How do you complete the square for $y=x^2 - 9x$ ?

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Answer 1 · 2015-07-10T15:45:28

$y=(x-9/2)^2-81/4$

$(x+a)^2=x^2+2ax+a^2$ => $x^2+bx=(x+b/2)^2-b^2/4$
Therefore substituting $b=-9$ we have:
$y=x^2-9x=(x-9/2)^2-81/4$

How do you complete the square for y=x^2 - 9xy=x^2 - 9x?