Let $f(x)$ be a quadratic expression which is positive for all real values of $x.$ If $g(x)=f(x)+f'(x)+f''(x)$ 
Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f'(x)+f''(x)$, then for any real $x$,
  $$(A)\, g(x)<0,\quad (B)\,g(x)>0,\quad (C)\, g(x)=0,\quad (D)\;g(x)\ge 0.$$

My work so far. 
Let $f(x)=ax^2+bx+c$ be the quadratic polynomial. Then
$b^2-4ac<0$ and
$$g(x)=ax^2+x(b+2a)+(c+b+2a).$$
Now I am stuck. I am not able to pick the right answer.

$f(x)=ax^2+bx+c$. $f>0 \Rightarrow a>0$.
$f(x)+f^\prime(x)+f^{\prime\prime}(x)=ax^2+(2a+b)x+c+b+2a=g(x)$
Reformulating $g$ in terms of $x+1$ gives 
$$a(x+1)^2+ (2a+b)(x+1)+c+b+2a-(2ax+a)-(2a+b) = $$
$$a(x+1)^2+ (2a+b)(x+1)+c-2a(x+1)+a = $$
$$a(x+1)^2+b(x+1)+c+a=g(x)$$.
So $g(x) = f(x+1)+a$, so $g(x)$ is $f(x)$ translated by 1 to the left and by $a$ upwards.
$f>0 \Longrightarrow g>0$
