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I was thinking graph A

yes

Alright thank you.

For this one ive gotten to y=- square root x +13 , y

The square root is only over x

Still need help on the second one.

? The second one is not multiple choice. I dont know where you just got C from.

I need help on the f(x)= (x-13)^2 , x

I dont want the answer told to me. Id like to be shown how to solve it.

oh is it the tyoe of question you will have to type an answer out for??

I can show my work up to where im stuck at if that will help.

Yes but my example lost me and i'm stuck at y= - square root x +13 , y

Im at the part where I found what the restriction and inverse equal..
Restriction = y

Sorry but I dont really understand all that. This is my second time going this type of problem but first with a square root. The picture attached is what I have written down

doing*

Do you know the domain and range of the original
y= (x-13)^2
?

Domain of f= Range of f&-1 = (- infinity sign,13]

and what is the range?

Following my example it should be [0, infinity sign) but I have to complete part A to be sure.

**(-14)^2= 196

Um okay.. That's part C.. I still need help completing part A.

Yes I know that.

you mean
\[f^{-1}(x)= -\sqrt{x}+13, \ \ x\ge 0\]

I followed the example.. f^-1 = - square root x +13 ; x>/0

I'm confused by one part.. The last step it squares both sides and flipped the sign /0

Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.

or perhaps I don't understand your question?

Um I'll type out what I mean ...

y/ 0
(square root x) ^2 >/ 0^2
x not <

Why is the sign for the final answer different from the one that it ends with x

If it does not show that, it is a typo

if you try to find the square root of -1 on your calculator is will complain

Okay so it was an error by Pearson.

Alright thanks