Q 14 - Precalculus Enhanced with Graphing Utilities

Question

Test each equation for symmetry with respect to the x-axis, the y-axis, and the origin. 

$x^2+x+y^2+2y=0$.

Step-by-Step Solution

Answer

The given equation is not symmetric anywhere.

Step 1. Given Information

We have given the following equation :- 

$x^2+x+y^2+2y=0$.

We have to check the symmetry of this equation with respect to x-axis, y-axis and the origin. 

Step 2. To check symmetry about x-axis.

The given equation is :-  

$x^2+x+y^2+2y=0$.

We know that a graph is symmetrical about x-axis, if a point $\left(x,y\right)$ lies on graph, then $\left(x,-y\right)$ is also lies on graph.

So to check symmetry about y-axis, change $y$ by $-y$ in the given equation, then we have :-

$$\begin{gathered} x^2+x+{(-y)}^2+2(-y)=0 \\ \Rightarrow x^2+x+y^2-2y=0 \end{gathered}$$

This equation is not same as the given equation.

So we can conclude that the given equation is not symmetric about x-axis.

Step 3. To check symmetry about y-axis.

The given equation is :- 

$x^2+x+y^2+2y=0$.

We know that a graph is symmetrical about y-axis, if a point $\left(x,y\right)$ lies on graph, then $\left(-x,y\right)$ is also lies on graph.

So to check symmetry about y-axis, change $x$ by $-x$ in the given equation, then we have :-

$$\begin{gathered} {(-x)}^2-x+y^2+2y=0 \\ \Rightarrow x^2-x+y^2+2y=0 \end{gathered}$$

This equation is not same as the given equation.

So we can conclude that the given equation is not symmetric about y-axis.

Step 4. To check symmetry about origin.

The given equation is :- 

$x^2+x+y^2+2y=0$.

We know that a graph is symmetrical about origin, if a point $\left(x,y\right)$ lies on graph, then $\left(-x,-y\right)$ is also lies on graph.

So to check symmetry about origin, change $x$ by $-x$ and $y$ by $-y$ in the given equation, then we have :-

$$\begin{gathered} {\left(-x\right)}^2-x+{(-y)}^2+2(-y)=0 \\ \Rightarrow x^2-x+y^2-2y=0 \end{gathered}$$

This equation is not same as the given equation.

So we can conclude that the given equation is not symmetric about origin.