SOLUTION: Given f(x) = -2x^2 + 12x - 5 I must complete the square and put in the form y = a(x - h)^2 + k
-2x^2 + 12x = 5
x^2 - 6x = -5/2
x^2 - 6x + 9 = 13/2
(x - 3)^2 = 13/2
x
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-> SOLUTION: Given f(x) = -2x^2 + 12x - 5 I must complete the square and put in the form y = a(x - h)^2 + k
-2x^2 + 12x = 5
x^2 - 6x = -5/2
x^2 - 6x + 9 = 13/2
(x - 3)^2 = 13/2
x
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Question 362381: Given f(x) = -2x^2 + 12x - 5 I must complete the square and put in the form y = a(x - h)^2 + k
-2x^2 + 12x = 5
x^2 - 6x = -5/2
x^2 - 6x + 9 = 13/2
(x - 3)^2 = 13/2
x - 3 = sqrt(13/2)
x = 3 +/- sqrt(13/2)
From here, how I would put this into the standard form? I have h = -b/2a = 3 and k = (c-b)^2/4a = -289/8 (?)
Not sure how I would enter x, wouldn't there be two of them?
Thanks for any help you can provide :-) Answer by Fombitz(32388) (Show Source):
