Answer
The given function is even, has symmetry about the y-axis only.
Work Step by Step
Step I: To check even or odd.
In the provided equation, if $h\left( -x \right)=h\left( x \right)$ then it is an even function, and if $h\left( -x \right)=-h\left( x \right)$ , then it is odd.
$\begin{align}
& h\left( x \right)={{x}^{2}}-{{x}^{4}} \\
& h\left( -x \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}} \\
& ={{x}^{2}}-{{x}^{4}} \\
& =h\left( x \right)
\end{align}$
It is an even 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.
$\begin{align}
& h\left( x \right)={{x}^{2}}-{{x}^{4}} \\
& h\left( -x \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}} \\
& ={{x}^{2}}-{{x}^{4}} \\
& =h\left( x \right)
\end{align}$
It is the same as given equation, hence it has symmetry 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.
$\begin{align}
& h\left( x \right)={{x}^{2}}-{{x}^{4}} \\
& y={{x}^{2}}-{{x}^{4}} \\
& -y={{x}^{2}}-{{x}^{4}} \\
& y=-{{x}^{2}}+{{x}^{4}}
\end{align}$
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.
$\begin{align}
& y={{x}^{2}}-{{x}^{4}} \\
& \left( -y \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}} \\
& -y={{x}^{2}}-{{x}^{4}} \\
& y=-{{x}^{2}}+{{x}^{4}}
\end{align}$
It is not the same as provided equation, hence it is not symmetric about the origin.
Therefore, the provided function is even, has symmetry about the y-axis only.
