# How do you identify all asymptotes or holes for f(x)=x/(-4x^2-16x)?

Jul 15, 2017

$\text{vertical asymptote at } x = - 4$
$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{simplify f(x) by factorising the denominator}$

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

$\text{the removal of the factor x from the numerator/denominator}$
$\text{indicates a hole at x = 0}$

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.

$\text{solve "4(x+4)=0rArrx=-4" is the asymptote}$

$\text{horizontal asymptotes occur as }$

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

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

$f \left(x\right) = - \frac{\frac{1}{x}}{\frac{4 x}{x} + \frac{16}{x}} = - \frac{\frac{1}{x}}{4 + \frac{16}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{4 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{x/(-4x^2-16x) [-10, 10, -5, 5]}