Graph rotation: explanation of equation

Question

Can someone please explain to me how in the attached image at the bottom right the author goes from $-y' = (-x')^2 -(-x')$ to $y = -x^2 - x$? Thanks.

Answer 1

$-y' = (-x')^2 -(-x')$

$-x'=x\quad\to\quad -y' = (x)^2 -(x)$

$-y'=y\quad\to\quad y = (x)^2 -(x)$

$y=x^2-x$

Answer 2

$$-y'=(-x')^2-(-x')$$ $$-y'=(x')^2+x'$$ $$y'=-(x')^2-x'$$ At this point we only drop the $'$s to indicate that we will be graphing on the same axes. $$y=-x^2-x$$

Answer 3

Rotation by $\pi$ radians is actually a central symmetry, hence if the center of rotation is the coordinate system's origin then each coordinate is transformed to its opposite:
$x$ to $-x$ and $y$ to $-y$