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

Question

misty1212 · Answer

HI!!

misty1212 · Answer

they want you to say "yes" even though the answer is "no"

misty1212 · Answer

you can rewrite it as 
$$y=1-x^2$$ whch makes$y$ dependent on $x$

misty1212 · Answer

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 · Answer

im soooo confusssed

misty1212 · Answer

of course you are, this question is ill posed

theloshua · Answer

Misty, you are not making any sense aha, my teacher cant be that stupid that he doesnt know the answer

misty1212 · Answer

think of it this way:
can you solve $$x^2+y=1$$ for $y$?

theloshua · Answer

no

anonymous · Answer

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 · Answer

so it is a function??

misty1212 · Answer

yes 
say yes

theloshua · Answer

what? yes? Misty, haha, make up your mind

lfreeman1285 · Answer

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 · Answer

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

anonymous · Answer

@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.