Square Root & Quadratic Inverses

$$f^{-1}\left(x\right)=\left(x+6\right)^2$$  

$$f^{-1}\left(x\right)=\left(x-6\right)^2$$  

$$f^{-1}\left(x\right)=x^2+6$$  

$$f^{-1}\left(x\right)=x^2+36$$  

$$y=\sqrt{x-3}+4$$  

$$y=\sqrt{x+3}-4$$  

$$y=\sqrt{x-3+4}$$  

$$y=\sqrt{3x+1}$$  

$$y=\sqrt{x+6}-10$$  

$$y=\sqrt{x-6}+10$$  

$$y=\sqrt{x-10}+6$$  

$$y=\sqrt{x+10}-6$$  

$$y=\left(x+2\right)^2-7$$  

$$y=\left(x-2\right)^2+7$$  

$$y=\left(x+7\right)^2-2$$  

$$y=\left(x-7\right)^2+2$$  

yes

no

$$f^{-1}\left(x\right)=\sqrt[]{x-6}+3$$  

$$f^{-1}\left(x\right)=\sqrt[]{x-3}+6$$  

$$f^{-1}\left(x\right)=\sqrt[]{x+3}-6$$  

$$f^{-1}\left(x\right)=\sqrt[]{x+6}-3$$  

Kind of... the graphs reflect over y = x; however the quadratic is invertible without restriction on the domain.

No, they do not reflect over the line y = x.

Yes, the functions reflect over the line y = x.

$$f\left(x\right)=\left(x-9\right)^2-2$$  

$$f\left(x\right)=\left(x+9\right)^2+2$$  

$$f\left(x\right)=\sqrt{x+9}+2$$  

$$f\left(x\right)=\sqrt{x-9}-2$$