The responses above are correct. I'm simply expanding their responses. It might help if there are any gaps needed to be filled in.
You have a square root function
This will yield a graph that looks like a regular "L" shape
Domain is on the horizontal axis labeled as 'x'
Range is on the vertical axis labeled as 'y'
f(x) is your 'y' or your range (just so you know)
Look at the equation in the denominator 6x+20 without the square root bar
What would make this equation equal zero? Because we need a benchmark in order to find the domain
We need to make 6*x equal to (-20) so that when we add it to (20) we get zero as our answer; this will establish our benchmark
x must equal (-10/3) in order to yield (-20)
Plug in
[6*(-10/3)] + 20
[-60/3] + 20
[-20] + 20 = 0
(-10/3) is our benchmark
#Remember the rule concerning fractions: the denominator cannot equal zero (this is considered undefined in algebra)
#Remember the rule concerning square roots: equations under the square root bar cannot yield a negative answer (this in turn yields an 'imaginary' number (yikes!), which we want to try and avoid here)
The domain can be any number greater than (-10/3) on the left--our benchmark; and since there are no limitations to x being greater than (-10/3)--that is to say any value greater than (-10/3) will yield a non-zero, non-negative square root in the denominator (woohoo!)
x can be any number to the right of (-10/3), which in turn means x goes to infinity to the right of (-10/3)
x>(-10/3)
or you can write it as
(-10/3,∞)
Your graph should have a regular "L" shape
As the graph runs right of (-10/3) it runs horizontally and seems to approach the x-axis but it never crosses it as it runs to infinity (hard to imagine--but there you have it)
As the graph runs left toward (-10/3) your graph runs up vertically to infinity but never crosses the boundary of (-10/3) (another hard to imagine--but there you have it moment)
