Pencil of singular affine hypersurfaces differing only in the constant term

Question--quick version: Does there exist a (nonconstant) polynomial $f \in \mathbb C[x,y,z]$ such that for all $c \in \mathbb C$, the affine hypersurface cut out by $f + c$ is singular?

Motivated version: Suppose you have a singular hypersurface $\mathbf V(f)$ in $\mathbb C^3$. Since a generic hypersurface is smooth, you can deform the hypersurface to be smooth by adding a bit of noise to every coefficient of $f$ (i.e., all coefficients of degree $\leq$ the degree of $f$). Can you achieve the same effect by adding noise only to the constant term? I would be surprised if there were no counterexamples, but I can't think of any.

Thoughts: You can't get an example just by putting e.g. a triple point in the base locus, because the pencil in question has empty affine base locus. (Perhaps this would give you singularities at infinity, but I only care about affine singularities.) The other approach that occurs to me is to make the pencil the image of a singular variety under a pencil of affine transformations, but I don't think it's ever possible to do this by shifting the constant term.

• This is Sard's Theorem, aka, generic smoothness. The critical locus of the morphism $f:\mathbb{A}^3\to \mathbb{A}^1$ is a proper closed subset of $\mathbb{A}^1$. – Jason Starr Jan 18 '16 at 2:02