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

1. misty1212

HI!!

2. misty1212

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

3. misty1212

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

4. 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"

5. Theloshua

im soooo confusssed

6. misty1212

of course you are, this question is ill posed

7. Theloshua

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

8. misty1212

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

9. Theloshua

no

10. 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.

11. Theloshua

so it is a function??

12. misty1212

yes say yes

13. Theloshua

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

14. 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.

15. 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 ".

16. 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.