How did we know that this distribution is triangular and symmetric about 0?

Question

I know that this distribution is symmetrical about y axis since f(-x)=f(x) but how did we know that it was symmetrical about 0 an dthat it is a triangular distribution? Since I know that a triangular distribution has the following expression:

And why did we infer from the symmetry that E(x)=μ=0 without computing any integral?

Answer 1

when $x$ is positive, $x \in (0, \tau)$, the graph is a line with negative slope.

Now, flip it along the $y$-axis, you get an increasing line.

Together, you get a triangle.