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

Apr 28, 2017

$\text{vertical asymptotes at "x=3/2" and } x = 7$

$\text{horizontal asymptote at } y = 0$

#### 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.

$\text{solve } \left(2 x - 3\right) \left(x - 7\right) = 0$

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

Horizontal asymptotes occur as

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

$f \left(x\right) = - \frac{x}{2 {x}^{2} - 17 x + 21}$

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

f(x)=-(x/x^2)/((2x^2)/x^2-(17x)/x^2+(21)/x^2)=-(1/x)/(2-17/x+(21)/x^2

as $x \to \pm \infty , f \left(x\right) \to - \frac{0}{2 - 0 + 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{-(x)/((2x-3)(x-7)) [-10, 10, -5, 5]}