Question 130330
Let's start with the definition of a mathematical function (courtesy of Wikipedia):

The mathematical concept of a function expresses dependence between two quantities, one of which is given (the independent variable, argument of the function, or its "input") and the other produced (the dependent variable, value of the function, or "output"). A function associates a single output to each input element drawn from a fixed set, such as the real numbers.


Let's look at your first equation:  {{{y=-x^2}}}.  First, let's discuss the domain of this relation over the set of real numbers.  Is there any real number or sub-set of the real numbers that would make this relationship undefined?  You can certainly square any real number and achieve another real number result.  And you can also take that positive real number result and take the negative of it.  Therefore the domain is the entire set of real numbers, in interval notation:  ({{{-infinity}}},{{{infinity}}}).


Next let's examine the function definition as it applies to this relation.  The key element of the definition is: "...associates a single output to each input element..."  Now, do you think you can find an element of the set of real numbers that could be substituted for x that would result in more than one value for y?  Don't stay up too late trying, because you will never find one.  Note that it doesn't matter that two different input (x) values have the same output (y) value.


{{{y=-x^2}}} is most assuredly a function.  Below you will find a graph of this function.  A visual test to determine whether a relation is a function or not is if you can draw a vertical line that intersects the graph of the relation in more than one place, the relation is NOT a function.  But as you can see, there is no such vertical line in the case of {{{y=-x^2}}}.


{{{drawing(600,600,-5,5,-5,5,
grid(1),
graph(600,600,-5,5,-5,5,-x^2))}}}


I'll give you the graphs of the other two relations and let you do the 'function/not a function' analysis yourself.


{{{xy=1}}}, which is the same thing as {{{y=1/x}}}
{{{drawing(600,600,-5,5,-5,5,
grid(1),
graph(600,600,-5,5,-5,5,1/x),
locate(2,2,y=f(x)=1/x))}}}



{{{y=-x^3}}}
{{{drawing(600,600,-5,5,-5,5,
grid(1),
graph(600,600,-5,5,-5,5,-x^3),
locate(2,2,y=f(x)=-x^3))}}}