Answer
The x-intercepts are (0,0) and (2,0)
The y-intercept are (0,-1), (0,0), and (0,1)
The equation has symmetry only with respect to the x-axis.
Work Step by Step
To find the x-intercept(s), we set y to 0 and solve for x:
$(x^2+0^2-x)^2=x^2+0^2$
$(x^2-x)^2=x^2$
$\sqrt{(x^2-x)^2}=\sqrt{x^2}$
$x^2-x=x$
$x^2-2x=0$
$x(x-2)=0$
$x_1=0$
$x_2-2=0\rightarrow x_2=2$
To find the y-intercept(s), we set x to 0 and solve for y:
$(0^2+y^2-0)^2=0^2+y^2$
$(y^2)^2=y^2$
$y^4-y^2=0$
$y^2(y^2-1)=0$
Two options:
$y_1^2=0\rightarrow y_1=0$
And:
$y^2-1=0$
$y^2=1$
$\sqrt{y^2}=\sqrt{1}$
$y=\pm1$
To test for symmetry with respect to the x-axis, we substitute y for -y and check if it equals the original equation:
$(x^2+(-y)^2-x)^2=x^2+(-y)^2$
$(x^2+y^2-x)^2=x^2+y^2 \checkmark$
To test for symmetry with respect to the y-axis, we substitute x for -x and check if it equals the original equation:
$((-x)^2+y^2-(-x))^2=(-x)^2+y^2$
$(x^2+y^2+x)^2=x^2+y^2$ nope
To test for symmetry with respect to the origin, we substitute x for -x, substitute y for -y and check if it equals the original equation:
$((-x)^2+(-y)^2-(-x))^2=(-x)^2+(-y)^2$
$(x^2+y^2+x)^2=x^2+y^2$ nope
