I have a question to ask, but before I do, let me just tell you that this might be a trivial - or even downright silly question. I only have a basic understanding of math, so please do try to keep that in mind when answering. Thank you so much.
So, essentially I have the following formula:
$$150 \cdot \sqrt{\frac{x}{10^{15}}}$$
Which would result in the following graph:
https://www.desmos.com/calculator/rfzjsgudsb
Now, what I'm trying to understand is, that with such a graph, can I determine which is the point before the curve becomes very flat? Given the above linked graph it would seem to me like somewhere between 5,000 - 15,000 but I'm not entirely sure.
Additionally, I know that the x value will be incremented at a constant rate - and given that, can I, knowing the rate of increase, calculate this 'ideal point'?
Apologies if my question is not so well-put together, and thank you all for your time.
Your question is indeed not well-formed; it really depends on what you mean by "very flat". As you've probably observed, the function increases without bound as $x\to\infty$, so the curve never becomes actually horizontal. However, the slope of the graph is continually (and continuously) decreasing towards zero. You can use calculus to determine the slope at any point: if $f(x) = 150\sqrt{x/10^{15}}$, then \begin{equation*} f'(x) = \frac{75}{10^{7.5}\sqrt{x}}, \end{equation*} so if you want to know, say, where the slope becomes less than $1$, you need only solve $f'(x) = 1$ for $x$.