Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `(x^3+4)/(2x^2+x-1)`?

Answer 1

The vertical asymptotes are `x=1/2` and `x=-1`
The oblique asymptote is `y=x/2-1/4`
No horizontal asymptote

Explanation:

Let start by factorising the denominator

`2x^2+x-1=(2x-1)(x+1)`

Let `f(x)=(x^3+4)/(2x^2+x-1)=(x^3+4)/((2x-1)(x+1))`

As, we cannot divide by `0`, `x!=1/2` and `x!=-1`

The vertical asymptotes are `x=1/2` and `x=-1`

As the degree of the numerator is `>` than the degree of the denominator, we have an oblique asymptote.

Let's do a long division

`color(white)(aaaa)` `x^3` `color(white)(aaaaaaaaaaaa)` `+4` `color(white)(aaaa)` `|` `2x^2+x-1`

`color(white)(aaaa)` `x^3+x^2/2-x/2` `color(white)(aaaa)` #color(white)(aaaaa)|# `1/2x-1/4`

`color(white)(aaaaa)` `0-x^2/2+x/2` `color(white)(aaaa)` `+4`

`color(white)(aaaaaaa)` `-x^2/2-x/4` `color(white)(a)` `+1/4`

`color(white)(aaaaaaa)` `-0+3/4x` `color(white)(aaaaa)` `+15/4`

Therefore,

`f(x)=x/2-1/4+(3/4x+15/4)/(2x^2+x-1)`

`lim_(x->-oo)f(x)-(x/2-1/4)=lim_(x->-oo)(3/4x)/(2x^2)=0^-`

`lim_(x->+oo)f(x)-(x/2-1/4)=lim_(x->+oo)(3/4x)/(2x^2)=0^+`

The oblique asymptote is `y=x/2-1/4`

graph{(y-(x^3+4)/(2x^2+x-1))(y-x/2+1/4)=0 [-18.02, 18.03, -9, 9.02]}