Table of Contents

Last modified on December 26th, 2022

A graph is said to have symmetry if it can divided into two equal parts along a line such that one part is an exact mirror image of the other. The concept is useful to graph an equation.

Graph symmetry is of 3 types:

A graph is symmetric about the x-axis when the points (x, y) and (x, -y) are present on the same graph. A graph will have x-axis symmetry if we get an **equivalent equation **to the original one when we **replace y with –y**.

Let us test the symmetry of a graph with the equation 6 + xy^{4} = 0

x-axis symmetry test: replacing y with –y:

6 + x(-y)^{4} = 0

⇒ 6 + xy^{4} = 0 (this is the original equation)

Thus the graph is symmetric about the x-axis.

A graph is symmetric about the y-axis when the points (x, y) and (-x, y) are present on the same graph. A graph will have y-axis symmetry if we get an **original equation **to the original one when we **replace x with –x.**

Let us test the symmetry of a graph with the equation y = x^{4} +3x^{2}

x-axis symmetry test: replacing y with –y:

-y = x^{4} +3x^{2}

Y = – (x^{4} +3x^{2}), this is not the equivalent equation.

y-axis symmetry test: replacing x with –x.

So, y = (-x)^{4} +3(-x)^{2}

y = x^{4} +3x^{2} (this is the original equation)

Thus the graph is symmetric about the y-axis.

A graph is symmetric about the origin when the points (x, y) and (-x,-y) are present on the same graph. A graph will have symmetry about the origin if we get an **equivalent equation** to the original one when we **replace y with –y** and also **x with –x**.

Let us test the symmetry of a graph with the equation xy = 4

x-axis symmetry test: replacing y with –y:

x(-y) = 4

-xy = 4, (this is not the original equation)

y-axis symmetry test: replacing x with –x.

(-x)y = 4

-xy = 4, (this is also not the original equation)

Origin symmetry test: replacing x with –x and y with -y.

(-x)(-y) = 4

xy = 4, (this is the original equation)

So, the graph is symmetric about the origin.

For a parabola the axis of symmetry always passes through the vertex. The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

For a quadratic function in standard form y = ax^{2} + bx + c, the axis of symmetry is x = -b/2a

Now, let us find the axis of symmetry of the given quadratic graph.

The given graph shows line symmetry about the y-axis.

So for a graph y = x^{2} – 4

Axis of symmetry is x = -b/2a, here a = 1, b = 0

∴ x = 0

**Prove the symmetry of the graph y = x ^{2}**

Solution:

x-axis symmetry test: Replacing y with –y,

-y = x^{2} (this is not equivalent to the original equation)

y-axis symmetry test: Replacing x with –x,

y = (-x)^{ 2}

= x^{2}

∴ The graph is symmetric about the **y-axis**

**Test the symmetry of the graph y = x ^{3}**

Solution:

x-axis symmetry test: Replacing y with –y,

-y = x^{3 }(this is not equivalent to the original equation)

(this is not equivalent to the original equation)

y = (-x)^{3}

= -x^{3 }(this is not equivalent to the original equation)

Origin symmetry test: replacing x with –x and y with -y.

-y = (-x)^{3}

-y = -x^{3}

Y = x^{3}, (this is the original equation)

∴ The graph is symmetric about the **origin**.

**Which graph shows rotational symmetry and which one shows line symmetry in the figure alongside?**

Solution:

Graph 2 shows rotational symmetry since the graph is a perfect circle about the origin.

Graph 1 shows point symmetry where the central point is the origin.

**Which graph shows line symmetry about the y-axis in the figure alongside?**

Solution:

Graph 2 shows line symmetry about the y-axis.

Graph 1 does not have the y-axis as its line of symmetry.

Last modified on December 26th, 2022