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"}
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The first function has a Real Number restriction, that , meaning , or and for the domain. It would be like one half of the circle, , .
The function itself will never become negative. Any real value of x in the domain will always give . This function will be the upper half of the circle.
Refer to the corresponding circle, , and understand that will be the part greater than or equal to zero:
(This graph is supposed to reach negative 3 on the left)