## College Algebra 7th Edition

$y$ is NOT a function of $x$.
Solve for $y$ in terms of $x$ to obtain: $\begin{array}{ccc} &x^2+(y-1)^2 &= &4 \\&(y-1)^2 &= &4-x^2 \\&\sqrt{(y-1)^2} &= &\pm\sqrt{4-x^2} \\&y-1 &= &\pm\sqrt{4-x^2} \\&y &= &1 \pm\sqrt{4-x^2} \end{array}$ The last equation above will give two different values of $y$ for each value of $x$. Thus, the equation does not represent $y$ as a function of $x$.