Can anyone help solve this taylor series?

Question

Can someone help find the taylor series for 1+x+x^2, a=2?

So far i have:

f(x)=1+x+x^2
f'=1+2x
f''=2
f'''=0

f(2)=7
f'(2)=5
f''(2)=2 ?
f'''(2)=0 ?

7+5+2+0

and... this is where I'm stuck. I don't know any function that correlates with this.

Answer 1

Now plug into the definition for Taylor Series:
f(x) = f(a) + (1/1!)f'(a)(x - a) + (1/2!)f''(a)(x - a)^2 + (1/3!)f''(a)(x - a)^3 + . . .
f(x) = f(2) + (1/1!)f'(2)(x - 2) + (1/2!)f''(2)(x - 2)^2 + (1/3!)f''(2)(x - 2)^3 + . . .
f(x) = 7 + (5)(x - 2) + (1/2)(2)(x - 2)^2 + (1/6!)'(0)(x - 2)^3 + . . .
f(x) = 7 + 5(x - 2) + (x - 2)^2

Expanding, you have
f(x) = 7 + 5x - 10 + x^2 - 4x + 4
f(x) = 1 + x + x^2

Answer 2

The taylor series at a=2 is
f(x) =f(2) +(x-2)f´(2 )+(x-2)^2 f´´(2)/2!
so f(2)=7
f´(2) =5
f´´(2) =2 so
f(x) = 7+5(x-2) +(x-2)^2