Looking for a function that approximates a parabola

I would suggest just approximating the $\max(0,\cdot)$ function, and then using that to implement $\max(0,1-x^2)$. This is a very well-studied problem, since $\max(0,\cdot)$ is the relu function which is currently ubiquitous in machine learning applications. One possibility is $$ \max(0,y) \approx \mu(w)(y) = \frac{y}2 + \sqrt{\frac{y^2}4 + w} $$

One way of deriving this formula: it's the positive-range inverse of $x\mapsto x-\tfrac{w}x$.

Then, composed with the quadratic, this looks thus:

Notice how unlike Calvin Khor's suggestions, this avoids ever going negative, and is easier to adopt to other parabolas.

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Nathaniel remarks that this approach does not preserve the height of the peak. I don't know if that matters at all – but if it does, the simplest fix is to just rescale the function with a constant factor. That requires however knowing the actual maximum of the parabola itself (in my case, 1).

To get an even better match to the original peak, you can define a version of $\mu$ whose first two Taylor coefficients around 1 are both 1 (i.e. like the identity), by rescaling both the result and the input (this exploits the chain rule): $$\begin{align} \mu_{1(1)}(w,y) :=& \frac{\mu(w,y)}{\mu(1)} \\ \mu_{2(1)}(w,y) :=& \mu_{1(1)}\left(w, 1 + \frac{y - 1}{\mu'_{1(1)}(w,1)}\right) \end{align}$$

And with that, this is your result:

The nice thing about this is that Taylor expansion around 0 will give you back the exact original (un-restricted) parabola.

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Source code (Haskell with dynamic-plot):

import Graphics.Dynamic.Plot.R2
import Text.Printf

μ₁₁, μ₁₁', μ₂₁ :: Double -> Double -> Double
μ₁₁ w y = (y + sqrt (y^2 + 4*w))
             / (1 + sqrt(1 + 4*w))
μ₁₁' w y = (1 + y / sqrt (y^2 + 4*w))
             / (1 + sqrt(1 + 4*w))
μ₂₁ w y = μ₁₁ w $ 1 + (y-1) / μ₁₁' w 1

q :: Double -> Double
q x = 1 - x^2

main :: IO ()
main = do
 plotWindow
  [ plotLatest [ plot [ legendName (printf "μ₂₍₁₎(w,q(x))")
                         . continFnPlot $ μ₂₁ w . q
                      , legendName (printf "w = %.2g" w) mempty
                      ]
               | w <- (^1500).recip<$>[1,1+3e-5..]]
  , legendName "max(0,q x)" . continFnPlot
       $ max 0 . q, xAxisLabel "x"
  , yInterval (0,1.5)
  , xInterval (-1.3,1.3) ]
 return ()

Here is a family of real analytic functions that does the job; making the parameter $N=1,2,3,4,\dots$ larger makes the approximation better. Lets just say the quadratic is $q(x) = 1-x^2$, so the function is $f(x)=\max(0,q(x))$. Its easy to do a more general case. Then set $$f_N(x) = \frac1{1+x^{2N}}q(x).$$ This works because for $N$ very big, when $|x|<1$, $x^{2N}\approx 0$, so $\frac1{1+x^{2N}}\approx 1$. But the moment $|x|>1$, $x^{2N}$ becomes huge, so then $\frac1{1+x^{2N}} \approx 0$. It's not exactly 0 outside $|x|<1$, but instead decays like $O(|x|^{-2N+2})$ (again, crank up $N$ to get a better approximation)

This is what it looks like for $N=10$:

The general case: the quadratic with height $h$ above $y=c$, base $d$ centered around $x_0$ is $$ q_{h,c,d,x_0}(x)=h q\left(\frac{x-x_0}{d/2}\right)+c.$$ The quadratic truncated at $y=c$ is then $$ f_{h,c,d,x_0}(x) = h f\left(\frac{x-x_0}{d/2}\right)+c,$$ and the corresponding approximation I am proposing is $$ f_{N,h,c,d,x_0}(x) = h f_N\left(\frac{x-x_0}{d/2}\right)+c.$$

You can adjust these parameters in this interactive Desmos graph.

A further variation is using $1/(1-x^{2N})$ instead to define $f_N$; this makes $f_N$ everywhere positive (because it changes sign exactly where $q$ does, and $f_N$ is differentiable at the truncation points by l'Hopital), which could be useful. But the approximation is worse near the boundary.

Here's a graph of this last variant-

The first variant is exactly 1 at $x=0$, and exactly 0 at $x=\pm 1$. It has the same sign as the original polynomial $1-x^2$ everywhere.

The second is exact only at $x=0$, (but even if $N$ is small), and is everywhere positive. (In particular, its a slight mischaracterisation to say that "unlike Calvin Khor's suggestions, this avoids ever going negative" :) but I didn't include a graph of this variant earlier, and their solution is great too!)

As an expansion of Calvin Khor's answer, you need to generate a suitable approximation to the square pulse function, which can be done in a number of different ways.

Consider that the error function goes from -1 to 1, with a relatively rapid transition from one side to the other. Using this function, we can write $$ f_a(x)=\frac{\text{erf}(ax)+1}2-\frac{\text{erf}(a(x-1))+1}2=\frac{\text{erf}(ax)-\text{erf}(a(x-1))}2 $$ Here is what that function looks like for $a=5$ (red), $a=50$ (blue), and $a=500$ green):

We can translate this as needed, then multiply by our required polynomial, in order to get our fit. Fitting $(1-x^2)$ leads us to $$ f(x)\approx (1-x^2)f_a(2x+1) $$ which, for the same values of $a$, gives

Any sequence that converges towards the square pulse will work. Instead of $\text{erf}$, you could use $\tan^{-1}$ (with appropriate rescaling), for example, or $\tanh$.