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find and label the vertex and line of symetry f(x)=3(x-1)^2 i have the vertex as(1,0) and line of symetry 0
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\[f(x) = 3(x-1)^{2} = 3(x ^{2}-2x+1)=3x ^{2} - 6x + 3\] Line of symmetry: -b/2a = -(-6)/2(3) = 6/6 = 1 ==> x = 1 f(1) = 0 and thus vertex is (1,0)
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