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

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

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.

Find $f (- x)$ .

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

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

Simplify the denominator.

Let $u = - x$ . Substitute $u$ for all occurrences of $- x$ .

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

Factor $u^{2} + 2 u - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$f (- x) = \frac{2}{(u - 1) (u + 3)}$

Replace all occurrences of $u$ with $- x$ .

$f (- x) = \frac{2}{(- x - 1) (- x + 3)}$

Simplify with factoring out.

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

$f (- x) = \frac{2}{(- (x) - 1) (- x + 3)}$

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

$f (- x) = \frac{2}{(- (x) - 1 \cdot 1) (- x + 3)}$

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

$f (- x) = \frac{2}{- (x + 1) (- x + 3)}$

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

$f (- x) = \frac{2}{- 1 (x + 1) (- x + 3)}$

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

$f (- x) = \frac{2}{- 1 (x + 1) (- (x) + 3)}$

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

$f (- x) = \frac{2}{- 1 (x + 1) (- (x) - 1 \cdot - 3)}$

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

$f (- x) = \frac{2}{- 1 (x + 1) (- (x - 3))}$

Simplify the expression.

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

$f (- x) = \frac{2}{- 1 (x + 1) (- 1 (x - 3))}$

Multiply $- 1$ by $- 1$ .

$f (- x) = \frac{2}{1 (x + 1) (x - 3)}$

Multiply $x + 1$ by $1$ .

$f (- x) = \frac{2}{(x + 1) (x - 3)}$

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

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

$\frac{2}{(x + 1) (x - 3)}$ $\neq$ $\frac{2}{x^{2} + 2 x - 3}$

Since $\frac{2}{(x + 1) (x - 3)}$ $\neq$ $\frac{2}{x^{2} + 2 x - 3}$ , the function is not even.

The function is not even

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

Factor $x^{2} + 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$- f (x) = - \frac{2}{(x - 1) (x + 3)}$

Since $\frac{2}{(x + 1) (x - 3)}$ $\neq$ $- \frac{2}{(x - 1) (x + 3)}$ , the function is not odd.

The function is not odd

The function is neither odd nor even

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

No origin symmetry

Since the function is not even, it is not symmetric about the y-axis.

No y-axis symmetry

Since the function is neither odd nor even, there is no origin / y-axis symmetry.

Function is not symmetric

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

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