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.
[tex]3^2+y^2=25[/tex]
[tex]9+y^2=25[/tex]
Subtract 9 on both sides:
[tex]y^2=25-9[/tex]
[tex]y^2=16[/tex]
Take the square root of both sides:
[tex]y=\pm \sqrt{16}[/tex]
[tex]y=\pm 4[/tex].
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:
[tex](\pm 5)^2+y^2=25[/tex]
[tex]25+y^2=25[/tex]
Subtract 25 on both sides:
[tex]y^2=25-25[/tex]
Simplify:
[tex]y^2=0[/tex]
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.
