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Identifying Graphs of Quadratic Functions
Contribució de: Palese
(Autor original: 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 + 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 + 4
f(x) = (x-4)2
f(x) = -x2 + 4
f(x) = -(x+4)2
f(x) = x2 - 2
f(x) = -x2 - 2
f(x) = -x2 + 2
f(x) = -(x+2)2
f(x) = 8x2
f(x) = -8x2
f(x) = -(⅛)x2
f(x) = (⅛)x2
f(x) = (⅛)x2 - 6
f(x) = -(⅛)x2 - 6
f(x) = -(⅛)x2 + 6
f(x) = (⅛)x2 + 6
f(x) = -(¼)x2 + 2
f(x) = (¼)x2 - 2
f(x) = -(¼)x2 - 2
f(x) = (¼)x2 + 2
f(x) = -(¼)x2 + 3
f(x) = (¼)x2 - 3
f(x) = -(¼)x2 - 3
f(x) = (¼)x2 + 3
f(x) = -(¼)x2
f(x) = (¼)x2
f(x) = -4x2
f(x) = 4x2
f(x) = -(¼)x2
f(x) = (¼)x2
f(x) = -4x2
f(x) = 4x2
f(x) = -8x2 - 2
f(x) = (⅛)x2 - 2
f(x) = -(⅛)x2 - 2
f(x) = -8x2 + 2
f(x) = -8x2 - 1
f(x) = 8x2 - 1
f(x) = -(⅛)x2 - 1
f(x) = -8x2 + 1
A
B
C
D
A
B
C
D
The vertex will be above the x-axis
The vertex will be below the x-axis
The vertex will be on the x-axis
None of the above
The parabola will not have a concavity
 The parabola will be concave down
 The parabola will be concave up
None of the above
The vertex will be above the x-axis
The vertex will be below the x-axis
The vertex will be on the x-axis
 None of the above
 The parabola will be wider than y=x2 
 The parabola will be narrower than y=x2 

 The parabola will be the same as y=x2 
None of the above
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