# What are the asymptotes of f(x)=-x/((2x-3)(7x-1)) ?

Mar 9, 2017

vertical asymptotes at $x = \frac{1}{7} \text{ and } x = \frac{3}{2}$
horizontal asymptote at y = 0

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve : $\left(2 x - 3\right) \left(7 x - 1\right) = 0$

$\Rightarrow x = \frac{1}{7} \text{ and "x=3/2" are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = - \frac{x}{14 {x}^{2} - 23 x + 3} = - \frac{\frac{x}{x} ^ 2}{\frac{14 {x}^{2}}{x} ^ 2 - \frac{23 x}{x} ^ 2 + \frac{3}{x} ^ 2}$

$= - \frac{\frac{1}{x}}{14 - \frac{23}{x} + \frac{3}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{14 - 0 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{-(x)/((2x-3)(x-1)) [-20, 20, -10, 10]}