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

Mar 17, 2016

( 1 , 0 )

#### Explanation:

The standard form of the quadratic function is $y = a {x}^{2} + b x + c$

The function $y = {x}^{2} - 2 x + 1 \text{ is in this form }$

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

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

x-coord of vertex $= - \frac{b}{2 a} = - \frac{- 2}{2} = 1$

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

$y = {\left(1\right)}^{2} - 2 \left(1\right) + 1 = 0$

thus coordinates of vertex = (1 , 0)
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Alternatively : factorise as $y = {\left(x - 1\right)}^{2}$

compare this to the vertex form of the equation

$y = {\left(x - h\right)}^{2} + k \text{ (h,k) being the vertex }$

now $y = {\left(x - 1\right)}^{2} + 0 \Rightarrow \text{ vertex } = \left(1 , 0\right)$
graph{x^2-2x+1 [-10, 10, -5, 5]}

Mar 17, 2016

Vertex$\to \left(x . y\right) \to \left(1 , 0\right)$

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

#### Explanation:

Compare to standard form of$\text{ } y = a {x}^{2} + b x + c$

Rewrite as: $y = a \left({x}^{2} + \frac{b}{a} x\right) + k$

In your case $a = 1$

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

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

Substitute for x=1

$\implies {y}_{\text{vertex}} = {\left(1\right)}^{2} - 2 \left(1\right) + 1 = 0$
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