta^2}
end{align*}
begin{align*}
y = frac{begin{array}{|cc|}
2 & -1
beta & -2
end{array}}{begin{array}{|cc|}
2 & beta
beta & 2
end{array}}=frac{-4+beta}{4-beta^2}=frac{beta -4}{4 – beta^2}
end{align*}

The second order necessary and sufficient conditions for optimality are based on the sign of the quadratic form $Q(mathbf{h})=mathbf{h}^T cdot Hf(mathbf{a}) cdot mathbf{h}$.

The Hessian of $f$ is given by,

$$Hf(mathbf{x})=begin{array}{|c c|}
2 & beta
beta & 2
end{array}$$

Thus, $d_1 = 2 > 0$ and $d_2 = 4 – beta^2$. Thus, $f$ has a local minimizer if and only if $4 – beta^2 > 0$. $g(beta) = 4 – beta^2$ is a downward facing parabola. So, the values of this expression positive, if and only if $-2 < beta < 2$. The function $f$ has no global maximum.

Question. How do I find the actual global minima?