Does the Taylor expansion and approximation centered about a point become more accurate at the point as more terms are used? I may have done a poor job of asking this question.  You can use the sine function as the Taylor expansion as an example. 
As you add more and more terms I can see how the Taylor representation of this function becomes more and more like the sine function!  I am OK with that. What I am asking is at the point 0 is the graph any more accurate past the 2nd term?
In other words assuming you are only interested in what the function looked like at the point 0 would an approximation with let us say using a hundred terms be more any more accurate than just using the first few terms.  Remember I am only interested at the point 0.  
Or is it the case that no matter what point I pick the line through that point becomes more and more accurate as the terms increase?
The reason I ask this is because I am trying to understand why the action in the principle of stationary action uses only the first order terms and the higher  order terms vanish!  This happens when the Lagrangian is expanded.  
All derivation examples end up using the Taylor series to write the expression and then they ignore the higher order terms.  
 A: Yes, your approximation becomes more accurate (or at least it does not become less accurate, if $f$ is already a polynomial in the first place).
There are many variants of Taylor's theorem, but in essence we have that if $P_{a, \,k}(x)$ is the $k$-th order Taylor polynomial about $a$, then the error term
$$\epsilon(x) = f(x) - P_{a, \,k}(x)$$
is $o\left({|x-a|}^k\right)$ as $x\to a$.
In other words, by increasing $k$ you ensure the error $\epsilon$ approaches $0$ faster as $x\to a$.
Taylor's theorem also assures us that $P_{a,\,k}$ is unique, in the following sense:

If $p$ is a $k$-th order polynomial and there is some $h:\Bbb R\longrightarrow \Bbb R$ with $\lim_{x\to a}\,h(x) = 0 $ and
$$f(x) = p(x) + h(x)\,{(x-a)}^k,$$
then $p=P_{a,\, k}$.

A: If $f(x)$ is expanded into a power series about $x = 0$, $$f(x) = \sum_{n=0}^{\infty} c_n x^n,$$ and hence $$f(0) = \sum_{n=0}^\infty c_n (0^n) = c_0.$$ Thus, if $f$ represented by power series centered at $0,$ then $f(0)$ is exactly the first coefficient of this expansion. So no terms beyond the first term are necessary to give the value of $f$ at $0$. Furthermore, near $0$, all terms of the form $c_nx^n$ for $n > 0$ vanish quickly. As $n$ increases, the speed the term vanishes near $0$ becomes faster, and low-order terms give a close approximation. This is why higher order terms are dropped in derivations in physics: in the limit as $x \rightarrow 0$, they vanish.
A: If the Taylor series of $f(x)$ about $x=a$ converges to the function, then we have 
$$ f(x)= f(a)+ f'(a)(x-a) + \frac {f''(a)}{2} (x-a)^2 +.....$$
As you see the first term $f(a)$ is a constant whose graph is a horizontal line.
The first two terms $f(a) +f'(a)(x-a)$ is a line with the same value and the sme slope as your function at $x=a$  
When you add the third term you get a parabola  with the same value, the same slope and the same curvature as $f(x)$ at $x=a$
Thus the more terms that you add, the more details you get out of your graph.
For most functions you do not have to go too far to get a reasonable graph which is very similar to the actual graph about $x=a$ 
