Question

*** Will give you a medal ***
Describe the vertical asymptote(s) and hole(s) for the graph of y = (x-5)/(x^2+4x+3)

Answer 1

@kropot72

Answer 2

\[\lim_{x \rightarrow \infty}\frac{x-5}{x^2+4x+3}=0\]
Therefore y=0 is an retricemptote.
\[y=\frac{x-5}{(x+3)(x+1)}\]
What value of x makes that equation become undefined? Undefined as in, what value of x makes the denominator become 0?

Answer 3

asymptote*

Answer 4

0?

Answer 5

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

These are my options

Answer 6

@jim_thompson5910

Answer 7

Dude...solve this quadratic for x.
\[(x+3)(x+1)=0\]
What values of x do you get from that?