Not recalling the theorem about the sufficient conditions for local extrema, show that no stationary point of the function $f:\Bbb R^2\to\Bbb R$ $$f(x)=\frac12\langle Ax,x\rangle+\langle b,x\rangle+c$$ where $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix},b=\begin{bmatrix}1\\2\end{bmatrix},c=100,$ is a local extremum.
My attempt:
Let $g:\Bbb R^2\to\Bbb R, g(x)=\langle Ax,Ix\rangle.$ Then, $$\begin{aligned}Dg(c)=(Ic)^TDA(c)+(Ac)^TDI(c)=c^TA+c^TA^TI=2c^TA^T=2(Ac)^T=2(c_1,-c_2)\end{aligned}$$
Let $h:\Bbb R^2\to\Bbb R,h(x)=b^Tx.$ Since $h$ is a linear functional, $Dh(c)=b^T=(1,2).$
So, the stationary point: $$Df(c)=(c_1+1,-c_2+2)=(0,0)\iff (c_1,c_2)=(-1,2).$$
Now, $f(x,y)$ can be rewritten as:
$$\begin{aligned}f(x,y)&=\frac12(x^2-y^2)+x+2y+100\\&=\frac12(x^2+2x+1)-\frac12-\frac12(y^2-4y+4)+2+100\\&=\frac{(x+1)^2-(y-2)^2}2+101\frac12.\end{aligned}$$
I was looking at the function $\overset{\sim}f:\Bbb R^2\to\Bbb R,$ $$\overset{\sim}f(x,y)=(x+1)^2-(y-2)^2.$$
And saw $\overset{\sim}f(x,2)>0,\forall x\in\Bbb R\setminus\{-1\}$ and $\overset{\sim}f(-1,y)<0,\forall y\in\Bbb R\setminus\{2\}.$
We should find some $(x_0,2),(-1,y_0)$ arbitrarily close to $(-1,2)$. Let $\varepsilon >0$. Then $$\|(x_0,2)-(-1,2)\|<\varepsilon\iff x_0\in(-1-\varepsilon,-1+\varepsilon)$$ and $$\|(-1,y_0)-(-1,2)\|<\varepsilon\iff y_0\in(2-\varepsilon,2+\varepsilon).$$
From above, it follows that for a given $\varepsilon>0,$ one can pick $(-1,y_0),(x_0,2)\in B((-1,2),\varepsilon)\setminus\{(-1,2)\}$ s. t $f(-1,y_0)<f(-1,2)$ and $f(x_0,2)>f(-1,2).$ Therefore, the only stationary point $(-1,2)$ of the given function $f$ isn't a point of a local extremum.
Could somebody check my answer?