Answer
The given function is neither an even nor odd function.
Work Step by Step
Step I: To check even or odd.
In the provided equation, if $g\left( -x \right)=g\left( x \right)$ then it is an even function, and if $g\left( -x \right)=-g\left( x \right)$ , then it is odd.
$\begin{align}
& g\left( x \right)={{x}^{2}}-x \\
& g\left( -x \right)={{\left( -x \right)}^{2}}-\left( -x \right) \\
& ={{x}^{2}}+x
\end{align}$
It is neither an even nor odd function.
Step II: To check symmetry about the y-axis:
In the provided equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y-axis.
$\begin{align}
& y={{x}^{2}}-x \\
& y={{\left( -x \right)}^{2}}-\left( -x \right) \\
& y={{x}^{2}}+x \\
\end{align}$
It is not the same as provided equation, hence it is not symmetric about the y-axis.
Step III: To check symmetry about the x-axis:
In the provided equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x- axis.
$\begin{align}
& y={{x}^{2}}-x \\
& \left( -y \right)={{x}^{2}}-x \\
& -y={{x}^{2}}-x
\end{align}$
It is not the same as the given equation, hence it is not symmetric about the x-axis.
Step IV: To check symmetry about the origin:
In the provided equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin.
$\begin{align}
& y={{x}^{2}}-x \\
& \left( -y \right)={{\left( -x \right)}^{2}}-\left( -x \right) \\
& -y={{x}^{2}}+x \\
& y=-{{x}^{2}}-x
\end{align}$
It is not the same as provided equation, hence it is not symmetric about the origin.
Therefore, the provided function is neither an even nor odd function.
