Question

describe the vertical asymptote and hole for the graph of

Answer 1

We will have the following:

$y=\frac{x-5}{(x+3)(x+1)}$

From this, we have that the graph will have vertical asymptotes when:

$x^2+4x+3=0\Rightarrow(x+3)(x+1)=0$

So, it will have asymptotes at x = -3 and x = -1.

We will also have that the expression has no holes.