Mark M. answered • 02/23/15

Mathematics Teacher - NCLB Highly Qualified

^{2}+ b

^{2}+ b

^{2}+ b

^{2}/27 + b

^{2}/27 + 26/3

Steven M.

asked • 02/23/15Took a pretest and this one stumped me. Any help would be greatful

http://i.gyazo.com/6703f24740c64115aae5d96dabaf09c6.png

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Mark M. answered • 02/23/15

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The relation could be a function. As it is presented it is not linear. The possible alternative is a parabola.

f(x) = ax^{2} + b

f(6) = a(6)^{2} + b

6 = 36a + b

f(3) = a(3)^{2} + b

8 = 9a + b

2 = -27a

-2/27 = a

f(x) = -2x/27 + b

f(3) = -2(3)^{2}/27 + b

8 = -2(9)/27 + b

8 = -2/3 + b

26/3 = b

f(x) = -2x/27 + 26/3

Testing this with another pair:

f(7) = -2(7)^{2}/27 + 26/3

8 = -98/27 + 26/3

8 = -98/27 + 234/27

8 = 136/27

8 ≠ 5.03

Assuming my arithmetic is correct, what is presented is neither linear nor parabolic.

Arthur D. answered • 02/23/15

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in a function, you can't have the same x-value go to 2 different y-values

"one to many" is not a function

in a function you can have 2 x-values go to the same y-value

"many to one" is a function

you have 3,6,7, and 9 for x-values

choose 8 for the x-value because you don't have 8 as an x-value *and...*

choose 9 for the y-value and you haven't broken either rule

Jon P. answered • 02/23/15

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Knowledgeable Math, Science, SAT, ACT tutor - Harvard honors grad

It's an odd question, because it looks to me that there is more than one correct answer.

First of all, they must mean "...so that it shows a function of x" because it's already not a function of y -- since there are two x values when y = 8.

Now, since there are already rows for x values 6, 3, 9, and 7 you can't use any of those numbers for x. That's because a function can only have one y value for each x value, so you can't repeat the same x value.

So either 8 or 12 for x should work, and then any y value can be used.

But maybe I'm not understanding the question?

Mark M.

Repetition of the input, the x-value, negates a relation from being a function. Repetition of the output, the v-value, has no relation to being a function.

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02/23/15

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Mark M.

02/24/15