$f(x)=2x^3+7x^2+x-6, a=2, f(2)=16+28+2-6=40\\ f'(x)=6x^2+14x+1, f'(2)=24+28+1=53\\ f''(x)=12x+14,f''(2)=24+14=38\\ f'''(x)=12\\ f(x)= f(2)+(x-2)f'(2)+\dfrac{(x-2)^2}{2!}.f''(2)+\dfrac{(x-2)^3}{3!}.f'''(2) \\=40+(x-2)53+\dfrac{(x-2)^2}{2!}.38+\dfrac{(x-2)^3}{3!}.12\\ =40+(x-2)53+{(x-2)^2}.19+{(x-2)^3}.2 $

$f(x)=53(x-2)+19(x-2)^2+2(x-3)^3+40\\ Thus \ f(x) is \ expanded \ in \ terms\ of \ (x-2)$