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Question 1062776: in the function f(x)=(x-2)^2+4, the minimum value occurs when x is
Found 4 solutions by rothauserc, math_helper, ikleyn, MathTherapy:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(x) = (x-2)^2 + 4
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f(x) = x^2 -2x +4 +4
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f(x) = x^2 -2x +8
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this is a parabola the curves upward
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the minimum value of f(x) occurs when x is the coordinate of the parabola's vertex
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x = -b/2a = -(-2) / (2*(1)) = 2 / 2 = 1
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the minimum value for f(x) occurs at x = 1
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here is a graph of f(x)
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Answer by math_helper(2461)  (Show Source):
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
in the function f(x)=(x-2)^2+4, the minimum value occurs when x is
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. . . x = 2. (answer).
Indeed, at x = 2 the function f(x) has the value of 4.
For any other value the function is greater than 4, because it adds the positive addend (x-2)^2 to 4.
It does not require any other/more explanations.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! in the function f(x)=(x-2)^2+4, the minimum value occurs when x is
Duplicate. See problem # Finance/1062778.
Vertex form: 
Minimum value occurs at h, and since - h = - 2, then that MINIMUM value occurs at: h = 2.
It's unbelievable!
Two people plotted graphs, one did the derivative, and one of those who graphed also FOILED, got an INCORRECT equation, and then an INCORRECT graph.
People, it's not that complicated! Why do you all have to go through so much to do a simple, simple problem? I just don't get it.
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