# How do you find the Vertical, Horizontal, and Oblique Asymptote given (x+3)/ (x^2+4x+3)?

Jul 24, 2016

vertical asymptote x = -1
horizontal asymptote y = 0

#### Explanation:

The first step is to factorise and simplify f(x).

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

The denominator of f(x) cannot be zero as this would be undefined. Equating the denominator and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by x

$\frac{\frac{1}{x}}{\frac{x}{x} + \frac{1}{x}} = \frac{\frac{1}{x}}{1 + \frac{1}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0 , denominator-degree 1 ) Hence there are no oblique asymptotes.
graph{(1)/(x+1) [-10, 10, -5, 5]}