What is the the vertex of $y = {(x - 1)}^{2} + 2 x + 16$ $y=(x -1)^2 + 2x +16$ ?

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Answer 1 · 2018-04-17T15:07:19

$(0, 17)$ $(0,17)$

Expand the brackets first

$y = x^{2} - 2 x + 1 + 2 x + 16$ $y=x^2-2x+1+2x+16$
$y = x^{2} + 17$ $y=x^2+17$

What is the the vertex of y=(x -1)^2 + 2x +16y=(x−1)2+2x+16y=(x -1)^2 + 2x +16?