# 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?

Asked on by lillyan

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)

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}.

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