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Identify the vertex and axis of symmetry f(x)= - 2x^2 + 2x-1
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If you arrange it into f(x) = a(x-h)^2 +k (h,k) is the coordinates of vertex and x=h is the axis of symmetry. In your case, f(x)= - 2x^2 + 2x-1 = -2 (x^2 -x) -1 = -2 ( x^2 -x +(1/2)^2 - (1/2)^2) -1 = -2 ( x- 1/2)^2 + 2(1/2)^2 -1 = -2 ( x- 1/2)^2 +1/2 -1 = -2 (x - 1/2)^2 -1/2 Can you get the answer from here?
I got 1/2 and -1/2
Complete the square, - 2x^2 + 2x-1 -2(x^2-x)-1 \[-2(x-1/2)^2-1/2(-2)-1\] \[-2(x-1/2)^2\] Symmetry at x=1/2
wait
−2(x−1/2)2−1/4(−2)−1 = -2 (x - 1/2)^2 -1/2 symmetry at x=1/2
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@dooley you're right for vertex
axis of symmetry is x=1/2 vertex is at (1/2, -1/2)
Thanks u guys!!1
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