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

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Answer 1 · 2015-03-12T22:48:24

The vertex form of an equation is in the form:
$y = {(x + a)}^{2} + b$ $y = (x + a)^2 + b$

${(x - a)}^{2}$ $(x-a)^2$ when expanded is $x^{2} - 2 a x + a^{2}$ $x^2 -2ax +a^2$

for the given equation, it follows that
$2 a x = - 6 x$ $2ax = -6x$
$\to a = - 3$ $rarr a=-3$

${(x - 3)}^{2} = x^{2} - 6 x + 9$ $(x-3)^2 = x^2 - 6x +9$

comparing this to the given equation, we see that
$b = - 3$ $b = -3$

So the vertex form of the given equation is
$y = {(x - 3)}^{2} - 3$ $y = (x-3)^2 - 3$

What is the vertex form of y=x^2-6x+6y=x2−6x+6y=x^2-6x+6?