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
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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.
in your case, it is the first condition that applies
what is the answer though cause i dont understand how to figure it out
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set the denominator of the function to 0 \[(x-6)(x-5)=0\] solve for x
so i would do like terms then divde omg idk i hate math
\[a\times b=0\] only if \(a=0\) or \(b=0\). so, \[x-6=0\qquad x-5=0\] get it?
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).