What equation best represents this set of data? Here is the graph. (It is the same as below.)

The points are symmetrical over the $y$-axis, but I cannot find an equation that accurately represents this graph.
 A: That looks suspiciously like $a \lvert x \rvert$ for some $a$... at least the branches look linear. If they aren't, some tweak should get you there.
Without any idea of the origin of the data, or the process that produces them (or they are representing), it is hard to hypothesize a matching functional form.
A: Well, observe that $y-x$ for $x>0$ are almost in arithmetical progression: $0, 3, 4.5, 6, 9, 12, 15, 30$. So we construct Lagrange polynomials for $y=y(t)$, $x=x(t)$ where $t=\frac{2}{3}(y-x)$:
$$x=t(t - 1),\ y=t(t + \frac{1}{2}).$$
We might leave this as an answer, but let's obtain $y(x)$.
$$t^2-t-x=0$$
$$t = \frac12 \left(1 \pm \sqrt{4 x + 1}\right)$$
$$t=\frac23(y-x)\Rightarrow y=\frac32 t+x$$
$$y=\frac34 \left(1 \pm \sqrt{4 x + 1}\right)+x$$
Now let's resolve the $\pm$.
\begin{array}{|l|c|c|c|c|c|c|c|c|}
\hline
x&0&2&6&12&30&56&90&380\\
\hline
y_1&0&1/2&3&15/2&45/2&91/2&153/2&703/2\\
\hline
y_2&3/2&5&21/2&18&39&68&105&410\\
\hline
\end{array}
$$\hbox{So }y(x)=\begin{cases}
\frac34 \left(1 + \sqrt{4 |x| + 1}\right)+|x|,&\hbox{for }x\ne 0,\\
0,&\hbox{for }x= 0.
\end{cases}$$

Edit: python script for computing the Lagrange polynomials
from sympy import *
R=Rational
from sympy.abc import x
xs=[0,2,6,12,30,56,90,380]
ys=[0,5,10+R(5,10),18,39,68,105,410]
def lp(xs,ys):
    monomes=[(x-R(i)) for i in xs]
    lps=[prod(monomes[:i]+monomes[i+1:]) for i in range(len(xs))]
    return simplify(sum(R(j)*f/f.subs({x:R(i)})
        for i,j,f in zip(xs,ys,lps)
                        ))
xs_=[(i-j)/R(3,2) for i,j in zip(ys,xs)]
print(lp(xs_,xs),lp(xs_,ys))

And it's output
x*(x - 1) x*(x + 1/2)
About $y-x$: observing $y-x$ is rather artificial and it is, but once you obtained it, you see that every $y-x$ is divisible by $3$, it's natural to divide by $3$. Also there's $1.5$ so it's natural to multiply by $2$ to get integer values. That's why $\frac 23$.
A: $y=|x|$ would be a good fit, although if you look at the data points carefully, the line actually has slope a little greater than $1$. To be as accurate as possible, take the average of all $\frac YX$ values for $X\gt 0$, which comes out to be $\approx 1.48$ and take this as the slope instead, giving the equation $$y=|1.48 x|$$
