Is #x^2 + y^2 = 7# a function?

2 Answers
Jul 3, 2015

No, it isn't.

Explanation:

You can see this the best by graphing the equation:
graph{x^2+y^2=7 [-10, 10, -5, 5]}
For a graph to be a function, every vertical line can only cross one (or zero) point(s). If you take the vertical line at #x=0#, it crosses the graph at #(0,sqrt(7))# and #(0,-sqrt(7))#. These are two points, so the equation can't be a function.

Jul 4, 2015

No it is not a function. (#y# is not a function of #x#.)

Explanation:

Graphing is a good way of deciding whether an equation defines a function.
Another way is to try to solve for #y#.

#x^2+y^2 = 7#

#y^2 = 7 - x^2#

#y = +- sqrt(7-x^2)#

"#y# equals plus or minus the square root of . . . "

Stop! Functions do not say "or". Functions do not give two answers. The give one or (if, we try to use an input that is not in the domain) they give no answer.