# How do you identify all asymptotes or holes and intercepts for f(x)=x/(3x^2+5x)?

Jan 26, 2018

$\text{see explanation}$

#### Explanation:

$\text{factorise the denominator}$

$f \left(x\right) = \frac{\cancel{x}}{\cancel{x} \left(3 x + 5\right)} = \frac{1}{3 x + 5}$

$\text{since the factor x has been cancelled from the numerator/}$
$\text{denominator this indicates a hole at} \left(0 , \frac{1}{5}\right)$

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.

$\text{solve "3x+5=0rArrx=-5/3" is the asymptote}$

$\text{horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{\frac{1}{x}}{\frac{3 x}{x} + \frac{5}{x}} = \frac{\frac{1}{x}}{3 + \frac{5}{x}}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0}{3 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{1/(3x+5) [-10, 10, -5, 5]}