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

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

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 $f (x)$ .

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

Simplify the numerator.

Apply the product rule to $- x$ .

$f (- x) = \frac{2 {(- 1)}^{2} x^{2}}{{(- x)}^{2} + 9}$

Raise $- 1$ to the power of $2$ .

$f (- x) = \frac{2 \cdot (1 x^{2})}{{(- x)}^{2} + 9}$

Multiply $2$ by $1$ .

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

Simplify the denominator.

Apply the product rule to $- x$ .

$f (- x) = \frac{2 x^{2}}{{(- 1)}^{2} x^{2} + 9}$

Raise $- 1$ to the power of $2$ .

$f (- x) = \frac{2 x^{2}}{1 x^{2} + 9}$

Multiply $x^{2}$ by $1$ .

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

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

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

$\frac{2 x^{2}}{x^{2} + 9} = \frac{2 x^{2}}{x^{2} + 9}$

Since $\frac{2 x^{2}}{x^{2} + 9} = \frac{2 x^{2}}{x^{2} + 9}$ , the function is even.

The function is even

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

No origin symmetry

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

Y-axis symmetry