Question

An equation is being tested for symmetry with respect to the $x$-axis, the $y$-axis, and the origin. Explain why, if two of these symmetries are present, the remaining one must also be present.

Short answer

If a graph is symmetry with respect to both the $x$-axis and the $y$-axis, by definition, any point $(x,y)$on the graph, the point $(x-y)$and $(-x,y)$are also on the graph. Therefore, for any point role="math" localid="1647402482927" $(x,y)$, role="math" localid="1647402489417" $(x,-y)$is on the graph since symmetry with respect to $x$-axis. role="math" localid="1647402495348" $(x,-y)$ is on the graph, role="math" localid="1647402501626" $(-x,-y)$ is also on the graph since symmetry on the $y$-axis, therefore, by definition, this graph is symmetric with respect to the origin. Similarly, we can prove other cases.

Step-by-step solution

Given information

An equation is being tested for symmetry with respect to the $x$-axis, the $y$-axis, and the origin.

Determine the symmetry.

Case 1: Suppose the equation is symmetric about the $x$and $y$-axes.

$$\begin{gathered} (x,y)=(-x,y)\text{y-axis symmetry: Given} \\ (x,y)=(x,-y)\mathrm{x}\text{-axis symmetry: Given} \end{gathered}$$

Therefore, $(-x,y)=(x,-y)$

$-x=x;y=-y$origin symmetry

Case 2: Suppose the equation is symmetric about the $x$-axis and origin.

$(x,y)=(-x,-y)\text{origin symmetry: Given}$

$(x,y)=(x,-y)\mathrm{x}\text{-axis symmetry: Given}$

Therefore, $(-x,-y)=(x,-y)$

$(x,y)=(-x,y)\text{y-axis symmetry}$

Case 3: Suppose the equation is symmetric about the $y$-axis and origin.

$(x,y)=(-x,-y)\text{origin symmetry: Given}$

$(x,y)=(-x,y)\text{y-axis symmetry: Given}$

Therefore, $(-x,-y)=(-x,y)$

$(x,y)=(x,-y)\mathrm{x}\text{-axis symmetry}$