State the domain and range of this function.

y=(x+1)^2-8

I think the domain is {D=x∈R} am i right?

Question

State the domain and range of this function.

y=(x+1)^2-8

I think the domain is {D=x∈R} am i right?

anonymous · Answer

Yup 

Domain = R   ( since There is no Restriction)

anonymous · Answer

thnx does that mean the range is the same

anonymous · Answer

For Range:

y =f(x)

y =(x+1)^2-8

y + 8 = (x + 1)^2

sqrt ( y+8)  - 1  = x

nw.....here comes the restriction the sqrt of the numebr should be Real so :

y + 8 > =0

y > = -8

Range = [-8 , infinity)

anonymous · Answer

hope this helps

anonymous · Answer

$$y=(x=1)^2-8$$
$$=\sqrt{y+8}=(x-1)$$
$$R={y \in R/y > -8}$$

i feel like im missing a step in between last and second last equations

anonymous · Answer

it is + there    (x + 1)

anonymous · Answer

oh yes sorry but im still a little confused