What is the the vertex of $y = x^{2} - 2 x - 16$ $y = x^2 - 2x - 16$ ?

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Answer 1 · 2017-04-03T05:09:12

minimum at $- 17$ $-17$

$y = x^{2} - 2 x - 16$ $y = x^2 - 2 x -1 6$

by using completing a square,

$y = {(x - 1)}^{2} - {(- 1)}^{2} - 16$ $y = (x - 1)^2 - (-1)^2 -1 6$

$y = {(x - 1)}^{2} - 17$ $y = (x - 1)^2 -17$

vertex is minimum at $- 17$ $-17$

What is the the vertex of y=x2−2x−16y = x^2 - 2x - 16?