Question

The function f(x) = x is translated using the rule (x, y) - (x - 6, y + 9) to create A(x).

Answer 1

It seems the correct function you need to mention is f(x) = sqrt x

so, I am assuming $f(x) = \sqrt{x}$ and will solve the question based on it.

Answer:

Check the explanation

Step-by-step explanation:

Given the function

$f(x) = \sqrt{x}$

As the function is translated according to the rule

• (x, y) → (x - 6, y + 9)

Translation of the function $f(x) = \sqrt{x}$ 6 units to the left will bring the function

$g(x) = \sqrt{x+6}$

Translation of the function  $g(x) = \sqrt{x+6}$  9 units up will bring the function

$A\left(x\right)=\sqrt{x+6}+9$

Determining the range of $A\left(x\right)=\sqrt{x+6}+9$

• As we know that the range is the set of dependent values for which the function is defined.

$\mathrm{The\:range\:of\:an\:radical\:function\:of\:the\:form}\:c\sqrt{ax+b}+k\:\mathrm{is}\:\:f\left(x\right)\ge \:k$

$k=9$

$f\left(x\right)\ge \:9$

so

$\mathrm{Range\:of\:}\sqrt{x+6}+9:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:9\:\\ \:\mathrm{Interval\:Notation:}&\:[9,\:\infty \:)\end{bmatrix}$

Therefore, the expression describes the range of A(x) will be:

$f\left(x\right)\ge \:\:9\:or\:y\:\ge \:\:\:9$