It seems the correct function you need to mention is f(x) = sqrt x
so, I am assuming [tex]f(x) = \sqrt{x}[/tex] and will solve the question based on it.
Answer:
Check the explanation
Step-by-step explanation:
Given the function
[tex]f(x) = \sqrt{x}[/tex]
As the function is translated according to the rule
Translation of the function [tex]f(x) = \sqrt{x}[/tex] 6 units to the left will bring the function
[tex]g(x) = \sqrt{x+6}[/tex]
Translation of the function [tex]g(x) = \sqrt{x+6}[/tex] 9 units up will bring the function
[tex]A\left(x\right)=\sqrt{x+6}+9[/tex]
Determining the range of [tex]A\left(x\right)=\sqrt{x+6}+9[/tex]
[tex]\mathrm{The\:range\:of\:an\:radical\:function\:of\:the\:form}\:c\sqrt{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k[/tex]
[tex]k=9[/tex]
[tex]f\left(x\right)\ge \:9[/tex]
so
[tex]\mathrm{Range\:of\:}\sqrt{x+6}+9:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:9\:\\ \:\mathrm{Interval\:Notation:}&\:[9,\:\infty \:)\end{bmatrix}[/tex]
Therefore, the expression describes the range of A(x) will be:
[tex]f\left(x\right)\ge \:\:9\:or\:y\:\ge \:\:\:9[/tex]
