Well, this is just the Taylor expansion of the exponential exp(x) around x=0:

f(x)=f'(x)=f''(x)=exp(x), so f(0)=f'(0)=f''(0)=1.

Now g(x)=ax²+bx+c, g'(x)=2ax+b, g''(x)=2a,

so g(0)=f(0) =>c=1

g'(0)=f'(0)=> b=1

g''(0)=f''(0)=> a=1/2,

and g(x)= 1+x+x²/2.