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?

Answer 1

Yup

Domain = R ( since There is no Restriction)

Answer 2

thnx does that mean the range is the same

Answer 3

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)

Answer 4

hope this helps

Answer 5

\[y=(x=1)^2-8\]
\[=\sqrt{y+8}=(x-1)\]
\[R={y \in R/y > -8}\]

i feel like im missing a step in between last and second last equations

Answer 6

it is + there (x + 1)

Answer 7

oh yes sorry but im still a little confused