Theloshua
  • Theloshua
is this equation a function? x^2 + y = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
they want you to say "yes" even though the answer is "no"
misty1212
  • misty1212
you can rewrite it as \[y=1-x^2\] whch makes\(y\) dependent on \(x\)

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More answers

misty1212
  • misty1212
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"
Theloshua
  • Theloshua
im soooo confusssed
misty1212
  • misty1212
of course you are, this question is ill posed
Theloshua
  • Theloshua
Misty, you are not making any sense aha, my teacher cant be that stupid that he doesnt know the answer
misty1212
  • misty1212
think of it this way: can you solve \[x^2+y=1\] for \(y\)?
Theloshua
  • Theloshua
no
anonymous
  • anonymous
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.
Theloshua
  • Theloshua
so it is a function??
misty1212
  • misty1212
yes say yes
Theloshua
  • Theloshua
what? yes? Misty, haha, make up your mind
lfreeman1285
  • lfreeman1285
Yes, it is a function. A function is defined as "a relationship or expression involving one or more variables," so this fits the bill.
IrishBoy123
  • IrishBoy123
"\(y(x)=−x^2+1\) is indeed a function', agreed but \(x(y) = \sqrt{1-y}\) isn't i would agree: the question is " ill posed ".
anonymous
  • anonymous
@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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