A silly question on range of a quadratic function. Let $f:\mathbb{R}\to\mathbb{R}\space|\space f(x) = x^2 + 3x + 2 \space\forall\space x 
\in\mathbb{R}$
We are asked to find the range of the function $f$
I begin as follows:
Assume $y=x^2 + 3x + 2$ for some $ x\in\mathbb{R}$
Collecting terms to one side,
$x^2 + 3x + 2-y=0$
Now as x is real, the discriminant of the above quadratic expression must be greater than or equal to zero. Hence,
$3^2 - 4(1)(2-y)\geqslant0$
Solving, we get,
$y\geqslant-1/4$
Hence the range of the function $f$ is the set $A=\{x|x\geq-1/4\}$
This answer is correct. The range of the function $f$ is indeed the set $A=\{x|x\geq-1/4\}$.
However, I have a problem with my last step. The final equality only tells me that $y$ is greater than OR equal to $-1/4$. It doesn't exactly say that $y$ WILL take all values greater than AND equal to $-1/4$. Is there a way to prove that $y$ will take all such values?
Any hint or help regarding this will be appreciated. Thank you

I would write this as $y = (x+\frac32)^2-\frac14$, from which it is obvious what the range would be, as the square term takes all nonnegative values.
