What is the vertex of $y = {(x - 2)}^{2} - 2 x - 6$ $y= (x-2)^2-2x-6$ ?

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Answer 1 · 2018-04-23T11:56:30

Vertex is at $(3, - 11)$ $(3,-11)$

$y = {(x - 2)}^{2} - 2 x - 6 or y = x^{2} - 4 x + 4 - 2 x - 6$ $y= (x-2)^2-2 x-6 or y= x^2-4 x +4-2 x-6$ or

$y = x^{2} - 6 x - 2 or y = (x^{2} - 6 x + 9) - 9 - 2$ $y= x^2-6 x -2 or y= (x^2-6 x +9)-9 -2$ or

$y = {(x - 3)}^{2} - 11$ $y= (x-3)^2-11$ . Comparing with vertex form of

equation $f (x) = a {(x - h)}^{2} + k; (h, k)$ $f(x) = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=3 , k=-11 :.$ Vertex is at $(3,-11)$

graph{(x-2)^2-2 x-6 [-40, 40, -20, 20]} [Ans]

What is the vertex of y= (x-2)^2-2x-6y=(x−2)2−2x−6 y= (x-2)^2-2x-6?