How to determine the center of symmetry of a function? I'm learning about functions and I came to a question that I'm struggling at:
$f(x) = \frac{x^2 -2}{x+1} $
The question is:

Show that the graph of $f$, has a center of symmetry to be determined in an orthonormal system.

What is the best approach to solve this question?
 A: Let's discover where the center of symmetry could be
Knowing that the map has a vertical asymptote for $x = -1$, the abscissa of the center of symmetry is $-1$. Let's find the ordinate. We have
$$\begin{aligned}f(x) &= \frac{x^2 -2}{x+1} = \frac{\left(x + 1\right)^2-2}{x + 1}\\
&=\frac{\left(x + 1\right)^2-2x-3}{x + 1}\\
&=x-1-\frac{1}{x+1}
\end{aligned}$$
Therefore $y = x-1$ is another asymptote of the map. The center of symmetric if it exists lie at the intersection of both asymptotes and therefore has for coordinates $C= (-1,-2)$.
Let's prove that $C$ is indeed a center of symmetry
For this, we need to notice that
$$-4-f(-2-x) = f(x).$$
Which I leave to the reader.
In general, for $(a,b)$ to be a center of symmetry you need to prove that $2b−f(2a−x)=f(x)$.
A: Since $f=x-1-\frac{1}{x+1}$, rotation around $(a,\,b)$ gives$$y=2b-f(2a-x)=2b-2a+x+1-\frac{1}{x-2a-1}.$$This is the original curve for $2b-2a+1=-1,\,-2a-1=1$, i.e. $a=-1,\,b=-2$.
A: Well, it is quite obviously $(-1,-2)$ if you draw a graph.
Notice that it lies in the intersection of both asymptots: $x=-1$ and $y=x-1$.


 A general rule simply involves shifting the function. Proving that $(a,b)$ is  a center of symmetry for $f(x)$ is the same as proving that the function
 $$g(x):=f(x+a)-b$$
 is symmetric around the origin. This means you have to prove $g(-x)=-g(x)$.


 So, in your case $g(x) ={x^2-1\over x}$ which is clearly odd.

