# What is the the vertex of y = (x - 3)^2 + x^2-4x+3?

Nov 14, 2017

$\left(2.5 , - 0.5\right) \min$

#### Explanation:

$y ' = 2 \left(x - 3\right) \cdot 1 + 2 x - 4$
$\implies 2 x - 6 + 2 x - 4$
$\implies 4 x - 10$
$\implies 2 \left(2 x - 5\right)$

$y ' = 0$
$\implies$
$2 \left(2 x - 5\right) = 0$
$\implies 2 x - 5 = 0$
$\implies 2 x = 5$
$\implies x = \frac{5}{2} = 2.5$

$y ' ' = 4 > 0 \implies \min$

${y}_{\left(2.5\right)} = {\left(2.5 - 3\right)}^{2} + {\left(2.5\right)}^{2} - 4 \left(2.5\right) + 3 =$
$= {\left(- 0.5\right)}^{2} + {\left(2.5\right)}^{2} - 10 + 3$
$= 0.25 + 6.25 - 7$
$= - 0.5$

$\left(2.5 , - 0.5\right) \min$