Which one of the following does not define y as a function of x?Question 16Answera.x^2 + y = 16b.y = (x - 3)^2 - 1c.x = y^2d.x + y = 3

To determine which option does not define $y$ as a function of $x$, we need to analyze each provided equation. A relation defines $y$ as a function of $x$ if for every value of $x$, there is at most one corresponding value of $y$.

1. Option a: $x^2 + y = 16$
   Rearranging gives $y = 16 - x^2$. This equation defines $y$ as a function of $x$ because for each $x$, there is exactly one $y$.

2. Option b: $y = (x - 3)^2 - 1$
   This is already in the form $y = f(x)$. It clearly defines $y$ as a function of $x$ because it assigns exactly one $y$ for each $x$.

3. Option c: $x = y^2$
   Rearranging gives $y = \pm \sqrt{x}$. This shows that for each $x$ (greater than or equal to 0), there are two possible values for $y$ (the positive and negative square roots). Thus, this equation does not define $y$ as a function of $x$.

4. Option d: $x + y = 3$
   Rearranging gives $y = 3 - x$. This is a linear equation that defines $y$ as a function of $x$ with exactly one corresponding $y$ for each $x$.

Conclusion

The option that does not define $y$ as a function of $x$ is c: $x = y^2$.