If I add a linear function to a nonlinear function, what happens?

It depends on the function. For quadratic functions the real effect of adding a linear term is expressing a new equation which is the expanded form of some square of a modification to x. Take your x^2 + 2x for example, which can be expressed as:

y = x^2 + 2x + 0

y = x^2 + 2x + 1 - 1

y = (x + 1)^2 - 1

As you can see, the only real change is a shift down and left in the graph of x^2. This is also seen in the derivative of the new function, which is 2x + 2. The derivative is exactly the same, and so must be the function that it's a derivative of - it's just a shift in position.

If you take cosine for instance though and add a linear term ax to it, you will get a graph that oscillates around y = ax. The residuals of those two graphs (y = coax + ax, and y = ax) will be the same as y = cosx and y = 0 (which makes sense when you think about it). The graph doesn't necessarily just shift the same as it does for a quadratic, however.

Edit: Well, the derivatives aren't really the same; the cyclical behavior of trigonometric functions doesn't allow such a broad statement based on derivatives to be made; the argument pertaining to the way the function is expressed though, and the additions to the y and x variables, should be enough to show that the effect is just a shift in position.

Edit2: I think the best way to test which will effectively result in a shift would be to take the old form and simply adjust one of the variables and isolate the old equation again to see what you have left. With quadratics this works:

y = x^2

into:

y = (x+a)^2 + c

y = x^2 + 2ax + a^2 + c

[y = x^2] + Kx + C

So, a shift in a linear term.

With cubics this does not happen:

y = x^3

y = (x + a)^3 + c

y = x^3 + 3ax^2 + 3xa^2 + a^3 + c

[y = x^3] + Kx^2 + Mx + C

so the only way that could come about (the shift, I mean) is through the addition of a quadratic. I assume this is the pattern as you increase in power - to shift you have to add a function a power less.

With exponentials, it's different too:

y = 2^x

y = 2^(x + a) + c

y = 2^x*(2^a) + c

[y = 2^x]*K + c

So to obtain a shift you must multiply by some scalar (this seems to work when I graph the results in Excel).

The general effect will always be, though, that when you add a linear term the graph will be slanted with the incline of the line defined by the linear term you added. That's almost stupidly tautological though.

I would think that it is just adding a constant to the parabola (intercept shift). Think about it as a linear regression. y=b0+b1x^2 is a parabola that crosses y at b1x^2 if b0=0. If b0=2x, it shifts the intercept but doesn't affect the slope or the shape of the function.