Find the Symmetry f(x)=-8/(x^2-9)

$f (x) = - \frac{8}{x^{2} - 9}$

Step 1

Determine if the function is odd, even, or neither in order to find the symmetry.

1. If odd, the function is symmetric about the origin.

2. If even, the function is symmetric about the y-axis.

Step 2

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$f (x) = - \frac{8}{x^{2} - 3^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$f (x) = - \frac{8}{(x + 3) (x - 3)}$

Step 3

Find $f (- x)$ .

Find $f (- x)$ by substituting $- x$ for all occurrence of $x$ in $f (x)$ .

$f (- x) = - \frac{8}{((- x) + 3) ((- x) - 3)}$

Factor $- 1$ out of $- x$ .

$f (- x) = - \frac{8}{(- (x) + 3) (- x - 3)}$

Rewrite $3$ as $- 1 (- 3)$ .

$f (- x) = - \frac{8}{(- (x) - 1 \cdot - 3) (- x - 3)}$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

$f (- x) = - \frac{8}{- (x - 3) (- x - 3)}$

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$f (- x) = - \frac{8}{- 1 (x - 3) (- x - 3)}$

Factor $- 1$ out of $- x$ .

$f (- x) = - \frac{8}{- 1 (x - 3) (- (x) - 3)}$

Rewrite $- 3$ as $- 1 (3)$ .

$f (- x) = - \frac{8}{- 1 (x - 3) (- (x) - 1 \cdot 3)}$

Factor $- 1$ out of $- (x) - 1 (3)$ .

$f (- x) = - \frac{8}{- 1 (x - 3) (- (x + 3))}$

Simplify the expression.

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$f (- x) = - \frac{8}{- 1 (x - 3) (- 1 (x + 3))}$

Multiply $- 1$ by $- 1$ .

$f (- x) = - \frac{8}{1 (x - 3) (x + 3)}$

Multiply $x - 3$ by $1$ .

$f (- x) = - \frac{8}{(x - 3) (x + 3)}$

Step 4

A function is even if $f (- x) = f (x)$ .

Check if $f (- x) = f (x)$ .

Since $- \frac{8}{(x - 3) (x + 3)} = - \frac{8}{(x + 3) (x - 3)}$ , the function is even.

The function is even

Step 5

Since the function is not odd, it is not symmetric about the origin.

No origin symmetry

Step 6

Since the function is even, it is symmetric about the y-axis.

Y-axis symmetry

Step 7

Find the Symmetry f(x)=-8/(x^2-9)

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