# How do you write F(x)=x^2-2x+5 in vertex form?

Jun 20, 2017

$y = {\left(x - 1\right)}^{2} + 4$

#### Explanation:

Given -

$F \left(x\right) = {x}^{2} - 2 x + 5$

$y = {x}^{2} - 2 x + 5$

Vertex

x - coordinate of the vertex

$x = \frac{- b}{2 a} = \frac{- \left(- 2\right)}{2 \times 1} = \frac{2}{2} = 1$

y - coordinate

$y = {\left(1\right)}^{2} - 2 \left(1\right) + 5 = 1 - 2 + 5 = 4$

Vertex $\left(1 , 4\right)$

Parabola in vertex form

$y = a {\left(x - h\right)}^{2} + k$

Where -

$a = 1$ ----- coefficient of ${x}^{2}$
$h = 1$ --- x - coordinate of the vertex
$k = 4$ --- y - coordinate of the vertex

Substitute these values into the equation.

$y = 1 {\left(x - 1\right)}^{2} + 4$
$y = {\left(x - 1\right)}^{2} + 4$
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