Let k be a constant. If y= -4(x^2) + 6x + (2k-3) < 0 for any real number x, find the range of values of k.
*解得詳細D.....thx
Quadratic function--F.4
2009-12-31 4:23 am
回答 (2)
2009-12-31 4:30 am
✔ 最佳答案
Because y = -4(x^2) + 6x + (2k-3) is smaller than zero, that means it is not possible to have y = -4(x^2) + 6x + (2k-3) = 0. The quadratric equation -4(x^2) + 6x + (2k-3) = 0 has no real root. Hence the discriminant < 0
Discriminant = 6^2 - 4(-4)(2k - 3) < 0
36 + 32k - 48 < 0
32k < 12
k < 3/8
2009-12-31 5:06 am
y-coordinate of vertex = (2k-3) - 6^2/4(-4) < 0
k < 3/8
k < 3/8
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