Algebra Examples

$y = x^{3} + 5 x$

Step 1

There are three types of symmetry:

1. X-Axis Symmetry

2. Y-Axis Symmetry

3. Origin Symmetry

Step 2

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

Step 3

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

$- y = x^{3} + 5 x$

Step 4

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

Not symmetric to the x-axis

Step 5

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

$y = {(- x)}^{3} + 5 (- x)$

Step 6

Simplify each term.

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Step 6.1

Apply the product rule to $- x$ .

$y = {(- 1)}^{3} x^{3} + 5 (- x)$

Step 6.2

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

$y = - x^{3} + 5 (- x)$

Step 6.3

Multiply $- 1$ by $5$ .

$y = - x^{3} - 5 x$

Step 7

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

Not symmetric to the y-axis

Step 8

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

$- y = {(- x)}^{3} + 5 (- x)$

Step 9

Simplify each term.

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Step 9.1

Apply the product rule to $- x$ .

$- y = {(- 1)}^{3} x^{3} + 5 (- x)$

Step 9.2

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

$- y = - x^{3} + 5 (- x)$

Step 9.3

Multiply $- 1$ by $5$ .

$- y = - x^{3} - 5 x$

Step 10

Multiply both sides by $- 1$ .

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Step 10.1

Multiply each term by $- 1$ .

$- - y = - - x^{3} - (- 5 x)$

Step 10.2

Multiply $- - y$ .

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Step 10.2.1

Multiply $- 1$ by $- 1$ .

$1 y = - - x^{3} - (- 5 x)$

Step 10.2.2

Multiply $y$ by $1$ .

$y = - - x^{3} - (- 5 x)$

Step 10.3

Simplify each term.

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Step 10.3.1

Multiply $- - x^{3}$ .

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Step 10.3.1.1

Multiply $- 1$ by $- 1$ .

$y = 1 x^{3} - (- 5 x)$

Step 10.3.1.2

Multiply $x^{3}$ by $1$ .

$y = x^{3} - (- 5 x)$

Step 10.3.2

Multiply $- 5$ by $- 1$ .

$y = x^{3} + 5 x$

Step 11

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

Symmetric with respect to the origin

Step 12