Question

How do you find the domain and range of `y =sqrt(x-2) / (6+x)`?

Answer 1

You need to ensure that:
1] the denominator is different from zero, so:
`6+x!=0`
`x!=-6`
2] the argument of the square root must be bigger or equal to zero, so:
`x-2>=0`
`x>=2`
These conditions are used to avoid points or intervals where your function cannot be calculated.
Finally the domain is all real `x` bigger or equal to `2`.

For the range I evaluated the maximum through the derivative:
`y'=(10-x)/(2*(sqrt(x-2))*(x+6)^2)`
The maximum is at `x=10` (by setting `y'=0`) corresponding to `y=sqrt(8)/16` giving a range for `y` from zero to `sqrt(8)/16`.