## anonymous 3 years ago find any points of discontinuity for the rational function y=(x+3)(x+7)(x+1)/(x-6)(x-5) a.)x = –3, x = –7, x = –1 b.)x = –6, x = –5 c.)x = 3, x = 7, x = 1 d.)x = 6, x = 5

1. anonymous

discontinuity will occur when you have vertical asymptotes.. i.e., when one of the following happens 1. a division by zero i.e., denomiator becomes "0" 2. there is a possibility of having square-root of a negative number.

2. anonymous

in your case, it is the first condition that applies

3. anonymous

what is the answer though cause i dont understand how to figure it out

4. anonymous

set the denominator of the function to 0 $(x-6)(x-5)=0$ solve for x

5. anonymous

so i would do like terms then divde omg idk i hate math

6. anonymous

$a\times b=0$ only if $$a=0$$ or $$b=0$$. so, $x-6=0\qquad x-5=0$ get it?

7. anonymous

OOOOOOOOOOOOO

8. anonymous

just observe that if x=6 or x=5, the denominator becomes zero. We don't no what happens when you divide something by zero (some say you get a black hole, but whatever.) So, we just say we don't know what happens and don't mark that point on a graph. So, the graph will look like a roller-coaster track that's missing a piece of the track (Oh! the horror).

9. anonymous

ty both of u i have more lol

10. anonymous

that's for another post :P

11. anonymous

lol yeha