Equation of a circle

gralla55 · #1 $Old$ November 14th, 2009, 05:04 PM

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!

adkinsjr · #2 $Old$ November 14th, 2009, 05:25 PM

Quote:

Originally Posted by gralla55 $View Post$

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 $x^2+y^2=4$ . If you solve this for $y$ you get $y=\pm\sqrt{4-x^2}$ . This isn't a function since there are going to be two values of $y$ for every $x$ . 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 $\sqrt{4-x^2}$ or $-\sqrt{4-x^2}$

gralla55 · #3 $Old$ November 14th, 2009, 05:26 PM

Thank, you very much! Makes perfect sense.