Why does this work in this type of function? I was just thinking about this: In a function like y = 3x you can find the correct slope even if you were to add two coordinates together.
Example: Find the slope by using x = 2 and x = 4 So, correct way is (12-6) / (4-2) which equals a slope of 3. 
But you will always get correct slope even if you were to add as follows: (12+6) / (4+2) = 3.
However this doesn't work in a function like y = 3x+1 (and most other functions I assume) So can someone clarify why it always works in the first function? 
 A: The equation $y=a x$ imposes a multiplicative relation on $x$ and $y$ that's uniquely compatible with the divisive computation of slope.
Of course, given two points $(x_1,y_1)$ and $(x_2,y_2)$ on any line, we define 
$$\text{slope} = \frac{\text{change in } y}{\text{change in }x} = \frac{y_1-y_2}{x_1-x_2}$$
For points on a line defined by $y=ax$ (which, note, is specifically a line through the origin), 
$$y_1 = a x_1 \qquad\text{and}\qquad y_2 = a x_2$$
so that
$$\text{slope} = \frac{a x_1 - a  x_2}{x_1 - x_2} = \frac{a(x_1-x_2)}{x_1-x_2}= a$$
Also, as you observe, you can add coordinates:
$$\frac{y_1+y_2}{x_1+x_2} = \frac{ax_1 + ax_2}{x_1+x_2}= \frac{a(x_1+x_2)}{x_1+x_2} = a$$
But, even more generally, you can take any "linear combination" of the coordinates:
$$\frac{Py_1+Qy_2}{Px_1+Qx_2} = \frac{Pax_1+Qax_2}{Px_1+Qx_2} = \frac{a(Px_1+Qx_2)}{Px_1+Qx_2} = a$$
The point here is that, because $y$-values are just multiples of the $x$-values, we can factor-out the multiplier in the numerator, and then cancel the denominator. The simple multiplicative nature of $y=ax$ just happens to get un-done by the divisive definition of slope. It's a nice property (and part of why we like when quantities are directly proportional), but it's too nice to expect to hold beyond this special case. 
A: It's because the function is odd.  It means that $f(-x)=-f(x)$.  For example $3(-x)=-3x$.  So, if $(a,b)$ is a point on the graph, so is $(-a,-b)$.  But, $3x+1$ is not odd, $3(-x)+1\neq -3x-1$.
