What is the vertex of $y = 1 - 4 x - x^{2}$ $y=1-4x-x^2$ ?

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Answer 1 · 2016-07-21T19:41:20

{-2,5}

$y = 1 - 4 x - x^{2}$ $y=1-4x-x^2$

$\frac{d y}{d x} = 0 - 4 - 2 x = 0$ $(d y)/(d x)=0-4-2x=0$

$- 4 - 2 x = 0$ $-4-2x=0$

$2 x = - 4; x = - \frac{4}{2} = - 2$ $2x=-4" ; "x=-4/2=-2$

$y = 1 - 4 (- 2) - {(- 2)}^{2}$ $y=1-4(-2)-(-2)^2$

$y = 1 + 8 - 4$ $y=1+8-4$ =5
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What is the vertex of y=1−4x−x2 y=1-4x-x^2 ?