Answer
$(\pm2,0), (0,-8)$, no symmetry
Work Step by Step
Step 1. For x-intercept(s), let $y=0$, we have $x^2(x+2)-4(x+2)=0\longrightarrow (x+2)(x^2-4)=0\longrightarrow (x+2)(x+2)(x-2)=0\longrightarrow (\pm2,0)$
Step 2. For y-intercept(s), let $x=0$, we have $y=-8\longrightarrow (0,-8)$,
Step 3. To test for x-axis symmetry, replace $(x,y)$ with $(x,-y)$, we have $-y=(x)^3+2(x)^2-4(x)-8$ which is different from the original equation, thus it is not symmetric with respect to the x-axis,
Step 4. To test for y-axis symmetry, replace $(x,y)$ with $(-x,y)$, we have $y=(-x)^3+2(-x)^2-4(-x)-8$ which is different from the original equation, thus it is not symmetric with respect to the y-axis,
Step 5. To test for origin symmetry, replace $(x,y)$ with $(-x,-y)$, we have $-y=(-x)^3+2(-x)^2-4(-x)-8$ which is different from the original equation, thus it is not symmetric with respect to the origin.
