If S is a circle with radius 2, why is the upper half of the circle the graph to the function f(x)= (root)(4-x^2) ? What would be the function to the lower half of the circle? (I'm guessing -(root)(4-x^2), but why?)
Thanks!
You're correct. The equation of a circle with radius 2 at the origin is $\displaystyle x^2+y^2=4$. If you solve this for $\displaystyle y$ you get $\displaystyle y=\pm\sqrt{4-x^2}$. This isn't a function since there are going to be two values of $\displaystyle y$ for every $\displaystyle x$. It doesn't pass the vertical line test. However, you can represent either the top or the bottom as a function by simply defining a function as either $\displaystyle \sqrt{4-x^2}$ or $\displaystyle -\sqrt{4-x^2}$