Question

Help please..Describe the vertical asymptote(s) and hole(s) for the graph of y = (x-5)/(x^2+4x+3)

asymptotes: x = –3, –1 and no holes.
asymptote: x = –3 and hole: x = –5
asymptotes: x = –3, –1 and hole: x = –5
asymptote: x = –5 and hole: x = –3

Answer 1

\[\frac{x-5}{(x+3)(x+1)}\] is a start

Answer 2

vertical asymptotes at the zeros of the denominator, namely at \(x=-3\) and at \(x=-1\)

no holes because nothing cancels

Answer 3

so the answer is the first choice? and if so how did you get that?

Answer 4

that is my best explanation above

Answer 5

thank you!!