Answer
$(\pm2,0),(0,-8)$
symmetric with respect to the y-axis,
Work Step by Step
Step 1. Let $y=f(x)=x^4-2x^2-8=(x^2+2)(x^2-4)=(x^2+2)(x+2)(x-2)$, we can find the x-intercept(s) $(\pm2,0)$ and y-intercept(s) $f(0)=-8$ or $(0,-8)$
Step 2. To test for x-axis symmetry, replace $(x,y)$ with $(x,-y)$, we have $-y=x^4-2x^2-8$ which is different from the original equation, thus it is not symmetric with respect to the x-axis,
Step 3. To test for y-axis symmetry, replace $(x,y)$ with $(-x,y)$, we have $y=x^4-2x^2-8$ which is the same as the original equation, thus it is symmetric with respect to the y-axis,
Step 4. To test for origin symmetry, replace $(x,y)$ with $(-x,-y)$, we have $-y=x^4-2x^2-8$ which is different from the original equation, thus it is not symmetric with respect to the origin.
