Show that x2−2x−1=(x−1)2−2x^2 - 2x - 1 = (x - 1)^2 - 2x2−2x−1=(x−1)2−2.
Hence, or otherwise, verify that the coordinates of the turning point of the graph of y=x2−2x−1y = x^2 - 2x - 1y=x2−2x−1 are (1,−2)(1, -2)(1,−2)
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