Higher To Lower Electric Potential The question I am working on is:

"An electron moving parallel to the x axis has an initial speed of $4.65 \cdot 10^6~m/s$ at the origin. Its speed is reduced to $1.27 \cdot 10^5 ~m/s$ at the point x = 2.00 cm.
(a) Calculate the electric potential difference between the origin and that point.
(b) Which point is at the higher potential?"

I am having difficulty visualizing this situation. Obviously, because of the reduction in speed, the kinetic energy is reducing; but, because of conservation of energy, the reduction of kinetic energy corresponds to an increase in potential energy.
Is something giving the electron an initial "push," and a positive source charge distribution is pulling the electron as it passes by, thereby reducing the speed of the electron?
Also, what confuses me is this paragraph:

We assume that the electron is moving in a vacuum. Its speed is reduced because it is >moving to a higher electric potential energy. Since its charge is negative, this means >that it is moving to a more negative potential; that is, its starting position is at a higher potential (a higher voltage) than its final position.

What do they mean by "more negative potential," what does a negative potential physically mean? I realize that at our initial position, we arbitrarily choose the initial electric potential the value zero, and so if we chose another point in the electric field, relative to our initial point our electric potential will either be positive or negative--assuming that our second point is not in the equipotential region that the first point is. I suppose I am also having trouble understanding what situation a negative electric potential corresponds to, and what situation a positive electric potential corresponds to.
I'd appreciate the help! Thanks in advance!
EDIT: Also, is it true that if a charge has negative electric potential energy, then it is not necessary that it has lost that energy. In other words, negative electric potential energy can mean a gain in that sort of energy?
I've heard electric potential being described as being "the measure of potential energy per unit charge." From this, I have developed the understanding that electric potential is a scalar that is assigned a different value at every point in an electric field.It almost seems like a conversion factor. It's how much potential energy a charged particle could possession if it were there. It's a potential of a potential. Again, sorry for all of the comments. I hope you can help me

Another question, if I push a negative test charge near a negative source charge, the test charge will be gaining potential energy, and that potential energy will be positive, right? Alternatively, if I move a negative test charge near a positive source charge, making sure it doesn't accelerate towards it, it will loss potential energy, showing up as a negative potential energy, right? And the mathematics will show that these are true? I guess the root of my problem is seeing a generalized situation, and the mathematics that accompany it.

Actually, I have one more question. I was able to answer the question I originally posed, except for part (b). I thought, because the electric potential energy was increasing at that point, that the electric potential at that point would be higher. After all, electric potential is defined one way as, "the amount of electric potential energy that a unitary point charge at that location would have." So, if the electron has more PE at the final position, wouldn't that mean the electric potential was greater at that final position too?
 A: First the math:
Given the electric potential $V(\mathbf a)$ at a point $\mathbf a$ in space and the potential $V(\mathbf b)$ at another point $\mathbf b$, the electric potential energy of a charge moving from point $\mathbf a$ to point $\mathbf b$ will change by
$$
  U(\mathbf b) - U(\mathbf a) = q[V(\mathbf b) - V(\mathbf a)]
$$
If the speed of an electron decreases in moving from a point $\mathbf a$ to a point $\mathbf b$, then that must mean that its potential energy has increased: $U(\mathbf b) - U(\mathbf a) >0$, so that since the electron has charge $-e$, we get
$$
  (-e)[V(\mathbf b) - V(\mathbf a)]>0
$$
which, upon dividing both sides by $-e$ (and noting to change the direction of the inequality when we do so) implies
$$
  V(\mathbf b) < V(\mathbf a)
$$
So we see that the starting position $\mathbf a$ is at a higher potential than the final position $\mathbf b$ as the second paragraph says.
The intuition here is that there are some other charges around (they could be positive or negative depending on where we place them) that setup the potential "landscape" V in such a way that the potential at the origin, where the charge began, is higher than the potential at the final point, and a negative charge that rolls "down this potential hill" will be going slower when it gets to the bottom of the hill.
Hope that wasn't confusing!
Cheers!
A: Consider this alternative approach: Since the electron is slowing down the point at $x=2$ is at lower potential than at 0, i.e. more negative (this is just physics.) Since it is the potential difference you are interested in, take $V-0=0.0$ and write the energy conservation equation 
$E_K (x=0)=E_K(x=2)+e(V_0-V_2)$
From this you get
$V_0-V_2=\frac {m_e(v_0^2-v_2^2)}{2e}$
and using $m_e=9.1\times 10^{-31}Kg$ you can calculate the potential difference.
Hope this helps?
