Question

Determine if the equation below is a function with independent variable x. If so, find the domain. If not, find a value of x to which there corresponds more than one value of y.
x^{2} +y^{2} =25

Answer 1

Answer:

No because it contains points (3,4) and (3,-4).  You cannot have a x assigned to more than one y-value if you want it to be a function.

Step-by-step explanation:

A function has one output per input.

If we are trying to determine if the given is a function of x, then x is the input.

However I can get two outputs from plugging in x=3.

$3^2+y^2=25$

$9+y^2=25$

Subtract 9 on both sides:

$y^2=25-9$

$y^2=16$

Take the square root of both sides:

$y=\pm \sqrt{16}$

$y=\pm 4$.

So input x=3 yields y=4 and y=-4.

Since this input has more than one output then the given is not a function of x.

----Also!

If you graph the equation, it is a circle with radius 5 and center (0,0). So I could I plug in any number for x between -5 and 5 excluding -5 and 5 which would yield only one output each. Since plugging in either one gives:

$(\pm 5)^2+y^2=25$

$25+y^2=25$

Subtract 25 on both sides:

$y^2=25-25$

Simplify:

$y^2=0$

There is only one value y such that when you square it gives you 0. That is 0.

x=5 only gives y=0 and x=-5 only gives y=0.

There is no circle, unless it is a circle with radius 0 which means it really wouldn't be a circle, that is a function.