Question

what is the domain and range of y=1/(x-2)^2? (in terms of infinity).

Answer 1

\[y=1\div(x-2)^{2}\]

Answer 2

what is the domain and range of the equation? @lucaz

Answer 3

you need to find the values of x that dosent make (x-2)^2 = 0

Answer 4

what do you mean? @lucaz

Answer 5

you have a fraction 1/(x-2)^2 the denominator cannot be 0, only positive or negative

Answer 6

if you set x=1 it would be (1-2)^2=1
if you set x=4 it would be (4-2)^2=4
but if x=2 (2-2)^2=0 x cannot be equal to 2
the domain is all real numbers except 2

Answer 7

got it?

Answer 8

so would it be \[[2,\infty)\]?

Answer 9

but then what would the range be?

Answer 10

no, it's R\[(x \in \mathbb{R}\2)\]

Answer 11

I think "except" is like that, I don't remember

Answer 12

it means the same thing because it's saying that you can all real numbers to infinity but the
"[" means it cannot be 2.

Answer 13

at least thats what my teacher is teaching us in PreCalc.

Answer 14

[2,∞) this means x>=2

Answer 15

but you can pick a negative value too

Answer 16

and a number less than 2

Answer 17

whatever you want, except 2, because the denominator would be 0

Answer 18

so would it be (2,∞) meaning the restriction of 2.

Answer 19

this is the domain

Answer 20

and the range?

Answer 21

I forgot how to write it, I'm trying to remember x__X

Answer 22

ops, wrong

Answer 23

I made a mistake

Answer 24

the range is only greater than 0 [0,infinity)

Answer 25

(0,infinity), sorry