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Identifying Graphs of Quadratic Functions
供稿人: Collins
(原作者: Scott)
f(x) = x2 + 1
f(x) = (x-1)2
f(x) = x2 - 1
f(x) = (x+1)2
f(x) = x2 + 1
f(x) = (x-1)2
f(x) = x2 - 1
f(x) = (x+1)2
f(x) = x2 + 1
f(x) = (x-1)2
f(x) = x2 - 1
f(x) = (x+1)2
f(x) = x2 + 1
f(x) = (x-1)2
f(x) = x2 - 1
f(x) = (x+1)2
f(x) = x2 + 2
f(x) = (x-2)2
f(x) = x2 - 2
f(x) = (x+2)2
f(x) = x2 + 2
f(x) = (x-2)2
f(x) = x2 - 2
f(x) = (x+2)2
f(x) = x2 + 2
f(x) = (x-2)2
f(x) = x2 - 2
f(x) = (x+2)2
f(x) = x2 + 2
f(x) = (x-2)2
f(x) = x2 - 2
f(x) = (x+2)2
f(x) = (x+4)2 + 3
f(x) = (x-4)2 + 3
f(x) = (x-4)2 - 3
f(x) = (x-4)2 - 3
f(x) = (x+4)2 + 3
f(x) = (x-4)2 + 3
f(x) = (x-4)2 - 3
f(x) = (x-4)2 - 3
f(x) = (x+4)2 + 3
f(x) = (x-4)2 + 3
f(x) = (x-4)2 - 3
f(x) = (x-4)2 - 3
f(x) = (x+4)2 + 3
f(x) = (x-4)2 + 3
f(x) = (x-4)2 - 3
f(x) = (x-4)2 - 3
f(x) = - (x+3)2 - 1
f(x) = - (x+3)2 + 1
f(x) = - (x-3)2 - 1
f(x) = - (x-3)2 + 1
f(x) = - (x+3)2 - 1
f(x) = - (x+3)2 + 1
f(x) = - (x-3)2 - 1
f(x) = - (x-3)2 + 1
f(x) =- (x+3)2 - 1
f(x) = -(x+3)2 + 1
f(x) = - (x-3)2 - 1
f(x) = - (x-3)2 + 1
f(x) =- (x+3)2 - 1
f(x) = -(x+3)2 + 1
f(x) = - (x-3)2 - 1
f(x) = - (x-3)2 + 1
f(x) = -2 (x-3)2 + 7
f(x) = -2 (x+3)2 - 7
f(x) = - 2 (x-3)2 - 7
f(x) =- 2(x+3)2 +7
f(x) =- 2(x+3)2 - 7
f(x) = - 2(x-3)2 + 7
f(x) = -2 (x-3)2 + 7
f(x) = -2(x+3)2 - 7
f(x) = 2 (x-3)2 + 7
f(x) = 2 (x+3)2 - 7
f(x) = 2 (x-3)2 - 7
f(x) = 2(x+3)2 +7
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Simplify radical expressions
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