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

Jul 26, 2017

$\text{vertical asymptote at } x = 3$
$\text{horizontal asymptote at } y = 0$

Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} - 9}$

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

$\text{factorise and simplify f(x)}$

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

$\text{the removal of the factor "(x+3)" from the numerator}$
$\text{and denominator indicates a hole at x = - 3}$

$\text{solve " x-3=0rArrx=3" is the asymptote}$

$\text{horizontal asymptotes occur as}$

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

$\text{divide terms on numerator/denominator by x}$

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

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 hence there are no oblique asymptotes.
graph{1/(x-3) [-10, 10, -5, 5]}