What is the vertex form of $y = x^{2} + 216 x + 432$ $y=x^2 +216x + 432$ ?

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Answer 1 · 2016-10-04T00:33:34

$y = {(x + 108)}^{2} - 11232$ $y=(x+108)^2-11232$

Complete the square with $x^{2} + 216 x$ $x^2+216x$
$y = x^{2} + 216 x + {(\frac{216}{2})}^{2} - {(\frac{216}{2})}^{2} + 432$ $y=x^2+216x+(216/2)^2-(216/2)^2+432$
Form a perfect square
$y = {(x + 108)}^{2} - {(\frac{216}{2})}^{2} + 432$ $y=(x+108)^2-(216/2)^2+432$
Simplify
$y = {(x + 108)}^{2} - 11664 + 432$ $y=(x+108)^2-11664+432$
$= {(x + 108)}^{2} - 11232$ $=(x+108)^2-11232$

What is the vertex form of y=x2+216x+432y=x^2 +216x + 432?