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anonymous
 5 years ago
Find the vertex of the parabola.
y=x^2+2x5
anonymous
 5 years ago
Find the vertex of the parabola. y=x^2+2x5

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0option 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(xh)^2 +k vertex: (h,k) factor out neg y = (x^2 2x +5) take half of b, square it and subtract from c y = ((x1)^2 +5  (1)^2) y = (x1)^2 4 vertex: (1,4)
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