Question

How do you find the domain and range of `(y^2)(x-4)=8`?

Answer 1

Given: `(y^2)(x-4)=8`

`y^2 = 8/(x-4)`

`y = +-(2sqrt2)/sqrt(x-4)`

Separate into two equations:

`y = -(2sqrt2)/sqrt(x-4)` and `y = (2sqrt2)/sqrt(x-4)`

In both cases, the expression under the radical cannot be negative and it must be greater than 0:

`x -4 > 0`

`x > 4 larr` this is the domain for both equations.

Find the range for the first equation:

`lim_(x to 4) -(2sqrt2)/sqrt(x-4) = -oo`

`lim_(x to oo) -(2sqrt2)/sqrt(x-4) = 0`

The range for the first equation is `-oo < y < 0`

`lim_(x to 4) (2sqrt2)/sqrt(x-4) = oo`

`lim_(x to oo) (2sqrt2)/sqrt(x-4) = 0`

The range for the second equation is `0 < y < oo`