= Solution
A <supersymmetric vacuum> satisfies $F=0$, equivalently $W'(v)=0$, and has $V(v)=0$. For $g\ne0$, the nonconstant quadratic $W'$ has at least one complex root, including a double root when $\Delta=0$. Since the scalar is complex, these roots are admissible physical vacua whether or not they lie on the real slice. Thus <supersymmetry> is unbroken at either zero of $W'$. The same is true for $g=0,m\ne0$, where $W'$ is linear and has the root $v=-\kappa/m$.
The exceptional pure-linear model $g=m=0,\kappa\ne0$ has $F^*=-\kappa$ everywhere and $V=|\kappa|^2>0$. It exhibits <flat F-term breaking with a linear superpotential>: all scalar values are flat vacua but no state annihilates every <supercharge>. If also $\kappa=0$, the potential is identically zero and <supersymmetry> is unbroken on a flat moduli space.
Consequently \b[the generic cubic or massive model does not spontaneously break <supersymmetry>; only the nonzero purely linear branch does in this single-field canonical model.] The answer is independent of $\alpha$.
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