How do you find the asymptotes for R(x) = (3x) / (x^2 - 9)?

Jul 4, 2016

Answer:

vertical asymptotes x = ± 3
horizontal asymptote y = 0

Explanation:

The denominator of R(x) cannot be zero.This would give division by zero which is undefined. Setting the denominator equal to zero and solving for x gives the values that x cannot be and if the numerator is non-zero for these values of x then they are the asymptotes.

solve: x^2-9=0rArr(x-3)(x+3)=0rArrx=±3

$\Rightarrow x = - 3 , x = 3 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

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

$\frac{\frac{3 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{9}{x} ^ 2} = \frac{\frac{3}{x}}{1 - \frac{9}{x} ^ 2}$

as $x \to \pm \infty , R \left(x\right) \to \frac{0}{1 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(3x)/(x^2-9) [-10, 10, -5, 5]}