Find the Symmetry f(x)=(x^2+3x-18)/(x-6)

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 ${\displaystyle -18}$ and whose sum is ${\displaystyle 3}$.

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 ${\displaystyle -3}$ 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 ${\displaystyle 6}$ as ${\displaystyle -1(-6)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1(-6)}$.

Simplify the expression.

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

Rewrite ${\displaystyle -6}$ as ${\displaystyle -1\left(6\right)}$.

Factor ${\displaystyle -1}$ out of ${\displaystyle -\left(x\right)-1\left(6\right)}$.

Rewrite negatives.

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

Since ${\displaystyle -\frac{(x+3)(x-6)}{x+6}}$${\displaystyle \ne }$${\displaystyle \frac{(x-3)(x+6)}{x-6}}$, the function is not even.

Multiply ${\displaystyle -1}$ by ${\displaystyle \frac{(x-3)(x+6)}{x-6}}$.

Since ${\displaystyle -\frac{(x+3)(x-6)}{x+6}}$${\displaystyle \ne }$${\displaystyle -\frac{(x-3)(x+6)}{x-6}}$, the function is not odd.