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Why can't we conclude extrema property of a function from its quadratic approximation when the discriminant is zero?

Suppose $f:\mathbf{R}^2\to\mathbf{R}$ is a smooth function and $P=(0,0)$ is a critical point of $f$. The [second-derivative test](https://en.wikipedia.org/wiki/Second_partial_derivative_test) is _inconclusive_ when the discriminant at $P$ is zero: 
$$f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^2=0\ .$$

For simplicity, assume further that $f(0,0)=0$ and $f_{xx}(0,0)\ne 0$. Then the quadratic approximation of $f$ at $P$ is of the form

\begin{align}
g(x,y)
=ax^2+bxy+cy^2=a\left[
(x+\frac{b}{2a}y)^2+(\frac{D}{4a^2})y^2\right\],
\end{align}
where $D=4ac-b^2$. In the case when $D\ne 0$, $f$ and $g$ have the *same* property at $P$ by the second-derivative test. 

But when $D=0$， we can no longer conclude the property of $f$ at $P$ from that of $g$ unless $f(x,y)=g(x,y)=a(x+\frac{b}{2a}y)^2$, where $f$ has a local min (resp. max) at $P$ when $a>0$ (resp. $a<0$). I would like to understand why. 

**Question:** why can we no longer conclude the local min/max property for $f$ from $g$ when $D=0$?