find the domain and range..x=(y-1)^2-2??

Question

anonymous · Answer

Domain: x>= -2

Range: all real y

anonymous · Answer

how did you find that?..i am a little lost, need a little more explanation..

anonymous · Answer

First you solve for y:

$$x=\left( y-1 \right)^{2}-2$$

$$x+2=\left( y-1 \right)^{2}$$

$$\sqrt{x+2}=y-1$$

$$y=1+\sqrt{x+2}$$

We can only have positive values under the square root sign, or our answer is no longer in the real numbers.

$$x+2\ge0$$

So our domain is:

$$x\ge-2$$


Now, for the range, our output from the square root sign is never going to be negative.  So the minimum value we can have for y is 1 (when x=0).

So our range is:

$$y \ge1$$

anonymous · Answer

ohhhhhhhhhhhhhhhh

anonymous · Answer

u sure mooby ?

anonymous · Answer

when you take sqrt you get +-

anonymous · Answer

$$y= 1+\sqrt{x+2} $$ is not the same as $$x= (y-1)^2 -2 $$

anonymous · Answer

range is all y