# 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?

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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.
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, $3(-x)+1\neq -3x-1$.