How do you find the parabola that best fits a function near a point?

Write the parabola as $q(x)=A+B(x-a)+C(x-a)^2$. At $x=a$ the linear and squared terms vanish, so $q(a)=A$, which forces $A=f(a)$. Differentiate once: $q'(x)=B+2C(x-a)$, so $q'(a)=B$, which forces $B=f'(a)$. Differentiate twice: $q''(x)=2C$, so $q''(a)=2C$. Matching the function's second rate gives $2C=f''(a)$, hence $C=f''(a)/2$. The half is not a convention; it is what makes the parabola's own second rate come out equal to $f''(a)$. Each condition also leaves the earlier ones intact, because $(x-a)^2$ and its first rate vanish at $a$.

Take $f(x)=e^x$ at $a=0$. Its value, first rate and second rate are all 1 at 0, so the matching line is $1+x$ and the matching parabola is $1+x+x^2/2$. At distance $d=x$ from 0 the true value is $e^d$, the line gives $1+d$, and the parabola gives $1+d+d^2/2$.

The natural guess is that any parabola through $(a,f(a))$ will do, or that a parabola can be chosen by eye once it touches the curve. Passing through the point is only one condition, and infinitely many parabolas satisfy it: in $q(x)=A+B(x-a)+C(x-a)^2$, $A$ is fixed but $B$ and $C$ are still free. Matching the slope fixes $B$, leaving a one-parameter family of parabolas that all touch the curve with the right direction but have different bending. Only the second-rate condition fixes $C$, and it fixes it as $f''(a)/2$. A parabola with any other $C$ still passes through the point and still has the right slope, but it bends differently, so its error near the point is of the same size as the distance squared rather than the distance cubed.

The error that remains has a sharp order. For $e^d$ at 0 the first omitted term is $d^3/6$, because the third rate of $e^x$ at 0 is 1 and $3!=6$, so the parabola's error is close to $d^3/6$: at $d=0.1$ that predicts 0.0001667 against the actual 0.000171, and at $d=0.5$ it predicts 0.020833 against 0.023721. The line's first omitted term is $d^2/2$, which predicts 0.005 at $d=0.1$ and 0.125 at $d=0.5$. The same factorial pattern continues: the cubed term has 6 underneath, the fourth has 24. This is why adding the squared term buys so much at small distances and why the gain is local.