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?

2. After the first transformation: $\displaystyle f_1(x)=-\sqrt{x}$
After the second transformation: $\displaystyle g(x)=-\sqrt{x-1}$

3. 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.

4. Originally Posted by s3a
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?
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.