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
 A: 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.
A: In your solution, you show that $x$ has a real value if $y\geq -1/4$ using discriminant of a quadratic. Well this works the other way too! If $y\geq -1/4$, then the quadratic has nonnegative discriminant and so you will always find $x$ that give that specific value of $y$.
Hope this helps. :)
A: Your calculation is correct and complete if you write it like this:
$$
\begin{align}
 &\text{$y$ is in the range of $f$} \\
\iff &x^2 + 3x + 2-y=0 \text{ for some } x \in \Bbb R \\
\iff &\text{the discriminant of $x^2 + 3x + 2-y$ is non-negative} \\
\iff &y \ge -1/4 \, .
\end{align}
$$
A: I think the OP’s doubt is that “f(x)$\geq\frac 14$“ states that any values taken by f(x) are greater than or equal to $\frac 14$, but not that f(x) takes all values greater than or equal to $\frac 14$.

Note that $y=f(x)$ is continuous and differentiable on all points in $\mathbb R$.
The minimum value of a quadratic function $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}$$ occurs at $\displaystyle x=-\frac{b}{2a}$ so there exists some x such that the minimum occurs.
Now, take an interval $\left[-\frac{b}{2a}, q\right]$ where $q\gt-\frac{b}{2a} $ is some real number.
Then, by the Intermediate Value Theorem, the function is guaranteed to take on all the values  between $\displaystyle -\frac{b}{2a}$ and $f(q)$. Notice that f(x) is increasing in $\left(-\frac{b}{2a},\infty\right)$ so $\displaystyle f(q)> -\frac{b}{2a}$. You can keep increasing q indefinitely and so we can conclude that f(x) assumes all values greater than and including $-\frac{b}{2a}$.
(P.S. a similar exercise can be carried out using $q’< -\frac{b}{2a}$.)
Thus, we have proved that the function is guaranteed to take on all the values  greater than and including its minimum.
