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Thread: Find the minimum value

20170218, 12:19 #1
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Find the minimum value
What is the minimum value of f(x) =  (x  h)² + k   q for each set of positive real numbers h, k, and q?
(A) q (B) k (C) k (D) k  q (E) k  q
While there's a 20% chance of guessing correctly, it's necessary to "explain" your answer.
This was a question on a recent standardized test for high school students.

20170218, 14:17 #2
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Possibly the highschool students had the benefit of mathematical typesetting  unfortunately I find the problem unintelligible as presented...
BATcher
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20170218, 14:53 #3
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Here's a graphic of f(x) =  (x  h)² + k   q
Clip0001.jpg

20170218, 16:29 #4
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The problem is therefore that I don't have a clue what the vertical bars represent!
And the other problem is that I don't know what
(A) q (B) k (C) k (D) k  q (E) k  q
means...
Perhaps I shouldn't have started this?BATcher
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20170218, 16:33 #5
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I assume that this is the modulus of the result  see Wikipedia. But like you Batcher I haven't a clue either
What do you mean nothing is impossible? I've been doing nothing for years.

20170218, 16:51 #6

20170219, 09:49 #7
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I take the vertical bars to mean "absolute value" (i.e., if the expression inside the vertical bars evaluates to a negative value, multiply it by 1). I'll defer posting my answer.

20170219, 11:00 #8
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The vertical bars mean ABSOLUTE VALUE. That's common mathematical notation.

20170220, 08:09 #9
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To minimize f(x) =  (x  h)² + k   q, try to set x such that taking the absolute value yields the lowest possible value, i.e., zero:
(x  h)² + k = 0,
(x  h)² = k,
(x  h)² = k,
x  h = +/sqrt(k), # "+/" represents "plus or minus"
x = +/sqrt(k) + h,
which is "real" because k is a positive real number.
So, the answer is (A) q.

20170220, 09:53 #10
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20170221, 08:50 #11
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I'm not sure I understand the question, but I believe I showed that the quadratic is zero if and only if x = +/sqrt(k) + h. Therefore, setting x = +/sqrt(k) + h will result in the quadratic's being zero.

20170223, 12:19 #12
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I'm not sure why you need to bother with what x might be: the absolute value part must be +ve, its minimum value must therefore be zero (for whatever x, h or k), and if we take q away from zero the minimum must be q.
What age might these students be?Last edited by mngerhold; 20170223 at 12:22.

20170223, 17:57 #13
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Mngerhold:
Standardized test(s) given to (generally) 11th grade students.
You are correct about your answer and approach. It's actually fairly straightforward.

20170228, 21:37 #14
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The solution is E kq. If x=h then (xh)=0>>0^2=0 >> 0+k  q = [abs val of k ] q = kq which is choice E.
Would you believe this is Algebra 1 ?

20170303, 07:22 #15
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I'm not seeing how setting x to h satisfies the problem of minimizing f(x). As I wrote earlier, that problem seems to require getting the entire expression within the absolute value symbols to equal zero.