How do you find the vertical, horizontal or slant asymptotes for (3x^2 - 4x + 2) /( 2x^3+3)?

Nov 27, 2016

The vertical asymptote is x=-(3/2)^(1/3
No slant asymptote.
The horizontal asymptote is $y = 0$

Explanation:

Let $f \left(x\right) = \frac{3 {x}^{2} - 4 x + 2}{2 {x}^{3} + 3}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- {\left(\frac{3}{2}\right)}^{\frac{1}{3}}\right\}$

As you cannot divide by $0$, x!=-(3/2)^(1/3

So a vertical asymptote is x=-(3/2)^(1/3

The degree of the numerator $<$ the degree of the denominator, there is

To calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{3 {x}^{2}}{2 {x}^{3}} = {\lim}_{x \to - \infty} \frac{3}{2 x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{3 {x}^{2}}{2 {x}^{3}} = {\lim}_{x \to + \infty} \frac{3}{2 x} = {0}^{+}$

So, the horizontal asymptote is $y = 0$

graph{(3x^2-4x+2)/(2x^3+3) [-8.89, 8.885, -4.444, 4.44]}