Question

Prove that if f is a quadratic polynomial function $f\left(x\right)=a{x}^2+bx+c$then the coefficient of f are completely determined by the values of f(x) and its derivatives at x=0 as follows

$a=\frac{f"\left(0\right)}{2};b=f\text{'}\left(0\right);c=f\left(0\right)$

Short answer

We determine the values of a, ,b, c in terms of f and its derivatives at x=0

Step-by-step solution

Given information

We are given a quadratic function$f\left(x\right)=a{x}^2+bx+c$

Differentiate the function

We have,

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\text{'}\left(x\right)=2ax+b(1stderivative) \\ f\text{'}\text{'}\left(x\right)=2a(2ndderivative) \end{gathered}$$

Compute the function and its derivative at x=0

We get,

$$\begin{gathered} f\left(0\right)=a\left(0\right)+b\left(0\right)+c \\ c=f\left(0\right)\left(1\right) \\ f\text{'}\left(0\right)=2a\left(0\right)+b \\ b=f\text{'}\left(0\right)\left(2\right) \\ f\text{'}\text{'}\left(0\right)=2a \\ a=\frac{f\text{'}\text{'}\left(0\right)}{2}\left(3\right) \end{gathered}$$

We determine the values of a, ,b, c in terms of f and its derivatives at x=0