Polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$ given some values what is $f(1)$? Let $f$ be a polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$, the minimum of f is attained at $3$, and $f(0)=3$, Then $f(1)$ equals. 
$(A) \ 1$
$(B) \ 2$
$(C) -1$
$(D) -2$
I am not sure how to deal with $f''(x) \rightarrow2$ as $x\rightarrow\infty$ . Give me a hint to try. 
EDIT : Work after hint 
Let $f(x)=ax^2+bx +c  $
$f''(x)=2\implies 2=2a \implies a=1 $
$f(0)=3=c$
$f(x)=x^2+bx +3  $
Using the fact that minima attained at 3 we have 
$f(3)=12+3b=3 \implies b=-3$
$f(x) = x^2-3x+3$
$f(1) = 1-3+3 = 1$

Dave showed that the degree of the polynomial is $2$. So we have $f(x)=a_2x^2+a_1x+a_0$.
We know that $f''(x)=2a_2=2\implies a_2=1\implies f(x)=x^2+a_1x+a_0$
We also know that there is a minima at $x=3\implies f'(3)=0\implies f'(3)=2\overbrace x^3+a_1=6+a_1=0\implies a_1=-6$ hence we have $f(x)=x^2-6x+a_0$.
We also know that $f(0)=3$ thus $f(0)={\overbrace x^0}^2-6\overbrace x^0+a_0=a_0=3$
And the final answer is $f(x)=x^2-6x+3$, with this we get $f(1)=1^2-6\times 1+3=\boxed{-2}$
