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

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)
#"--------------------------------------------------------------------"#

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]}