Question

How do you find the vertical asymptotes and holes of `f(x)=(x^2+4x+3)/(x+3)`?

Answer 1

There are no asymptotes, only a hole at `x=-3`

Explanation:

Let's factorise the numerator

`x^2+4x+3=(x+1)(x+3)`

Therefore,

`f(x)=(x^2+4x+3)/(x+3)=((x+1)cancel(x+3))/cancel(x+3)`

There we have a hole at `x=-3`

graph{(x^2+4x+3)/(x+3) [-7.316, 6.73, -5.28, 1.743]}