How to find symmetry of a function y=f(x) if it is not centered on the y-axis?

Question

I'm thinking about the example $f(x)=(x-1)^2$ which is clearly symmetric about the line $x=1$. The question is really how do you show that it is symmetric about $x=1$ algebraically? I notice that if you plug in $-x+2$ you end up with $y=(-x+1)^2$ which is equivalent to the original function. I'm looking for this answer because I'm curious how one would show a similar property for a function that is more tricky to graph.

Answer 1

If $x=a$ is the axis of symmetry, then we have

$$f(a-x)=f(a+x), \forall x\in \mathbb{R}$$

let $z=a-x$,

then $$f(z)=f(2a-z), \forall z \in \mathbb{R}$$

For your chosen function,

$$(z-1)^2=(2a-z-1)^2$$

$$(2a-z-1)^2-(z-1)^2=0$$

$$(2a-2)(2a-2z)=0$$

$a$ can't be equal to every $z$, hence $a=1$.

Answer 2

If $f(x)$ is symmetrical around $x=a$, the symmetrical point of $x=x$ at the other side of $x=a$ is $x=y$ . $x=a$ must be the mid point, with mid point formula: \begin{align}\frac{x+y}{2}=a, y= 2a-x \to f(x)=f(2a-x)\end{align}.

In this particular case, $f(x)=(x-1)^2, a=1, f(x)=f(2-x)=f(2-x-1)^2=f(1-x)^2$