No, domain and range finding techniques
aren't involved here.

Since 0 is on the right side,



We begin by finding all critical values, by
setting the numerator = 0 and solving, then
setting the denominator = 0.

Setting the numerator = 0, x+8=0 gives x=-8,
so -8 is one critical value.

Setting the denominator = 0, x-1=0 gives x=1,
so 1 is the other critical value.

Plot these on a number line.  First make all
the circles open, we may have to close some of
them, but for now just use open circles:
  
-------------------o-----------------------------------o-------------
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4 

Now we need to find out what part of the line to shade.
First we pick any test value in the region to the left of -8,
say -9, and substitute it into the inequality







That is true, so we shade the number line left of -8

<==================o-----------------------------------o-------------
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Next we pick any test value in the between -8 and 1,
say 0, and substitute it into the inequality







That is false, so we do not shade the number line between
-8 and 1

Thirdly we pick any test value in the region to the right of 1,
say 2, and substitute it into the inequality







That is true, so we shade the number line right of 1

<==================o-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Now we must test the critical values to see if we can
shade them or not.

Test critical value -8 by substituting it







This is true, so we darken the circle at -8

<==================@-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Test critical value 1 by substituting it






Division by zero is always undefined. So we must leave
the circle at 1 open, so we still have

<==================@-----------------------------------o============>
-12  -11 -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4

Now we want to get the interval notation.  On the far left of the 
shading we have negative infinity, -oo and the shading stops at
-8, so we write the left shaded part as

(-oo,-8]

We use a "(" at -oo because infinity is never included.  We use a ] at 
at -8, because -8 is included and has a darkened circle.


Now we put a "union" symbol "U"

(-oo,-8] U

Then the shaded part on the right going
left to right is from 1 to infinity, or (1,oo)

We use "(" at 1 because 1 is not included, so the
final answer is:

(-oo,-8] U (1,oo)

Edwin