How do i prove this is symmetric

I know that this function is neither even nor odd in symmetry, but it is symmetric at x=3.

But how do I prove this?

the function is...

abs((x^2)-6x+5)

You need to show that f(3 - x) = f(x + 3) for all x (do you see why? - draw a picture).

f(3 - x)
= |(3 - x)^2 - 6(3 - x) + 5|
= |(x^2 - 6x + 9) - (18 - 6x) + 5|
= |x^2 - 4|

f(x + 3)
= |(x + 3)^2 - 6(x + 3) + 5|
= |(x^2 + 6x + 9) - (6x + 18) + 5|
= |x^2 - 4|.

Hence, f is symmetric about the line x = 3.

I hope this helps!