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  • 5 years ago

Find the vertex of the parabola. y=-x^2+2x-5

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  1. dumbcow
    • 5 years ago
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    option 1: get line of symmetry x = -b/2a, where a=-1 b=2, c=-5 x = -2/-2 = 1 substitute that in for x to find y y = -(1)^2 +2(1) -5 = -4 vertex: (1,-4) option 2: use completing the square to put it in vertex form y=a(x-h)^2 +k vertex: (h,k) factor out neg y = -(x^2 -2x +5) take half of b, square it and subtract from c y = -((x-1)^2 +5 - (-1)^2) y = -(x-1)^2 -4 vertex: (1,-4)

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