Question 844460
You did not say exactly what you want from these two Q's.


The first function has a Real Number restriction, that {{{9-x^2>=0}}}, meaning {{{x^2<=9}}}, or {{{x>=-3}}} and {{{x<=3}}} for the domain.  It would be like one half of the circle, {{{y^2=9-x^2}}},
{{{x^2+y^2=9}}}.
The function itself will never become negative.  Any real value of x in the domain will always give {{{sqrt(9-x^2)>=0}}}.   This function will be the upper half of the circle.  


Refer to the corresponding circle, {{{x^2+y^2=9}}}, and understand that {{{f(x)=sqrt(9-x^2)}}} will be the part greater than or equal to zero:


(This graph is supposed to reach negative 3 on the left)
{{{graph(275,275,-5,5,-5,5,sqrt(9-x^2))}}}