Question

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

Answer 1

Domain: x>= -2

Range: all real y

Answer 2

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

Answer 3

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:

\[y \ge1\]

Answer 4

ohhhhhhhhhhhhhhhh

Answer 5

u sure mooby ?

Answer 6

when you take sqrt you get +-

Answer 7

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

Answer 8

range is all y