I get g(x)=sqrt(-(x+1)) and the solution guide says g(x)=sqrt(-(x-1)). Is my answer wrong or is the solution guide wrong?
If the negative is outside the square root, you're changing the a parameter (vertical change). Both the solution guide and my answer has the negative inside the square root. In high school, I know that you can't have a negative numerical value in a square root but the letters (unknowns) can also be negative which could result in a positive value. But, I still need help understanding the minor difference in answers between me and my solutions guide.
Hello s3a,
$\displaystyle f(x)=\sqrt{x}$ reflected about the y-axis will yield
$\displaystyle f_1(x)=\sqrt{-x}$.
Then, if you translate (slide) 1 unit to the right, you end up with
$\displaystyle \boxed{g(x)=\sqrt{-x+1}=\sqrt{-(x-1)}}$, which is what your answer key says.
Your solution, $\displaystyle g(x)=\sqrt{-(x+1)}$ will translate the graph 1 unit to the left.