What is the vertex of $y= 2x^2-x-3-4(x-1)^2$ ?

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Answer 1 · 2018-05-01T14:32:44

$"Vertex": (7/4, -7/8)$

$y= 2x^2-x-3-4(x-1)^2$
$y = 2x^2-x-3-4x^2+8x-4$
$y = -2x^2 + 7x - 7$

$f(x) = ax^2 + bx + c " : x vertex" = (-b)/(2a)$

$(-b)/(2a) = (-7)/(2(-2) = 7/4$

$y = -2(7/4)^2 + 7(7/4) - 7 = (-7)/8$

What is the vertex of y= 2x^2-x-3-4(x-1)^2 y= 2x^2-x-3-4(x-1)^2?