jbriggs444 said:
I see no hint in the problem statement about a direction for the coordinate axis. A positive field strength would point in the direction of the force on a positive test charge. Which would be the direction of increasing kinetic energy and decreasing electrical potential energy.
Line your coordinate axis up so that the force on a positive test charge points in the positive direction and you have a positive-pointing field strength.
The main problem with such apparently simplified treatments (it's in fact oversimplified) is that you get easily confused. It's an example, where misunderstood didactics is in the way of understanding. One should start with clear concepts and treat scalars as scalars and vectors as vectors. Then together with a bit intuition in which direction the vectors (here the electric field and forces on a point charge) should point all the trouble with signs is gone.
Take the example with the constant electric field. You have
$$\vec{E}=E_1 \vec{e}_1+ E_2 \vec{e}_2 + E_3 \vec{e}_3,$$
where ##\vec{e}_1##, ##\vec{e}_2##, and ##\vec{e}_3## are three vectors of length 1 which are perpendicular to each other and form in this order a right-handed Cartesian coordinate system (if you point with the thumb of your right hand in direction of ##\vec{e}_1## with the index finger in direction of ##\vec{e}_2##, then your middle finger points in direction of ##\vec{e}_3##).
With the given basis you can now also just use the components of the electric field to uniquely describe it. Usually it's written as a column vector:
$$\vec{E}=\begin{pmatrix} E_1 \\ E_2 \\ E_3 \end{pmatrix}.$$
Now, if you have a scalar field ##\phi(\vec{x})## by definition the components of its gradient wrt. to this basis is
$$\vec{\nabla} \phi = \begin{pmatrix} \partial_1 \phi \\ \partial_2 \phi \\ \partial_3 \phi \end{pmatrix},$$
where
$$\partial_1 \phi=\frac{\partial \phi}{\partial x_1},\; \ldots.$$
Now for our constant field we can easily check that it has a potential, i.e., we look for a scalar field ##\phi## such that ##\vec{E}=-\vec{\nabla} \phi##. The minus sign here is a convention, but it's a useful one such that for forces in the corresponding energy-conservation law, the potential energy is added to the kinetic energy. That's why usually one writes this minus sign in the relation between a vector field and its potential (if it exists).
To find ##\phi## we start with the first component. From ##\vec{E}=-\vec{\nabla} \phi## we get
$$\partial_1 \phi=-E_1.$$
Since ##E_1=\text{const}## (by assumption, this integrates to
$$\phi(x_1,x_2,x_3)=-x_1 E_1 + \phi_2(x_2,x_3),$$
because in integrating with respect to ##x_1## you can always add some function which is constant wrt. ##x_1##. For the 2nd component we find
$$\partial_2 \phi=\partial_2 \phi_2=-E_2 \; \Rightarrow \; \phi_2 = -x_2 E_2 + \phi_3(x_3).$$
And once more for the 3rd component you finally get
$$\phi_3=-x_3 E_3+\phi_0,$$
where ##\phi_0## is an arbitrary constant. Plugging everything together we get
$$\phi=-x_1 E_1 - x_2 E_2 -x_3 E_3+\phi_0=-\vec{x} \cdot \vec{E} + \phi_0.$$
The constant ##\phi_0## is arbitrary. Potentials are usually only defined up to such an arbitrary additive constant. What's really physical in the description here is ##\vec{E}## and because ##\vec{E}=-\vec{\nabla} \phi## the constant drops out again.