[Math] Function or relation: $y^4 = 8x^2$

algebra-precalculusfunctionsrelations

The question under debate:
Is $y^4=8x^2$ a function or a relation?

We disagree. I say, it's a relation because when I substitute in points

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I get two outputs for most inputs, and thus it's not a function (fails the vertical line test). I also used Wolfram Alpha and Desmos to check the graph, and both of those yield vertical-line-failing horizontal parabolas…
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which fail the vertical line test.

Another person says "…because the original problem did not include a root… it was an exponent – we quad-rooted the other side to solve for y only… but that wasn't the original question/problem. So the way the original problem is set up, there would be no negative answers… " which means each x has one y, it passes the vertical line test, and it’s a function.

I'll admit, I don't quite understand his reasoning, but there's a lot I don't understand, like how the square root function yields only a positive result but $x^2=9$ yields $+3$ and $-3$, so I can't argue convincingly.

Is there some way to rule either in/out? The answer given by the system indicates my answer is correct, that it's a relation and not a function, but for a host of reasons I won't describe, we don't fully trust that system.

Thank you for any light you can shed that will resolve our debate!

Best Answer

You are definitely correct. You can regard the equation $y^4 = 8x^2$ as the relation $$xRy \iff y^4 = 8x^2,$$ and $R$ is a function of $x$ if and only if, for each $x$, there is at most one $y$ such that $xRy$. But that is clearly false in this case. For example, let $x=2$; then $y^4 = 32$ so that $y = \pm\sqrt[4]{32} = \pm 2\sqrt[4]{2}$ (as well as some complex values). Thus this relation does not define a function of $x$.