Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the potential at any point is the superposition of the potentials due to all charges.
So the potential at point $\rm p$ due to the two charges is given by
$$V_{\rm p}=V_1+V_2$$
$$V_{\rm p}=\dfrac{k_eq}{r}+\dfrac{k_eq}{r}=\dfrac{2k_eq}{r}$$
From the geometry of figure below, $r=\sqrt{x^2+(s/2)^2}$
$$\boxed{V_{\rm p}=\dfrac{2k_eq}{\sqrt{x^2+(s/2)^2}} }$$
$$\color{blue}{\bf [b]}$$
The graph of $V$ versus $x$ is shown below.
The potential of $2q$ that is at the origin is given by
$$V_{\rm 2q}=\dfrac{2k_eq}{x }$$
which is the same case when $x\rightarrow\infty$ of the two charges above,
$$V_{\rm p}=\dfrac{2k_eq}{x\sqrt{1+\left(\dfrac{s}{4x}\right)^2}}$$
where $\left(\dfrac{s}{4x}\right)\approx 0$
$$V_{\rm p}=\dfrac{2k_eq}{x }$$
Note that when $x=0$, $V_{\rm p}=\dfrac{k_eq}{s}$ which is the top point of our figure, while $V_{\rm 2q}=\infty$.
The dashed curve for $V_{\rm 2q}$.
