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

Aug 4, 2018

$\text{vertical asymptote at } x = - 2$
$\text{horizontal asymptote at } y = 0$

Explanation:

$\text{let } f \left(x\right) = \frac{2 x - 4}{{x}^{2} - 4}$

$\text{factor numerator/denominator}$

$f \left(x\right) = \frac{2 \cancel{\left(x - 2\right)}}{\cancel{\left(x - 2\right)} \left(x + 2\right)} = \frac{2}{x + 2}$

$\text{the removal of the factor "(x-2)" indicates a removable}$
$\text{discontinuity (hole) at } x = 2$

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 numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve "x+2=0rArrx=-2" is the asymptote}$

$\text{Horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{\frac{2}{x}}{\frac{x}{x} + \frac{2}{x}} = \frac{\frac{2}{x}}{1 + \frac{2}{x}}$

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

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

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(2x-4)/(x^2-4) [-10, 10, -5, 5]}