Graphing with functions and inverse functions

[fattymangoes]

I need to show my work for the whole problem and the answer is B. ( ive attached it). My only problem is that I don't know how to do anything (big ik). Any help would be much appreciated!
Here's the work that i did (i tried to find the inverse first):
y=(x-5)^2
square root X = square root (y-5)^2
square root X +5= y

 

[blamocur]

square root X = square root (y-5)^2
square root X +5= y

I cannot figure out what you mean. Can you use parenthesis and a single word for square roots? E.g. "y-5 = sqrt (x)".

 

[Dr.Peterson]

I need to show my work for the whole problem and the answer is B. ( ive attached it). My only problem is that I don't know how to do anything (big ik). Any help would be much appreciated!
Here's the work that i did (i tried to find the inverse first):
y=(x-5)^2
square root X = square root (y-5)^2
square root X +5= y

Your work presumably means this:
[math]y=(x-5)^2\\x=(y-5)^2\\\sqrt{x}=\sqrt{(y-5)^2}\\\sqrt{x}=y-5\\y=\sqrt{x}+5[/math]
If so, then you have the correct inverse, except that when you take the square root, you need to consider both positive and negative signs:
[math]y=(x-5)^2\\x=(y-5)^2\\\pm\sqrt{x}=\sqrt{(y-5)^2}\\\pm\sqrt{x}=y-5\\y=\pm\sqrt{x}+5[/math]
But that is not a function. Look back at the problem, and tell me which sign you need to use.

Then you can graph the two functions (that is, confirm that the graph you show agrees with yours), and use that to answer the domain and range questions.

Then there will be more to discuss.