How is the Taylor expansion for $f(x + h)$ derived?

It looks as if the notation you are accustomed to for the Taylor expansion is something like $$f(x)\approx f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+ \frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots.$$

Now write $x$ instead of $x_0$, and $x\pm h$ instead of $x$.

This has already been answered, but I have seen some of the comments to the answer from Andre Nicolas and I thought I could make it clear with a step by step explanation.

Instead of just doing a substitution, as Andre mentioned, think about the meaning of it. Start with the standard Taylor series expansion, $$ f(x) \approx f(x_0)+f′(x_0)(x−x_0)+\frac{f′′(x_0)}{2!}(x−x_0)^2+\frac{f′′′(x_0)}{3!}(x−x_0)^3+⋯. \qquad (*)$$ Now what does $x-x_0$ mean? For convergence, we usually need this to be small, so we can call this $h$. Now substitute $x-x_0=h$ (and obviously $x=x_0+h$) into $(*)$ to get: $$ f(x_0+h) \approx f(x_0)+f′(x_0)h+\frac{f′′(x_0)}{2!}h^2+\frac{f′′′(x_0)}{3!}h^3+⋯.$$ This is in the required form but with $x_0$ instead of $x$ and since this is just a variable we can just substitute it for $x$.