Question

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

Answer 1

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

Explanation:

Let `f(x)=(3x^2-4x+2)/(2x^3+3)`

The domain of `f(x)` is `D_f(x) =RR-{-(3/2)^(1/3)}`

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->+-oo`, we take the terms of highest degree in the numerator and the denominator.

`lim_(x->-oo)f(x)=lim_(x->-oo)(3x^2)/(2x^3)=lim_(x->-oo)3/(2x)=0^(-)`

`lim_(x->+oo)f(x)=lim_(x->+oo)(3x^2)/(2x^3)=lim_(x->+oo)3/(2x)=0^(+)`

So, the horizontal asymptote is `y=0`

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