How do you write $y=x^2-6x+11$ in vertex form?

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Answer 1 · 2016-07-19T00:59:41

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

x-coordinate of vertex:
$x = -b/(2a) = 6/2 = 3$
y-coordinate of vertex:
y(3) = 9 - 18 + 11 = 2
Vertex (3, 2)
Vertex form of y:
$y = (x - 3)^2 + 2$

How do you write y=x^2-6x+11y=x^2-6x+11 in vertex form?