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Theloshua
 one year ago
is this equation a function? x^2 + y = 1
Theloshua
 one year ago
is this equation a function? x^2 + y = 1

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.1they want you to say "yes" even though the answer is "no"

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1you can rewrite it as \[y=1x^2\] whch makes\(y\) dependent on \(x\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1the reason it is not really a function is that it is not written as one say like \[f(x)=1x^2\] but nvm you math teacher does not know any better, just say "yes"

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1of course you are, this question is ill posed

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0Misty, you are not making any sense aha, my teacher cant be that stupid that he doesnt know the answer

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1think of it this way: can you solve \[x^2+y=1\] for \(y\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 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
 one year ago
Best ResponseYou've already chosen the best response.0so it is a function??

Theloshua
 one year ago
Best ResponseYou've already chosen the best response.0what? yes? Misty, haha, make up your mind

lfreeman1285
 one year ago
Best ResponseYou've already chosen the best response.0Yes, it is a function. A function is defined as "a relationship or expression involving one or more variables," so this fits the bill.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0"\(y(x)=−x^2+1\) is indeed a function', agreed but \(x(y) = \sqrt{1y}\) isn't i would agree: the question is " ill posed ".

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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 \(y=\sqrt{1x}\) is not a function (I've swapped the variables for clarity). By definition, \(\sqrt{1x}\) 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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