# Qual é o vértice de  y = x ^ 2-2x + 1 ?

(1, 0)

#### Explicação:

The standard form of the quadratic function is y =ax^2+bx+c

The function  y = x^2 - 2x + 1 " is in this form "

with a = 1 , b = -2 and c = 1

the x-coordinate of the vertex can be found as follows

x-coord of vertex  = - b/(2a )= -(-2)/2 = 1

substitute x = 1 into equation to obtain y-coord.

 y = (1)^2 -2(1) + 1 = 0

thus coordinates of vertex = (1 , 0)
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Alternatively : factorise as y = (x - 1 )^2

compare this to the vertex form of the equation

 y = (x - h )^2 + k " (h,k) being the vertex "

now y = (x-1)^2 + 0 rArr " vertex " = (1,0)
graph{x^2-2x+1 [-10, 10, -5, 5]}