Question

How do you decide whether the relation `x^2 + y = 81` defines a function?

Answer 1

`y = f(x) = -x^2+81` is a function

`x = f(y) = \pm\sqrt(-y+81)` is not.

Explanation:

A function is a law of association between elements in two different sets.

So, to define a function, you must have a "starting" set `A` (the domain), a "landing" set `B` (the codomain), and a rule that associates to every element of `A` one and only one element of `B`.

In most cases, students deal with numerical functions, i.e. functions in which both domain and codomain are the set of real numbers `\mathbb{R}`.

So, does your equation describes a function? It depends on the variable we choose and independent:

Case 1: `x` is independent

In this case, we're looking for a rule that assigns a value to `y`, given a value for `x` as input. Subtracting `x^2` from both sides of the equation, we get

`y = -x^2+81`

which is indeed a function: for every value `x` we may choose, the correponding `y` will be `-x^2+81`

Case 1: `y` is independent

This case is the other way around: we're looking for a rule that assigns a value to `x`, given a value for `y` as input. Subtracting `y` from both sides of the equation, we get

`x^2 = -y+81`

which is not a function: for every value `y` we may choose, assuming `-y+81` is positive, we have ambiguity on the correponding `x`.

Assume, for example, `y=0`. The equation would be `x^2 = 81`, which yields `x=\pm9`.

So, in this case, we're not able to associate one and only one output value to our input value.