Successive approximations of a function (taylor formula derivation) I am reading about the ways of function approximation which led to Taylor's formula.
I understand that in a very close neighbourhood of the point a we can approximate f(x) by f(a). However this is very rough approximation and just by looking at the function graph we see that a tangent line can do a much better job of approximation. So we can find the tangent line formula by calculating the derivative of f(x) and rearranging terms a bit:
$$\frac{f(x)-f(a)}{x-a} = f'(a) $$
thus
$$ y = f(a) + f'(a)(x-a)$$
so far so good but just by looking at the graph of the tangent it's obvious that this approximation will get worst and worst the further away x is from a. 
So now what can we do next to approximate f(x) even better? Author of the book I read suddenly, without any derivation shows this expression:
$$\frac{f(x) - [f(a) + f'(a)(x-a)]}{(x-a)^2}$$
It closely resembles second derivative but it isn't that. I am especially confused about the denominator $$(x-a)^2$$ what is its origin? How can we prove / explain it geometrically, base on the function graph I linked below?
My explanation of the next step on a way to better approximation after the tangent line is we want the tangent to travel the graph according to our delegation from a described by factor $$(x-a)$$ Is it correct thinking?
Function and its "tangential" approximation

We have that 
$$
f(x) = f(a) + f'(a)\left( {x - a} \right) + {1 \over 2}f''(a)\left( {x - a} \right)^{\,2}  + O\left( {\left( {x - a} \right)^{\,3} } \right)
$$
so that we can write
$$
{1 \over 2}f''(a) \approx {{f(x) - f(a) - f'(a)\left( {x - a} \right)} \over {\left( {x - a} \right)^{\,2} }}
$$
i.e., (probably ) they are telling you that, once you have $f(a), f'(a)$ you can estimate $1/2f''(a)$ in that way instead of directly differentiate twice.
