The equation provided is:

$y^2 + x^2 = 9$

This represents the equation of a circle with a radius of 3, centered at the origin $(0, 0)$.

To determine whether it defines a function, we can use the vertical line test, which checks whether any vertical line crosses the graph more than once. For an equation to represent a function, each value of $x$ should correspond to exactly one value of $y$.

For the equation of a circle, when solving for $y$, we get:

$y^2 = 9 - x^2$ $y = \pm \sqrt{9 - x^2}$

This shows that for a given $x$, there are generally two corresponding values of $y$ (one positive and one negative). Since one value of $x$ can give two values of $y$, the equation does not represent a function.

Conclusion:

The equation is NF (not a function).

Would you like any more details or have any further questions? Here are some related questions for practice:

1. How would you graph the equation of this circle?
2. How can you convert this equation to parametric form?
3. What would the equation look like if it were a function?
4. How would this equation change if the radius were doubled?
5. How do you define a function algebraically?

Tip: A function must pass the vertical line test, meaning no vertical line should intersect the graph at more than one point.