How to find the integral value of $a$ for which $f(x) = x^2 - 6ax + 3 - 2a + 9a^2$ is surjective

Question

Let $f:\mathbb{R} \to [1, \infty)$ be defined by $f(x)=x^2-6ax+3-2a+9a^2$. The integral value of $a$ for which $f(x)$ is surjective is equal to

I tried putting $f(x)=1$. Is this the right approach?

Answer 1

$f(x) = (x-3a)^2+3-2a\Rightarrow f_\text{min}= f(3a) = 1\Rightarrow 3-2a = 1\Rightarrow a = 1$.

Answer 2

As it is quadratic equation and its range defines from [-D/4a,infinity] (a>0 in equation) so we can find D in from this equation and can equate to 1 because it is given that it is surjective .(as range is 1 to infinity given) -D/4a=-(36a^2-4(3-2a+9a^2))/4*1 =12-8a/4 1=12-8a/4 1=a