I have this equation:
$$ f(x) = \frac{2x^2+3}{x-2} $$
and I have to prove it has half-turn symmetry around the point (2,8). I know that for a function to have half-turn symmetry, it needs to have the property:
$$ f(-x) = -f(x) $$
So I need to substitute (2,8) into the equation, so
$$ -\frac{2(x-2)^2+3}{(x-2)-2}+8 = \frac{2(2-x)^2+3}{(2-x)-2}+8 $$
But after solving it out and/or graphing it, I don't get equality. In desmos, the two graphs are reflections of each other, in the line $x=2$, but they're not equal. What am I doing wrong?
edit:
Updated equation, but I'm still not getting it. After simplifying I reached $$-\frac{2x^2-8x+11}{x-4} = \frac{2x^2-8x+11}{-x} $$ but I don't know where to go from there.
Hint:
Symmetry w.r.t. the point $(a,b)$ translates as $$f(2a-x)=2b-f(x).$$ Can you check it?
Justification:
If the points $M=(x, f(x))$ and $M'=(x', f(x'))$ are symmetric w.r.t. $I=(a,b)$, this means $I$ is the midpoint of $[MM']$, i.e. $$a=\frac{x+x'}2, \quad b=\frac{f(x)+f(x')}2,$$ whence $$x'=2a-x,\quad f(x')=f(2a-x)=2b-f(x).$$