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:

$$\begin{array}{rcl}f\left(x\right)& =& 2+\sqrt{-{(x-2)}^{2}+1}\\ g\left(x\right)& =& 2-\sqrt{-{(x-2)}^{2}+1}\\ h\left(x\right)& =& 3x-3\\ i\left(x\right)& =& -3x+9\\ j\left(x\right)& =& \mathrm{0,2}x+\mathrm{1,7}\end{array}$$

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.

## The quadric

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

$$-{(x-2)}^{2}+1\ge 0,$$

that is

$$1\ge {(x-2)}^{2},$$

which would more commonly written as

$${(x-2)}^{2}\le 1,$$

whose solution is

$$-1\le x-2\le 1,$$

or

$$1\le x\le 3.$$

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:

$$y=2\pm \sqrt{-{(x-2)}^{2}+1},$$

which can be rearranged to

$$y-2=\pm \sqrt{-{(x-2)}^{2}+1}.$$

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 $(2,2)$, and with radius 1. $f$ represents the upper half, $g$ the lower half.

## 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

$$\begin{array}{rcl}h\left(x\right)& =& \phantom{-}3x-3\\ i\left(x\right)& =& -3x+9\end{array}$$

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:

$$j\left(x\right)=\frac{x}{5}+\frac{17}{10}$$

or inline using the solidus:

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

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:

So, an ‘A’ shape.

## Putting it all together

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