Explanation:
Step 1. We need to find the option for which the domain is:
![\lbrack a,\infty)](https://img.qammunity.org/2023/formulas/mathematics/college/27gwy29l6r9jyxpsfja2src713n7mgaki9.png)
and the range
![(-\infty,b\rbrack](https://img.qammunity.org/2023/formulas/mathematics/college/4sk9rpw0hu8f5st12b3vr06sf6pvzlv1wa.png)
Step 2. Let's start with the domain. The domain is the set of possible x values.
when we have a square root, the value inside the square root has to be greater than or equal to 0, for options B and D:
Solving for x on each inequality:
Step 3. That means that for B, the domain is:
![\begin{gathered} \lbrack a,\infty) \\ and\text{ for B:} \\ \lbrack-a,\infty) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3mf6e5iy2hl1z7cnbgxyj86krcumvn6qvq.png)
Only B meets the condition for the domain, since the other two options A, and C has third roots, the values inside of them can be negative and the domain is not [a, infinity).
Step 4. If we graph the function for option B, here we are using a=1 and b=2 but these could be any values since it is just for demonstration
We can see that the values possible for the y-axis (the range) go from minus infinity and stop at value b.
Therefore, option B also has the correct range.
Answer.
B
![f(x)=-√(x-a)+b](https://img.qammunity.org/2023/formulas/mathematics/college/lf4xfir3xse6az5rbc7laiygjabl8696gl.png)