Which of the following is true for the rational function f(x) = x-16/x-4 ?

 First of all, F is true because (16 - 16)/(16 - 4) = 0. It just means when you plug in that number (16) you'll get 0. This will NOT happen for 4 because the 0 is in the denominator. 

Look what happens if you have x = 4. This gives you 12/0, which is undefined. For some graphs, this is a hole, but let's look closer. 

What happens if you have x = 3.9? You'll have 12.1/0.1. 3.9999? 12.0001/0.0001. The closer you get to 4, the closer you'll get to y = infinity. 

But what if you have 4.1? 11.9/-0.1. You'll get the same results, but NEGATIVE infinity. So it is NOT a hole in the graph. 

If you draw it out, you'll see that there is a vertical asymptote at x = 4. 

B and F are true. 

As for horizontal asymptotes, look at it like this: y = 16-16/16 - 4 means y = 0. There is no asymptote here. Try subbing in 1 = (x -16)/(x - 4). 
Multiply by x - 4 on both sides 
x - 4 = x - 16 
There is no solution here; there will be an asymptote, so D is also true. 

B, D, and F are true. ,"yahoo answers"if i get busted at least i put down my main source (you know what I'm saying)

The answer should be look like this, there will be underline on f=x2-4x-16

f=‌x2−4x−16
      x2−4x

Hope it helps