What's the equation of the graph $y=x^3-x^2+x-2$ after it is reflected in the x axis, and then the y axis as well? I don't fully understand the steps to reach the answer. 
I've plotted the graph and I found out that the graph is an odd graph and that after reflecting in both axes, the only thing that changes on the graph are the points where it intersects the x & y axis. 
I would like to see the algebraic steps one would take on how to solve this equation to better grasp what is going on, thanks.
 A: Reflecting wrt $x$ axis, inverts the sign of $y$ coordinate. So After reflecting wrt $x$ axis, the equation is:
$$
-y = x^3-x^2+x-2
$$
Now reflecting wrt $y$ axis, inverts the sign of $x$. This gives:
$$
-y = (-x)^3-(-x)^2+(-x)-2
$$
$$
-y = -x^3-x^2-x-2
$$
$$
y = x^3+x^2+x+2
$$
A: you wanted to understand fully so here is an explanation.
what happens when you reflect a graph about the x-axis. All the points (x,y)on the graph 
changes their y value with -y because distance still remains the same thus only sign is changing you can also see in the graph below
$\space \space $
same happens when you reflect a graph w.r.t. to the y-axis   sign of x changes with -x  
thus first change the sign of y   so your equation becomes $$-y=x^3-x^2+x-2$$
now change the sign of x so your equation becomes $$-y=(-x)^3-(-x)^2-x-2$$ 
can be written in simplified form as $$y=x^3+x^2+x+2$$ 
and the graph of the final equation looks like this 

A: 1)Reflection about $x-$axis:
$(x,y) \rightarrow (x,-y)$;
2)Reflection about $y$-axis:
$(u,v) \rightarrow (-u,v)$;
Combining:
3) $(x,y) \rightarrow (x,-y) \rightarrow (-x,-y)$.
Given : $y=f(x) =x^3-x^2+x-2$;
We have $(x,f(x)) \rightarrow (-x,-f(x))$, the reflected graph .
$(-x)$ is mapped to $-f(x)$, or
$x = -(-x)$ is mapped to $-f(-x)$.
The reflected graph is $(x,-f(-x))$, or
$y = -( (-x)^3 -(-x)^2+(-x) -2)$.
