Apparently graffiti things such as this and this have appeared in Bruxelles (and who knows where else).

Note: this article makes use of MathML, the standard XML markup for math formulas. Sadly, this is not properly supported on some allegedly ‘modern’ and ‘feature-rich’ browsers. If the formulas don't make sense in your browser, consider reporting the issue to the respective developers and/or switching to a standard-compliant browser.

I must say it's not too common seeing mathematical graffiti, so let's have a look at them a little bit closer. Let's start with a transcription:

$f(x) = 2+ -(x-2)2 + 1 g(x) = 2- -(x-2)2 + 1 h(x) = 3x-3 i(x) = -3x+9 j(x) = 0,2x+1,7$

There's a few things I don't like about some choices made (such as the choice of decimal separator —which could be avoided altogether, as we'll see later), but let's first try to understand what we have, as-is.

What we're looking at is the definition of five distinct functions of a single variables. One of the nice things about real-valued functions of real-valued variables (which is the assumption we make here) is that, thanks to the brilliant intuition of Réne Descartes to associate algebra and geometry, we can visualize these functions.

Thus, another way to look at this is that we have an algebraic description of five curves. The obvious implication here would be that, if we were to plot these curves, we'd get another picture, the actual, hidden, graffiti.

Can we reason about the functions to get an idea about what to expect from the visualization? Indeed, we can.

For example, the functions $f$ and $g$ are obviously closely related, as they differ just for the sign of radical. We can thus expect them to be the two possible solution for a second-order equation, which we're going to discover soon. Additionally, the form these functions have also give as a domain of existence: since the argument to the radical must be non-negative for the functions to have real values (and thus be plottable), we must have

$-(x-2)2+1≥0,$

that is

$1≥(x-2)2,$

which would more commonly written as

$(x-2)2≤1,$

whose solution is

$-1≤x-2≤1,$

or

$1≤x≤3.$

Now we know that, for the functions $f$ and $g$ to exist, $x$ must be in the interval between 1 and 3. Since the other functions do not have restrictions on the domain of existence, we can take this as domain for the whole plot.

The next step is to find out what are $f$ and $g$ the two halves of. To do this, we can “combine” them using $y$ as placeholder for either, and writing:

$y=2±-(x-2)2+1,$

which can be rearranged to

$y-2=±-(x-2)2+1.$

Since we're taking both the positive and negative solutions for the square root, we can square both sides without worrying about introducing spurious solutions, obtaining:

$(y-2)2=-(x-2)2+1,$

which we can again rearrange to get

$(x-2)2+(y-2)2=1.$

This is the equation of a circle, centered at the point with coordinates $\left(2,2\right)$, and with radius 1. $f$ represents the upper half, $g$ the lower half.

The circle represented by the first two equations

## The lines

The next three functions ($h$, $i$, $j$) are much simpler, and anybody that remembers their analytical geometry from high school should recognize them for the equations of straight lines.

If we look at the first two of these more closely

$h(x) = -3x-3 i(x) = -3x+9$

we notice that they are rather steep (the absolute value of the $x$ coefficient is $3>1$), and symmetrical with respect to the vertical axis (the $x$ coefficient has opposite sign). They intersect for $x=2$ which is conveniently placed halfway through our domain (derived from the circle equations). Note that the resulting ordinate $y=3$ places the intersection point on the circle we've seen, as:

$(2-2)2+(3-2)2=1$

is satisfied.

The last equation, as I mentioned at the beginning meets my full displeasure due to the choice of decimal separator —in fact, the most annoying thing about is that it uses one at all, as the same values could be written in a more universal way by using fractions:

$j(x)=x5+1710$

or inline using the solidus:

$j(x)=x/5+17/10.$

This straight line much less steep (in fact, one could say it's barely sloping at all). It also intersects the other two lines in places with some funky values which I'm not even going to bother computing, as we are only interested in the visualization:

The straight lines

So, an ‘A’ shape.

## Putting it all together

If we put it all together now, we obtain the quite famous anarchist logo:

Anarchy