= Solution
Eliminating the <auxiliary field> gives $F^*=-W'(\varphi)$ and the nonnegative <F-term scalar potential>
$$
\boxed{V(\varphi)=\left|\kappa+m\varphi+\frac g2\varphi^2\right|^2.}
$$
Use $u=\varphi-s$ after the critical-point shift, so that
$$
\boxed{V(u)=\left|Mu+\frac g2u^2\right|^2.}
$$
This notation keeps the original parameters distinct from the shifted mass coefficient $M$. For real $M,g$ with $Mg\ne0$, the real slice is $V(u)=u^2(M+gu/2)^2$. It has zero-energy minima at $u=0$ and $u=-2M/g$, a maximum between them at $u=-M/g$ with height $M^4/(4g^2)$, and grows quartically toward both ends. These statements follow by differentiating: $V'(u)=2u(M+gu/2)(M+gu)$.
When $M=0$ it becomes $|g|^2u^4/4$, with one degenerate zero minimum. When $g=0$, it is $|M|^2u^2$ after a permissible shift. For the real parameters with negative <discriminant>, the original real slice can have a positive minimum even though the full complex <scalar potential> has zero-energy vacua away from that slice. The sketch distinguishes these regimes:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-45-scalar-potentials.png]
{title=Real scalar-potential slices: two supersymmetric minima, a double root, complex vacua off the slice, and flat F-term breaking}
The plots use illustrative normalized coefficients, not values fixed by the exam. \b[A positive minimum on a chosen real slice is not evidence for <supersymmetry breaking> in the full complex field space.]
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