To find the range of the new function [tex]\( A(x) \)[/tex] which is derived from translating the original function [tex]\( f(x) = \sqrt{x} \)[/tex], we can follow these steps:
1. Understand the original function: The given function [tex]\( f(x) = \sqrt{x} \)[/tex]. The square root function [tex]\(\sqrt{x}\)[/tex] is defined for [tex]\( x \geq 0 \)[/tex]. It outputs values [tex]\( y \geq 0 \)[/tex]. Thus, the range of [tex]\( f(x) \)[/tex] is [tex]\( y \geq 0 \)[/tex].
2. Translation of the function: The translation rule given is [tex]\((x, y) \rightarrow (x-6, y+9)\)[/tex]. This indicates two transformations:
- A horizontal shift to the right by 6 units, transforming [tex]\( x \)[/tex] to [tex]\( x - 6 \)[/tex].
- A vertical shift upwards by 9 units, transforming [tex]\( y \)[/tex] to [tex]\( y + 9 \)[/tex].
3. Applying translation to the original function:
- Horizontally shifting [tex]\( f(x) \)[/tex] by 6 units to [tex]\((x-6)\)[/tex], transforms [tex]\( f(x) \)[/tex] to [tex]\( \sqrt{x-6} \)[/tex]. This keeps the domain of [tex]\( x \geq 6 \)[/tex] since we need [tex]\( x-6 \geq 0 \)[/tex].
- Vertically shifting [tex]\( f(x) \)[/tex] by 9 units to [tex]\( (y+9) \)[/tex], the new function becomes [tex]\( A(x) = \sqrt{x-6} + 9 \)[/tex].
4. Determine the new range: Initially, the range of [tex]\( f(x) = \sqrt{x} \)[/tex] was [tex]\( y \geq 0 \)[/tex]. By shifting this range upwards by 9 units, every value in the range increases by 9:
- If [tex]\( y \geq 0 \)[/tex] becomes [tex]\( y \geq 0 + 9 \)[/tex].
- Hence, the new range for [tex]\( A(x) \)[/tex] is [tex]\( y \geq 9 \)[/tex].
Thus, the correct expression that describes the range of [tex]\( A(x) = \sqrt{x-6} + 9 \)[/tex] is:
[tex]\[ y \geq 9 \][/tex]
So, the correct choice from the given options is:
[tex]\[ \boxed{y \geq 9} \][/tex]
