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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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HI!!
they want you to say "yes" even though the answer is "no"
you can rewrite it as \[y=1-x^2\] whch makes\(y\) dependent on \(x\)

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