2x^2-4x-6=0 in vertex form and in standard form?
回答 (4)
2017-01-03 9:24 am
x^2-2x=3
x^2-2x+1 = 4
(x-1)^2 = 4
(x-1)^2-4 = 0
x^2-2x+1 = 4
(x-1)^2 = 4
(x-1)^2-4 = 0
2017-01-03 10:21 am
Standard form (US):
.. x^2 -2x -3 = 0 . . . . descending powers, relatively prime coefficients, =0.
Vertex form (standard form UK):
.. (x -1)^2 -4 = 0 . . . . a(x-h)^2 +k for vertex (h, k). Here, a=1.
_____
.. (x +1)(x -3) = 0 . . . . factored form
.. x^2 -2x -3 = 0 . . . . descending powers, relatively prime coefficients, =0.
Vertex form (standard form UK):
.. (x -1)^2 -4 = 0 . . . . a(x-h)^2 +k for vertex (h, k). Here, a=1.
_____
.. (x +1)(x -3) = 0 . . . . factored form
2017-01-03 8:49 am
2x^2-4x-6=0
General form: Y= 2*X^2-4*X-6
Standard form: Y = ( 2x² -4x +2) -2 -6 = 2(x -1)² -8
Transformational form: 0.5(Y +8) = (x -1)²
General form: Y= 2*X^2-4*X-6
Standard form: Y = ( 2x² -4x +2) -2 -6 = 2(x -1)² -8
Transformational form: 0.5(Y +8) = (x -1)²
2017-01-03 8:10 am
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收錄日期: 2021-05-01 21:24:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170103000832AA1M675