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

Question

Theodore Amidon · Accepted Answer

#x=-1/3#

Set each factor in the denominator to #0# and solve for #x#. Since the factor, #-3x-1#, appears twice, you only have to solve for #x# in one of the factors.

#-3x-1=0#

#-3x=1#

#color(green)(|bar(ul(color(white)(a/a)x=-1/3color(white)(a/a)|)))#

If you graph the function, you can see that the #x# values approach, but never touch #x=-1/3#.

graph{(2x^2)/(-3x-1)^2 [-7.97, 7.834, -3, 4.9]}

Anna Allen · Answer

undefined To find the asymptotes of the function $ \frac{2x^2}{(-3x-1)^2} $:

1. **Vertical Asymptotes**: Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is not zero.

Set the denominator, $(-3x-1)^2$, equal to zero and solve for $x$:

$$ (-3x-1)^2 = 0 $$

Taking the square root of both sides:

$$ -3x - 1 = 0 $$
$$ -3x = 1 $$
$$ x = -\frac{1}{3} $$

Thus, the vertical asymptote is $x = -\frac{1}{3}$.

2. **Horizontal Asymptotes**: Horizontal asymptotes can be found by comparing the degrees of the numerator and the denominator.

Given that both the numerator and the denominator are polynomials of degree 2, divide each term of the numerator by the corresponding term of the denominator to find the horizontal asymptote:

$$ \frac{2x^2}{(-3x-1)^2} $$

As $ x $ approaches positive or negative infinity, the highest degree terms dominate the function. Therefore, the horizontal asymptote is determined by the ratio of the leading coefficients:

$$ \text{Horizontal Asymptote} = \frac{2}{1} = 2 $$

Thus, the horizontal asymptote is $y = 2$.

In summary:
- Vertical asymptote: $x = -\frac{1}{3}$
- Horizontal asymptote: $y = 2$