Find the Symmetry y = cube root of x

$y = \sqrt[3]{x}$

There are three types of symmetry:

1. X-Axis Symmetry

2. Y-Axis Symmetry

3. Origin Symmetry

If $(x, y)$ exists on the graph, then the graph is symmetric about the:

1. X-Axis if $(x, - y)$ exists on the graph

2. Y-Axis if $(- x, y)$ exists on the graph

3. Origin if $(- x, - y)$ exists on the graph

Check if the graph is symmetric about the x-axis by plugging in $- y$ for $y$ .

$- y = \sqrt[3]{x}$

Simplify the left side.

Remove parentheses.

$- y = \sqrt[3]{x}$

Since the equation is not identical to the original equation, it is not symmetric to the x-axis.

Not symmetric to the x-axis

Check if the graph is symmetric about the y-axis by plugging in $- x$ for $x$ .

$y = \sqrt[3]{- x}$

Simplify.

Simplify the left side.

Remove parentheses.

$y = \sqrt[3]{- x}$

Simplify the right side.

Simplify $\sqrt[3]{- x}$ .

Rewrite $- x$ as ${({(- 1)}^{3})}^{3} x$ .

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = \sqrt[3]{{(- 1)}^{3} x}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = \sqrt[3]{{({(- 1)}^{3})}^{3} x}$

Pull terms out from under the radical.

$y = {(- 1)}^{3} \sqrt[3]{x}$

Raise $- 1$ to the power of $3$ .

$y = - \sqrt[3]{x}$

Since the equation is not identical to the original equation, it is not symmetric to the y-axis.

Not symmetric to the y-axis

Check if the graph is symmetric about the origin by plugging in $- x$ for $x$ and $- y$ for $y$ .

$- y = \sqrt[3]{- x}$

Solve for $y$ .

Simplify the left side.

Simplify $- y$ .

Remove parentheses.

$(- y) = \sqrt[3]{- x}$

Remove parentheses.

$- y = \sqrt[3]{- x}$

Simplify the right side.

Simplify $\sqrt[3]{- x}$ .

Rewrite $- x$ as ${({(- 1)}^{3})}^{3} x$ .

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$- y = \sqrt[3]{{(- 1)}^{3} x}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$- y = \sqrt[3]{{({(- 1)}^{3})}^{3} x}$

Pull terms out from under the radical.

$- y = {(- 1)}^{3} \sqrt[3]{x}$

Raise $- 1$ to the power of $3$ .

$- y = - \sqrt[3]{x}$

Simplify the left side.

Simplify $- (- y)$ .

Remove parentheses.

$- (- y) = - (\sqrt[3]{- x})$

Multiply $- (- y)$ .

Multiply $- 1$ by $- 1$ .

$1 y = - (\sqrt[3]{- x})$

Multiply $y$ by $1$ .

$y = - (\sqrt[3]{- x})$

Simplify the right side.

Simplify $- (\sqrt[3]{- x})$ .

Rewrite $- x$ as ${({(- 1)}^{3})}^{3} x$ .

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = - \sqrt[3]{{(- 1)}^{3} x}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$y = - \sqrt[3]{{({(- 1)}^{3})}^{3} x}$

Pull terms out from under the radical.

$y = - ({(- 1)}^{3} \sqrt[3]{x})$

Multiply $- 1$ by ${(- 1)}^{3}$ by adding the exponents.

Move ${(- 1)}^{3}$ .

$y = {(- 1)}^{3} \cdot - 1 \sqrt[3]{x}$

Multiply ${(- 1)}^{3}$ by $- 1$ .

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

$y = {(- 1)}^{3} \cdot {(- 1)}^{1} \sqrt[3]{x}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = {(- 1)}^{3 + 1} \sqrt[3]{x}$

Add $3$ and $1$ .

$y = {(- 1)}^{4} \sqrt[3]{x}$

Raise $- 1$ to the power of $4$ .

$y = 1 \sqrt[3]{x}$

Multiply $\sqrt[3]{x}$ by $1$ .

$y = \sqrt[3]{x}$

Since the equation is identical to the original equation, it is symmetric to the origin.

Symmetric with respect to the origin

Find the Symmetry y = cube root of x

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