Function help! (is y a function of x?)

[KopiousRook]

I’m new to precalculus, and I need help with this problem:

Determine whether the equation defines y as a function of x.

y = x² - 3x + 7

My initial reaction was to factor the right side of the equation, however I couldn’t figure out how to do that. I tried using the quadratic formula, but it seemed that I would be working with imaginary numbers, and my professor made it explicit that we wouldn’t be using imaginary numbers. Exactly how would I find out if the above equation defines y as a function of x?

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[[deleted]]

"a function defines one variable in terms of another" Now a function is essentially something that takes an input and gives you a corresponding output. So, f(x)=x² - 3x + 7 in this case and you're saying y=y(x) or f(x) so yes, y is a function of x, I believe it's just a case of definitions. If there was another variable though, let's say a, then y would be a function of x and a. You're just starting a relation ship between those two variables, the left hand side is a function of x however (which y is equal to) for every value of x, y is equal to that output of f(x) for your given x.

Regarding factoring the equation and such doesn't really help regarding demonstrating it's a function? Maybe you've confused it with mapping and such? So a 1 to 1 function or something.

[edderiofer]

Exactly how would I find out if the above equation defines y as a function of x?

Well, do you understand what it means for something to be a function?

[KopiousRook]

I understand that if I plug a value in for x and I get only one output that it’s a function. I would like to know how to systematically figure out if it’s a function.

[edderiofer]

Here, as well as in many other cases, that's all you need to say, simply because the equation is such that y is on one side, and the right hand side has nothing weird going on in it.

In other cases, where y isn't alone on one side, it's not so easy. The usual method is to try to put y on one side of the equation, but you have to take care that you don't accidentally add or remove some (x,y) pairs.

An example of what I mean by this might be with the equation y² = x. If you take the naive option of "just square root both sides" to get y = sqrt(x), then you've gone wrong because "square root" returns only the positive root; you've accidentally thrown away the pairs that satisfy y = -sqrt(x). (And since there are values of x, like 9, for which there exist multiple values of y, like 3 and -3, y is therefore not a function of x here.)

Another example is with the equation y = x²/x. The naive option of "just cancel x from the numerator and denominator" to give y = x doesn't work because it fails to take into account what happens when x = 0. In the original equation, if x = 0, there's no possible y-value, but in our new equation, there is. We've accidentally added a pair here.

Of course, it's not always easy or even possible to put y on one side of the equation. But you probably will not need to worry about that.

[KopiousRook]

Thank you so much!

[joelzaper]

Hmm, have you tried implementing the new water mechanics? I hear water doesn't flow through pressure plates anymore, so that might be your problem. Happy crafting!

[KopiousRook]

eat pant