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

Nov 16, 2016

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

#### Explanation:

Let's factorise the numerator

${x}^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right)$

Therefore,

$f \left(x\right) = \frac{{x}^{2} + 4 x + 3}{x + 3} = \frac{\left(x + 1\right) \cancel{x + 3}}{\cancel{x + 3}}$

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

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