How do you find the asymptotes for  (2x^2)/((-3x-1)^2)?

Sep 20, 2016

vertical asymptote at $x = - \frac{1}{3}$
horizontal asymptote at $y = \frac{2}{9}$

Explanation:

The denominator of the function cannot be zero as this would make the function 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.

solve: ${\left(- 3 x - 1\right)}^{2} = 0 \Rightarrow - 3 x - 1 = 0 \Rightarrow x = - \frac{1}{3}$

$\Rightarrow x = - \frac{1}{3} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

$f \left(x\right) = \frac{2 {x}^{2}}{{\left(- 3 x - 1\right)}^{2}} = \frac{2 {x}^{2}}{9 {x}^{2} + 6 x + 1}$

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

$f \left(x\right) = \frac{\frac{2 {x}^{2}}{{x}^{2}}}{\frac{9 {x}^{2}}{x} ^ 2 + \frac{6 x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{2}{9 + \frac{6}{x} + \frac{1}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{2}{9 + 0 + 0}$

$\Rightarrow y = \frac{2}{9} \text{ is the asymptote}$
graph{(2x^2)/(9x^2+6x+1) [-20, 20, -10, 10]}