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 
$f(3)=12+3b=3 \implies b=-3$
$f(x) = x^2-3x+3$
$f(1) = 1-3+3 = 1$

Suppose $f$ has degree $n$, then we can write $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$$
We are given that $f''(x)\to 2$ as $x\to\infty$, so what could $n$ possibly be? 
Suppose $n>2$, then $\deg(f''(x))=n-2>0$, so $f''(x)\to\pm\infty$ as $x\to\infty$.
Suppose $n<2$, then $f''(x)\equiv 0$ , so clearly $f''(x)\to 0\neq 2$ as $x\to\infty$.
This should get you pretty close to a solution.
