Can you explain why there is no local minima for the following function when $b \neq 0$:
$$ f(x,y) = ax^2 + by + c, \text{where}\ f:\mathbb R^2 \rightarrow \mathbb R $$
In this case a global minimum does not exist.
$f(x,y) = c$ is the local minima when $a > 0$ and $b = 0$.
This equation in Wolfram Alpha
The gradient is $\nabla f(x,y) = (2ax, b)$.
I am going to make the function arbitrarily negative. Take $x=0$ to simplify the problem. Let $M<0$ be as negative a number as you want.
If $b>0$, you can increase $y$ to make the function as negative as desired by making $y$ negative. In particular, since $x=0$, $by+c <M$ can be achieved by taking $y < (M-c)/b$.
If $b<0$, you can decrease $y$ to make the function as negative as desired by making $y$ positive. In particular, since $x=0$, $by+c <M$ can be achieved by taking $y>(M-c)/b$.
So the function is unbounded below, so it has no global minimum when $b \neq 0$.