find the optimum value for each of the following quadratic functions and the axis of symmetry of their graphs by completing the square.
y=-X^2+4X
y=x^2-2X+3
y=X^2+6X+36
y=-X^2-8X+7
呢條數點做(20點)
2007-10-14 10:54 pm
回答 (2)
2007-10-14 11:07 pm
✔ 最佳答案
y = -x^2 + 4x= -(x^2 - 4x)
= -(x^2 - 4x + 4) + 4
= -(x - 2)^2 + 4
optimum value = max value = 4
axis of symmetry : x = 2
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y = x^2 - 2x + 3
= x^2 - 2x + 1 - 1 + 3
= (x - 1)^2 + 2
optimum value = min value = 2
axis of symmetry : x = 1
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y = x^2 + 6x + 36
= x^2 + 6x + 9 - 9 + 36
= (x + 3)^2 + 27
optimum value = min value = 27
axis of symmetry : x = -3
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y = -x^2 - 8x + 7
= -(x^2 + 8x) + 7
= -(x^2 + 8x + 16) + 16 + 7
= -(x + 4)^2 + 23
optimum value = max value = 23
axis of symmetry : x = -4
2007-10-14 11:11 pm
for y=-X^2+4X
y=-1(x^2-4x)
=-1[(x^2-4x+4-4)]
=-1[(x-2)^2-4]
=-1(x-2)^2+4
i.e. the maximum point is (2,4) .
i.e. axis of symmetry is x=2 ,the value is y=4 .
y=-1(x^2-4x)
=-1[(x^2-4x+4-4)]
=-1[(x-2)^2-4]
=-1(x-2)^2+4
i.e. the maximum point is (2,4) .
i.e. axis of symmetry is x=2 ,the value is y=4 .
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