Find the Symmetry f(x)=(x+2)/(x^2+x-12)

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle -12}$ and whose sum is ${\displaystyle 1}$.

Write the factored form using these integers.

Find ${\displaystyle f(-x)}$ by substituting ${\displaystyle -x}$ for all occurrence of ${\displaystyle x}$ in ${\displaystyle f\left(x\right)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle 2}$ as ${\displaystyle -1(-2)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1(-2)}$.

Rewrite ${\displaystyle -(x-2)}$ as ${\displaystyle -1(x-2)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle -3}$ as ${\displaystyle -1\left(3\right)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1\left(3\right)}$.

Rewrite ${\displaystyle -(x+3)}$ as ${\displaystyle -1(x+3)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -x}$.

Rewrite ${\displaystyle 4}$ as ${\displaystyle -1(-4)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1(-4)}$.

Simplify the expression.

Check if ${\displaystyle f(-x)=f\left(x\right)}$.

Since ${\displaystyle -\frac{x-2}{(x+3)(x-4)}}$${\displaystyle \ne }$${\displaystyle \frac{x+2}{(x-3)(x+4)}}$, the function is not even.

Multiply ${\displaystyle -1}$ by ${\displaystyle \frac{x+2}{(x-3)(x+4)}}$.

Since ${\displaystyle -\frac{x-2}{(x+3)(x-4)}}$${\displaystyle \ne }$${\displaystyle -\frac{x+2}{(x-3)(x+4)}}$, the function is not odd.