Geometric equation of center of symmetry

Question

$ f(x) = \frac{x^2 + mx + 6}{x-m} $ what is the geometric equation of center of symmetry ?

I tried to pick random numbers for m and then finding roots of f''(x) but it didn't work..

Answer 1

The relation $$ y = \frac{x^2 + mx + 6}{x-m} $$ is a hyperbola with center of symmetry at $$ x = m \; \; , y = 3m \; . $$ There are no inflection points involved.

Example with $m=1$

Answer 2

Note that

$$f(x) = \frac{x^2 + mx + 6}{x-m}= \frac{(x+2m)(x-m)+2m^2+ 6}{x-m}=x+2m+\frac{2m^2+ 6}{x-m}$$

thus

$$f'(x)=1-\frac{2m^2+ 6}{(x-m)^2}\implies f''(x)=\frac{4m^2+ 12}{(x-m)^3}\neq 0$$

thus we have not inflection points.