Q. 86 - Precalculus Enhanced with Graphing Utilities

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.

Step-by-Step Solution

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 $(x, y)$, $(x,-y)$ is on the graph since symmetry with respect to $x$-axis. $(x,-y)$ is on the graph, $(-x,-y)$ 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 1: Given information

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

Step 2: 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, $(-x,-y)=(-x, y)$

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