# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (x^2+4x+3)/(x^2 - 9)?

##### 2 Answers
Jun 4, 2016

using lim calculus

#### Explanation:

vertical asymptote:
first find domain: ${x}^{2} - 9 \ne 0$ then $x \ne \pm 3$
then calculate
$\lim f \left(x\right)$when $x \to + 3$ and it is $\frac{24}{0} = \infty$
but is not the same for $x \to - 3$
so you have vertical asymptote x=3

horizontal asymptote;
calculate
$\lim f \left(x\right)$ when $x \to \infty$ that's 1
so you have horizontal asymptote for y=1

when you have horizontal asymptote you haven't oblique ones

Jun 4, 2016

vertical asymptote x = 3
horizontal asymptote y = 1

#### Explanation:

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

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

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 3 = 0 → x = 3 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

$\frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x} - \frac{3}{x}} = \frac{1 + \frac{1}{x}}{1 - \frac{3}{x}}$

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

$\Rightarrow y = 1 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ). Hence there are no oblique asymptotes.
graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}