Question

the functio f(x)=square root is translated using the rule (x,y) --> (x-6,y+9) to create A(x). which expression describes the range of A(x)?

Answer 1

Answer:  Range is $y \ge 9$

Explanation:

The range of $f(x) = \sqrt{x}$ is  $y \ge 0$

In other words, it's the set of nonnegative real numbers.

We only focus on the y values. We go from y to y+9, which means we shift each (x,y) point 9 unit sup.

Since we shifted 9 units up, we add 9 to each y value and that effectively moves the lowest point in the range (0) up nine units.

So we go from $y \ge 0$ to $y \ge 9$

If you need the range in interval notation, then you would write $[9, \infty)$. Note the square bracket next to the 9 so we include this as part of the range.