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

May 29, 2018

$y = f \left(x\right) = - {x}^{2} + 81$ is a function

$x = f \left(y\right) = \setminus \pm \setminus \sqrt{- y + 81}$ 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 $\setminus m a t h \boldsymbol{R}$.

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 $x = \setminus \pm 9$.

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