Question

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

Answer 1

( 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]}

Answer 2

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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