Answer
The given function is even, has symmetry about the y-axis only.
Work Step by Step
Step I: To check if the given function is even or odd.
In the given equation, if $f\left( -x \right)=f\left( x \right)$ then it is an even function, and if $f\left( -x \right)=-f\left( x \right)$ , then it is odd.
$\begin{align}
& f\left( x \right)={{x}^{2}}-{{x}^{4}}+1 \\
& f\left( -x \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}}+1 \\
& ={{x}^{2}}-{{x}^{4}}+1 \\
& =f\left( x \right)
\end{align}$
It is an even function.
Step II: To check if the given function has symmetry about the y-axis:
In the given equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y-axis.
$\begin{align}
& f\left( x \right)={{x}^{2}}-{{x}^{4}}+1 \\
& f\left( -x \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}}+1 \\
& ={{x}^{2}}-{{x}^{4}}+1 \\
& =f\left( x \right)
\end{align}$
It is the same as the provided equation, hence it has symmetry about the y-axis.
Step III: To check if the given function has symmetry about the x-axis:
In the given equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x-axis.
$\begin{align}
& f\left( x \right)={{x}^{2}}-{{x}^{4}}+1 \\
& y={{x}^{2}}-{{x}^{4}}+1 \\
& -y={{x}^{2}}-{{x}^{4}}+1 \\
& y=-{{x}^{2}}+{{x}^{4}}-1
\end{align}$
It is not the same as the provided equation, hence it is not symmetric about the x-axis.
Step IV: To check if the given function has symmetry about the origin:
In the given equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin.
$\begin{align}
& y={{x}^{2}}-{{x}^{4}}+1 \\
& \left( -y \right)={{\left( -x \right)}^{2}}-{{\left( -x \right)}^{4}}+1 \\
& -y={{x}^{2}}-{{x}^{4}}+1 \\
& y=-{{x}^{2}}+{{x}^{4}}-1
\end{align}$
It is not the same as the provided equation, hence it is not symmetric about the origin.
The graph of the function is symmetric about the origin and thus the definition of even function is fulfilled.
