# Quasiconvexity of a specific quartic bivariate polynomial family

I have a quartic polynomial of shape $$(a\cdot x+b+c\cdot xy-d\cdot y+y^2\big)^2$$ where $$a,b,c,d\in\mathbb Z_{>0}$$ are coefficients.

The polynomial is not convex.

1. However is it quasiconvex for various values of $$(a,b,c,d)$$?

2. My problem is to find the regions of $$(a,b,c,d)\in\mathbb Z_{>0}^4$$ on which the polynomial is quasiconvex?

Note this polynomial is always non-negative and expressible as sum of squares. If so may be explicit representation will help. Note $$0$$ level is always the lower bound. It might be easy to handle.

Let $$g(x,y) = ax+b+cxy-dy+y^2$$ so that $$f(x,y) = (ax+b+cxy-dy+y^2)^2 = g(x,y)^2$$. If $$g$$ has a non-convex zero set, then $$f$$ won't be quasi-convex - for small enough $$\epsilon >0$$, $$f^{-1}((-\infty,\epsilon))$$ will be a neighborhood of $$g^{-1}(0)$$ which will again be nonconvex. So our only hope of $$f$$ being quasiconvex is when $$g=0$$ is convex.
Our goals are for this conic to be a single line, a single point, or empty, as these are the only conics which are convex. We can read off when $$ax+b+cxy-dy+y^2=0$$ is of the appropriate form by examining the matrix form of our conic, $$G =\begin{pmatrix} 0 & c/2 & a/2 \\ c/2 & 1 & -d/2 \\ a/2 & -d/2 & b \end{pmatrix}$$
Let $$U$$ be the upper left $$2\times 2$$ minor.
In order to get a single line, a single point, or an empty conic, we must have $$\det G =0$$ and $$\det U \geq 0$$. On the other hand, $$\det U = \frac{-c^2}{4}$$, which as $$c\neq 0$$ by the problem statement, is always negative, so $$f$$ is never quasiconvex.