I have recently had a thought to construct a parabola by zooming in all the way on a pixel grid, then drawing it in rectangles and zooming out afterwards.
For example, we start by drawing $1\times1$ box in the center of the coordinate system. Then lets make the second stack of blocks bigger by $+1$ on both sides and join them next to the previous block, and repeat. Then join the outer points of the stacks to get a parabola. Image:
We can work out the formula for this parabola.
The $x$s of the points are $0+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}, 3+\frac{1}{2}, 4+\frac{1}{2} \dots$
The $y$s of the points are $-1+\frac{1}{2}, 0+\frac{1}{2}, 2+\frac{1}{2}, 5+\frac{1}{2},9+\frac{1}{2}\dots$
Meaning the $x$ coordinates have a sequence formula $n-\frac{1}{2}$
And the $y$ coordinates have a sequence formula $\frac{n(n-1)}{2}-\frac{1}{2}$
From that we can work out that $f(x) = \frac{1}{2}x^2-\frac{5}{8}$
But what if our stacks are the prime numbers?
First stack centered at $(0,0)$ then is $1\times2$ followed by $1\times3, 1\times5,1\times7,1\times11 \dots$
Since we can't have a sequence that will produce primes, can we at least estimate the $f(x)$ for such a "parabola" for all primes?
Or how can we find such a function that'll be correct for at least first $n$ primes?
"parabola" $\to$ I'm thinking that the construction won't yield a parabola, but a curve that looks similarly to a parabola.


As you noted, there is no simple precise formula. But there are estimates.
For example:
(For sources and more precise versions of these formulas, see https://en.wikipedia.org/wiki/Prime_number_theorem)
So you may take something like $f(x)=\frac{x^2}{2}\ln(x)$, which is indeed similar to a parabola.