What is the vertex of $y = {(x + 3)}^{2} - 6 x$ $y=(x +3)^2 -6x$ ?

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Answer 1 · 2017-07-01T14:00:27

$V = (0, 9)$ $V = (0, 9)$

If $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ then $V = (-b/(2a), - Delta/(4a))$

$y = x^2 + 6x + 9 - 6x$
$a = 1, b = 0, c = 9$
$Delta = 0^2 - 4 * 1 * 9 = -36$

What is the vertex of y=(x +3)^2 -6x y=(x+3)2−6xy=(x +3)^2 -6x ?