What is the vertex of # y=x^2-2x+1 #?

2 Answers
Mar 17, 2016

( 1 , 0 )

Explanation:

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]}

Mar 17, 2016

Vertex#->(x.y)->(1,0)#

Look at https://socratic.org/s/aMzfZyB2 for detailed determination of the vertex by 'completing the square'.

Explanation:

Compare to standard form of#" "y=ax^2+bx+c#

Rewrite as: #y=a(x^2+b/ax) +k#

In your case #a=1#

#x_("vertex")" "= (-1/2)xxb/a#

#x_("vertex")" " =" " (-1/2)xx(-2)" " =" " +1#

Substitute for x=1

#=>y_("vertex")=(1)^2-2(1)+1 = 0#
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Tony B