What is the vertex of $y= -(2x-1)^2+x^2-x+3$ ?

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Answer 1 · 2016-04-13T14:02:54

$"vertex "->(x,y)->(1/2,11/4)$

Multiply out the brackets giving:

$y=-(4x^2-4x+1)+x^2-x+3$

Multiply everything inside the bracket by $(-1)$ giving

$y=-4x^2+4x-1+x^2-x+3$

$y=-3x^2+3x+2$

Write as: $y=-3(x^2+3/(-3)x)+2$

$=>y=-3(x^2-x)+2$

Consider the coefficient $-1$ from $-x$ inside the brackets

$color(blue)(x_("vertex")=(-1/2)xx(-1)=+1/2)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for #x_("vertex") in the equation

$color(brown)(y=-3x^2+3x+2" "->" " y=-3(color(blue)(1/2))^2+3(color(blue)(1/2) )+2$

$color(blue)(y_("vertex")= 2 3/4 = 11/4)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("vertex "->(x,y)->(1/2,11/4)$

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What is the vertex of y= -(2x-1)^2+x^2-x+3 y= -(2x-1)^2+x^2-x+3?