Explore function of two variables for extremas at given points

Question

The task is to explore the function of two variables $$f(x,y)=-x^2(y-2)^3$$ for extremas in two points $$A=(5,0), \qquad B=(0,5)$$ I tried to use the second derivative test, but it turns inconclusive: $$\frac{\partial f}{\partial x} = -2 x (y - 2)^3, \qquad \frac{\partial f}{\partial y} = -3 x^2 (y - 2)^2$$ so the critical points are: $\{(0,y) \mid y \in \Bbb{R}\}$ and $\{(x,2) \mid x \in \Bbb{R}\}$, that gives us that $B=(0,5)$ is critical point, but $$\frac{\partial^2 f}{\partial x^2} = -2 (y - 2)^3, \qquad \frac{\partial^2 f}{\partial x \partial y} = -6 x (y - 2)^2 \qquad \frac{\partial^2 f}{\partial y^2} = -6 x^2 (y - 2)$$ and the Hessian turns to zero at this point: $$ \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - (\frac{\partial^2 f}{\partial x \partial y})^2 = -24 x^2 (-2 + y)^4 = -24 \cdot (0)^2 \cdot (3)^4 = 0 $$ So, the question is, how can I explore these types of functions for extremas? (for example, for this function the test also doesn't give a conclusion: $f(x,y)=-xy^3$)

Answer 1

As regards the point $B=(0,5)$ consider the function $f$ in a neighbourhood of the point: note that for $4<y<6$ (and for all $x$), $$f(x,y)=-x^2(y-2)^3\leq -x^2(4-2)^3\leq 0=f(0,5).$$ What kind of extrema is $B$?

Using the same approach (or by exploring the sign of $f$), what kind of extrema are the other critical points?