Identifying Graphs of Quadratic Functions @ State High
 f(x) = x2 + 1f(x) = (x-1)2f(x) = x2 - 1f(x) = (x+1)2 f(x) = x2 + 1f(x) = (x-1)2f(x) = x2 - 1f(x) = (x+1)2 f(x) = x2 + 1f(x) = (x-1)2f(x) = x2 - 1f(x) = (x+1)2 f(x) = x2 + 1f(x) = (x-1)2f(x) = x2 - 1f(x) = (x+1)2 f(x) = x2 + 2f(x) = (x-2)2f(x) = x2 - 2f(x) = (x+2)2 f(x) = x2 + 2f(x) = (x-2)2f(x) = x2 - 2f(x) = (x+2)2 f(x) = x2 + 2f(x) = (x-2)2f(x) = x2 - 2f(x) = (x+2)2 f(x) = x2 + 2f(x) = (x-2)2f(x) = x2 - 2f(x) = (x+2)2 f(x) = (x+4)2 + 3f(x) = (x+4)2 - 3f(x) = (x-4)2 - 3f(x) = (x-4)2 + 3 f(x) = (x+4)2 + 3f(x) = (x-4)2 + 3f(x) = (x-4)2 - 3f(x) = (x+4)2 - 3 f(x) = (x+4)2 + 3f(x) = (x+4)2 - 3f(x) = (x-4)2 + 3f(x) = (x-4)2 - 3 f(x) = (x+4)2 + 3f(x) = (x-4)2 - 3f(x) = (x-4)2 + 3f(x) = (x+4)2 - 3 f(x) = - (x+3)2 - 1f(x) = - (x+3)2 + 1f(x) = - (x-3)2 - 1f(x) = - (x-3)2 + 1 f(x) = - (x+3)2 - 1f(x) = - (x+3)2 + 1f(x) = - (x-3)2 - 1f(x) = - (x-3)2 + 1 f(x) =- (x+3)2 - 1f(x) = -(x+3)2 + 1f(x) = - (x-3)2 - 1f(x) = - (x-3)2 + 1 f(x) =- (x+3)2 - 1f(x) = -(x+3)2 + 1f(x) = - (x-3)2 - 1f(x) = - (x-3)2 + 1 f(x) = -2 (x-3)2 + 7f(x) = -2 (x+3)2 - 7f(x) = - 2 (x-3)2 - 7f(x) =- 2(x+3)2 +7 f(x) =- 2(x+3)2 - 7f(x) = - 2(x+3)2 + 7f(x) = -2(x-3)2 + 7f(x) = -2(x-3)2 - 7 f(x) = 2 (x-3)2 + 7f(x) = 2 (x+3)2 - 7f(x) = 2 (x-3)2 - 7f(x) = 2(x+3)2 +7
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