By definition of the Taylor polynomials, T 2 ( x ) = ∑ i = 0 2 f ( i ) ( a ) i ! ( x − a ) i = f ( 0 ) ( a ) 0 ! ( x − a ) 0 + f ( 1 ) ( a ) 1 ! ( x − a ) 1 + f ( 2 ) ( a ) 2 ! ( x − a ) 2 = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ( x − a ) 2 {\displaystyle {\begin{aligned}T_{2}(x)&=\sum _{i=0}^{2}{\frac {f^{(i)}(a)}{i!}}(x-a)^{i}\\&={\frac {f^{(0)}(a)}{0!}}(x-a)^{0}+{\frac {f^{(1)}(a)}{1!}}(x-a)^{1}+{\frac {f^{(2)}(a)}{2!}}(x-a)^{2}\\&=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}\end{aligned}}}
Computing the values of the derivatives of f {\displaystyle f} at 1, we find that f ( 1 ) = 4 , f ′ ( 1 ) = 12 , f ″ ( 1 ) = 24 {\displaystyle f(1)=4,\,f'(1)=12,\,f''(1)=24} . Therefore T 2 ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) 2 ( x − a ) 2 = 4 + 12 ( x − 1 ) + 12 ( x − 1 ) 2 = 12 x 2 − 12 x + 4 {\displaystyle {\begin{aligned}T_{2}(x)&=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2}}(x-a)^{2}\\&={\color {blue}4+12(x-1)+12(x-1)^{2}}\\&={\color {blue}12x^{2}-12x+4}\end{aligned}}}
