SOLUTION: Q 1 f(x) = sqroot(9 - x^2) Q 2 f(x) = - |x| I am a slow and poor grasper. Kindly include all steps.617419"}

Algebra ->  Absolute-value -> SOLUTION: Q 1 f(x) = sqroot(9 - x^2) Q 2 f(x) = - |x| I am a slow and poor grasper. Kindly include all steps.617419"}      Log On


   



Question 844460: Q 1 f(x) = sqroot(9 - x^2)
Q 2 f(x) = - |x|
I am a slow and poor grasper. Kindly include all steps.617419"}

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
You did not say exactly what you want from these two Q's.

The first function has a Real Number restriction, that 9-x%5E2%3E=0, meaning x%5E2%3C=9, or x%3E=-3 and x%3C=3 for the domain. It would be like one half of the circle, y%5E2=9-x%5E2,
x%5E2%2By%5E2=9.
The function itself will never become negative. Any real value of x in the domain will always give sqrt%289-x%5E2%29%3E=0. This function will be the upper half of the circle.

Refer to the corresponding circle, x%5E2%2By%5E2=9, and understand that f%28x%29=sqrt%289-x%5E2%29 will be the part greater than or equal to zero:

(This graph is supposed to reach negative 3 on the left)
graph%28275%2C275%2C-5%2C5%2C-5%2C5%2Csqrt%289-x%5E2%29%29