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

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

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Answer 1 · 2015-07-10T15:45:28

$y = {(x - \frac{9}{2})}^{2} - \frac{81}{4}$ $y=(x-9/2)^2-81/4$

Explanation:

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

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

1 Answer

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