Consider the real-valued function of a real variable $f(x) = \sqrt {ax^2 + bx + c}$ with $a$, $b$ and $c$ given and $a>0$.
When $\Delta = 0$, the function is equal to $|\sqrt{a} (x+\frac{b}{2a})|$. In other words, it is equal to the absolute value of some linear function. But when $\Delta \neq 0$ the function does not behave quite so nicely. Then, $\sqrt {ax^2 + bx + c}$ is not equal to $|\sqrt{a} (x+\frac{b}{2a})|$. However, when $|x|$ is very large, it is approximately equal to it regardless of how large you $|\Delta|$ is. In other words it "approaches" linearity - the graph of $f(x)$ can be made arbitrarily close to the graph of $|\sqrt{a} (x+\frac{b}{2a})|$ by making $x$ sufficiently large.
My question is, simply, why does this happen?
As $|x| \to \infty$ with $a, b, c$ constant and $a > 0$: $$a x^2 + b x + c = a \left(x + \dfrac{b}{2a}\right)^2 + O(1) = a \left(x + \dfrac{b}{2a}\right)^2 (1 + O(1/x^2))$$ so $$\sqrt{a x^2 + bx + c} = \pm \sqrt{a} \left(x + \dfrac{b}{2a}\right) (1 + O(1/x^2)) = \pm \sqrt{a} \left(x + \dfrac{b}{2a}\right) + O(1/x)$$