What is the vertex of $y = - 2 {(x + 3)}^{2} + 12 x + 4$ $y=-2(x+3)^2+12x+4$ ?

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What is the vertex of $y = - 2 {(x + 3)}^{2} + 12 x + 4$ $y=-2(x+3)^2+12x+4$ ?

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Answer 1 · 2017-04-05T07:09:47

vertex $(0, - 14)$ $(0, -14)$

Explanation:

Given -

$y = - 2 {(x + 3)}^{2} + 12 x + 4$ $y=-2(x+3)^2+12x+4$

$y = - 2 (x^{2} + 6 x + 9) + 12 x + 4$ $y=-2(x^2+6x+9)+12x+4$

$y = - 2 x^{2} - 12 x - 18 + 12 x + 4$ $y=-2x^2-12x-18+12x+4$

$y = - 2 x^{2} - 14$ $y=-2x^2-14$

$x$ $x$ term is missing in the expression $- 2 x^{2} - 14$ $-2x^2-14$
Let us supply it.

$y = - 2 x^{2} + 0 x - 14$ $y=-2x^2+0x-14$

$x = \frac{- b}{2 \times a} = \frac{0}{2 \times (- 2)} = 0$ $x=(-b)/(2xxa)=0/(2xx (-2))=0$

At $x = 0$ $x=0$

$y = - 2 {(0)}^{2} - 14 = - 14$ $y=-2(0)^2-14=-14$

vertex $(0, - 14)$ $(0, -14)$

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What is the vertex of $y = - 2 {(x + 3)}^{2} + 12 x + 4$ $y=-2(x+3)^2+12x+4$ ?

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What is the vertex of y=-2(x+3)^2+12x+4 y=−2(x+3)2+12x+4y=-2(x+3)^2+12x+4 ?

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Explanation:

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What is the vertex of $y = - 2 {(x + 3)}^{2} + 12 x + 4$ $y=-2(x+3)^2+12x+4$ ?