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Theloshua

  • one year ago

is this equation a function? x^2 + y = 1

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    they want you to say "yes" even though the answer is "no"

  3. misty1212
    • one year ago
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    you can rewrite it as \[y=1-x^2\] whch makes\(y\) dependent on \(x\)

  4. misty1212
    • one year ago
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    the reason it is not really a function is that it is not written as one say like \[f(x)=1-x^2\] but nvm you math teacher does not know any better, just say "yes"

  5. Theloshua
    • one year ago
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    im soooo confusssed

  6. misty1212
    • one year ago
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    of course you are, this question is ill posed

  7. Theloshua
    • one year ago
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    Misty, you are not making any sense aha, my teacher cant be that stupid that he doesnt know the answer

  8. misty1212
    • one year ago
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    think of it this way: can you solve \[x^2+y=1\] for \(y\)?

  9. Theloshua
    • one year ago
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    no

  10. anonymous
    • one year ago
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    Sorry to interrupt, but \(y=-x^2+1\) is indeed a function. It is a parabola opening downward with its vertex at (0, 1). Don't mean to mess up what you're doing.

  11. Theloshua
    • one year ago
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    so it is a function??

  12. misty1212
    • one year ago
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    yes say yes

  13. Theloshua
    • one year ago
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    what? yes? Misty, haha, make up your mind

  14. lfreeman1285
    • one year ago
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    Yes, it is a function. A function is defined as "a relationship or expression involving one or more variables," so this fits the bill.

  15. IrishBoy123
    • one year ago
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    "\(y(x)=−x^2+1\) is indeed a function', agreed but \(x(y) = \sqrt{1-y}\) isn't i would agree: the question is " ill posed ".

  16. anonymous
    • one year ago
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    @IrishBoy123 and @misty1212 , I think you'd have to agree that, in 100% of high school mathematics courses, \(x\) is the independent variable and \(y\) is the dependent variable. Otherwise, we'd have this same argument for every quadratic function, regardless of how it's written. I've never seen the notation \(y(x)\) or \(x(y)\) used in a high school textbook - it's always \(f(x)\), \(g(y)\), or something similar. And I disagree that \(y=\sqrt{1-x}\) is not a function (I've swapped the variables for clarity). By definition, \(\sqrt{1-x}\) is the principal (positive) square root only, making this a function. It goes back to the difference between solving \(x^2 = 4\) and solving \(x=\sqrt{4}\). The answer to the former is \(\pm2\) while the answer to the latter is \(2\) only.

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