Where does the normal line to the parabola y = x − x2 at the point (1, 0) intersect the parabola a second time?

Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)

we just need the slope

take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1

the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation

solve for where y=x-1 and y=x-x² intersect

set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero

x-1=0
x=1
we got this point already

x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2

it intersects at (-1,-2)