Find the minimum value

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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.

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Possibly the high-school students had the benefit of mathematical typesetting – unfortunately I find the problem unintelligible as presented…

BATcher

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Here’s a graphic of f(x) = | -(x – h)² + k | – q

46719-Clip0001

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Here’s a graphic of f(x) = | -(x – h)² + k | – q

46719-Clip0001

it still makes no sense
but i only did 5 years of graduate school as a math major

clearly they are inventing new meanings for symbols in HS
or they are clueless about how to properly write the question

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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

Plethora means a lot to me.

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And the other problem is that I don’t know what
(A) -q (B) -k (C) k (D) -k – q (E) k – q
means…

Multiple choice answers

A. -q
B. -k
C. k
D. -k-q
E. k-q

The clue was in the 1 in 5 chance of getting it right!

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The vertical bars mean ABSOLUTE VALUE. That’s common mathematical notation.

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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:blink:

Eliminate spare time: start programming PowerShell

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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.

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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.

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While you have found the correct answer, how do you know for sure that the absolute value of the quadratic is = 0?

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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.

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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?

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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 straight-forward.

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The solution is E k-q. If x=h then (x-h)=0>>0^2=0 >>| -0+k | -q = [abs val of -k ] -q = k-q which is choice E.

Would you believe this is Algebra 1 ?

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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.

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DHL, you are right.

FYI: students are permitted to use a graphing calculator on the test. Since the constants are real, one might set them all = 1 and graph. I would look like this, where the minimum value is -q since q = 1.

46856-Clip0006

Even if h and k are changed to other values and q remains = 1, the graph shifts and would widen or get narrower.

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This is the minimum value or lowest point on a parabola not 0. the lowest point of this parabola is at k-q.
This is when the abs value is k and f(x) = k-q.
This is the vertex of the curve.
Go to KhanAcademy.org and watch the videos on Parabolas in the algebra 1 section. I have been doing this for the past 4 months. It is a great way to brush up on your math up thru calculus.

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philomel: are you questioning if the answer is A which is -q? That’s the correct answer. As you can see from the graph, the min value is -1. If q = 1, the min is -q.

Khan is a good source if you can stand to listen to him…I can’t stand him, frankly. Also, his grammar leave much to be challenged and some of this math statements are actually INCORRECT! But, for most, he’s a good source.

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the way that i read the problem
if there is a minimum
it occurs at x=h and k=0 ; for all values of q

that is a bit advanced for 11th grade
even if it was not stated well

but if the idea is to use graphing calculators to do the solution
it is more a how to enter a formula into your calculator problem
than a real math problem

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The vertical bars have always meant absolute value (well, since 1841, anyway).
That’s not new notation!
The problem can be solve algebraically as has been stated or can be solved by graphing.
Students may use graphing calculators on the standardized test(s).

This is NOT advanced for 11th grade. Many students in high school today take precalculus as a sophomore and their first year of calculus as a junior.

https://en.wikipedia.org/wiki/Absolute_value

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