Is this the correct voltage graph of a conducting sphere in the presence of another charge? This grading rubric show an example of an acceptable answer, however I wonder if the slopes at the surface of the second sphere are wrong. This is a previous AP Physics test question so I'm inclined to trust the key, however, the graph is not presented as "the correct answer" but just as an "acceptable answer" and I want to dig deeper into it.

The sphere on the left is a conductor with a net charge; the sphere on the right is an uncharged conductor.
The graph shows "curved segment 1" between the two spheres ending with a slope of zero and "curved segment 2" to the right of the second sphere has a steep slope.  It's almost like two separate 1/r graphs.
My thought was that it should look like a single 1/r graph with a flat "pause" in the middle that picks up with the same slope it left off with - that the slopes immediately to the left and right of the second sphere should both be shallow negative and match one another.
Which graph is correct and why?
 A: The problem is complicated (or simplified) by considering how the charges of the sphere on the left induce a charge distribution on the uncharged sphere.  I cannot see how either situation results in the graph you have shown being correct.


*

*I don't see a good reason why the potential should be flat immediately on the left of the uncharged sphere, as this would imply $0$ field there, which I can't see being realistic.  In this scenario, the potential curve would be discontinuous at the edge of the neutral sphere on both sides, and this does not correspond to the figure as given.  The discontinuit would be proportional to the local surface density of induced charges.

*The flat potential on the left of the uncharged sphere could be correct if you ignore induced charges on the neutral sphere: the dielectric-conductor boundary conditions give a field proportional to the surface charge density, which would presumably be $0$ without induced charges.  If you believe this, the potential should also be flat immediately on the right of the neutral sphere, so this does not correspond to the graph presented either.

*Thus I suspect the solution you propose, with non-zero slope of $V$'on both sides of the neutral sphere, is likely correct.  
Hopefully I'm not missing something obvious that will make may answer look silly...
A: The existing answer by ZeroTheHero is correct.  The neutral sphere has nonzero surface charge density where it is polarized by the external field.  Therefore there is a discontinuity of $\sigma/\epsilon$ in the field normal to the surface, which would appear in a more careful sketch as a discontinuity in the slope of the potential.
From a pedagogical standpoint, I think the error is forgivable.  There will be some modification of the $1/r$ potential around the neutral sphere due to its nonzero dipole moment, so a pure $1/r$ with a flat segment isn't quite right either.  The important parts for the students to understand are that the potential is constant (but not necessarily zero) within the conductors and decreasing nonlinearly as you move away from the charged conductor.  My AP-level students who are not yet either physicists or artists sometimes struggle to freehand curves with just these features.  I have some competence as a physicist and some competence as an artist, and I think it would take me two or three tries to show the curvature in the segment of the potential between the spheres without making the curve go "too flat," as in your solution, where it meets the polarized sphere.
