The graph of a function $y(x)$ is a straight line in a regular coordinate system if and only if $y(x) = ax + b$? How does one prove: 
The graph of a function $y(x)$ is a straight line (no curves) in a regular coordinate system ($x,y$ coordinate system with axis' having interval $c$) if and only if $y(x) = ax + b$ ?

This question might be odd. By looking at graphs of functions one can be convinced that the implication holds both ways, but how is this proven rigoursly ? 
 A: Consider the generic case of a line that intersects the $x$-axis and the $y$-axis in two distinct points  $A=(a,0)$ and $B=(0,b)$ with $a,b\ne 0$. (The exceptional cases of $a=b=0$ or a line parallel to the $x$-axis [but not to the $y$-axis, why?]) are left as an exercise.
Let $P$ be any point on that line. And let $Q=(x,0)$ be the point of intersection of the $x$-axis with the line parallel to the $y$-axis through $P$. Let $R=(0,y)$ be the point of intersection of the $y$-axis with the line parallel to the $x$-axis through $P$. Then by definition, $P$ has coordinates $(x,y)$.
By the intercept theorem (once with center $A$, once with center $B$)
$$ \frac{x-a}{a}=\frac{AQ}{OA} = \frac{AP}{BA}=\frac{OR}{BO}=\frac{y}{-b}$$
and hence 
$$ y = m x+b $$
with $m:=-\frac ba$.
The converse (i.e. that $y=mx+b$ descibes a line) is easily verified with the converse of the intercept theorem.
A: I don't know if this is correct or rigorous enough (I guess it is not) but, every function $y = y(x)$ defined over $x \in \mathbb{R}$ is a straight line in the $(x,y)$ plane if and only if its curvature is zero (see MJD's comment below):
$$y''(x) = 0, \quad x \in \mathbb{R},$$
then it must hold:
$$y(x) = a  x + b,$$
for $a$ and $b$ constants of integration. 
Cheers!
A: I think your questions has been adequately answered by Dmoreno's post elsewhere in this thread, but I would like to generalize it a little bit.  You asked about functions where $y = y(x)$, which includes lines that are not vertical.  It is possible to generalize this situation.
Let $t$ be some parameter and let $x=x(t), y=y(t)$ be functions of $t$.  We can plot the set of all points $(x(t), y(t))$ for all $t$ in some range, say for all real $t$.  This gives us a curve of some sort.  The case where $y$ depends directly on $x$ is included as a special case of this situation, where $x(t)$ is the identity function $t$; then we are plotting $(t, y(t))$.  But the parametric case also includes the special case where $x(t)$ is a constant, in which case the graph is a vertical line, as well as many other curves where $y$ is not a function of $x$.  For example, the graph of $x(t) = \cos t, y(t) = \sin t$ is a circle, which is not the graph of any function $y(x)$.
We can compute the curvature of the graph for each $t$; it comes out to $$\kappa(t) = \frac{\left\lvert\dot x\ddot y - \dot y \ddot x\right\rvert}{\left(\dot x^2 + \dot y^2\right)^{3/2}}$$
where $\dot x$ and $\ddot x$ are the first and second derivatives of $x(t)$ with respect to $t$, and similarly $\dot y$ and $\ddot y$.
We would like to know when $\kappa(t)$ is identically zero.  Assuming that the denominator never vanishes (which can only occur when $\dot x(t) = \dot y(t) = 0$, which for smooth curves we can avoid by changing the parameter) we want $$\dot x\ddot y - \dot y \ddot x = 0.$$
If $\dot x = 0$ then $y$ can be anything, but then $x(t)$ is constant and the graph is  vertical line.  Similarly if $\dot y=0$ we have a horizontal line.  So suppose neither $\dot x$ nor $\dot y$ is identically zero.  Then we can divide by $\dot x \dot y$ to get:
$$\frac{\ddot x}{\dot x} - \frac{\ddot y}{\dot y} = 0.$$
Integrating both sides with respect to $t$ gives $$\ln{\dot x} - \ln{\dot y} = C\\ \ln\left(\frac{\dot x}{\dot y}\right) = C\\ \dot x = a\dot y$$
  where $a = e^C$.
Integrating again, we have $$x = ay+b$$ as the only solution when neither $\dot x$ nor $\dot y$ is zero.  So the only solutions are lines.
