What is the the vertex of $y = - x^{2} - 12 x - 4$ $y = -x^2 - 12x - 4$ ?

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Answer 1 · 2017-08-25T11:32:54

Vertex is at $(- 6, 32)$ $(-6,32)$

$y = - x^{2} - 12 x - 4 or y = - (x^{2} + 12 x) - 4$ $y= -x^2-12x-4 or y = -(x^2+12x) -4$

$y = - (x^{2} + 12 x + 36) + 36 - 4$ $y = -(x^2+12x+36) +36 - 4$

$y = - {(x + 6)}^{2} + 36 - 4 = - {(x + 6)}^{2} + 32$ $y = -(x+6)^2 +36 - 4 = -(x+6)^2 +32$ . Comparing with

vertex form of equation $y = a {(x - h)}^{2} + k; (h, k)$ $y = a(x-h)^2+k ; (h,k)$ being vertex ,

we find here $h= -6 , k =32 :.$ Vertex is at $(-6,32)$ [Ans]

What is the the vertex of y = -x^2 - 12x - 4 y=−x2−12x−4y = -x^2 - 12x - 4 ?