How would i graph the sqrt of (a^2)-(x^2)??
Hello, Godfather!
How would i graph: .$\displaystyle y \;=\;\sqrt{a^2 -x^2}$ ?
Square both sides: .$\displaystyle y^2 \:=\:a^2-x^2 \quad\Rightarrow\quad x^2+y^2 \:=\:a^2$
This is a circle: center at the origin, with radius $\displaystyle a.$
Then: .$\displaystyle y \;=\;^+ \sqrt{a^2-x^2}$ .is the upper semicircle.
The domain consists of all legal values for x given that the function must be a real number. So, $\displaystyle a^2 - x^2$ must be nonnegative. For what values of x is this true?
The range consists of all the values the function can take on its domain. To start, consider what the largest possible value for $\displaystyle \sqrt{a^2 - x^2}$ would be. Can the function give a negative result?
The zeroes of the function will be the same as the zeroes of $\displaystyle a^2 - x^2$. For what values of x is the function equal to zero?
Whether or not the function is even or odd depends on whether or not $\displaystyle f(x) = f(-x)$ or $\displaystyle f(x) = -f(-x)$, respectively.
Given that $\displaystyle f(x) = \sqrt{a^2 - x^2}$, is either of these statements true?
If the function is periodic with period p, then $\displaystyle f(x) = f(x + np)$ for every integer n. Is a semicircle periodic?
If the function is one-to-one, then every y value must be matched by exactly one x value. Is this the case for a semicircle?