91Ó°ÊÓ

For $x$-intercept, substitute $0$for $y$and then solve for $x$.

$\begin{array}{rcl}{\left(x^2+0^2-x\right)}^2& =& x^2+0^2\\ {\left(x^2-x\right)}^2& =& x^2\\ x^4-2x^3+x^2& =& x^2\\ x^4-2x^3& =& 0\\ x^3\left(x-2\right)& =& 0\\ x& =& 0,2\end{array}$

For $y$-intercept, substitute $0$for $x$and then solve for $y$.

$\begin{array}{rcl}{\left(0^2+y^2-0\right)}^2& =& 0^2+y^2\\ y^4& =& y^2\\ y^4-y^2& =& 0\\ y^2\left(y^2-1\right)& =& 0\\ y^2\left(y-1\right)\left(y+1\right)& =& 0\\ y& =& 0,1,-1\end{array}$

Perform the test for symmetry with respect to the x-axis.

Replace $y$with $-y$.

$\begin{array}{rcl}{\left(x^2+(-y{)}^2-x\right)}^2& =& x^2+(-y{)}^2\\ {\left(x^2+y^2-x\right)}^2& =& x^2+y^2\end{array}$

The above equation is same as the given equation.

Perform the test for symmetry with respect to the$y$-axis.

Replace $x$with $-x$.

$\begin{array}{rcl}{\left({\left(-x\right)}^2+y^2-\left(-x\right)\right)}^2& =& {\left(-x\right)}^2+y^2\\ {\left(x^2+y^2+x\right)}^2& =& x^2+y^2\end{array}$

The above equation is not same as the given equation.

Perform the test for symmetry with respect to the origin.

Replace $x$with $-x$and $y$with $-y$.

$\begin{array}{rcl}{\left({\left(-x\right)}^2+(-y{)}^2-\left(-x\right)\right)}^2& =& {\left(-x\right)}^2+(-y{)}^2\\ {\left(x^2+y^2+x\right)}^2& =& x^2+y^2\end{array}$

The above equation is not same as the given equation.

The $x$-intercept is $\left(0,0\right),\left(2,0\right)$.

The $y$-intercept is $\left(0,1\right),\left(0,-1\right)$.

The equation is symmetric with respect to the $x$-axis,