<s>[INST] How to change the value of $a$ in a Taylor series yet still get the correct approximation? Set up
For example, if $f(x) = x^3 + 4x + 6$, then the second degree Taylor polynomial
is
$$ 
P_2(a) = (a^3 + 4a + 6) + (3a + 4)( x - a) + \left(\frac{6a}{2!}(x - a)^2\right)
$$ 
If I want to approximate $f(1.2)$ then I can use $P_2 (1.2)$. 
You can see that this is close by graphing the functions, and noting that about
$f(1.2)$ they're very close.

Question
From other questions I get the idea that I'm able to change the value of $a$ ,
by setting it to zero or whatever, and still approximate the function as needed. 
But looking at this example, I don't know how I could change the value of $a$ ,
and still approximate the original function at $x = 1.2$.
Sometimes people say that you're able to change the value which the function is being evaluated about, but I don't understand how this is so. 

definitions
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation} 
  \begin{aligned}
    P_n(x) = f(a)
    + \frac{f'(a)}{1!}(x - a)
    + \frac{f''(a)}{2!}(x - a)^2
    & + \ldots
    + \frac{f^{(n)}(a)}{n!}(x - a)^n \\
  \end{aligned}
\end{equation}
And $R_n (x) $ is  (where $\xi$ is between $a$ and $x$)
\begin{equation}
  \begin{aligned}
    R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
  \end{aligned}
\end{equation}
 [/INST] I think you've gotten a little confused between your $x$ and your $a$. The Taylor series is a function of $x$, parameterised by $a$. So, for example, the 2nd-order Taylor polynomial of your cubic is:
$$T_{2;a}(x) = (a^3 + 4a + 6) + (x - a)(3a^2 + 4) + 6a\frac{(x - a)^2}{2}$$
If we evaluate this series at $a = 0$, we get the (linear) polynomial:
$$\begin{eqnarray}T_{2;0}(x) & = & (0^3 + 4\cdot0 + 6) + (x - 0)(3\cdot0^2 + 4) + 6\cdot 0 \frac{(x - a)^2}{2} \\
& = & 6 + 4x\end{eqnarray}$$
Using this function, the approximation for $f(1.2)$ is $T_{2;0}(1.2) = 6 + 4\cdot 1.2 = 10.8$.
If we evaluate the series at $a = 1$, we get the polynomial:
$$\begin{eqnarray}T_{2;1}(x) & = & (1^3 + 4 \cdot 1 + 6) + (x - 1)(3 \cdot 1^2 + 4) + 6 \cdot 1 \frac{(x - 1)^2}{2} \\
& = & 11 + 7(x - 1) + 3(x - 1)^2 \\
& = & 7 + x + 3x^2\end{eqnarray}$$
Using this function, the approximation for $f(1.2)$ is $T_{2;1}(1.2) = 11 + 7(0.2) + 3(0.2^2) = 12.52$. Note that this is a lot closer to the true value of $f(1.2) = 12.528$.
If we take $a = 1.2$, then we'll get a Taylor polynomial that is exactly right for $f(1.2)$, but whose approximation gets worse further away from it.
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