What are the values of parameter m if the vertex of parabola y = x^2-2(m-1)*x+m-1 is on the first bisector line?

Firts, we'll determine the coordinates of the vertex of
the parabola.

V(-b/2a ;
-delta/4a)

deta = b^2 -
4ac

a,b,c, are the coefficients of the
quadratic.

a = 1 , b = -2(m-1) , c =
m-1

xV = 2(m-1)/2

xV = 2m -
2

delta = 4(m-1)^2 -
4(m-1)

delta =
4(m-1)(m-1-1)

delta =
4(m-1)(m-2)

yV =
-4(m-1)(m-2)/4

yV =
-(m-1)(m-2)

The coordinates of the vertex are: V(2m - 2
;  -(m-1)(m-2)).

We'll impose the constraint of enumciation
that the vertex has to be located on the bisectrix of the first
quadrant.

yV = xV

2m - 2 =
-(m-1)(m-2)

2(m-1) =
-(m-1)(m-2)

We'll add
(m-1)(m-2):

2(m-1) + (m-1)(m-2) =
0

We'll factorize by
(m-1):

(m-1)(2 +m - 2) =
0

m(m-1) = 0

m1 =
0

m-1 = 0

m2 =
1

The values of m for the vertex of the
parabola to be on the bisector of the 1st quadrant are: {0 ;
1}.