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

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Answer 1 · 2017-11-14T21:32:42

$(2.5,-0.5) min$

$y'=2(x-3)*1+2x-4$
$=> 2x-6+2x-4$
$=> 4x-10$
$=> 2(2x-5)$

$y'=0$
$=>$
$2(2x-5)=0$
$=> 2x-5=0$
$=> 2x=5$
$=> x=5/2=2.5$

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

$y_((2.5))=(2.5-3)^2+(2.5)^2-4(2.5)+3=$
$=(-0.5)^2+(2.5)^2-10+3$
$=0.25+6.25-7$
$=-0.5$

$(2.5,-0.5) min$

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